Re: [R] Graph many points without hiding some

[Peter Langfelder]

On Wed, Mar 30, 2011 at 10:04 PM, Samuel Dennis  wrote:
> I have a very large dataset with three variables that I need to graph using
> a scatterplot. However I find that the first variable gets masked by the
> other two, so the graph looks entirely different depending on the order of
> variables. Does anyone have any suggestions how to manage this?
>
> This code is an illustration of what I am dealing with:
>
> x <- 1
> plot(rnorm(x,mean=20),rnorm(x),col=1,xlim=c(16,24))
> points(rnorm(x,mean=21),rnorm(x),col=2)
> points(rnorm(x,mean=19),rnorm(x),col=3)
>
> gives an entirely different looking graph to:
>
> x <- 1
> plot(rnorm(x,mean=19),rnorm(x),col=3,xlim=c(16,24))
> points(rnorm(x,mean=20),rnorm(x),col=1)
> points(rnorm(x,mean=21),rnorm(x),col=2)
>
> despite being identical in all respects except for the order in which the
> variables are plotted.
>
> I have tried using pch=".", however the colours are very difficult to
> discern. I have experimented with a number of other symbols with no real
> solution.
>
> The only way that appears to work is to iterate the plot with a for loop,
> and progressively add a few numbers from each variable, as below. However
> although I can do this simply with random numbers as I have done here, this
> is an extremely cumbersome method to use with real datasets.
>
> plot(1,1,xlim=c(16,24),ylim=c(-4,4),col="white")
> x <- 100
> for (i in 1:100) {
> points(rnorm(x,mean=19),rnorm(x),col=3)
> points(rnorm(x,mean=20),rnorm(x),col=1)
> points(rnorm(x,mean=21),rnorm(x),col=2)
> }
>
> Is there some function in R that could solve this through automatically
> iterating my data as above, using transparent symbols, or something else? Is
> there some other way of solving this issue that I haven't thought of?

Assume you are plotting variables y1, y2, y3 of the same length
against a common x, and you would like to assign colors say c(1,2,3).
You can automate the randomization of order as follows:

n = length(y1);
y = c(y1, y2, y3);
xx = rep(x, 3);
colors = rep(c(1,2,3), c(n, n, n));

order = sample(c(1:(3*n)));

plot(xx[order], y[order], col= colors[order])

I basically turn the y's into a single vector y with the corresponding
values of x stored in xx and the plotting colors, then randomize the
order using the sample function.

HTH,

Peter

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[R] ANCOVA for linear regressions without intercept

[Yusuke Fukuda]

Hello R experts

I have two linear regressions for sexes (Male, Female, Unknown). All have a 
good correlation between body length (response variable) and head length 
(explanatory variable). I know it is not recommended, but for a good practical 
reason (the purpose of study is to find a single conversion factor from head 
length to body length), the regressions need to go through the origin (0 
intercept). 

Is it possible to do ANCOVA for these regressions without intercepts? When I do

summary(lm(body length ~ sex*head length))

this will include the intercepts as below


Coefficients:
   Estimate Std. Error t value Pr(>|t|)
(Intercept)-6.496971.68497  -3.856 0.000118 ***
sexMale-9.393401.97760  -4.750 2.14e-06 ***
sexUnknown -1.337912.35453  -0.568 0.569927
head_length 7.123070.05503 129.443  < 2e-16 ***
sexMale:head_length 0.316310.06246   5.064 4.37e-07 ***
sexUnknown:head_length  0.199370.07022   2.839 0.004556 ** 
--- 

Is there any way I can remove the intercepts so that I can simply compare the 
slopes with no intercept taken into account?

Thanks for help in advance.

Yusuke Fukuda

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Re: [R] Graph many points without hiding some

[Dennis Murphy]

Hi:

I can think of a couple: (1) size reduction of the points; (2) alpha
transparency; (3)  (1) + (2)

>From your original plot in base graphics, I reduced cex to 0.2 and it didn't
look too bad:

plot(rnorm(x,mean=19),rnorm(x),col=3,xlim=c(16,24), cex = 0.2)
points(rnorm(x,mean=20),rnorm(x),col=1, cex = 0.2)
points(rnorm(x,mean=21),rnorm(x),col=2, cex = 0.2)

AFAIK, base graphics doesn't have alpha transparency available, but the
ggplot2 package does. One approach is to adjust the alpha transparency on
default size points; another is to combine reduced point size with alpha
transparency. Here is your example rehashed for ggplot2.

require(ggplot2)
d <- data.frame(x1 = rnorm(1, mean = 19), x2 = rnorm(1, mean = 20),
x3 = rnorm(1, mean = 21), x = rnorm(1))
# Basically stacking x1 - x3, creating two new vars named variable and value
dm <- melt(d, id = 'x')   # from reshape package, loads with ggplot2
# Alpha transparency is set to a low level with default point size,
# but the colors in the legend are muted by the level of transparency
ggplot(dm, aes(x = x, y = value, colour = variable)) + theme_bw() +
   geom_point(alpha = 0.05) +
   scale_colour_manual(values = c('x1' = 'black',
  'x2' = 'red', 'x3' = 'green'))

# A tradeoff is to reduce the point size and increase alpha a bit, but these
changes will
# also be reflected in the legend.

ggplot(dm, aes(x = x, y = value, colour = variable)) + theme_bw() +
   geom_point(alpha = 0.15, size = 1) +
   scale_colour_manual(values = c('x1' = 'black',
  'x2' = 'red', 'x3' = 'green'))

You may well find the legend to be useless for this example, so to get rid
of it,

ggplot(dm, aes(x = x, y = value, colour = variable)) + theme_bw() +
   geom_point(alpha = 0.15, size = 1) +
   scale_colour_manual(values = c('x1' = 'black',
  'x2' = 'red', 'x3' = 'green')) +
   opts(legend.position = 'none')

The nice thing about the ggplot2 graph is that you can adjust the point size
and alpha transparency to your tastes. The default point size is 2 and the
default alpha = 1 (no transparency).

HTH,
Dennis

On Wed, Mar 30, 2011 at 10:04 PM, Samuel Dennis  wrote:

> I have a very large dataset with three variables that I need to graph using
> a scatterplot. However I find that the first variable gets masked by the
> other two, so the graph looks entirely different depending on the order of
> variables. Does anyone have any suggestions how to manage this?
>
> This code is an illustration of what I am dealing with:
>
> x <- 1
> plot(rnorm(x,mean=20),rnorm(x),col=1,xlim=c(16,24))
> points(rnorm(x,mean=21),rnorm(x),col=2)
> points(rnorm(x,mean=19),rnorm(x),col=3)
>
> gives an entirely different looking graph to:
>
> x <- 1
> plot(rnorm(x,mean=19),rnorm(x),col=3,xlim=c(16,24))
> points(rnorm(x,mean=20),rnorm(x),col=1)
> points(rnorm(x,mean=21),rnorm(x),col=2)
>
> despite being identical in all respects except for the order in which the
> variables are plotted.
>
> I have tried using pch=".", however the colours are very difficult to
> discern. I have experimented with a number of other symbols with no real
> solution.
>
> The only way that appears to work is to iterate the plot with a for loop,
> and progressively add a few numbers from each variable, as below. However
> although I can do this simply with random numbers as I have done here, this
> is an extremely cumbersome method to use with real datasets.
>
> plot(1,1,xlim=c(16,24),ylim=c(-4,4),col="white")
> x <- 100
> for (i in 1:100) {
> points(rnorm(x,mean=19),rnorm(x),col=3)
> points(rnorm(x,mean=20),rnorm(x),col=1)
> points(rnorm(x,mean=21),rnorm(x),col=2)
> }
>
> Is there some function in R that could solve this through automatically
> iterating my data as above, using transparent symbols, or something else?
> Is
> there some other way of solving this issue that I haven't thought of?
>
> Thankyou,
>
> Samuel Dennis
>
>[[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

[[alternative HTML version deleted]]

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Re: [R] VECM with UNRESTRICTED TREND

[Pfaff, Bernhard Dr.]

Hello Greg,
 
you include your trend as a (Nx1) matrix and use this for 'dumvar'. The matrix 
'dumvar' is just added to the VECM as deterministic regressors and while you 
are referring to case 5, this is basically what you are after, if I am not 
mistaken. But we aware that this implies a quadratic trend for the levels.
 
Best,
Bernhard




Von: Grzegorz Konat [mailto:grzegorz.ko...@ibrkk.pl] 
Gesendet: Mittwoch, 30. März 2011 20:50
An: Pfaff, Bernhard Dr.; r-help@r-project.org
Betreff: Re: [R] VECM with UNRESTRICTED TREND


Hello Bernhard, 

Thank You very much. Unfortunately I'm still not really sure how should 
I use dummy vars in this context...
If I have a system of three variables (x, y, z), lag order = 2 and 1 
cointegrating relation, what should I do? I mean, what kind of 'pattern' should 
be used to create those dummy variables, what should they represent and how 
many of them do I need?

Many thanks in advance!

Best,
Greg


2011/3/30 Pfaff, Bernhard Dr. 


Hello Greg,

you can exploit the argument 'dumvar' for this. See ?ca.jo

Best,
Bernhard

> -Ursprüngliche Nachricht-
> Von: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] Im Auftrag von Grzegorz 
Konat
> Gesendet: Mittwoch, 30. März 2011 16:46
> An: r-help@r-project.org
> Betreff: [R] VECM with UNRESTRICTED TREND

>
> Dear All,
>
> My question is:
>
> how can I estimate VECM system with "unrestricted trend" (aka
> "case 5") option as a deterministic term?
>
> As far as I know, ca.jo in urca package allows for 
"restricted trend"
> only [vecm
> <- ca.jo(data, type = "trace"/"eigen", ecdet = "trend", K =
> n, spec = "transitory"/"longrun")].
> Obviously, I don't have to do this in urca, so if another
> package gives the possibility, please let me know too!
>
> Thanks in advance!
>
> Greg
>

>   [[alternative HTML version deleted]]
>
> __
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> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible 
code.
>


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Re: [R] Graph many points without hiding some

[Nick Sabbe]

Hi.
You could also turn it into a 3D plot with some variation on the function
below:
plot4d<-function(x,y,z, u, main="", xlab="", ylab="", zlab="", ulab="")
{
require(rgl)#may need to install this package first

#standard trick to get some intensity colors
uLim<-range(u)
uLen<-uLim[2] - uLim[1] + 1
colorlut<-terrain.colors(uLen)
col<-colorlut[u - uLim[1] + 1]

open3d()#Open new device
points3d(x=x, y=y, z=z,  col=col)
aspect3d(x=1, y=1, z=1) #ensure bounding box is in cube-form
(scaling variables)
#note: if you want to flip an axis, use -1 in the statement above

axes3d() #Show axes
title3d(main = main, sub=paste("Green is low", ulab, ", red is
high")
xlab = xlab, ylab = ylab, zlab = zlab)
}

HTH,


Nick Sabbe
--
ping: nick.sa...@ugent.be
link: http://biomath.ugent.be
wink: A1.056, Coupure Links 653, 9000 Gent
ring: 09/264.59.36

-- Do Not Disapprove




-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Peter Langfelder
Sent: donderdag 31 maart 2011 9:26
To: Samuel Dennis
Cc: R-help@r-project.org
Subject: Re: [R] Graph many points without hiding some

On Wed, Mar 30, 2011 at 10:04 PM, Samuel Dennis  wrote:
> I have a very large dataset with three variables that I need to graph
using
> a scatterplot. However I find that the first variable gets masked by the
> other two, so the graph looks entirely different depending on the order of
> variables. Does anyone have any suggestions how to manage this?
>
> This code is an illustration of what I am dealing with:
>
> x <- 1
> plot(rnorm(x,mean=20),rnorm(x),col=1,xlim=c(16,24))
> points(rnorm(x,mean=21),rnorm(x),col=2)
> points(rnorm(x,mean=19),rnorm(x),col=3)
>
> gives an entirely different looking graph to:
>
> x <- 1
> plot(rnorm(x,mean=19),rnorm(x),col=3,xlim=c(16,24))
> points(rnorm(x,mean=20),rnorm(x),col=1)
> points(rnorm(x,mean=21),rnorm(x),col=2)
>
> despite being identical in all respects except for the order in which the
> variables are plotted.
>
> I have tried using pch=".", however the colours are very difficult to
> discern. I have experimented with a number of other symbols with no real
> solution.
>
> The only way that appears to work is to iterate the plot with a for loop,
> and progressively add a few numbers from each variable, as below. However
> although I can do this simply with random numbers as I have done here,
this
> is an extremely cumbersome method to use with real datasets.
>
> plot(1,1,xlim=c(16,24),ylim=c(-4,4),col="white")
> x <- 100
> for (i in 1:100) {
> points(rnorm(x,mean=19),rnorm(x),col=3)
> points(rnorm(x,mean=20),rnorm(x),col=1)
> points(rnorm(x,mean=21),rnorm(x),col=2)
> }
>
> Is there some function in R that could solve this through automatically
> iterating my data as above, using transparent symbols, or something else?
Is
> there some other way of solving this issue that I haven't thought of?

Assume you are plotting variables y1, y2, y3 of the same length
against a common x, and you would like to assign colors say c(1,2,3).
You can automate the randomization of order as follows:

n = length(y1);
y = c(y1, y2, y3);
xx = rep(x, 3);
colors = rep(c(1,2,3), c(n, n, n));

order = sample(c(1:(3*n)));

plot(xx[order], y[order], col= colors[order])

I basically turn the y's into a single vector y with the corresponding
values of x stored in xx and the plotting colors, then randomize the
order using the sample function.

HTH,

Peter

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[R] nls.profile

[Daniel Kaschek]

Hello,

I use nls.profile to compute confidence intervals of parameter estimates
of a non-linear model. When computing the profiles, the model function
produces an error for certain parameter combinations. Therefore nls
fails and so does nls.profile.

Is there a way to tell nls.profile to ignore errors and to proceed with
the next parameter value?

Cheers,
Daniel.


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Re: [R] summing values by week - based on daily dates - but with somedates missing

[Martyn Byng]

Hi,

Yep, that was what it was doing. For a sum across week, try something like

get.week.flag <- function(dd) {
  ## get weekday from the date in dd and code it as Monday = 1, Tuesday = 2 etc
  idd = 
factor(weekdays(dd),levels=c("Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"))

  ## convert to numeric
  ndd = as.numeric(idd)

  ## flag entries where weekday code gets less (this will flag changes in week)
  wflag = c(FALSE,(ndd[-length(idd)] > ndd[-1]))

  ## cumulative sum to get the week flag
  cumsum(wflag) + 1
}

to get a week flag (this is assuming that your data is sorted by date, if not 
you'll have to sort it first). If you want the week to start on a different 
day, just change the ordering of the weekdays in the levels statement.

data.frame(date=myframe$date,day=weekdays(myframe$date),week=get.week.flag(myframe$date))

seems to indicate that the function is doing what it should, so you can then 
amend the previous code to use get.week.flag instead of weekdays, as in

sum.by.week <- function(ff) {
 by.day <- split(ff$value,get.week.flag(ff$dates))
 lapply(by.day,sum)
}

by.grp <- split(myframe,myframe$group)
lapply(by.grp,sum.by.week)

Martyn

-Original Message-
From: Dimitri Liakhovitski [mailto:dimitri.liakhovit...@gmail.com] 
Sent: 30 March 2011 18:03
To: Martyn Byng
Cc: r-help
Subject: Re: [R] summing values by week - based on daily dates - but with 
somedates missing

Thank you, Martyn.
But it looks like this way we are getting sums by day - i.e., across
all Mondays, all Tuesdays, etc.
Maybe I did not explain well, sorry! The desired output would contain
sums for each WHOLE week - across all days that comprise that week -
Monday through Sunday.
Makes sense?
Dimitri

On Wed, Mar 30, 2011 at 12:53 PM, Martyn Byng  wrote:
> Hi,
>
> How about something like:
>
> sum.by.day <- function(ff) {
>  by.day <- split(ff$value,weekdays(ff$dates))
>  lapply(by.day,sum)
> }
>
> by.grp <- split(myframe,myframe$group)
>
> lapply(by.grp,sum.by.day)
>
>
> Martyn
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Dimitri Liakhovitski
> Sent: 30 March 2011 15:23
> To: r-help
> Subject: [R] summing values by week - based on daily dates - but with
> somedates missing
>
> Dear everybody,
>
> I have the following challenge. I have a data set with 2 subgroups,
> dates (days), and corresponding values (see example code below).
> Within each subgroup: I need to aggregate (sum) the values by week -
> for weeks that start on a Monday (for example, 2008-12-29 was a
> Monday).
> I find it difficult because I have missing dates in my data - so that
> sometimes I don't even have the date for some Mondays. So, I can't
> write a proper loop.
> I want my output to look something like this:
> group   dates   value
> group.1 2008-12-29  3.0937
> group.1 2009-01-05  3.8833
> group.1 2009-01-12  1.362
> ...
> group.2 2008-12-29  2.250
> group.2 2009-01-05  1.4057
> group.2 2009-01-12  3.4411
> ...
>
> Thanks a lot for your suggestions! The code is below:
> Dimitri
>
> ### Creating example data set:
> mydates<-rep(seq(as.Date("2008-12-29"), length = 43, by = "day"),2)
> myfactor<-c(rep("group.1",43),rep("group.2",43))
> set.seed(123)
> myvalues<-runif(86,0,1)
> myframe<-data.frame(dates=mydates,group=myfactor,value=myvalues)
> (myframe)
> dim(myframe)
>
> ## Removing same rows (dates) unsystematically:
> set.seed(123)
> removed.group1<-sample(1:43,size=11,replace=F)
> set.seed(456)
> removed.group2<-sample(44:86,size=11,replace=F)
> to.remove<-c(removed.group1,removed.group2);length(to.remove)
> to.remove<-to.remove[order(to.remove)]
> myframe<-myframe[-to.remove,]
> (myframe)
>
>
>
> --
> Dimitri Liakhovitski
> Ninah Consulting
> www.ninah.com
>
> __
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
> 
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R-

[R] statistical question

[Anna Lee]

Dear List!

I want to compare medians of  non normal distributed data. Is it
possible and usefull to calculate 95% confidence intervals for
medians? And if so - how can this be achieved in R?

Thanks a lot!
Anna

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[R] dfsane arguments

[Kristian Lind]

Hi there,

I'm trying to solve 2 nonlinear equations in 2 unknowns using the BB
package.

The first part of my program solves 3 ODEs using the deSolve package. This
part works. The output is used as parameter values in the functions I need
to solve.

The second part is to solve 2 equations in 2 unknowns. This does not work. I
get the error message "unexpected end of input". So what inputs am I missing
here? As I understand it the arguments I have excluded from dfsane(), such
as control, are set to default?

parameters <- c(K_vv= 0.0047,
K_rv=-0.0268,
K_rr=0.3384,
theta_v=107.4039,
theta_r =5.68,
Sigma_rv=0.0436,
Sigma_rr=0.1145,
lambda_v=0,
lambda_r=-0.0764
)
state <- c(b_1 = 0,
b_2 = 0,
a = 0)
Kristian <- function(t, state, parameters){
with(as.list(c(state, parameters)),{
db_1 =
-((K_vv+lambda_v)*b_1+(K_rv+Sigma_rv*lambda_v+Sigma_rr*lambda_r)*b_2+0.5*(b_1)^2+Sigma_rv*b_1*b_2+0.5*((Sigma_rv)^2+(Sigma_rr)^2)*(b_2)^2
)
db_2 = -K_rr*b_2+1
da = K_vv*theta_v*b_1+(K_rv*theta_v+K_rr*theta_r)*b_2
})
}
times <- seq(0, 10, by = 0.5)
library(deSolve)
out <- ode(y = state, times = times, func = Kristian, parms = parameters)
# constructing output as a matrix
outmat <- as.matrix(out)
library(BB) #loading BB package
Bo <- function(x, s){
f <- rep(NA, length(x))
f[1] <- (1-exp(outmat[4,4]-outmat[4,3]*x[1]-outmat[4,2]*x[2]))/
( exp(-outmat[1,4]-outmat[1,3]*x[1]-outmat[1,2]*x[2])
+ exp(-outmat[2,4]-outmat[2,3]*x[1]-outmat[2,2]*x[2])
+ exp(-outmat[3,4]-outmat[3,3]*x[1]-outmat[3,2]*x[2])
+ exp(-outmat[4,4]-outmat[4,3]*x[1]-outmat[4,2]*x[2]))-s[1]

f[2] <-(1-exp(-outmat[20,4]-outmat(20,3)*x[1]-outmat[20,2]*x[2]))/
( exp(-outmat[1,4]-outmat[1,3]*x[1]-outmat[1,2]*x[2])
+ exp(-outmat[2,4]-outmat[2,3]*x[1]-outmat[2,2]*x[2])
+ exp(-outmat[3,4]-outmat[3,3]*x[1]-outmat[3,2]*x[2])
+ exp(-outmat[4,4]-outmat[4,3]*x[1]-outmat[4,2]*x[2])
+ exp(-outmat[5,4]-outmat[5,3]*x[1]-outmat[5,2]*x[2])
+ exp(-outmat[5,4]-outmat[5,3]*x[1]-outmat[5,2]*x[2])
+ exp(-outmat[6,4]-outmat[6,3]*x[1]-outmat[6,2]*x[2])
+ exp(-outmat[7,4]-outmat[7,3]*x[1]-outmat[7,2]*x[2])
+ exp(-outmat[8,4]-outmat[8,3]*x[1]-outmat[8,2]*x[2])
+ exp(-outmat[9,4]-outmat[9,3]*x[1]-outmat[9,2]*x[2])
+ exp(-outmat[10,4]-outmat[10,3]*x[1]-outmat[10,2]*x[2])
+ exp(-outmat[11,4]-outmat[11,3]*x[1]-outmat[11,2]*x[2])
+ exp(-outmat[12,4]-outmat[12,3]*x[1]-outmat[12,2]*x[2])
+ exp(-outmat[13,4]-outmat[13,3]*x[1]-outmat[13,2]*x[2])
+ exp(-outmat[14,4]-outmat[14,3]*x[1]-outmat[14,2]*x[2])
+ exp(-outmat[15,4]-outmat[15,3]*x[1]-outmat[15,2]*x[2])
+ exp(-outmat[16,4]-outmat[16,3]*x[1]-outmat[16,2]*x[2])
+ exp(-outmat[17,4]-outmat[17,3]*x[1]-outmat[17,2]*x[2])
+ exp(-outmat[18,4]-outmat[18,3]*x[1]-outmat[18,2]*x[2])
+ exp(-outmat[19,4]-outmat[19,3]*x[1]-outmat[19,2]*x[2])
+ exp(-outmat[20,4]-outmat[20,3]*x[1]-outmat[20,2]*x[2])) -s[2]
f
s <- c(0.03, 0.045)
p<-c(0.5, 0.5)
ans <- dfsane(par=p, fn=Bo, s=s)
ans$par

Any help will be much appreciated!

Thank you,

Kristian Lind

[[alternative HTML version deleted]]

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Re: [R] statistical question

[Roger Koenker]

The default rank test in the quantreg package would look like this

summary(rq(y ~ d, tau = .5))

where d is a factor variable indicating which sample the elements of
y belonged to.  Summary returns a confidence interval for the coef
of the factor variable -- if this interval excludes zero at the chosen
alpha level then the difference in medians is "significant."


Roger Koenker
rkoen...@illinois.edu



On Mar 31, 2011, at 4:15 AM, Anna Lee wrote:

> Dear List!
> 
> I want to compare medians of  non normal distributed data. Is it
> possible and usefull to calculate 95% confidence intervals for
> medians? And if so - how can this be achieved in R?
> 
> Thanks a lot!
> Anna
> 
> -- 
> 
> 
> 
> Der Inhalt dieser E-Mail ist vertraulich. Sollte Ihnen die E-Mail
> irrtümlich zugesandt worden sein, bitte ich Sie, mich unverzüglich zu
> benachrichtigen und die E-Mail zu löschen.
> 
> This e-mail is confidential. If you have received it in error, please
> notify me immediately and delete it from your system.
> 
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

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[R] Simple lattice question

[Rubén Roa]

DeaR ComRades,

require(lattice)
data <- data.frame(SP=sort(rep(as.factor(c('A','B','C','D','E')),12)),
   x=rpois(60,10),
   y=rep(c(rep(0,4),rep(10,4),rep(20,4)),5),
   z=rep(1:4,15))
xyplot(x~y|SP,data=data,groups=z,layout=c(2,3),pch=1:4,lty=1:4,col='black',type='b')

How do I put a legend for the grouping variable in the empty upper-right panel?

TIA

Rubén


 

Dr. Rubén Roa-Ureta
AZTI - Tecnalia / Marine Research Unit
Txatxarramendi Ugartea z/g
48395 Sukarrieta (Bizkaia)
SPAIN

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Re: [R] how about a "

[Kenn Konstabel]

In addition to %...% operators one can define := (I haven't seen this
possibility documented anywhere but it's used in a package) which
seems to have different precedence.

> `:=`<-`%<-%`   # the %<-% by John Fox
> a
[1] -1
> a := 1000 + 1
[1] 1001
> a
[1] 1001
> a %<-% 1000 + 1
[1] 1001
> a
[1] 1000

Regards,
Kenn

Kenn Konstabel
National Institute for Health Development
Hiiu 42
Tallinn

On Thu, Mar 31, 2011 at 2:50 AM, William Dunlap  wrote:
> The %...% operators are not a panacea.
> they have the same precedence as `*`
> and `/` (I think) so you get things like:
>
>  > x %<-% 10 - 8  # %<-% has higher precedence than -
>  [1] 2
>  > x # not what you thought it would be
>  [1] 10
>
>  > x %<-% 10 ^3 # but lower than ^
>  [1] 1000
>  > x # this is what you expected
>  [1] 1000
>
> It isn't that hard to write a package with your
> own parser in it.  Just have it generate the
> call tree from your input text and call eval()
> on it.
>
> Bill Dunlap
> Spotfire, TIBCO Software
> wdunlap tibco.com
>
>> -Original Message-
>> From: r-help-boun...@r-project.org
>> [mailto:r-help-boun...@r-project.org] On Behalf Of John Fox
>> Sent: Wednesday, March 30, 2011 4:34 PM
>> To: 'Carl Witthoft'
>> Cc: r-help@r-project.org
>> Subject: Re: [R] how about a ">
>> Dear Carl,
>>
>> I think that the following does what you want:
>>
>> > `%<-%` <- function(e1, e2){
>> +   e1 <- deparse(substitute(e1))
>> +   env <- parent.frame()
>> +   assign(e1, e2, envir=env)
>> +   e2
>> + }
>>
>> > x %<-% 10
>> [1] 10
>>
>> > x
>> [1] 10
>>
>> But, as has been pointed out, it's probably easier just to
>> parenthesize the
>> usual assignment command.
>>
>> Regards,
>>  John
>>
>> 
>> John Fox
>> Senator William McMaster
>>   Professor of Social Statistics
>> Department of Sociology
>> McMaster University
>> Hamilton, Ontario, Canada
>> http://socserv.mcmaster.ca/jfox
>>
>>
>>
>>
>> > -Original Message-
>> > From: r-help-boun...@r-project.org
>> [mailto:r-help-boun...@r-project.org]
>> > On Behalf Of Carl Witthoft
>> > Sent: March-30-11 7:00 PM
>> > To: r-help@r-project.org
>> > Subject: [R] how about a "> >
>> > I was cursing Matlab again today (what else is new) because
>> the default
>> > action for every Matlab command is to spew the result to
>> the console, and
>> > one must remember to put that darn ";"  at the end of every line.
>> >
>> > So I just wondered:  was there ever a discussion as to
>> providing some
>> > modified version of the "<-" and "->" operators in R to do
>> the reverse?
>> >   That is, since R does not print the values of a command
>> to the console,
>> > what if there were  an operator such that
>> >
>> >
>> >   newobject > >
>> > would do the same as
>> >
>> > print(newobject <- somefunction())
>> >
>> >
>> > Any thoughts?
>> > Carl
>> >
>> > __
>> > R-help@r-project.org mailing list
>> > https://stat.ethz.ch/mailman/listinfo/r-help
>> > PLEASE do read the posting guide http://www.R-project.org/posting-
>> > guide.html
>> > and provide commented, minimal, self-contained, reproducible code.
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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[R] That dreaded floating point trap

[Alexander Engelhardt]

Hi,
I had a piece of code which looped over a decimal vector like this:


for( i in where ){
  thisdata <- subset(herde, herde$mlr >= i)
  # do stuff with thisdata..
}

'where' is a vector like seq(-1, 1, by=0.1)

My problem was: 'nrow(thisdata)' in loop repetition 0.4 was different if 
'where' was seq(-1, 1, by=0.1) than when 'where' was seq(-0.8, 1, by=0.1)

It went away after I changed the first line to:

  thisdata <- subset(herde, herde$mlr >= round(i, digits=1))

This is that "floating point trap" the R inferno pdf talked about, 
right? That file talked about the problem, but didn't offer a solution.


Similar things happened when I created a table() from a vector with 
values in seq(-1, 1, by=0.1)


Do I really have to round every float at every occurence from now on, or 
is there another solution? I only found all.equal() and identical(), but 
I want to subset for observations with a value /greater/ than something.


Thanks in advance,
 Alex

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Re: [R] VECM with UNRESTRICTED TREND

[Grzegorz Konat]

Hello Bernhard,

thank You so much one again! Now I (more or less) understand the idea, but
still have problem with its practical application.

I do (somewhat following example 8.1 in your textbook):

library(urca)
data(my.data)
names(my.data)
attach(my.data)
dat1 <- my.data[, c("dY", "X", "dM")]
dat2 <- cbind(time)
args('ca.jo')
yxm.vecm <- ca.jo(dat1, type = "trace", ecdet = "trend", K = 2, spec =
"longrun", dumvar=dat2)

The above code produces following output:

Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), , drop =
FALSE] :
  non-numeric argument to binary operator

What does that mean? Should I use cbind command to dat1 as well? And doesn't
it transform the series into series of integer numbers?

Thank you once again (especially for your patience).

Best,
Greg



2011/3/31 Pfaff, Bernhard Dr. 

>  Hello Greg,
>
> you include your trend as a (Nx1) matrix and use this for 'dumvar'. The
> matrix 'dumvar' is just added to the VECM as deterministic regressors and
> while you are referring to case 5, this is basically what you are after, if
> I am not mistaken. But we aware that this implies a quadratic trend for the
> levels.
>
> Best,
> Bernhard
>
>  --
> *Von:* Grzegorz Konat [mailto:grzegorz.ko...@ibrkk.pl]
> *Gesendet:* Mittwoch, 30. März 2011 20:50
> *An:* Pfaff, Bernhard Dr.; r-help@r-project.org
> *Betreff:* Re: [R] VECM with UNRESTRICTED TREND
>
> Hello Bernhard,
>
> Thank You very much. Unfortunately I'm still not really sure how should I
> use dummy vars in this context...
> If I have a system of three variables (x, y, z), lag order = 2 and 1
> cointegrating relation, what should I do? I mean, what kind of 'pattern'
> should be used to create those dummy variables, what should they represent
> and how many of them do I need?
>
> Many thanks in advance!
>
> Best,
> Greg
>
> 2011/3/30 Pfaff, Bernhard Dr. 
>
>> Hello Greg,
>>
>> you can exploit the argument 'dumvar' for this. See ?ca.jo
>>
>> Best,
>> Bernhard
>>
>> > -Ursprüngliche Nachricht-
>> > Von: r-help-boun...@r-project.org
>> > [mailto:r-help-boun...@r-project.org] Im Auftrag von Grzegorz Konat
>> > Gesendet: Mittwoch, 30. März 2011 16:46
>> > An: r-help@r-project.org
>> > Betreff: [R] VECM with UNRESTRICTED TREND
>>  >
>> > Dear All,
>> >
>> > My question is:
>> >
>> > how can I estimate VECM system with "unrestricted trend" (aka
>> > "case 5") option as a deterministic term?
>> >
>> > As far as I know, ca.jo in urca package allows for "restricted trend"
>> > only [vecm
>> > <- ca.jo(data, type = "trace"/"eigen", ecdet = "trend", K =
>> > n, spec = "transitory"/"longrun")].
>> > Obviously, I don't have to do this in urca, so if another
>> > package gives the possibility, please let me know too!
>> >
>> > Thanks in advance!
>> >
>> > Greg
>> >
>> >   [[alternative HTML version deleted]]
>> >
>> > __
>> > R-help@r-project.org mailing list
>> > https://stat.ethz.ch/mailman/listinfo/r-help
>> > PLEASE do read the posting guide
>> > http://www.R-project.org/posting-guide.html
>> > and provide commented, minimal, self-contained, reproducible code.
>> >
>> *
>> Confidentiality Note: The information contained in this message,
>> and any attachments, may contain confidential and/or privileged
>> material. It is intended solely for the person(s) or entity to
>> which it is addressed. Any review, retransmission, dissemination,
>> or taking of any action in reliance upon this information by
>> persons or entities other than the intended recipient(s) is
>> prohibited. If you received this in error, please contact the
>> sender and delete the material from any computer.
>> *
>>
>>
>

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Re: [R] That dreaded floating point trap

[Ted Harding]

On 31-Mar-11 11:24:01, Alexander Engelhardt wrote:
> Hi,
> I had a piece of code which looped over a decimal vector like this:
> 
> for( i in where ){
>thisdata <- subset(herde, herde$mlr >= i)
># do stuff with thisdata..
> }
> 
> 'where' is a vector like seq(-1, 1, by=0.1)
> 
> My problem was: 'nrow(thisdata)' in loop repetition 0.4 was different
> if 'where' was seq(-1, 1, by=0.1) than when 'where' was
> , when you wan
> It went away after I changed the first line to:
> 
>thisdata <- subset(herde, herde$mlr >= round(i, digits=1))
> 
> This is that "floating point trap" the R inferno pdf talked about, 
> right? That file talked about the problem, but didn't offer a solution.
> 
> Similar things happened when I created a table() from a vector with 
> values in seq(-1, 1, by=0.1)
> 
> Do I really have to round every float at every occurence from now on,
> or is there another solution? I only found all.equal() and identical(),
> but I want to subset for observations with a value /greater/ than
> something.
> 
> Thanks in advance,
>   Alex

A very straightforward way to avoid this problem is to construct the
sequence by multiplying a sequence of integers by an approriate
constant. E.g. for your first example:

  for( i in where ){
 thisdata <- subset(herde, herde$mlr >= i)
 # do stuff with thisdata..
  }
 
'where' is a vector like 0.1*((-10):10)
[ instead of seq(-1, 1, by=0.1) ]

and then, when you want to change to seq(-0.8, 1, by=0.1),
use instead 0.1*(-80,10).

Hoping this helps,
Ted.


E-Mail: (Ted Harding) 
Fax-to-email: +44 (0)870 094 0861
Date: 31-Mar-11   Time: 12:52:35
-- XFMail --

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Re: [R] how about a "

[John Fox]

Dear Ken and Bill,

Thanks for the comments. I obviously didn't think about the precedence issue
and was unaware that one could define :=.

Best,
 John

> -Original Message-
> From: Kenn Konstabel [mailto:lebats...@gmail.com]
> Sent: March-31-11 7:12 AM
> To: William Dunlap
> Cc: John Fox; Carl Witthoft; r-help@r-project.org
> Subject: Re: [R] how about a " 
> In addition to %...% operators one can define := (I haven't seen this
> possibility documented anywhere but it's used in a package) which seems to
> have different precedence.
> 
> > `:=`<-`%<-%`   # the %<-% by John Fox
> > a
> [1] -1
> > a := 1000 + 1
> [1] 1001
> > a
> [1] 1001
> > a %<-% 1000 + 1
> [1] 1001
> > a
> [1] 1000
> 
> Regards,
> Kenn
> 
> Kenn Konstabel
> National Institute for Health Development Hiiu 42 Tallinn
> 
> On Thu, Mar 31, 2011 at 2:50 AM, William Dunlap  wrote:
> > The %...% operators are not a panacea.
> > they have the same precedence as `*`
> > and `/` (I think) so you get things like:
> >
> >  > x %<-% 10 - 8  # %<-% has higher precedence than -
> >  [1] 2
> >  > x # not what you thought it would be
> >  [1] 10
> >
> >  > x %<-% 10 ^3 # but lower than ^
> >  [1] 1000
> >  > x # this is what you expected
> >  [1] 1000
> >
> > It isn't that hard to write a package with your own parser in it.
> > Just have it generate the call tree from your input text and call
> > eval() on it.
> >
> > Bill Dunlap
> > Spotfire, TIBCO Software
> > wdunlap tibco.com
> >
> >> -Original Message-
> >> From: r-help-boun...@r-project.org
> >> [mailto:r-help-boun...@r-project.org] On Behalf Of John Fox
> >> Sent: Wednesday, March 30, 2011 4:34 PM
> >> To: 'Carl Witthoft'
> >> Cc: r-help@r-project.org
> >> Subject: Re: [R] how about a " >>
> >> Dear Carl,
> >>
> >> I think that the following does what you want:
> >>
> >> > `%<-%` <- function(e1, e2){
> >> +   e1 <- deparse(substitute(e1))
> >> +   env <- parent.frame()
> >> +   assign(e1, e2, envir=env)
> >> +   e2
> >> + }
> >>
> >> > x %<-% 10
> >> [1] 10
> >>
> >> > x
> >> [1] 10
> >>
> >> But, as has been pointed out, it's probably easier just to
> >> parenthesize the usual assignment command.
> >>
> >> Regards,
> >>  John
> >>
> >> 
> >> John Fox
> >> Senator William McMaster
> >>   Professor of Social Statistics
> >> Department of Sociology
> >> McMaster University
> >> Hamilton, Ontario, Canada
> >> http://socserv.mcmaster.ca/jfox
> >>
> >>
> >>
> >>
> >> > -Original Message-
> >> > From: r-help-boun...@r-project.org
> >> [mailto:r-help-boun...@r-project.org]
> >> > On Behalf Of Carl Witthoft
> >> > Sent: March-30-11 7:00 PM
> >> > To: r-help@r-project.org
> >> > Subject: [R] how about a " >> >
> >> > I was cursing Matlab again today (what else is new) because
> >> the default
> >> > action for every Matlab command is to spew the result to
> >> the console, and
> >> > one must remember to put that darn ";"  at the end of every line.
> >> >
> >> > So I just wondered:  was there ever a discussion as to
> >> providing some
> >> > modified version of the "<-" and "->" operators in R to do
> >> the reverse?
> >> >   That is, since R does not print the values of a command
> >> to the console,
> >> > what if there were  an operator such that
> >> >
> >> >
> >> >   newobject  >> >
> >> > would do the same as
> >> >
> >> > print(newobject <- somefunction())
> >> >
> >> >
> >> > Any thoughts?
> >> > Carl
> >> >
> >> > __
> >> > R-help@r-project.org mailing list
> >> > https://stat.ethz.ch/mailman/listinfo/r-help
> >> > PLEASE do read the posting guide http://www.R-project.org/posting-
> >> > guide.html and provide commented, minimal, self-contained,
> >> > reproducible code.
> >>
> >> __
> >> R-help@r-project.org mailing list
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> PLEASE do read the posting guide
> >> http://www.R-project.org/posting-guide.html
> >> and provide commented, minimal, self-contained, reproducible code.
> >>
> >
> > __
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> > http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >

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Re: [R] Italicized title from index

[Henrique Dallazuanna]

Try this:

plot(1, main = bquote("Yield  for " ~ italic(.(spp)) ~ "in management
region" ~ .(region)))

On Thu, Mar 31, 2011 at 2:35 AM, Jeremy Newman  wrote:
> Hi all!
>
> I've written a handy script that uses a for loop to allow me to generate a
> large number of figures and statistical outputs for a large dataset.
>
> I am using indexing to retrieve a species name for the title of my graphs-
> which worked fine. However, I need to italicize these species names.
>
> I originally used the paste function, and had no problems with indexing:
>
> *main=paste("Yield for ", testsub[1,3], " in management region ",
> testsub[1,2])*
>
> The title looks like: "*Yield for Gadus morhua in management region 4X5Y*"
>
> I tried bquote from a related help thread, I tried to emulate it:
>
> *spp<-testsub[1,3]*
> *region<-testsub[1,2]*
> *
> *
> *main=bquote(Yield ~ for ~ italic(.(spp)) ~ in ~ management ~ region ~
> .(region))*
>
> Which doesn't seem to work at all, but when I try not putting anything after
> the italic:
>
> *main=bquote(Yield ~ for ~ italic(.(spp)))*
>
> I get: "*Yield for 1*"
> While *spp=Gadus morhua*
>
> I'm at wit's end, I tried to read about substitute, expression, and eval
> functions in the hopes I can figure it out, but I am lost!
>
> Thanks for any help!
>
> Cheers,
>
> -Jeremy N
>
> Undergraduate Researcher in Macroecology
> University of Ottawa
> Department of Biology
>
> Ad astra per alia porci!
>
>        [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

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Re: [R] Italicized title from index

[Henrique Dallazuanna]

Better:

plot(1, main = bquote("Yield  for " ~ italic(.(as.character(spp))) ~
"in management region" ~ .(region)))

You have a factor, so you need convert it to character


On Thu, Mar 31, 2011 at 8:59 AM, Henrique Dallazuanna  wrote:
> Try this:
>
> plot(1, main = bquote("Yield  for " ~ italic(.(spp)) ~ "in management
> region" ~ .(region)))
>
> On Thu, Mar 31, 2011 at 2:35 AM, Jeremy Newman  wrote:
>> Hi all!
>>
>> I've written a handy script that uses a for loop to allow me to generate a
>> large number of figures and statistical outputs for a large dataset.
>>
>> I am using indexing to retrieve a species name for the title of my graphs-
>> which worked fine. However, I need to italicize these species names.
>>
>> I originally used the paste function, and had no problems with indexing:
>>
>> *main=paste("Yield for ", testsub[1,3], " in management region ",
>> testsub[1,2])*
>>
>> The title looks like: "*Yield for Gadus morhua in management region 4X5Y*"
>>
>> I tried bquote from a related help thread, I tried to emulate it:
>>
>> *spp<-testsub[1,3]*
>> *region<-testsub[1,2]*
>> *
>> *
>> *main=bquote(Yield ~ for ~ italic(.(spp)) ~ in ~ management ~ region ~
>> .(region))*
>>
>> Which doesn't seem to work at all, but when I try not putting anything after
>> the italic:
>>
>> *main=bquote(Yield ~ for ~ italic(.(spp)))*
>>
>> I get: "*Yield for 1*"
>> While *spp=Gadus morhua*
>>
>> I'm at wit's end, I tried to read about substitute, expression, and eval
>> functions in the hopes I can figure it out, but I am lost!
>>
>> Thanks for any help!
>>
>> Cheers,
>>
>> -Jeremy N
>>
>> Undergraduate Researcher in Macroecology
>> University of Ottawa
>> Department of Biology
>>
>> Ad astra per alia porci!
>>
>>        [[alternative HTML version deleted]]
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
>
> --
> Henrique Dallazuanna
> Curitiba-Paraná-Brasil
> 25° 25' 40" S 49° 16' 22" O
>



-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

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Re: [R] That dreaded floating point trap

[Alexander Engelhardt]

A very straightforward way to avoid this problem is to construct the
sequence by multiplying a sequence of integers by an approriate
constant. E.g. for your first example:

   for( i in where ){
  thisdata<- subset(herde, herde$mlr>= i)
  # do stuff with thisdata..
   }

'where' is a vector like 0.1*((-10):10)
[ instead of seq(-1, 1, by=0.1) ]

and then, when you want to change to seq(-0.8, 1, by=0.1),
use instead 0.1*(-80,10).



Hi,

this helps, thank you.
But if this code is in a function, and some user supplies a vector, I 
will still have to round it in the function, I guess.


It's weird how 0.1 is different from round(0.1, digits=1) , but I don't 
want to read that 90 page long floating point paper which was referenced 
somewhere :)


Thanks,
 Alex

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Re: [R] VECM with UNRESTRICTED TREND

[Pfaff, Bernhard Dr.]

 

Hello Bernhard, 


thank You so much one again! Now I (more or less) understand the idea, 
but still have problem with its practical application.


I do (somewhat following example 8.1 in your textbook):


library(urca)
data(my.data)
names(my.data)
attach(my.data)
dat1 <- my.data[, c("dY", "X", "dM")]
dat2 <- cbind(time)
 
What is 'time'? Just employ matrix(seq(1:nrow(dat1)), ncol = 1) for 
creating the trend variable.
 
Best,
Bernhard
 
 
 args('ca.jo')
yxm.vecm <- ca.jo(dat1, type = "trace", ecdet = "trend", K = 2, spec = 
"longrun", dumvar=dat2)

The above code produces following output:

Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), , 
drop = FALSE] : 
  non-numeric argument to binary operator

What does that mean? Should I use cbind command to dat1 as well? And 
doesn't it transform the series into series of integer numbers?

Thank you once again (especially for your patience).

Best,
Greg



2011/3/31 Pfaff, Bernhard Dr. 


Hello Greg,
 
you include your trend as a (Nx1) matrix and use this for 
'dumvar'. The matrix 'dumvar' is just added to the VECM as deterministic 
regressors and while you are referring to case 5, this is basically what you 
are after, if I am not mistaken. But we aware that this implies a quadratic 
trend for the levels.
 
Best,
Bernhard




Von: Grzegorz Konat [mailto:grzegorz.ko...@ibrkk.pl] 
Gesendet: Mittwoch, 30. März 2011 20:50
An: Pfaff, Bernhard Dr.; r-help@r-project.org
Betreff: Re: [R] VECM with UNRESTRICTED TREND


Hello Bernhard, 

Thank You very much. Unfortunately I'm still not really 
sure how should I use dummy vars in this context...
If I have a system of three variables (x, y, z), lag 
order = 2 and 1 cointegrating relation, what should I do? I mean, what kind of 
'pattern' should be used to create those dummy variables, what should they 
represent and how many of them do I need?

Many thanks in advance!

Best,
Greg


2011/3/30 Pfaff, Bernhard Dr. 



Hello Greg,

you can exploit the argument 'dumvar' for this. 
See ?ca.jo

Best,
Bernhard

> -Ursprüngliche Nachricht-
> Von: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] Im 
Auftrag von Grzegorz Konat
> Gesendet: Mittwoch, 30. März 2011 16:46
> An: r-help@r-project.org
> Betreff: [R] VECM with UNRESTRICTED TREND

>
> Dear All,
>
> My question is:
>
> how can I estimate VECM system with 
"unrestricted trend" (aka
> "case 5") option as a deterministic term?
>
> As far as I know, ca.jo in urca package 
allows for "restricted trend"
> only [vecm
> <- ca.jo(data, type = "trace"/"eigen", ecdet 
= "trend", K =
> n, spec = "transitory"/"longrun")].
> Obviously, I don't have to do this in urca, 
so if another
> package gives the possibility, please let me 
know too!
>
> Thanks in advance!
>
> Greg
>

>   [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
   

Re: [R] VECM with UNRESTRICTED TREND

[Grzegorz Konat]

'time' was a trend variable from my.data set. Equivalent to the output of
the command 'matrix' you just gave me.

So now I did:

library(urca)
data(my.data)
names(my.data)
attach(my.data)
dat1 <- my.data[, c("dY", "X", "dM")]
mat1 <- matrix(seq(1:nrow(dat1)), ncol = 1)
args('ca.jo')
yxm.vecm <- ca.jo(dat1, type = "trace", ecdet = "const", K = 2, spec =
"longrun", dumvar=mat1)

and the output is:

Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), , drop =
FALSE] :
  non-numeric argument to binary operator
In addition: Warning message:
In ca.jo(dat1, type = "trace", ecdet = "const", K = 2, spec = "longrun",  :
No column names in 'dumvar', using prefix 'exo' instead.

What do I do wrong?

Best,
Greg


2011/3/31 Pfaff, Bernhard Dr. 

>
>
>
> Hello Bernhard,
>
> thank You so much one again! Now I (more or less) understand the idea, but
> still have problem with its practical application.
>
> I do (somewhat following example 8.1 in your textbook):
>
>  library(urca)
> data(my.data)
> names(my.data)
> attach(my.data)
> dat1 <- my.data[, c("dY", "X", "dM")]
> dat2 <- cbind(time)
>
> What is 'time'? Just employ matrix(seq(1:nrow(dat1)), ncol = 1) for
> creating the trend variable.
>
> Best,
> Bernhard
>
>
>  args('ca.jo')
> yxm.vecm <- ca.jo(dat1, type = "trace", ecdet = "trend", K = 2, spec =
> "longrun", dumvar=dat2)
>
> The above code produces following output:
>
>  Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), , drop
> = FALSE] :
>   non-numeric argument to binary operator
>
> What does that mean? Should I use cbind command to dat1 as well? And
> doesn't it transform the series into series of integer numbers?
>
> Thank you once again (especially for your patience).
>
> Best,
> Greg
>
>
>
> 2011/3/31 Pfaff, Bernhard Dr. 
>
>>  Hello Greg,
>>
>> you include your trend as a (Nx1) matrix and use this for 'dumvar'. The
>> matrix 'dumvar' is just added to the VECM as deterministic regressors and
>> while you are referring to case 5, this is basically what you are after, if
>> I am not mistaken. But we aware that this implies a quadratic trend for the
>> levels.
>>
>> Best,
>> Bernhard
>>
>>  --
>> *Von:* Grzegorz Konat [mailto:grzegorz.ko...@ibrkk.pl]
>> *Gesendet:* Mittwoch, 30. März 2011 20:50
>> *An:* Pfaff, Bernhard Dr.; r-help@r-project.org
>> *Betreff:* Re: [R] VECM with UNRESTRICTED TREND
>>
>>   Hello Bernhard,
>>
>> Thank You very much. Unfortunately I'm still not really sure how should I
>> use dummy vars in this context...
>> If I have a system of three variables (x, y, z), lag order = 2 and 1
>> cointegrating relation, what should I do? I mean, what kind of 'pattern'
>> should be used to create those dummy variables, what should they represent
>> and how many of them do I need?
>>
>> Many thanks in advance!
>>
>> Best,
>> Greg
>>
>> 2011/3/30 Pfaff, Bernhard Dr. 
>>
>>> Hello Greg,
>>>
>>> you can exploit the argument 'dumvar' for this. See ?ca.jo
>>>
>>> Best,
>>> Bernhard
>>>
>>> > -Ursprüngliche Nachricht-
>>> > Von: r-help-boun...@r-project.org
>>> > [mailto:r-help-boun...@r-project.org] Im Auftrag von Grzegorz Konat
>>> > Gesendet: Mittwoch, 30. März 2011 16:46
>>> > An: r-help@r-project.org
>>> > Betreff: [R] VECM with UNRESTRICTED TREND
>>>  >
>>> > Dear All,
>>> >
>>> > My question is:
>>> >
>>> > how can I estimate VECM system with "unrestricted trend" (aka
>>> > "case 5") option as a deterministic term?
>>> >
>>> > As far as I know, ca.jo in urca package allows for "restricted trend"
>>> > only [vecm
>>> > <- ca.jo(data, type = "trace"/"eigen", ecdet = "trend", K =
>>> > n, spec = "transitory"/"longrun")].
>>> > Obviously, I don't have to do this in urca, so if another
>>> > package gives the possibility, please let me know too!
>>> >
>>> > Thanks in advance!
>>> >
>>> > Greg
>>> >
>>> >   [[alternative HTML version deleted]]
>>> >
>>> > __
>>> > R-help@r-project.org mailing list
>>> > https://stat.ethz.ch/mailman/listinfo/r-help
>>> > PLEASE do read the posting guide
>>> > http://www.R-project.org/posting-guide.html
>>> > and provide commented, minimal, self-contained, reproducible code.
>>> >
>>> *
>>> Confidentiality Note: The information contained in this message,
>>> and any attachments, may contain confidential and/or privileged
>>> material. It is intended solely for the person(s) or entity to
>>> which it is addressed. Any review, retransmission, dissemination,
>>> or taking of any action in reliance upon this information by
>>> persons or entities other than the intended recipient(s) is
>>> prohibited. If you received this in error, please contact the
>>> sender and delete the material from any computer.
>>> *
>>>
>>>
>>
>

[[alternative HTML version deleted]]

__
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Re: [R] That dreaded floating point trap

[Sarah Goslee]

On Thu, Mar 31, 2011 at 8:14 AM, Alexander Engelhardt
> this helps, thank you.
> But if this code is in a function, and some user supplies a vector, I will
> still have to round it in the function, I guess.
>
> It's weird how 0.1 is different from round(0.1, digits=1) , but I don't want
> to read that 90 page long floating point paper which was referenced
> somewhere :)


Or you could try the much shorter R FAQ 7.31. Turns out it
isn't weird at all, if you are a computer.

Sarah

-- 
Sarah Goslee
http://www.functionaldiversity.org

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Re: [R] fonts in mosaic

[Michael Friendly]

If you are doing multiple plots like this, there is no way I know of
to specify the fonts for labeling *once* for all such plots.  However,
you can do something like this to save typing and keep things consistent:

my.largs <- list(
gp_labels = gpar(fontsize = 12, fontfamily = "calibri"),
gp_varnames = gpar(fontsize = 16, fontfamily = "calibri"))

mosaic(UCBAdmissions, labeling_args=my.largs)
mosaic(UCBAdmissions, labeling_args=my.largs, shade=TRUE)

etc.


On 3/30/2011 2:42 PM, Henrique Dallazuanna wrote:

Try this:

windowsFonts(calibri = windowsFont("Calibri"))

mosaic(UCBAdmissions, labeling_args = list(
   gp_labels = gpar(fontsize = 12, fontfamily = "calibri"),
   gp_varnames = gpar(fontsize = 16, fontfamily = "calibri")
))

On Wed, Mar 30, 2011 at 3:25 PM, Erich Neuwirth
  wrote:

Achim
I simply want to replace the font R uses on mosaic (whatever it is)
by a font of my choice (say Calibri or Arial)
because I need to embed the R charts in a PowerPoint
presentation and want the fonts to match.
And I want the most simple way of accomplishing this.
I worked my way through the strucplot vignette,
but I could not extract enough information there.
Is there some information about the proper font names
to use in R?



Personally, I simply change the size of the device I'm plotting on. When
I plot on a large device, the fonts will be relatively smaller, and vice

versa. This is what I do when including graphics in PDF files (papers,
slides, reports, etc.).

For fine control, you can set the arguments of the labeling function
employed. ?strucplot shows that the default is ?labeling_border which
has several arguments. For example you can set the graphical parameters
of the labels (gp_labels) or the graphical parameters of the variable
names (gp_varnames). Both arguments take ?gpar lists ("grid" graphical
parameters). For example you may do



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--
Michael Friendly Email: friendly AT yorku DOT ca
Professor, Psychology Dept.
York University  Voice: 416 736-5115 x66249 Fax: 416 736-5814
4700 Keele StreetWeb:   http://www.datavis.ca
Toronto, ONT  M3J 1P3 CANADA

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Re: [R] VECM with UNRESTRICTED TREND

[Pfaff, Bernhard Dr.]

Well, without further information, I do not know, but try the following
 
library(urca)
example(ca.jo)
trend <- matrix(1:nrow(sjf), ncol = 1)
colnames(trend) <- "trd"
ca.jo(sjf, type = "trace", ecdet = "const", K = 2, spec = "longrun", dumvar = 
trend)
 
Best,
Bernhard
 
 




Von: Grzegorz Konat [mailto:grzegorz.ko...@ibrkk.pl] 
Gesendet: Donnerstag, 31. März 2011 14:40
An: Pfaff, Bernhard Dr.; r-help@r-project.org
Betreff: Re: [R] VECM with UNRESTRICTED TREND


'time' was a trend variable from my.data set. Equivalent to the output 
of the command 'matrix' you just gave me. 


So now I did:


library(urca)
data(my.data)
names(my.data)
attach(my.data)
dat1 <- my.data[, c("dY", "X", "dM")]
mat1 <- matrix(seq(1:nrow(dat1)), ncol = 1)
args('ca.jo')
yxm.vecm <- ca.jo(dat1, type = "trace", ecdet = "const", K = 2, spec = 
"longrun", dumvar=mat1)

and the output is:

Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), , 
drop = FALSE] : 
  non-numeric argument to binary operator
In addition: Warning message:
In ca.jo(dat1, type = "trace", ecdet = "const", K = 2, spec = 
"longrun",  :
No column names in 'dumvar', using prefix 'exo' instead.

What do I do wrong?

Best,
Greg


2011/3/31 Pfaff, Bernhard Dr. 



 

Hello Bernhard, 


thank You so much one again! Now I (more or less) 
understand the idea, but still have problem with its practical application.


I do (somewhat following example 8.1 in your textbook):


library(urca)
data(my.data)
names(my.data)
attach(my.data)
dat1 <- my.data[, c("dY", "X", "dM")]
dat2 <- cbind(time)
 
What is 'time'? Just employ matrix(seq(1:nrow(dat1)), 
ncol = 1) for creating the trend variable.
 
Best,
Bernhard
 
 
 args('ca.jo')
yxm.vecm <- ca.jo(dat1, type = "trace", ecdet = 
"trend", K = 2, spec = "longrun", dumvar=dat2)

The above code produces following output:

Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - 
lag + 1L), , drop = FALSE] : 
  non-numeric argument to binary operator

What does that mean? Should I use cbind command to dat1 
as well? And doesn't it transform the series into series of integer numbers?

Thank you once again (especially for your patience).

Best,
Greg



2011/3/31 Pfaff, Bernhard Dr. 



Hello Greg,
 
you include your trend as a (Nx1) matrix and 
use this for 'dumvar'. The matrix 'dumvar' is just added to the VECM as 
deterministic regressors and while you are referring to case 5, this is 
basically what you are after, if I am not mistaken. But we aware that this 
implies a quadratic trend for the levels.
 
Best,
Bernhard




Von: Grzegorz Konat 
[mailto:grzegorz.ko...@ibrkk.pl] 
Gesendet: Mittwoch, 30. März 2011 20:50
An: Pfaff, Bernhard Dr.; 
r-help@r-project.org
Betreff: Re: [R] VECM with UNRESTRICTED 
TREND


Hello Bernhard, 

Thank You very much. Unfortunately I'm 
still not really sure how should I use dummy vars in this context...
If I have a system of three variables 
(x, y, z), lag order = 2 and 1 cointegrating relation, what should I do? I 
mean, what kind of 'pattern' should be used to create those dummy variables, 
what should they represent and how many of them do I need?

Many thanks in advance!

Best,
Greg
   

Re: [R] That dreaded floating point trap

[Alexander Engelhardt]

Am 31.03.2011 14:41, schrieb Sarah Goslee:

On Thu, Mar 31, 2011 at 8:14 AM, Alexander Engelhardt

this helps, thank you.
But if this code is in a function, and some user supplies a vector, I will
still have to round it in the function, I guess.

It's weird how 0.1 is different from round(0.1, digits=1) , but I don't want
to read that 90 page long floating point paper which was referenced
somewhere :)



Or you could try the much shorter R FAQ 7.31. Turns out it
isn't weird at all, if you are a computer.


You're a computer! :)

But yes.. the FAQ entry was where I found all.equal and the referenced 
90-page-paper. But I didn't find out how to do a subset with 'somevector 
> 0.4'.


I think I'll have to round the numbers every time now.. or use some 
other not-so-pretty workaround like 'somevector > 0.4 - 0.05' for 
0.1-binned data.


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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Lists of tables and conditional statements

[Herbert, Alan G]

Thanks Henrique - that worked  like a charm - I had tried lots of other 
combinations before seeing your reply - wished I had asked sooner!

Alan 

-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net] 
Sent: Wednesday, March 30, 2011 10:09 PM
To: Herbert, Alan G
Cc: r-help@r-project.org Help
Subject: Re: [R] Lists of tables and conditional statements


On Mar 30, 2011, at 7:27 PM, Henrique Dallazuanna wrote:

> Try this:
>
> lapply(l, function(x)x[x[,'Sum'] == 3,])

If this is the right answer, you should send a "solved" message. The dput 
extract was incomplete.

--
David.
>
> On Wed, Mar 30, 2011 at 7:38 PM, Herbert, Alan G 
> wrote:
>> Hi R-users,
>>
>> I have a list containing numeric tables of differing row length. I 
>> want to make a new list that contains only rows from tables with a 
>> "Sum" greater than 3, plus the names of each table. I was wondering 
>> whether there is an elegant way to do this using apply of related 
>> functions as this list has many thousands of such tables.
>>
>> Here is an example of the list
>>
>>
>> $AACS
>>
>>POOL
>>
>> INFO pool1 pool2 pool6 pool7 pool8 pool.all Sum
>>
>>  12:125561133:novel 0 0 0 0 10   1
>>
>>  12:125570904:novel 0 0 0 0 10   1
>>
>>  12:125571014:novel 0 1 0 0 00   1
>>
>>  12:125571038:novel 0 0 0 1 00   1
>>
>>  12:125575996:novel 0 0 0 1 00   1
>>
>>  12:125591844:rs2297478 1 0 1 0 01   3
>>
>>  12:125599114:novel 0 0 0 1 00   1
>>
>>  12:125612668:novel 0 0 0 0 10   1
>>
>>  12:125612839:rs900411  1 0 1 0 11   4
>>
>>  12:125626650:novel 0 0 0 0 10   1
>>
>>  12:125626737:novel 0 0 0 1 00   1
>>
>>
>>
>> $AADAC
>>
>>POOL
>>
>> INFO pool1 pool2 pool5 pool6 pool7 pool8  
>> pool.all Sum
>>
>>  3:151542411:novel  0 0 0 0 1 0 
>> 1   2
>>
>>  3:151542412:novel  0 0 0 0 1 0 
>> 1   2
>>
>>  3:151542643:novel  0 1 0 0 0 0 
>> 0   1
>>
>>  3:151545322:rs2410836  0 1 0 0 0 0 
>> 1   2
>>
>>  3:151545323:rs62272918 0 1 0 0 0 0 
>> 1   2
>>
>>  3:151545509:novel  0 0 1 0 0 0 
>> 1   2
>>
>>  3:151545601:rs1803155  1 1 1 1 1 1 
>> 1   7
>>
>>  3:151545721:novel  0 0 1 0 0 0 
>> 0   1
>>
>>  3:151545802:novel  0 0 0 0 1 0 
>> 0   1
>>
>>  3:151545824:novel  0 1 0 0 0 0 
>> 0   1
>>
>>
>> Thanks for your help
>>
>>[[alternative HTML version deleted]]
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide 
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
>
> --
> Henrique Dallazuanna
> Curitiba-Paraná-Brasil
> 25° 25' 40" S 49° 16' 22" O
>
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> http://www.R-project.org/posting-guide.html
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David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] That dreaded floating point trap

[Kenn Konstabel]

n Thu, Mar 31, 2011 at 3:56 PM, Alexander Engelhardt
 wrote:
> Am 31.03.2011 14:41, schrieb Sarah Goslee:
>>
>> On Thu, Mar 31, 2011 at 8:14 AM, Alexander Engelhardt
>>>
>>> this helps, thank you.
>>> But if this code is in a function, and some user supplies a vector, I
>>> will
>>> still have to round it in the function, I guess.
>>>
>>> It's weird how 0.1 is different from round(0.1, digits=1) , but I don't
>>> want
>>> to read that 90 page long floating point paper which was referenced
>>> somewhere :)
>>
>>
>> Or you could try the much shorter R FAQ 7.31. Turns out it
>> isn't weird at all, if you are a computer.
>
> You're a computer! :)
>
> But yes.. the FAQ entry was where I found all.equal and the referenced
> 90-page-paper. But I didn't find out how to do a subset with 'somevector >
> 0.4'.
>
> I think I'll have to round the numbers every time now.. or use some other
> not-so-pretty workaround like 'somevector > 0.4 - 0.05' for 0.1-binned data.

You could define your own %>% and %<% to do rounding and comparison in
one step.

Kenn

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Re: [R] VECM with UNRESTRICTED TREND

[Grzegorz Konat]

The code you gave me works fine with Finland, but the same for my data -
does not!
I do:

library(urca)
data(my.data)
dat1 <- my.data[, c("dY", "X", "dM")]
trend <- matrix(1:nrow(dat1), ncol = 1)
colnames(trend) <- "trd"
yxm.vecm <- ca.jo(dat1, type = "trace", ecdet = "const", K = 2, spec =
"longrun", dumvar = trend)

and the result is again:

Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), , drop =
FALSE] :
  non-numeric argument to binary operator

I attach my dataset in xls format. If you have 5 minutes and wish to check
it out, I'd be extremely grateful!

Best,
Greg



2011/3/31 Pfaff, Bernhard Dr. 

>  Well, without further information, I do not know, but try the following
>
> library(urca)
> example(ca.jo)
> trend <- matrix(1:nrow(sjf), ncol = 1)
> colnames(trend) <- "trd"
> ca.jo(sjf, type = "trace", ecdet = "const", K = 2, spec = "longrun",
> dumvar = trend)
>
> Best,
> Bernhard
>
>
>
>  --
> *Von:* Grzegorz Konat [mailto:grzegorz.ko...@ibrkk.pl]
> *Gesendet:* Donnerstag, 31. März 2011 14:40
>
> *An:* Pfaff, Bernhard Dr.; r-help@r-project.org
> *Betreff:* Re: [R] VECM with UNRESTRICTED TREND
>
> 'time' was a trend variable from my.data set. Equivalent to the output of
> the command 'matrix' you just gave me.
>
> So now I did:
>
>  library(urca)
> data(my.data)
> names(my.data)
> attach(my.data)
> dat1 <- my.data[, c("dY", "X", "dM")]
> mat1 <- matrix(seq(1:nrow(dat1)), ncol = 1)
> args('ca.jo')
> yxm.vecm <- ca.jo(dat1, type = "trace", ecdet = "const", K = 2, spec =
> "longrun", dumvar=mat1)
>
> and the output is:
>
>  Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), , drop
> = FALSE] :
>   non-numeric argument to binary operator
> In addition: Warning message:
> In ca.jo(dat1, type = "trace", ecdet = "const", K = 2, spec = "longrun",
>  :
> No column names in 'dumvar', using prefix 'exo' instead.
>
> What do I do wrong?
>
> Best,
> Greg
>
>
> 2011/3/31 Pfaff, Bernhard Dr. 
>
>>
>>
>>
>>  Hello Bernhard,
>>
>> thank You so much one again! Now I (more or less) understand the idea, but
>> still have problem with its practical application.
>>
>> I do (somewhat following example 8.1 in your textbook):
>>
>>  library(urca)
>> data(my.data)
>> names(my.data)
>> attach(my.data)
>> dat1 <- my.data[, c("dY", "X", "dM")]
>> dat2 <- cbind(time)
>>
>> What is 'time'? Just employ matrix(seq(1:nrow(dat1)), ncol = 1) for
>> creating the trend variable.
>>
>> Best,
>> Bernhard
>>
>>
>>  args('ca.jo')
>> yxm.vecm <- ca.jo(dat1, type = "trace", ecdet = "trend", K = 2, spec =
>> "longrun", dumvar=dat2)
>>
>> The above code produces following output:
>>
>>  Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), ,
>> drop = FALSE] :
>>   non-numeric argument to binary operator
>>
>> What does that mean? Should I use cbind command to dat1 as well? And
>> doesn't it transform the series into series of integer numbers?
>>
>> Thank you once again (especially for your patience).
>>
>> Best,
>> Greg
>>
>>
>>
>> 2011/3/31 Pfaff, Bernhard Dr. 
>>
>>>  Hello Greg,
>>>
>>> you include your trend as a (Nx1) matrix and use this for 'dumvar'. The
>>> matrix 'dumvar' is just added to the VECM as deterministic regressors and
>>> while you are referring to case 5, this is basically what you are after, if
>>> I am not mistaken. But we aware that this implies a quadratic trend for the
>>> levels.
>>>
>>> Best,
>>> Bernhard
>>>
>>>  --
>>> *Von:* Grzegorz Konat [mailto:grzegorz.ko...@ibrkk.pl]
>>> *Gesendet:* Mittwoch, 30. März 2011 20:50
>>> *An:* Pfaff, Bernhard Dr.; r-help@r-project.org
>>> *Betreff:* Re: [R] VECM with UNRESTRICTED TREND
>>>
>>>   Hello Bernhard,
>>>
>>> Thank You very much. Unfortunately I'm still not really sure how should I
>>> use dummy vars in this context...
>>> If I have a system of three variables (x, y, z), lag order = 2 and 1
>>> cointegrating relation, what should I do? I mean, what kind of 'pattern'
>>> should be used to create those dummy variables, what should they represent
>>> and how many of them do I need?
>>>
>>> Many thanks in advance!
>>>
>>> Best,
>>> Greg
>>>
>>> 2011/3/30 Pfaff, Bernhard Dr. 
>>>
 Hello Greg,

 you can exploit the argument 'dumvar' for this. See ?ca.jo

 Best,
 Bernhard

 > -Ursprüngliche Nachricht-
 > Von: r-help-boun...@r-project.org
 > [mailto:r-help-boun...@r-project.org] Im Auftrag von Grzegorz Konat
 > Gesendet: Mittwoch, 30. März 2011 16:46
 > An: r-help@r-project.org
 > Betreff: [R] VECM with UNRESTRICTED TREND
  >
 > Dear All,
 >
 > My question is:
 >
 > how can I estimate VECM system with "unrestricted trend" (aka
 > "case 5") option as a deterministic term?
 >
 > As far as I know, ca.jo in urca package allows for "restricted trend"
 > only [vecm
 > <- ca.jo(data, type = "trace"/"eigen", ecdet = "trend", K =
 > n, spec = "transitory"/"longrun")].
 > Obvio

Re: [R] That dreaded floating point trap

[ONKELINX, Thierry]

Dear Alexander,

Instead of testing 'somevector > 0.4', test 'abs(somevector - 0.4) < 
some.small.number'

Best regards,

Thierry


ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek
team Biometrie & Kwaliteitszorg
Gaverstraat 4
9500 Geraardsbergen
Belgium

Research Institute for Nature and Forest
team Biometrics & Quality Assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium

tel. + 32 54/436 185
thierry.onkel...@inbo.be
www.inbo.be

To call in the statistician after the experiment is done may be no more than 
asking him to perform a post-mortem examination: he may be able to say what the 
experiment died of.
~ Sir Ronald Aylmer Fisher

The plural of anecdote is not data.
~ Roger Brinner

The combination of some data and an aching desire for an answer does not ensure 
that a reasonable answer can be extracted from a given body of data.
~ John Tukey
  

> -Oorspronkelijk bericht-
> Van: r-help-boun...@r-project.org 
> [mailto:r-help-boun...@r-project.org] Namens Alexander Engelhardt
> Verzonden: donderdag 31 maart 2011 14:57
> Aan: r-help@r-project.org
> Onderwerp: Re: [R] That dreaded floating point trap
> 
> Am 31.03.2011 14:41, schrieb Sarah Goslee:
> > On Thu, Mar 31, 2011 at 8:14 AM, Alexander Engelhardt
> >> this helps, thank you.
> >> But if this code is in a function, and some user supplies 
> a vector, I 
> >> will still have to round it in the function, I guess.
> >>
> >> It's weird how 0.1 is different from round(0.1, digits=1) , but I 
> >> don't want to read that 90 page long floating point paper 
> which was 
> >> referenced somewhere :)
> >
> >
> > Or you could try the much shorter R FAQ 7.31. Turns out it 
> isn't weird 
> > at all, if you are a computer.
> 
> You're a computer! :)
> 
> But yes.. the FAQ entry was where I found all.equal and the 
> referenced 90-page-paper. But I didn't find out how to do a 
> subset with 'somevector  > 0.4'.
> 
> I think I'll have to round the numbers every time now.. or 
> use some other not-so-pretty workaround like 'somevector > 
> 0.4 - 0.05' for 0.1-binned data.
> 
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide 
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
> 
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[R] how to do t-test in r for difference of mean

[arkajyoti jana]

I am trying to do t-test to test whether the mean of one one column of the
data frame is greater then another. please help me out.

-- 
Arkajyoti Jana
M. Phil/ 2nd semester
Centre for Economic Studies and planning
School of Social Sciences
Jawaharlal Nehru University
New Delhi-67

[[alternative HTML version deleted]]

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Re: [R] summing values by week - based on daily dates - but with somedates missing

[Dimitri Liakhovitski]

Thank you so much, everyone, for your help.
Extremely valuable suggestions and extremely valuable learnings!
Dimitri

On Thu, Mar 31, 2011 at 5:03 AM, Martyn Byng  wrote:
> Hi,
>
> Yep, that was what it was doing. For a sum across week, try something like
>
> get.week.flag <- function(dd) {
>  ## get weekday from the date in dd and code it as Monday = 1, Tuesday = 2 etc
>  idd = 
> factor(weekdays(dd),levels=c("Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"))
>
>  ## convert to numeric
>  ndd = as.numeric(idd)
>
>  ## flag entries where weekday code gets less (this will flag changes in week)
>  wflag = c(FALSE,(ndd[-length(idd)] > ndd[-1]))
>
>  ## cumulative sum to get the week flag
>  cumsum(wflag) + 1
> }
>
> to get a week flag (this is assuming that your data is sorted by date, if not 
> you'll have to sort it first). If you want the week to start on a different 
> day, just change the ordering of the weekdays in the levels statement.
>
> data.frame(date=myframe$date,day=weekdays(myframe$date),week=get.week.flag(myframe$date))
>
> seems to indicate that the function is doing what it should, so you can then 
> amend the previous code to use get.week.flag instead of weekdays, as in
>
> sum.by.week <- function(ff) {
>  by.day <- split(ff$value,get.week.flag(ff$dates))
>  lapply(by.day,sum)
> }
>
> by.grp <- split(myframe,myframe$group)
> lapply(by.grp,sum.by.week)
>
> Martyn
>
> -Original Message-
> From: Dimitri Liakhovitski [mailto:dimitri.liakhovit...@gmail.com]
> Sent: 30 March 2011 18:03
> To: Martyn Byng
> Cc: r-help
> Subject: Re: [R] summing values by week - based on daily dates - but with 
> somedates missing
>
> Thank you, Martyn.
> But it looks like this way we are getting sums by day - i.e., across
> all Mondays, all Tuesdays, etc.
> Maybe I did not explain well, sorry! The desired output would contain
> sums for each WHOLE week - across all days that comprise that week -
> Monday through Sunday.
> Makes sense?
> Dimitri
>
> On Wed, Mar 30, 2011 at 12:53 PM, Martyn Byng  wrote:
>> Hi,
>>
>> How about something like:
>>
>> sum.by.day <- function(ff) {
>>  by.day <- split(ff$value,weekdays(ff$dates))
>>  lapply(by.day,sum)
>> }
>>
>> by.grp <- split(myframe,myframe$group)
>>
>> lapply(by.grp,sum.by.day)
>>
>>
>> Martyn
>>
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
>> On Behalf Of Dimitri Liakhovitski
>> Sent: 30 March 2011 15:23
>> To: r-help
>> Subject: [R] summing values by week - based on daily dates - but with
>> somedates missing
>>
>> Dear everybody,
>>
>> I have the following challenge. I have a data set with 2 subgroups,
>> dates (days), and corresponding values (see example code below).
>> Within each subgroup: I need to aggregate (sum) the values by week -
>> for weeks that start on a Monday (for example, 2008-12-29 was a
>> Monday).
>> I find it difficult because I have missing dates in my data - so that
>> sometimes I don't even have the date for some Mondays. So, I can't
>> write a proper loop.
>> I want my output to look something like this:
>> group   dates   value
>> group.1 2008-12-29  3.0937
>> group.1 2009-01-05  3.8833
>> group.1 2009-01-12  1.362
>> ...
>> group.2 2008-12-29  2.250
>> group.2 2009-01-05  1.4057
>> group.2 2009-01-12  3.4411
>> ...
>>
>> Thanks a lot for your suggestions! The code is below:
>> Dimitri
>>
>> ### Creating example data set:
>> mydates<-rep(seq(as.Date("2008-12-29"), length = 43, by = "day"),2)
>> myfactor<-c(rep("group.1",43),rep("group.2",43))
>> set.seed(123)
>> myvalues<-runif(86,0,1)
>> myframe<-data.frame(dates=mydates,group=myfactor,value=myvalues)
>> (myframe)
>> dim(myframe)
>>
>> ## Removing same rows (dates) unsystematically:
>> set.seed(123)
>> removed.group1<-sample(1:43,size=11,replace=F)
>> set.seed(456)
>> removed.group2<-sample(44:86,size=11,replace=F)
>> to.remove<-c(removed.group1,removed.group2);length(to.remove)
>> to.remove<-to.remove[order(to.remove)]
>> myframe<-myframe[-to.remove,]
>> (myframe)
>>
>>
>>
>> --
>> Dimitri Liakhovitski
>> Ninah Consulting
>> www.ninah.com
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>> 
>> This e-mail has been scanned for all viruses by Star.
>> 
>>
>> 
>> The Numerical Algorithms Group Ltd is a company registered in England
>> and Wales with company number 1249803. The registered office is:
>> Wilkinson House, Jordan Hill Road, Oxford OX2 8DR, United Kingdom.
>>
>> This e-mail has been scanned for all viruses by Star. The service is
>> powere

Re: [R] how about a "

[Duncan Murdoch]

On 11-03-30 7:00 PM, Carl Witthoft wrote:

I was cursing Matlab again today (what else is new) because the default
action for every Matlab command is to spew the result to the console,
and one must remember to put that darn ";"  at the end of every line.

So I just wondered:  was there ever a discussion as to providing some
modified version of the "<-" and "->" operators in R to do the reverse?
   That is, since R does not print the values of a command to the
console,  what if there were  an operator such that


   newobject

Others have given alternatives.  Just some comments on this particular 
proposal:


We already have a problem that

x<-3

is slightly ambiguous:  does it mean

x <- 3   # yes

or

x < -3   # no

I think the compound operator mean?  If the amount of typing in print(x <- 3) is a problem, that's a 
user interface issue, so it should be addressed at the user interface 
level, not by changing the language.  A front-end could make it easy to 
transform x <- 3 into the longer expression (or an equivalent), or just 
make it easy to examine the .Last.value variable.


Duncan Murdoch

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Re: [R] That dreaded floating point trap

[Duncan Murdoch]

On 11-03-31 7:24 AM, Alexander Engelhardt wrote:

Hi,
I had a piece of code which looped over a decimal vector like this:


for( i in where ){
thisdata<- subset(herde, herde$mlr>= i)
# do stuff with thisdata..
}

'where' is a vector like seq(-1, 1, by=0.1)


The solution to this problem is to take steps by representable numbers, 
not by numbers like 0.1 that can't be represented exactly.  For example, 
seq(-1, 1, by=0.25) has exact entries, because fractions with small 
powers of 2 in the denominator are all exactly representable.  ("Small" 
depends on the numerator, but for fractions between 0 and 1 it's about 
52, so not really so small.)


Duncan Murdoch




My problem was: 'nrow(thisdata)' in loop repetition 0.4 was different if
'where' was seq(-1, 1, by=0.1) than when 'where' was seq(-0.8, 1, by=0.1)
It went away after I changed the first line to:

thisdata<- subset(herde, herde$mlr>= round(i, digits=1))

This is that "floating point trap" the R inferno pdf talked about,
right? That file talked about the problem, but didn't offer a solution.

Similar things happened when I created a table() from a vector with
values in seq(-1, 1, by=0.1)

Do I really have to round every float at every occurence from now on, or
is there another solution? I only found all.equal() and identical(), but
I want to subset for observations with a value /greater/ than something.

Thanks in advance,
   Alex

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Re: [R] Simple lattice question

[Peter Ehlers]

On 2011-03-31 03:39, Rubén Roa wrote:


DeaR ComRades,

require(lattice)
data<- data.frame(SP=sort(rep(as.factor(c('A','B','C','D','E')),12)),
x=rpois(60,10),
y=rep(c(rep(0,4),rep(10,4),rep(20,4)),5),
z=rep(1:4,15))
xyplot(x~y|SP,data=data,groups=z,layout=c(2,3),pch=1:4,lty=1:4,col='black',type='b')

How do I put a legend for the grouping variable in the empty upper-right panel?


See the help page for xyplot for an example using the iris data.
You just need to add something like

 auto.key = list(x = .6, y = .7, corner = c(0, 0))

to your lattice call. See the 'key' entry on the ?xyplot page.

Peter Ehlers



TIA

Rubén



Dr. Rubén Roa-Ureta
AZTI - Tecnalia / Marine Research Unit
Txatxarramendi Ugartea z/g
48395 Sukarrieta (Bizkaia)
SPAIN

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[R] another statistical question

[Anna Lee]

Dear List!

I have a unverse (basic population) which is not normally distributed.
Now from this universe I take some subsets. Each subset is normally
distributed within itself. I now want to compare the subsets and see
if they differ significantly. So what is my assumption - normal
distributed data and therefore comparison of means, or non normal
distributed data and therefore comparison of medians?


Cheers, Anna
-- 



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Re: [R] Simple lattice question

[Rubén Roa]

Thanks Peters!

Just a few minor glitches now:

require(lattice)
data <- data.frame(SP=sort(rep(as.factor(c('A','B','C','D','E')),12)),
   x=rpois(60,10),
   y=rep(c(rep(0,4),rep(10,4),rep(20,4)),5),
   z=rep(1:4,15))
xyplot(x~y|SP,data=data,groups=z,layout=c(2,3),pch=1:4,lty=1:4,col='black',type='b',
   key=list(x = .65, y = .75, corner = c(0, 0), points=TRUE,
  lines=TRUE, pch=1:4, lty=1:4, type='b',
  text=list(lab = as.character(unique(data$z)

I have an extra column of symbols on the legend,

and,

would like to add a title to the legend. Such as 'main' for plots.

Any suggestions?

Rubén


 

Dr. Rubén Roa-Ureta
AZTI - Tecnalia / Marine Research Unit
Txatxarramendi Ugartea z/g
48395 Sukarrieta (Bizkaia)
SPAIN



> -Mensaje original-
> De: Peter Ehlers [mailto:ehl...@ucalgary.ca] 
> Enviado el: jueves, 31 de marzo de 2011 15:41
> Para: Rubén Roa
> CC: r-help@r-project.org
> Asunto: Re: [R] Simple lattice question
> 
> On 2011-03-31 03:39, Rubén Roa wrote:
> >
> > DeaR ComRades,
> >
> > require(lattice)
> > data<- 
> data.frame(SP=sort(rep(as.factor(c('A','B','C','D','E')),12)),
> > x=rpois(60,10),
> > y=rep(c(rep(0,4),rep(10,4),rep(20,4)),5),
> > z=rep(1:4,15))
> > 
> xyplot(x~y|SP,data=data,groups=z,layout=c(2,3),pch=1:4,lty=1:4,col='bl
> > ack',type='b')
> >
> > How do I put a legend for the grouping variable in the 
> empty upper-right panel?
> 
> See the help page for xyplot for an example using the iris data.
> You just need to add something like
> 
>   auto.key = list(x = .6, y = .7, corner = c(0, 0))
> 
> to your lattice call. See the 'key' entry on the ?xyplot page.
> 
> Peter Ehlers
> 
> >
> > TIA
> >
> > Rubén
> >
> > 
> __
> > __
> >
> > Dr. Rubén Roa-Ureta
> > AZTI - Tecnalia / Marine Research Unit Txatxarramendi Ugartea z/g
> > 48395 Sukarrieta (Bizkaia)
> > SPAIN
> >
> > __
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide 
> > http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> 
> 

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] another statistical question

[Alexander Engelhardt]

Am 31.03.2011 15:46, schrieb Anna Lee:

Dear List!

I have a unverse (basic population) which is not normally distributed.
Now from this universe I take some subsets. Each subset is normally
distributed within itself. I now want to compare the subsets and see
if they differ significantly. So what is my assumption - normal
distributed data and therefore comparison of means, or non normal
distributed data and therefore comparison of medians?


If you want to do an ANOVA, the assumption is normality /within/ the 
groups. That is, Y given X=x is normal distributed. Also, you want the 
same variance within each group (group = your subset = factor value).


Short answer: Means.
(I am only 95% certain)

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[R] choosing best 'match' for given factor

[Murali.Menon]

Folks,

I have a 'matching' matrix between variables A, X, L, O:

> a <- structure(c(1, 0.41, 0.58, 0.75, 0.41, 1, 0.6, 0.86, 0.58, 
0.6, 1, 0.83, 0.75, 0.86, 0.83, 1), .Dim = c(4L, 4L), .Dimnames = list(
c("A", "X", "L", "O"), c("A", "X", "L", "O")))

> a
  A X L O
A  1.00  0.41  0.58  0.75
X  0.41  1.00  0.60  0.86
L  0.58  0.75  1.00  0.83
O  0.60  0.86  0.83  1.00

And I have a search vector of variables

> v <- c("X", "O")

I want to write a function bestMatch(searchvector, matchMat) such that for each 
variable in searchvector, I get the variable that it has the highest match to - 
but searching only among variables to the left of it in the 'matching' matrix, 
and not matching with any variable in searchvector itself.

So in the above example, although "X" has the highest match (0.86) with "O", I 
can't choose "O" as it's to the right of X (and also because "O" is in the 
searchvector v already); I'll have to choose "A".

For "O", I will choose "L", the variable it's best matched with - as it can't 
match "X" already in the search vector.

My function bestMatch(v, a) will then return c("A", "L")

My matrix a is quite large, and I have a long list of search vectors v, so I 
need an efficient method.

I wrote this:

bestMatch <- function(searchvector,  matchMat) {
sapply(searchvector, function(cc) {
 y <- matchMat[!(rownames(matchMat) %in% 
searchvector) & (index(rownames(matchMat)) < match(cc, rownames(matchMat))), 
cc, drop = FALSE];
 rownames(y)[which.max(y)]
})   
}

Any advice?

Thanks,

Murali

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[R] error in recode.defalt ....object '.data' not found

[Simon Kiss]

Dear colleagues, working with the data frame below, trying to reverse two 
variables I the error message below.
i searched through the help list but could not find any postings which could 
help me solve the situation. I tried attaching and detaching the data frame to 
no avail.
Yours, Simon Kiss

*DATA FRAME
'data.frame':   1569 obs. of  9 variables:
 $ equal : num  3 4 3 2 3 4 2 3 2 2 ...
 $ disc  : num  3 2 3 3 2 2 3 3 3 3 ...
 $ family: num  3 2 2 2 3 2 2 1 2 1 ...
 $ special   : num  3 3 4 4 3 3 4 4 3 4 ...
 $ immigrants: num  3 8 3 8 3 3 4 1 1 2 ...
 $ wedlock   : num  3 3 3 3 3 2 2 8 2 3 ...
 $ crime : num  3 2 2 1 2 3 1 8 2 1 ...
 $ breakdown : num  3 3 3 2 2 4 8 2 2 4 ...
 $ nonwhites : num  2 4 3 3 2 2 3 4 3 3 ...

*RECODE
social$nonwhites<-recode(social$nonwhites, "1=4; 2=3; 3=2; 4=1; 8=NA; -9=NA")

*ERROR
Error in recode.default(social$nonwhites, "1=4; 2=3; 3=2; 4=1; 8=NA; -9=NA") : 
  object '.data' not found


*
Simon J. Kiss, PhD
Assistant Professor, Wilfrid Laurier University
73 George Street
Brantford, Ontario, Canada
N3T 2C9
Cell: +1 519 761 7606

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Re: [R] error in recode.defalt ....object '.data' not found

[John Fox]

Dear Simon,

Your intention appears to be to call the recode() function in the car package. 
That function is not generic, and consequently has no default method, leading 
me to believe that you've called a recode() function in some other package.

I hope this helps,
 John


John Fox
Sen. William McMaster Prof. of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox/

On Thu, 31 Mar 2011 11:01:42 -0400
 Simon Kiss  wrote:
> Dear colleagues, working with the data frame below, trying to reverse two 
> variables I the error message below.
> i searched through the help list but could not find any postings which could 
> help me solve the situation. I tried attaching and detaching the data frame 
> to no avail.
> Yours, Simon Kiss
> 
> *DATA FRAME
> 'data.frame': 1569 obs. of  9 variables:
>  $ equal : num  3 4 3 2 3 4 2 3 2 2 ...
>  $ disc  : num  3 2 3 3 2 2 3 3 3 3 ...
>  $ family: num  3 2 2 2 3 2 2 1 2 1 ...
>  $ special   : num  3 3 4 4 3 3 4 4 3 4 ...
>  $ immigrants: num  3 8 3 8 3 3 4 1 1 2 ...
>  $ wedlock   : num  3 3 3 3 3 2 2 8 2 3 ...
>  $ crime : num  3 2 2 1 2 3 1 8 2 1 ...
>  $ breakdown : num  3 3 3 2 2 4 8 2 2 4 ...
>  $ nonwhites : num  2 4 3 3 2 2 3 4 3 3 ...
> 
> *RECODE
> social$nonwhites<-recode(social$nonwhites, "1=4; 2=3; 3=2; 4=1; 8=NA; -9=NA")
> 
> *ERROR
> Error in recode.default(social$nonwhites, "1=4; 2=3; 3=2; 4=1; 8=NA; -9=NA") 
> : 
>   object '.data' not found
> 
> 
> *
> Simon J. Kiss, PhD
> Assistant Professor, Wilfrid Laurier University
> 73 George Street
> Brantford, Ontario, Canada
> N3T 2C9
> Cell: +1 519 761 7606
> 
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

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[R] AEM package

[Fernanda Melo Carneiro]

I'm trying to do AEM eigenfunctions, but I have a different number of points 
into each reservoir (e.g.: reservoir 1-4 points, reservoir 2-6 points, 
reservoir 3-2 points). Can someone help me?Besides this in the Legendre's 
book exemple in the part:>length.edge<-vector(length=nrow(edges.b))>for (i in 
1:nrow(edges.b)){length.edge[i]<-D1.mat[edges.b[i,1], edges.b[i,2]]}There is 
some problem with the "lenght" argument.Thanks 
Fernanda Melo Carneiro contato: (62) 3521-1480 e 8121-7374www.ecoevol.ufg.br
Laboratório de Ecologia Teórica e Síntese (UFG)  
[[alternative HTML version deleted]]

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] fonts in mosaic

[Erich Neuwirth]

The easiest way of changing the font used for labels by the windows
graphics device (opened by a call to windows()) seems to be the following:

Let us assume we want to use the font Consolas for all labels:

windowsFonts(myfont="Consolas")
par(family="myfont")

If one later on wants to change the default font for graphics,
one has to run both commands again. Just issueing a new windowsFonts
command will not change the font used by the graphics device.

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Re: [R] ANCOVA for linear regressions without intercept

[Bert Gunter]

If you haven't already received an answer, a careful reading of

?formula

will provide it.

-- Bert

On Wed, Mar 30, 2011 at 11:42 PM, Yusuke Fukuda wrote:

>
> Hello R experts
>
> I have two linear regressions for sexes (Male, Female, Unknown). All have a
> good correlation between body length (response variable) and head length
> (explanatory variable). I know it is not recommended, but for a good
> practical reason (the purpose of study is to find a single conversion factor
> from head length to body length), the regressions need to go through the
> origin (0 intercept).
>
> Is it possible to do ANCOVA for these regressions without intercepts? When
> I do
>
> summary(lm(body length ~ sex*head length))
>
> this will include the intercepts as below
>
>
> Coefficients:
>   Estimate Std. Error t value Pr(>|t|)
> (Intercept)-6.496971.68497  -3.856 0.000118 ***
> sexMale-9.393401.97760  -4.750 2.14e-06 ***
> sexUnknown -1.337912.35453  -0.568 0.569927
> head_length 7.123070.05503 129.443  < 2e-16 ***
> sexMale:head_length 0.316310.06246   5.064 4.37e-07 ***
> sexUnknown:head_length  0.199370.07022   2.839 0.004556 **
> ---
>
> Is there any way I can remove the intercepts so that I can simply compare
> the slopes with no intercept taken into account?
>
> Thanks for help in advance.
>
> Yusuke Fukuda
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
"Men by nature long to get on to the ultimate truths, and will often be
impatient with elementary studies or fight shy of them. If it were possible
to reach the ultimate truths without the elementary studies usually prefixed
to them, these would not be preparatory studies but superfluous diversions."

-- Maimonides (1135-1204)

Bert Gunter
Genentech Nonclinical Biostatistics

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] how to do t-test in r for difference of mean

[Ista Zahn]

Hi Arkajyoti,
Please learn to help yourself. Start with

help.search("t-test")

Also read one of the many introductory R tutorials on the web.

Best,
Ista

On Thu, Mar 31, 2011 at 9:07 AM, arkajyoti jana  wrote:
> I am trying to do t-test to test whether the mean of one one column of the
> data frame is greater then another. please help me out.
>
> --
> Arkajyoti Jana
> M. Phil/ 2nd semester
> Centre for Economic Studies and planning
> School of Social Sciences
> Jawaharlal Nehru University
> New Delhi-67
>
>        [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Ista Zahn
Graduate student
University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.org

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Re: [R] choosing best 'match' for given factor

[Nick Sabbe]

Hi Murali.
I haven't compared, but this is what I would do:

bestMatch<-function(searchVector, matchMat)
{
searchRow<-unique(sort(match(searchVector, colnames(matchMat #if
you're sure, you could drop unique
cat("Original row indices:")
print(searchRow)
matchMat<-matchMat[, -searchRow, drop=FALSE] #avoid duplicates
altogether
cat("Corrected Matrix:\n")
print(matchMat)
correctedRows<-searchRow - seq_along(searchRow) + 1 #works because
of the sort above
cat("Corrected row indices:")
print(correctedRows)
sapply(correctedRows, function(cr){
lookWhere<-matchMat[cr, seq(cr-1)]
cat("Will now look into:\n")
print(lookWhere)
cc<-which.max(lookWhere)
cat("Max at position", cc, "\n")
colnames(matchMat)[cc]
})
}
I don't think there's that much difference. Depending on specific sizes, it
may be more or less costly to first shrink the search matrix like I do. And
similarly depending, I may be better still if you remove the rows that
you're not interested in as well (some more but similar index trickery
required then.

HTH,


Nick Sabbe
--
ping: nick.sa...@ugent.be
link: http://biomath.ugent.be
wink: A1.056, Coupure Links 653, 9000 Gent
ring: 09/264.59.36

-- Do Not Disapprove





-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of murali.me...@avivainvestors.com
Sent: donderdag 31 maart 2011 16:46
To: r-help@r-project.org
Subject: [R] choosing best 'match' for given factor

Folks,

I have a 'matching' matrix between variables A, X, L, O:

> a <- structure(c(1, 0.41, 0.58, 0.75, 0.41, 1, 0.6, 0.86, 0.58, 
0.6, 1, 0.83, 0.75, 0.86, 0.83, 1), .Dim = c(4L, 4L), .Dimnames = list(
c("A", "X", "L", "O"), c("A", "X", "L", "O")))

> a
  A X L O
A  1.00  0.41  0.58  0.75
X  0.41  1.00  0.60  0.86
L  0.58  0.75  1.00  0.83
O  0.60  0.86  0.83  1.00

And I have a search vector of variables

> v <- c("X", "O")

I want to write a function bestMatch(searchvector, matchMat) such that for
each variable in searchvector, I get the variable that it has the highest
match to - but searching only among variables to the left of it in the
'matching' matrix, and not matching with any variable in searchvector
itself.

So in the above example, although "X" has the highest match (0.86) with "O",
I can't choose "O" as it's to the right of X (and also because "O" is in the
searchvector v already); I'll have to choose "A".

For "O", I will choose "L", the variable it's best matched with - as it
can't match "X" already in the search vector.

My function bestMatch(v, a) will then return c("A", "L")

My matrix a is quite large, and I have a long list of search vectors v, so I
need an efficient method.

I wrote this:

bestMatch <- function(searchvector,  matchMat) {
sapply(searchvector, function(cc) {
 y <- matchMat[!(rownames(matchMat) %in%
searchvector) & (index(rownames(matchMat)) < match(cc, rownames(matchMat))),
cc, drop = FALSE];
 rownames(y)[which.max(y)]
})   
}

Any advice?

Thanks,

Murali

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and provide commented, minimal, self-contained, reproducible code.

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] another statistical question

[Peter Ehlers]

On 2011-03-31 07:44, Alexander Engelhardt wrote:

Am 31.03.2011 15:46, schrieb Anna Lee:

Dear List!

I have a unverse (basic population) which is not normally distributed.
Now from this universe I take some subsets. Each subset is normally
distributed within itself. I now want to compare the subsets and see
if they differ significantly. So what is my assumption - normal
distributed data and therefore comparison of means, or non normal
distributed data and therefore comparison of medians?


If you want to do an ANOVA, the assumption is normality /within/ the
groups. That is, Y given X=x is normal distributed. Also, you want the
same variance within each group (group = your subset = factor value).


The equal-variances assumption is not strictly required.
R has the multiple-groups equivalent of the two-sample t-test.
See ?oneway.test.

Peter Ehlers



Short answer: Means.
(I am only 95% certain)

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and provide commented, minimal, self-contained, reproducible code.


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and provide commented, minimal, self-contained, reproducible code.

Re: [R] How to put line linking two plots

[Greg Snow]

Here is a way to do it using just base graphics:

layout(matrix(c(1,1,2,3), 2, 2, byrow = TRUE))
plot(runif(10), type='b', ylim=c(0,1))
x.tmp <- grconvertX(4, to='ndc')
y.tmp <- grconvertY(0.9, to='ndc')
plot(runif(20), type='l', ylim=c(0,1))
par(xpd=NA)
segments( 10, 1,
 grconvertX(x.tmp,  from='ndc'), grconvertY(y.tmp, from='ndc'),
col='red' )
plot(runif(20), type='l')



-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111


> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Paul Murrell
> Sent: Wednesday, March 30, 2011 4:29 PM
> To: Mario Valle
> Cc: R-help@r-project.org
> Subject: Re: [R] How to put line linking two plots
> 
> Hi
> 
> On 30/03/2011 10:54 p.m., Mario Valle wrote:
> > Hello!
> > Suppose I have three charts like below. The top chart is a general
> > overview and the bottom charts are related so some point of this
> chart.
> > To make clear this relationship I want to draw a line between (4,0.9)
> in
> > the top chart and (10,1) in the bottom-left one.
> > Currently I add it manually using Inkscape on the resulting pdf file.
> > Is it possible to add it inside R? Should I switch to other charting
> > packages?
> 
> You'll have your work cut out using traditional graphics, but this is
> doable in grid-based graphics.  For example, ...
> 
> library(grid)
> library(lattice)
> 
> set.seed(123)
> print(xyplot(runif(10)~1:10, type="b"),
>   position=c(0, .5, 1, 1),
>   prefix="top",
>   more=TRUE)
> print(xyplot(runif(20)~1:20, type="l"),
>   position=c(0, 0, .5, .5),
>   prefix="left",
>   more=TRUE)
> print(xyplot(runif(20)~1:20, type="l"),
>   position=c(.5, 0, 1, .5),
>   prefix="right")
> trellis.focus("panel", 1, 1, prefix="top")
> grid.move.to(unit(4, "native"), unit(.9, "native"))
> trellis.unfocus()
> trellis.focus("panel", 1, 1, prefix="left", clip.off=TRUE)
> grid.line.to(unit(10, "native"), unit(1, "native"))
> trellis.unfocus()
> 
> Paul
> 
> > Thanks for the advice!
> >   mario
> >
> > set.seed(123)
> > pdf("test.pdf", width=14, height=7)
> > layout(matrix(c(1,1,2,3), 2, 2, byrow = TRUE))
> > plot(runif(10), type='b')
> > plot(runif(20), type='l')
> > plot(runif(20), type='l')
> > dev.off()
> >
> > R 2.12.2 on Windows 7 (32bits)
> >
> 
> --
> Dr Paul Murrell
> Department of Statistics
> The University of Auckland
> Private Bag 92019
> Auckland
> New Zealand
> 64 9 3737599 x85392
> p...@stat.auckland.ac.nz
> http://www.stat.auckland.ac.nz/~paul/
> 
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-
> guide.html
> and provide commented, minimal, self-contained, reproducible code.

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Re: [R] Italicized title from index

[Jeremy Newman]

Hi,

Thank you *very* much. I'll remember that in the future!

Cheers,

-Jeremy N

Undergraduate Researcher in Macroecology
University of Ottawa
Department of Biology

Ad astra per alia porci!



On Thu, Mar 31, 2011 at 8:01 AM, Henrique Dallazuanna wrote:

> Better:
>
> plot(1, main = bquote("Yield  for " ~ italic(.(as.character(spp))) ~
> "in management region" ~ .(region)))
>
> You have a factor, so you need convert it to character
>
>
> On Thu, Mar 31, 2011 at 8:59 AM, Henrique Dallazuanna 
> wrote:
> > Try this:
> >
> > plot(1, main = bquote("Yield  for " ~ italic(.(spp)) ~ "in management
> > region" ~ .(region)))
> >
> > On Thu, Mar 31, 2011 at 2:35 AM, Jeremy Newman 
> wrote:
> >> Hi all!
> >>
> >> I've written a handy script that uses a for loop to allow me to generate
> a
> >> large number of figures and statistical outputs for a large dataset.
> >>
> >> I am using indexing to retrieve a species name for the title of my
> graphs-
> >> which worked fine. However, I need to italicize these species names.
> >>
> >> I originally used the paste function, and had no problems with indexing:
> >>
> >> *main=paste("Yield for ", testsub[1,3], " in management region ",
> >> testsub[1,2])*
> >>
> >> The title looks like: "*Yield for Gadus morhua in management region
> 4X5Y*"
> >>
> >> I tried bquote from a related help thread, I tried to emulate it:
> >>
> >> *spp<-testsub[1,3]*
> >> *region<-testsub[1,2]*
> >> *
> >> *
> >> *main=bquote(Yield ~ for ~ italic(.(spp)) ~ in ~ management ~ region ~
> >> .(region))*
> >>
> >> Which doesn't seem to work at all, but when I try not putting anything
> after
> >> the italic:
> >>
> >> *main=bquote(Yield ~ for ~ italic(.(spp)))*
> >>
> >> I get: "*Yield for 1*"
> >> While *spp=Gadus morhua*
> >>
> >> I'm at wit's end, I tried to read about substitute, expression, and eval
> >> functions in the hopes I can figure it out, but I am lost!
> >>
> >> Thanks for any help!
> >>
> >> Cheers,
> >>
> >> -Jeremy N
> >>
> >> Undergraduate Researcher in Macroecology
> >> University of Ottawa
> >> Department of Biology
> >>
> >> Ad astra per alia porci!
> >>
> >>[[alternative HTML version deleted]]
> >>
> >> __
> >> R-help@r-project.org mailing list
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> >> and provide commented, minimal, self-contained, reproducible code.
> >>
> >
> >
> >
> > --
> > Henrique Dallazuanna
> > Curitiba-Paraná-Brasil
> > 25° 25' 40" S 49° 16' 22" O
> >
>
>
>
> --
> Henrique Dallazuanna
> Curitiba-Paraná-Brasil
> 25° 25' 40" S 49° 16' 22" O
>

[[alternative HTML version deleted]]

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Re: [R] Graph many points without hiding some

[Greg Snow]

Just a note, Base graphics does support transparency as long as the device 
plotting to supports it.

-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111


> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Dennis Murphy
> Sent: Thursday, March 31, 2011 1:36 AM
> To: Samuel Dennis
> Cc: R-help@r-project.org
> Subject: Re: [R] Graph many points without hiding some
> 
> Hi:
> 
> I can think of a couple: (1) size reduction of the points; (2) alpha
> transparency; (3)  (1) + (2)
> 
> >From your original plot in base graphics, I reduced cex to 0.2 and it
> didn't
> look too bad:
> 
> plot(rnorm(x,mean=19),rnorm(x),col=3,xlim=c(16,24), cex = 0.2)
> points(rnorm(x,mean=20),rnorm(x),col=1, cex = 0.2)
> points(rnorm(x,mean=21),rnorm(x),col=2, cex = 0.2)
> 
> AFAIK, base graphics doesn't have alpha transparency available, but the
> ggplot2 package does. One approach is to adjust the alpha transparency
> on
> default size points; another is to combine reduced point size with
> alpha
> transparency. Here is your example rehashed for ggplot2.
> 
> require(ggplot2)
> d <- data.frame(x1 = rnorm(1, mean = 19), x2 = rnorm(1, mean =
> 20),
> x3 = rnorm(1, mean = 21), x = rnorm(1))
> # Basically stacking x1 - x3, creating two new vars named variable and
> value
> dm <- melt(d, id = 'x')   # from reshape package, loads with ggplot2
> # Alpha transparency is set to a low level with default point size,
> # but the colors in the legend are muted by the level of transparency
> ggplot(dm, aes(x = x, y = value, colour = variable)) + theme_bw() +
>geom_point(alpha = 0.05) +
>scale_colour_manual(values = c('x1' = 'black',
>   'x2' = 'red', 'x3' = 'green'))
> 
> # A tradeoff is to reduce the point size and increase alpha a bit, but
> these
> changes will
> # also be reflected in the legend.
> 
> ggplot(dm, aes(x = x, y = value, colour = variable)) + theme_bw() +
>geom_point(alpha = 0.15, size = 1) +
>scale_colour_manual(values = c('x1' = 'black',
>   'x2' = 'red', 'x3' = 'green'))
> 
> You may well find the legend to be useless for this example, so to get
> rid
> of it,
> 
> ggplot(dm, aes(x = x, y = value, colour = variable)) + theme_bw() +
>geom_point(alpha = 0.15, size = 1) +
>scale_colour_manual(values = c('x1' = 'black',
>   'x2' = 'red', 'x3' = 'green')) +
>opts(legend.position = 'none')
> 
> The nice thing about the ggplot2 graph is that you can adjust the point
> size
> and alpha transparency to your tastes. The default point size is 2 and
> the
> default alpha = 1 (no transparency).
> 
> HTH,
> Dennis
> 
> On Wed, Mar 30, 2011 at 10:04 PM, Samuel Dennis 
> wrote:
> 
> > I have a very large dataset with three variables that I need to graph
> using
> > a scatterplot. However I find that the first variable gets masked by
> the
> > other two, so the graph looks entirely different depending on the
> order of
> > variables. Does anyone have any suggestions how to manage this?
> >
> > This code is an illustration of what I am dealing with:
> >
> > x <- 1
> > plot(rnorm(x,mean=20),rnorm(x),col=1,xlim=c(16,24))
> > points(rnorm(x,mean=21),rnorm(x),col=2)
> > points(rnorm(x,mean=19),rnorm(x),col=3)
> >
> > gives an entirely different looking graph to:
> >
> > x <- 1
> > plot(rnorm(x,mean=19),rnorm(x),col=3,xlim=c(16,24))
> > points(rnorm(x,mean=20),rnorm(x),col=1)
> > points(rnorm(x,mean=21),rnorm(x),col=2)
> >
> > despite being identical in all respects except for the order in which
> the
> > variables are plotted.
> >
> > I have tried using pch=".", however the colours are very difficult to
> > discern. I have experimented with a number of other symbols with no
> real
> > solution.
> >
> > The only way that appears to work is to iterate the plot with a for
> loop,
> > and progressively add a few numbers from each variable, as below.
> However
> > although I can do this simply with random numbers as I have done
> here, this
> > is an extremely cumbersome method to use with real datasets.
> >
> > plot(1,1,xlim=c(16,24),ylim=c(-4,4),col="white")
> > x <- 100
> > for (i in 1:100) {
> > points(rnorm(x,mean=19),rnorm(x),col=3)
> > points(rnorm(x,mean=20),rnorm(x),col=1)
> > points(rnorm(x,mean=21),rnorm(x),col=2)
> > }
> >
> > Is there some function in R that could solve this through
> automatically
> > iterating my data as above, using transparent symbols, or something
> else?
> > Is
> > there some other way of solving this issue that I haven't thought of?
> >
> > Thankyou,
> >
> > Samuel Dennis
> >
> >[[alternative HTML version deleted]]
> >
> > __
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> > http://www.R-project.o

Re: [R] Simple lattice question

[Peter Ehlers]

On 2011-03-31 06:58, Rubén Roa wrote:

Thanks Peters!

Just a few minor glitches now:

require(lattice)
data<- data.frame(SP=sort(rep(as.factor(c('A','B','C','D','E')),12)),
x=rpois(60,10),
y=rep(c(rep(0,4),rep(10,4),rep(20,4)),5),
z=rep(1:4,15))
xyplot(x~y|SP,data=data,groups=z,layout=c(2,3),pch=1:4,lty=1:4,col='black',type='b',
key=list(x = .65, y = .75, corner = c(0, 0), points=TRUE,
   lines=TRUE, pch=1:4, lty=1:4, type='b',
   text=list(lab = as.character(unique(data$z)

I have an extra column of symbols on the legend,

and,

would like to add a title to the legend. Such as 'main' for plots.

Any suggestions?


for key(), make 'lines' into a list:

 xyplot(x~y|SP,data=data,groups=z,layout=c(2,3),
pch=1:4,lty=1:4,col='black',type='b',
key=list(x = .65, y = .75, corner = c(0, 0),
 title="title here", cex.title=.9, lines.title=3,
 lines=list(pch=1:4, lty=1:4, type='b'),
 text=list(lab = as.character(unique(data$z)

Peter Ehlers



Rubén



Dr. Rubén Roa-Ureta
AZTI - Tecnalia / Marine Research Unit
Txatxarramendi Ugartea z/g
48395 Sukarrieta (Bizkaia)
SPAIN




-Mensaje original-
De: Peter Ehlers [mailto:ehl...@ucalgary.ca]
Enviado el: jueves, 31 de marzo de 2011 15:41
Para: Rubén Roa
CC: r-help@r-project.org
Asunto: Re: [R] Simple lattice question

On 2011-03-31 03:39, Rubén Roa wrote:


DeaR ComRades,

require(lattice)
data<-

data.frame(SP=sort(rep(as.factor(c('A','B','C','D','E')),12)),

 x=rpois(60,10),
 y=rep(c(rep(0,4),rep(10,4),rep(20,4)),5),
 z=rep(1:4,15))


xyplot(x~y|SP,data=data,groups=z,layout=c(2,3),pch=1:4,lty=1:4,col='bl

ack',type='b')

How do I put a legend for the grouping variable in the

empty upper-right panel?

See the help page for xyplot for an example using the iris data.
You just need to add something like

   auto.key = list(x = .6, y = .7, corner = c(0, 0))

to your lattice call. See the 'key' entry on the ?xyplot page.

Peter Ehlers



TIA

Rubén



__

__

Dr. Rubén Roa-Ureta
AZTI - Tecnalia / Marine Research Unit Txatxarramendi Ugartea z/g
48395 Sukarrieta (Bizkaia)
SPAIN

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Re: [R] choosing best 'match' for given factor

[Bert Gunter]

Folks:

I think the following may be somewhat faster, as it avoids sorting:

bmat <- function(mx,vec)
{
  nm <- colnames(mx)
  ivec <- match(vec,nm)
  sapply(ivec,function(k){
   if(k==1)NA  else {
lookat <- setdiff(seq_len(k-1),ivec) ## only those to left and not
in search vector ##
nm[lookat[which.max(mx[lookat,k] )]]
   }
  }
 )
}

-- Bert

On Thu, Mar 31, 2011 at 8:30 AM, Nick Sabbe  wrote:
>
> Hi Murali.
> I haven't compared, but this is what I would do:
>
> bestMatch<-function(searchVector, matchMat)
> {
>        searchRow<-unique(sort(match(searchVector, colnames(matchMat #if
> you're sure, you could drop unique
>        cat("Original row indices:")
>        print(searchRow)
>        matchMat<-matchMat[, -searchRow, drop=FALSE] #avoid duplicates
> altogether
>        cat("Corrected Matrix:\n")
>        print(matchMat)
>        correctedRows<-searchRow - seq_along(searchRow) + 1 #works because
> of the sort above
>        cat("Corrected row indices:")
>        print(correctedRows)
>        sapply(correctedRows, function(cr){
>                        lookWhere<-matchMat[cr, seq(cr-1)]
>                        cat("Will now look into:\n")
>                        print(lookWhere)
>                        cc<-which.max(lookWhere)
>                        cat("Max at position", cc, "\n")
>                        colnames(matchMat)[cc]
>                })
> }
> I don't think there's that much difference. Depending on specific sizes, it
> may be more or less costly to first shrink the search matrix like I do. And
> similarly depending, I may be better still if you remove the rows that
> you're not interested in as well (some more but similar index trickery
> required then.
>
> HTH,
>
>
> Nick Sabbe
> --
> ping: nick.sa...@ugent.be
> link: http://biomath.ugent.be
> wink: A1.056, Coupure Links 653, 9000 Gent
> ring: 09/264.59.36
>
> -- Do Not Disapprove
>
>
>
>
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of murali.me...@avivainvestors.com
> Sent: donderdag 31 maart 2011 16:46
> To: r-help@r-project.org
> Subject: [R] choosing best 'match' for given factor
>
> Folks,
>
> I have a 'matching' matrix between variables A, X, L, O:
>
> > a <- structure(c(1, 0.41, 0.58, 0.75, 0.41, 1, 0.6, 0.86, 0.58,
> 0.6, 1, 0.83, 0.75, 0.86, 0.83, 1), .Dim = c(4L, 4L), .Dimnames = list(
>    c("A", "X", "L", "O"), c("A", "X", "L", "O")))
>
> > a
>      A     X     L     O
> A  1.00  0.41  0.58  0.75
> X  0.41  1.00  0.60  0.86
> L  0.58  0.75  1.00  0.83
> O  0.60  0.86  0.83  1.00
>
> And I have a search vector of variables
>
> > v <- c("X", "O")
>
> I want to write a function bestMatch(searchvector, matchMat) such that for
> each variable in searchvector, I get the variable that it has the highest
> match to - but searching only among variables to the left of it in the
> 'matching' matrix, and not matching with any variable in searchvector
> itself.
>
> So in the above example, although "X" has the highest match (0.86) with "O",
> I can't choose "O" as it's to the right of X (and also because "O" is in the
> searchvector v already); I'll have to choose "A".
>
> For "O", I will choose "L", the variable it's best matched with - as it
> can't match "X" already in the search vector.
>
> My function bestMatch(v, a) will then return c("A", "L")
>
> My matrix a is quite large, and I have a long list of search vectors v, so I
> need an efficient method.
>
> I wrote this:
>
> bestMatch <- function(searchvector,  matchMat) {
>        sapply(searchvector, function(cc) {
>                             y <- matchMat[!(rownames(matchMat) %in%
> searchvector) & (index(rownames(matchMat)) < match(cc, rownames(matchMat))),
> cc, drop = FALSE];
>                             rownames(y)[which.max(y)]
>        })
> }
>
> Any advice?
>
> Thanks,
>
> Murali
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.



--
"Men by nature long to get on to the ultimate truths, and will often
be impatient with elementary studies or fight shy of them. If it were
possible to reach the ultimate truths without the elementary studies
usually prefixed to them, these would not be preparatory studies but
superfluous diversions."

-- Maimonides (1135-1204)

Bert Gunter
Genentech Nonclinical Biostatistics

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PLEASE do read the posting guide http://www.R-project

[R] Sequential multiple regression

[Tyler Rinker]

Hello,
 
In the past I have tended to reside more in the ANOVA camp but am trying to 
become more familiar with regression techniques in R.  I would like to get the 
F change from a model as I take away factors:
 
SO...
 
mod1<-lm(y~x1+x2+x3)...mod2<-lm(y~x1,x2)...mod3<-lm(y~x1)
 
 
I can do this by hand by running several models in R and taking the MSr1/MSe1, 
MSr2/MSe2...  This is slow and I know there's a better way.
 
In SPSS (which I no longer use) I could easily obtain these results (F-change) 
as documented by Professor Andy Fields:
http://www.statisticshell.com/multireg.pdf
 
You can see the F changes for his two IV model yielding 2 F changes.  Maybe 
it's the language I'm using (sequential multiple regression) that yields me 
poor results in searching the archives and Rseek.  The results tend to be 
around hierarchal regression (I'm not familiar with this terminology being in 
the ANOVA camp).  When I look at the hier.part package and run the examples it 
doesn't seem to give me the F change I'm looking for.  The step function in the 
base program reduces the model but takes away the non sig. IV's (which is a 
great approach but I'm really after those F changes).  As is the usually the 
case I'm sure R does this simply and beautifully, I'm just not experienced with 
the statistical vocabulary and techniques around regression to find what I'm 
looking for.
 
F-change values with R:  Any help would be appreciated.
 
Thank you in advance,
Tyler 
[[alternative HTML version deleted]]

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Sequential multiple regression

[Bert Gunter]

?drop1

-- Bert

On Thu, Mar 31, 2011 at 9:24 AM, Tyler Rinker  wrote:
>
> Hello,
>
> In the past I have tended to reside more in the ANOVA camp but am trying to 
> become more familiar with regression techniques in R.  I would like to get 
> the F change from a model as I take away factors:
>
> SO...
>
> mod1<-lm(y~x1+x2+x3)...mod2<-lm(y~x1,x2)...mod3<-lm(y~x1)
>
>
> I can do this by hand by running several models in R and taking the 
> MSr1/MSe1, MSr2/MSe2...  This is slow and I know there's a better way.
>
> In SPSS (which I no longer use) I could easily obtain these results 
> (F-change) as documented by Professor Andy Fields:
> http://www.statisticshell.com/multireg.pdf
>
> You can see the F changes for his two IV model yielding 2 F changes.  Maybe 
> it's the language I'm using (sequential multiple regression) that yields me 
> poor results in searching the archives and Rseek.  The results tend to be 
> around hierarchal regression (I'm not familiar with this terminology being in 
> the ANOVA camp).  When I look at the hier.part package and run the examples 
> it doesn't seem to give me the F change I'm looking for.  The step function 
> in the base program reduces the model but takes away the non sig. IV's (which 
> is a great approach but I'm really after those F changes).  As is the usually 
> the case I'm sure R does this simply and beautifully, I'm just not 
> experienced with the statistical vocabulary and techniques around regression 
> to find what I'm looking for.
>
> F-change values with R:  Any help would be appreciated.
>
> Thank you in advance,
> Tyler
>        [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
"Men by nature long to get on to the ultimate truths, and will often
be impatient with elementary studies or fight shy of them. If it were
possible to reach the ultimate truths without the elementary studies
usually prefixed to them, these would not be preparatory studies but
superfluous diversions."

-- Maimonides (1135-1204)

Bert Gunter
Genentech Nonclinical Biostatistics

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] statistical question

[Greg Snow]

You could do bootstrapping for confidence intervals or permutation tests for 
just comparing the medians.  There are other tools as well, just be sure that 
you understand what you are testing and what assumptions are being made in 
whatever method you choose.

-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111


> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Anna Lee
> Sent: Thursday, March 31, 2011 3:15 AM
> To: r-help@r-project.org
> Subject: [R] statistical question
> 
> Dear List!
> 
> I want to compare medians of  non normal distributed data. Is it
> possible and usefull to calculate 95% confidence intervals for
> medians? And if so - how can this be achieved in R?
> 
> Thanks a lot!
> Anna
> 
> --
> 
> 
> 
> Der Inhalt dieser E-Mail ist vertraulich. Sollte Ihnen die E-Mail
> irrtümlich zugesandt worden sein, bitte ich Sie, mich unverzüglich zu
> benachrichtigen und die E-Mail zu löschen.
> 
> This e-mail is confidential. If you have received it in error, please
> notify me immediately and delete it from your system.
> 
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-
> guide.html
> and provide commented, minimal, self-contained, reproducible code.

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] choosing best 'match' for given factor

[Henrique Dallazuanna]

Try this:

bestMatch <- function(search, match) {
colnames(match)[pmax(apply(match[,search], 2, which.max) - 1, 1)]
}


On Thu, Mar 31, 2011 at 11:46 AM,   wrote:
> Folks,
>
> I have a 'matching' matrix between variables A, X, L, O:
>
>> a <- structure(c(1, 0.41, 0.58, 0.75, 0.41, 1, 0.6, 0.86, 0.58,
> 0.6, 1, 0.83, 0.75, 0.86, 0.83, 1), .Dim = c(4L, 4L), .Dimnames = list(
>    c("A", "X", "L", "O"), c("A", "X", "L", "O")))
>
>> a
>      A     X     L     O
> A  1.00  0.41  0.58  0.75
> X  0.41  1.00  0.60  0.86
> L  0.58  0.75  1.00  0.83
> O  0.60  0.86  0.83  1.00
>
> And I have a search vector of variables
>
>> v <- c("X", "O")
>
> I want to write a function bestMatch(searchvector, matchMat) such that for 
> each variable in searchvector, I get the variable that it has the highest 
> match to - but searching only among variables to the left of it in the 
> 'matching' matrix, and not matching with any variable in searchvector itself.
>
> So in the above example, although "X" has the highest match (0.86) with "O", 
> I can't choose "O" as it's to the right of X (and also because "O" is in the 
> searchvector v already); I'll have to choose "A".
>
> For "O", I will choose "L", the variable it's best matched with - as it can't 
> match "X" already in the search vector.
>
> My function bestMatch(v, a) will then return c("A", "L")
>
> My matrix a is quite large, and I have a long list of search vectors v, so I 
> need an efficient method.
>
> I wrote this:
>
> bestMatch <- function(searchvector,  matchMat) {
>        sapply(searchvector, function(cc) {
>                             y <- matchMat[!(rownames(matchMat) %in% 
> searchvector) & (index(rownames(matchMat)) < match(cc, rownames(matchMat))), 
> cc, drop = FALSE];
>                             rownames(y)[which.max(y)]
>        })
> }
>
> Any advice?
>
> Thanks,
>
> Murali
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

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Re: [R] Sequential multiple regression

[Tyler Rinker]

Bert and anyone else with info,
 
First, Bert thank you for your quick reply.  drop1 gives the results as a type 
II anova.  Is there a way to make drop1 give you type I anova (the args don't 
appear to have a way to do so)?  Another package/function perhaps?
 
Tyler 
 
> Date: Thu, 31 Mar 2011 09:32:02 -0700
> Subject: Re: [R] Sequential multiple regression
> From: gunter.ber...@gene.com
> To: tyler_rin...@hotmail.com
> CC: r-help@r-project.org
> 
> ?drop1
> 
> -- Bert
> 
> On Thu, Mar 31, 2011 at 9:24 AM, Tyler Rinker  
> wrote:
> >
> > Hello,
> >
> > In the past I have tended to reside more in the ANOVA camp but am trying to 
> > become more familiar with regression techniques in R.  I would like to get 
> > the F change from a model as I take away factors:
> >
> > SO...
> >
> > mod1<-lm(y~x1+x2+x3)...mod2<-lm(y~x1,x2)...mod3<-lm(y~x1)
> >
> >
> > I can do this by hand by running several models in R and taking the 
> > MSr1/MSe1, MSr2/MSe2...  This is slow and I know there's a better way.
> >
> > In SPSS (which I no longer use) I could easily obtain these results 
> > (F-change) as documented by Professor Andy Fields:
> > http://www.statisticshell.com/multireg.pdf
> >
> > You can see the F changes for his two IV model yielding 2 F changes.  Maybe 
> > it's the language I'm using (sequential multiple regression) that yields me 
> > poor results in searching the archives and Rseek.  The results tend to be 
> > around hierarchal regression (I'm not familiar with this terminology being 
> > in the ANOVA camp).  When I look at the hier.part package and run the 
> > examples it doesn't seem to give me the F change I'm looking for.  The step 
> > function in the base program reduces the model but takes away the non sig. 
> > IV's (which is a great approach but I'm really after those F changes).  As 
> > is the usually the case I'm sure R does this simply and beautifully, I'm 
> > just not experienced with the statistical vocabulary and techniques around 
> > regression to find what I'm looking for.
> >
> > F-change values with R:  Any help would be appreciated.
> >
> > Thank you in advance,
> > Tyler
> >[[alternative HTML version deleted]]
> >
> > __
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
> 
> 
> 
> -- 
> "Men by nature long to get on to the ultimate truths, and will often
> be impatient with elementary studies or fight shy of them. If it were
> possible to reach the ultimate truths without the elementary studies
> usually prefixed to them, these would not be preparatory studies but
> superfluous diversions."
> 
> -- Maimonides (1135-1204)
> 
> Bert Gunter
> Genentech Nonclinical Biostatistics
  
[[alternative HTML version deleted]]

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Re: [R] nls.profile

[Douglas Bates]

On Thu, Mar 31, 2011 at 4:02 AM, Daniel Kaschek
 wrote:
> Hello,

> I use nls.profile to compute confidence intervals of parameter estimates
> of a non-linear model. When computing the profiles, the model function
> produces an error for certain parameter combinations. Therefore nls
> fails and so does nls.profile.

> Is there a way to tell nls.profile to ignore errors and to proceed with
> the next parameter value?

Not easily.  What I tend to do is to adjust (increase) the alphamax
parameter (which is a misnomer, it should have been alphamin but I got
confused) until the profile can succeed.

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[R] one question about bioconductor

[wang peter]

dear lady and gentalmen:
  i am gaoshan from kansas university.
i used such coding to deal with gel data

data <- ReadAffy()
 Warning messages:
1: In file(out, "wt") :
  cannot open file
'C:\Users\gaoshan\AppData\Local\Temp\RtmpvsyXOV\Rhttpd3f0b2e85': No such
file or directory
2: In file(out, "wt") :
  cannot open file
'C:\Users\gaoshan\AppData\Local\Temp\RtmpvsyXOV\Rhttpd1d7427a'
eset <- rma(data)
Error in gzfile(file, "r") : cannot open the connection
In addition: Warning message:
In gzfile(file, "r") :
  cannot open compressed file
'C:\Users\gaoshan\AppData\Local\Temp\RtmpvsyXOV\file5f6a6cb5', probable
reason 'No such file or directory'
 can you help me to fix it?
thank u very much
gao shan

[[alternative HTML version deleted]]

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[R] How to update R?

[Li, Yunfei]

Hi,

My R version is old, and I would like to update it. How can I update R and keep 
all the libraries? Thanks

Best,

Yunfei Li
--
Research Assistant
Department of Statistics &
School of Molecular Biosciences
Biotechnology Life Sciences Building 427
Washington State University
Pullman, WA 99164-7520
Phone: 509-339-5096
http://www.wsu.edu/~ye_lab/people.html


[[alternative HTML version deleted]]

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Re: [R] one question about bioconductor

[Philipp Pagel]

On Thu, Mar 31, 2011 at 09:32:06AM -0700, wang peter wrote:
> dear lady and gentalmen:
>   i am gaoshan from kansas university.
> i used such coding to deal with gel data
> 
> data <- ReadAffy()
>  Warning messages:
> 1: In file(out, "wt") :
>   cannot open file
> 'C:\Users\gaoshan\AppData\Local\Temp\RtmpvsyXOV\Rhttpd3f0b2e85': No such
> file or directory

As the message says: there is something wrong with the path. 

In order to get more helpful replies, you should show the actual code
you used and also give a hint about the spoecific packages you were
using. E.g. ReadAffy most certainly requires at least a filename which
seems to be missing from your comamnd above. 

In addition, I recommend to post your question on the bioconductor
mailing list.

cu
Philipp


-- 
Dr. Philipp Pagel
Lehrstuhl für Genomorientierte Bioinformatik
Technische Universität München
Wissenschaftszentrum Weihenstephan
Maximus-von-Imhof-Forum 3
85354 Freising, Germany
http://webclu.bio.wzw.tum.de/~pagel/

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[R] Cluster analysis, factor variables, large data set

[Hans Ekbrand]

Dear R helpers,

I have a large data set with 36 variables and about 50.000 cases. The
variabels represent labour market status during 36 months, there are 8
different variable values (e.g. Full-time Employment, Student,...)

Only cases with at least one change in labour market status is
included in the data set.

To analyse sub sets of the data, I have used daisy in the
cluster-package to create a distance matrix and then used pam (or pamk
in the fpc-package), to get a k-medoids cluster-solution. Now I want
to analyse the whole set.

clara is said to cope with large data sets, but the first step in the
cluster analysis, the creation of the distance matrix must be done by
another function since clara only works with numeric data.

Is there an alternative to the daisy -> clara route that does not
require as much RAM?

What functions would you recommend for a cluster analysis of this kind
of data on large data set?


regards,

Hans Ekbrand

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[R] read password-protected files

[Shi, Tao]

Hi list,

I have a bunch of .csv files that are password-protected.  I wonder if there is 
a way to read them in in R without manually removing the password protection 
for 
each file?

Thank you very much!


...Tao
[[alternative HTML version deleted]]

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] How to update R?

[Shi, Tao]

This question has been asked by many people already.  The easiest way is:

1) install the new version
2) copy all or the libraries that you installed later from the "library" folder 
of older version to the new version
3) uninstall the old version
4) do a library update in the new version

Done!

...Tao






From: "Li, Yunfei" 
To: r-help@r-project.org
Sent: Thu, March 31, 2011 10:16:23 AM
Subject: [R] How to update R?

Hi,

My R version is old, and I would like to update it. How can I update R and keep 
all the libraries? Thanks

Best,

Yunfei Li
--

Research Assistant
Department of Statistics &
School of Molecular Biosciences
Biotechnology Life Sciences Building 427
Washington State University
Pullman, WA 99164-7520
Phone: 509-339-5096
http://www.wsu.edu/~ye_lab/people.html


[[alternative HTML version deleted]]

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and provide commented, minimal, self-contained, reproducible code.

[[alternative HTML version deleted]]

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] How to update R?

[Joshua Wiley]

Hi Yunfei,

It really depends to some extent on the version of R (and your OS) you
were using.  Here is a link to a blog post that discusses some
potential strategies:
http://www.r-statistics.com/2010/04/changing-your-r-upgrading-strategy-and-the-r-code-to-do-it-on-windows/

There are also a number of messages if you search the R-help archive.
However, old version of libraries may not be portable to the most
recent R, so you may need to update R and update all your libraries
too.

Cheers,

Josh

On Thu, Mar 31, 2011 at 10:16 AM, Li, Yunfei  wrote:
> Hi,
>
> My R version is old, and I would like to update it. How can I update R and 
> keep all the libraries? Thanks
>
> Best,
>
> Yunfei Li
> --
> Research Assistant
> Department of Statistics &
> School of Molecular Biosciences
> Biotechnology Life Sciences Building 427
> Washington State University
> Pullman, WA 99164-7520
> Phone: 509-339-5096
> http://www.wsu.edu/~ye_lab/people.html
>
>
>        [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

__
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Cluster analysis, factor variables, large data set

[Christian Hennig]

Dear Hans,

clara doesn't require a distance matrix as input (and therefore doesn't 
require you to run daisy), it will work with the raw data matrix using

Euclidean distances implicitly.
I can't tell you whether Euclidean distances are appropriate in this 
situation (this depends on the interpretation and variables and 
particularly on how they are scaled), but they may be fine at least after 
some transformation and standardisation of your variables.


Hope this helps,
Christian

On Thu, 31 Mar 2011, Hans Ekbrand wrote:


Dear R helpers,

I have a large data set with 36 variables and about 50.000 cases. The
variabels represent labour market status during 36 months, there are 8
different variable values (e.g. Full-time Employment, Student,...)

Only cases with at least one change in labour market status is
included in the data set.

To analyse sub sets of the data, I have used daisy in the
cluster-package to create a distance matrix and then used pam (or pamk
in the fpc-package), to get a k-medoids cluster-solution. Now I want
to analyse the whole set.

clara is said to cope with large data sets, but the first step in the
cluster analysis, the creation of the distance matrix must be done by
another function since clara only works with numeric data.

Is there an alternative to the daisy -> clara route that does not
require as much RAM?

What functions would you recommend for a cluster analysis of this kind
of data on large data set?


regards,

Hans Ekbrand

__
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and provide commented, minimal, self-contained, reproducible code.



*** --- ***
Christian Hennig
University College London, Department of Statistical Science
Gower St., London WC1E 6BT, phone +44 207 679 1698
chr...@stats.ucl.ac.uk, www.homepages.ucl.ac.uk/~ucakche

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Re: [R] How to update R?

[Shi, Tao]

Forgot to mention that my way of updating is for Windows only.

...Tao






From: Joshua Wiley 
To: "Li, Yunfei" 
Cc: r-help@r-project.org
Sent: Thu, March 31, 2011 11:05:29 AM
Subject: Re: [R] How to update R?

Hi Yunfei,

It really depends to some extent on the version of R (and your OS) you
were using.  Here is a link to a blog post that discusses some
potential strategies:
http://www.r-statistics.com/2010/04/changing-your-r-upgrading-strategy-and-the-r-code-to-do-it-on-windows/


There are also a number of messages if you search the R-help archive.
However, old version of libraries may not be portable to the most
recent R, so you may need to update R and update all your libraries
too.

Cheers,

Josh

On Thu, Mar 31, 2011 at 10:16 AM, Li, Yunfei  wrote:
> Hi,
>
> My R version is old, and I would like to update it. How can I update R and 
> keep 
>all the libraries? Thanks
>
> Best,
>
> Yunfei Li
>--
>-
> Research Assistant
> Department of Statistics &
> School of Molecular Biosciences
> Biotechnology Life Sciences Building 427
> Washington State University
> Pullman, WA 99164-7520
> Phone: 509-339-5096
> http://www.wsu.edu/~ye_lab/people.html
>
>
>[[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

__
R-help@r-project.org mailing list
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[[alternative HTML version deleted]]

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Re: [R] read password-protected files

[Joshua Wiley]

Hi Tao,

You will need to give us more details.  CSV files are pure text.  How
are they password protected?  Are they Excel files?  In a password
protected compressed format?  Encrypted?

Best Regards,

Josh

On Thu, Mar 31, 2011 at 11:00 AM, Shi, Tao  wrote:
> Hi list,
>
> I have a bunch of .csv files that are password-protected.  I wonder if there 
> is
> a way to read them in in R without manually removing the password protection 
> for
> each file?
>
> Thank you very much!
>
>
> ...Tao
>        [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] read password-protected files

[Prof Brian Ripley]

On Thu, 31 Mar 2011, Shi, Tao wrote:


Hi list,

I have a bunch of .csv files that are password-protected.  I wonder if there is


There is nothing in the CSV standard about password protection.  I 
very much doubt that these actually CSV files.  So please follow the 
posting guide [*]



a way to read them in in R without manually removing the password protection for
each file?

Thank you very much!


...Tao
[[alternative HTML version deleted]]

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.


[*] as above: and as you sent HTML you are either full of contempt for 
the helpeRs or failed to do your homework.



--
Brian D. Ripley,  rip...@stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

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Re: [R] read password-protected files

[Philipp Pagel]

On Thu, Mar 31, 2011 at 11:00:48AM -0700, Shi, Tao wrote:
> Hi list,
> 
> I have a bunch of .csv files that are password-protected.  I wonder if there 
> is 
> a way to read them in in R without manually removing the password protection 
> for 
> each file?

I doubt that there is such a thing as a password protected csv file.
They are just text files, after all. So I guess you have something
else. How or what did the presumed pasword protection?

cu
Philipp

-- 
Dr. Philipp Pagel
Lehrstuhl für Genomorientierte Bioinformatik
Technische Universität München
Wissenschaftszentrum Weihenstephan
Maximus-von-Imhof-Forum 3
85354 Freising, Germany
http://webclu.bio.wzw.tum.de/~pagel/

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] read password-protected files

[Shi, Tao]

Hi Josh and Prof. Ripley,

Thanks for the quick replies!

Sorry about the HTML email.  It's the default for my yahoo account and forgot 
to 
switch.

The question was asked by a colleague of mine.  After double-checking, yes, 
they 
are actually Excel files.  Any idea on how to approach this?

Thanks!

...Tao




From: Prof Brian Ripley 

Cc: r-help@r-project.org
Sent: Thu, March 31, 2011 11:15:35 AM
Subject: Re: [R] read password-protected files

On Thu, 31 Mar 2011, Shi, Tao wrote:

> Hi list,
>
> I have a bunch of .csv files that are password-protected.  I wonder if there 
is

There is nothing in the CSV standard about password protection.  I 
very much doubt that these actually CSV files.  So please follow the 
posting guide [*]

> a way to read them in in R without manually removing the password protection 
>for
> each file?
>
[[elided Yahoo spam]]
>
>
> ...Tao
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

[*] as above: and as you sent HTML you are either full of contempt for 
the helpeRs or failed to do your homework.


-- 
Brian D. Ripley,  rip...@stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] read password-protected files

[Joshua Wiley]

Hi Tao,

There are a number of R packages to work with reading/writing Excel
files (e.g., xlsReadWrite).  If Excel is on the system in question, as
long as you have opened the file in Excel, you should be able to use
RODBC with  RDCOMClient or rcom, but I am not sure those will work if
it is password protected unless it is currently unlocked in Excel.

HTH,

Josh

On Thu, Mar 31, 2011 at 11:33 AM, Shi, Tao  wrote:
> Hi Josh and Prof. Ripley,
>
> Thanks for the quick replies!
>
> Sorry about the HTML email.  It's the default for my yahoo account and forgot 
> to
> switch.
>
> The question was asked by a colleague of mine.  After double-checking, yes, 
> they
> are actually Excel files.  Any idea on how to approach this?
>
> Thanks!
>
> ...Tao
>
>
>
> 
> From: Prof Brian Ripley 
> To: "Shi, Tao" 
> Cc: r-help@r-project.org
> Sent: Thu, March 31, 2011 11:15:35 AM
> Subject: Re: [R] read password-protected files
>
> On Thu, 31 Mar 2011, Shi, Tao wrote:
>
>> Hi list,
>>
>> I have a bunch of .csv files that are password-protected.  I wonder if there
> is
>
> There is nothing in the CSV standard about password protection.  I
> very much doubt that these actually CSV files.  So please follow the
> posting guide [*]
>
>> a way to read them in in R without manually removing the password protection
>>for
>> each file?
>>
>> Thank you very much!
>>
>>
>> ...Tao
>>     [[alternative HTML version deleted]]
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
> [*] as above: and as you sent HTML you are either full of contempt for
> the helpeRs or failed to do your homework.
>
>
> --
> Brian D. Ripley,                  rip...@stats.ox.ac.uk
> Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
> University of Oxford,             Tel:  +44 1865 272861 (self)
> 1 South Parks Road,                     +44 1865 272866 (PA)
> Oxford OX1 3TG, UK                Fax:  +44 1865 272595

__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] read password-protected files

[Shi, Tao]

Thanks, Josh.  I use xlsReadWrite routinely.  It would be nice if it has a 
password option.

...Tao




- Original Message 
> From: Joshua Wiley 
> To: "Shi, Tao" 
> Cc: r-help@r-project.org
> Sent: Thu, March 31, 2011 11:50:27 AM
> Subject: Re: [R] read password-protected files
> 
> Hi Tao,
> 
> There are a number of R packages to work with reading/writing  Excel
> files (e.g., xlsReadWrite).  If Excel is on the system in  question, as
> long as you have opened the file in Excel, you should be able to  use
> RODBC with  RDCOMClient or rcom, but I am not sure those will work  if
> it is password protected unless it is currently unlocked in  Excel.
> 
> HTH,
> 
> Josh
> 
> On Thu, Mar 31, 2011 at 11:33 AM, Shi, Tao   wrote:
> > Hi  Josh and Prof. Ripley,
> >
> > Thanks for the quick  replies!
> >
> > Sorry about the HTML email.  It's the default for my  yahoo account and 
>forgot to
> > switch.
> >
> > The question was  asked by a colleague of mine.  After double-checking, 
> > yes, 
>they
> > are  actually Excel files.  Any idea on how to approach this?
> >
> >  Thanks!
> >
> > ...Tao
> >
> >
> >
> >  
> > From: Prof Brian Ripley 
> > To:  "Shi, Tao" 
> > Cc: r-help@r-project.org
> > Sent: Thu,  March 31, 2011 11:15:35 AM
> > Subject: Re: [R] read password-protected  files
> >
> > On Thu, 31 Mar 2011, Shi, Tao wrote:
> >
> >>  Hi list,
> >>
> >> I have a bunch of .csv files that are  password-protected.  I wonder if 
>there
> > is
> >
> > There is  nothing in the CSV standard about password protection.  I
> > very much  doubt that these actually CSV files.  So please follow the
> > posting guide  [*]
> >
> >> a way to read them in in R without manually removing the  password 
>protection
> >>for
> >> each  file?
> >>
> >> Thank you very  much!
> >>
> >>
> >> ...Tao
> >> [[alternative  HTML version deleted]]
> >>
> >>  __
> >> R-help@r-project.org mailing  list
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >>  PLEASE do read the posting guide  
>http://www.R-project.org/posting-guide.html
> >> and provide commented,  minimal, self-contained, reproducible code.
> >
> > [*] as above: and as  you sent HTML you are either full of contempt for
> > the helpeRs or failed  to do your homework.
> >
> >
> > --
> > Brian D. Ripley,   rip...@stats.ox.ac.uk
> > Professor  of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
> > University of  Oxford, Tel:  +44 1865 272861 (self)
> > 1 South Parks Road,  +44 1865 272866 (PA)
> > Oxford OX1 3TG, UK Fax:  +44 1865 272595
>

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Re: [R] read password-protected files

[Henrique Dallazuanna]

Using RDCOMCliente:

library(RDCOMClient)
eApp <- COMCreate("Excel.Application")
wk <- eApp$Workbooks()$Open(Filename="your_file",Password="your_password")
tf <- tempfile()
wk$Sheets(1)$SaveAs(tf, 3)

DF <- read.table(sprintf("%s.txt", tf), header = TRUE, sep = "\t")

On Thu, Mar 31, 2011 at 3:33 PM, Shi, Tao  wrote:
> Hi Josh and Prof. Ripley,
>
> Thanks for the quick replies!
>
> Sorry about the HTML email.  It's the default for my yahoo account and forgot 
> to
> switch.
>
> The question was asked by a colleague of mine.  After double-checking, yes, 
> they
> are actually Excel files.  Any idea on how to approach this?
>
> Thanks!
>
> ...Tao
>
>
>
> 
> From: Prof Brian Ripley 
>
> Cc: r-help@r-project.org
> Sent: Thu, March 31, 2011 11:15:35 AM
> Subject: Re: [R] read password-protected files
>
> On Thu, 31 Mar 2011, Shi, Tao wrote:
>
>> Hi list,
>>
>> I have a bunch of .csv files that are password-protected.  I wonder if there
> is
>
> There is nothing in the CSV standard about password protection.  I
> very much doubt that these actually CSV files.  So please follow the
> posting guide [*]
>
>> a way to read them in in R without manually removing the password protection
>>for
>> each file?
>>
> [[elided Yahoo spam]]
>>
>>
>> ...Tao
>>     [[alternative HTML version deleted]]
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
> [*] as above: and as you sent HTML you are either full of contempt for
> the helpeRs or failed to do your homework.
>
>
> --
> Brian D. Ripley,                  rip...@stats.ox.ac.uk
> Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
> University of Oxford,             Tel:  +44 1865 272861 (self)
> 1 South Parks Road,                     +44 1865 272866 (PA)
> Oxford OX1 3TG, UK                Fax:  +44 1865 272595
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

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Re: [R] Cluster analysis, factor variables, large data set

[Hans Ekbrand]

On Thu, Mar 31, 2011 at 07:06:31PM +0100, Christian Hennig wrote:
> Dear Hans,
> 
> clara doesn't require a distance matrix as input (and therefore
> doesn't require you to run daisy), it will work with the raw data
> matrix using
> Euclidean distances implicitly.
> I can't tell you whether Euclidean distances are appropriate in this
> situation (this depends on the interpretation and variables and
> particularly on how they are scaled), but they may be fine at least
> after some transformation and standardisation of your variables.

The variables are unordered factors, stored as integers 1:9, where 

1 means "Full-time employment"
2 means "Part-time employment"
3 means "Student"
4 means "Full-time self-employee"
...

Does euclidean distances make sense on unordered factors coded as
integers?

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[R] transparent grays?

[Frostygoat]

Is there a grayscale equivalent to alpha levels in rgb?

Example:  I have the following to make red transparent circles overlap
with previously plotted blue symbols.

symbols(x=sites$long,y=sites$lat,circles=log(sites$prop.nem
+1),add=T,inches=F,bg=rgb(red=1,green=0,blue=0,
alpha=0.5),fg=rgb(red=1,green=0,blue=0, alpha=0.5))

I'm having a hard time coming up with a grayscale equivalent.

Thanks!

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Re: [R] Cluster analysis, factor variables, large data set

[Hans Ekbrand]

On Thu, Mar 31, 2011 at 08:48:02PM +0200, Hans Ekbrand wrote:
> On Thu, Mar 31, 2011 at 07:06:31PM +0100, Christian Hennig wrote:
> > Dear Hans,
> > 
> > clara doesn't require a distance matrix as input (and therefore
> > doesn't require you to run daisy), it will work with the raw data
> > matrix using
> > Euclidean distances implicitly.
> > I can't tell you whether Euclidean distances are appropriate in this
> > situation (this depends on the interpretation and variables and
> > particularly on how they are scaled), but they may be fine at least
> > after some transformation and standardisation of your variables.
> 
> The variables are unordered factors, stored as integers 1:9, where 
> 
> 1 means "Full-time employment"
> 2 means "Part-time employment"
> 3 means "Student"
> 4 means "Full-time self-employee"
> ...
> 
> Does euclidean distances make sense on unordered factors coded as
> integers?

To be clear, here is an extract

> my.df.full[900:910, 16:19]
PL210F.first.year PL210G.first.year PL210H.first.year PL210I.first.year
900 2 2 1 2
901 1 1 1 1
902 1 1 1 1
903 2 2 2 2
904 1 1 1 1
905 2 2 2 2
906 7 8 2 7
907 5 5 5 5
908 1 1 1 1
909 1 1 1 1
910 1 1 1 1

> class(my.df.full[,16])
[1] "integer"

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Re: [R] Cluster analysis, factor variables, large data set

[Peter Langfelder]

On Thu, Mar 31, 2011 at 11:48 AM, Hans Ekbrand  wrote:
>
> The variables are unordered factors, stored as integers 1:9, where
>
> 1 means "Full-time employment"
> 2 means "Part-time employment"
> 3 means "Student"
> 4 means "Full-time self-employee"
> ...
>
> Does euclidean distances make sense on unordered factors coded as
> integers?

It probably doesn't. You said you have some 36 observations for each
case, correct? You can turn these 36 observations into a vector of
length 36 * 9 on which Euclidean distance will make some sense, namely
k changes will produce a distance of sqrt(2*k). For each observation
with value p (p between 1 and 9), create a vector r = c(0,0,1,0,...0)
where the entry 1 is in the p-th component. Hence, if values p1 and p2
are the same, euclidean distance between r1 and r2 is zero; if they
are not the same, Euclidan distance is sqrt(2).

Here's some possible R code:


transform = function(obsVector, maxVal)
{
  templateMat = matrix(0, maxVal, maxVal);
  diag(templateMat) = 1;

  return(as.vector(templateMat[, obsVector]));
}

set.seed(10)
n = 4;
m = 5;
max = 4;
data = matrix(sample(c(1:max), n*m, replace = TRUE), m, n);

> data
 [,1] [,2] [,3] [,4]
[1,]3312
[2,]1332
[3,]3324
[4,]1242
[5,]4141


trafoData = apply(data, 2, transform, maxVal = max);

> trafoData
  [,1] [,2] [,3] [,4]
 [1,]0010
 [2,]0001
 [3,]1100
 [4,]0000
 [5,]1000
 [6,]0001
 [7,]0110
 [8,]0000
 [9,]0000
[10,]0010
[11,]1100
[12,]0001
[13,]1000
[14,]0101
[15,]0000
[16,]0010
[17,]0101
[18,]0000
[19,]0000
[20,]1010



The code assumes that cases are in columns and observations in rows of
data. Examine data and trafoData to see how the transformation works.
Once you have the transformed data, simply apply your favorite
clustering method that uses Euclidean distance.

HTH,

Peter

>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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[R] rank of Matrix

[Carson Farmer]

Dear list,

Can anyone tell me how to obtain the rank of a sparse Matrix, for
example from package Matrix (class dgCMatrix)? Here is an example of
QR decomposition of a sparse matrix (from the sparseQR class help).

library(Matrix)
data(KNex)
mm <- KNex$mm
str(mmQR <- qr(mm))

Similarly, using the functions/classes from the relatively new
MatrixModels package:

library(MatrixModels)
str(trial <- data.frame(counts=c(18,17,15,20,10,20,25,13,12),
outcome=gl(3,1,9,labels=LETTERS[1:3]),
treatment=gl(3,3,labels=letters[1:3])))
glmS <- glm4(counts ~ 0+outcome + treatment, poisson, trial,
  verbose = TRUE, sparse = TRUE)
str(glmS)
str(X <- glmS@pred@X)
str(QR <- qr(X))

I'm not very familiar with matrix decomposition, but is the 'p' slot
from a sparseQR object the same as $pivot from base qr? Additionally,
how do I obtain the rank of the input matrix? qr from the base package
produces an object with several components, two of which are rank, and
pivot, is there an equivalent way of getting at these values/vectors
for sparse matrices?
Basically, I am trying to get at the variance-covariance matrix in
order to compute t-values for the coefficients of a (sparse) GLM (i.e.
a summary function for class glpModel). Am I going about this the
right way? Are they perhaps more efficient ways of doing this when
working with sparse matrices. Is there a preferred way of calculating
t-statistics? Note that 'rankMatrix' from package Matrix appears to
densify the matrix before use, so this is not a viable option.

I have asked a similar question before [1], though I am now trying to
'answer' it myself by digging in and actually coding a bit more
myself, however, as you can see I'm still a bit stuck :-(

Thanks for any suggestions,

Carson

[1] http://www.mail-archive.com/r-help@r-project.org/msg130145.html

-- 
Carson J. Q. Farmer
ISSP Doctoral Fellow
National Centre for Geocomputation
National University of Ireland, Maynooth,
http://www.carsonfarmer.com/

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Re: [R] transparent grays?

[Henrique Dallazuanna]

Try this:

with(trees, symbols(Height, Volume, circles=Girth/24, inches=FALSE, fg
= "gray30",
 bg = sapply(apply(rbind(col2rgb(gray(seq(0,1, l = 10)))/255, alpha =
0.5), 2, as.list), do.call, what = rgb)))


On Thu, Mar 31, 2011 at 2:52 PM, Frostygoat  wrote:
> Is there a grayscale equivalent to alpha levels in rgb?
>
> Example:  I have the following to make red transparent circles overlap
> with previously plotted blue symbols.
>
> symbols(x=sites$long,y=sites$lat,circles=log(sites$prop.nem
> +1),add=T,inches=F,bg=rgb(red=1,green=0,blue=0,
> alpha=0.5),fg=rgb(red=1,green=0,blue=0, alpha=0.5))
>
> I'm having a hard time coming up with a grayscale equivalent.
>
> Thanks!
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

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[R] generate random numbers

[Anna Lee]

Hey List,

does anyone know how I can generate a vector of random numbers from a
given distribution? Something like "rnorm" just for non normal
distributions???

Thanks a lot!
Anna

-- 



Der Inhalt dieser E-Mail ist vertraulich. Sollte Ihnen die E-Mail
irrtümlich zugesandt worden sein, bitte ich Sie, mich unverzüglich zu
benachrichtigen und die E-Mail zu löschen.

This e-mail is confidential. If you have received it in error, please
notify me immediately and delete it from your system.

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Sequential multiple regression

[Bert Gunter]

See

http://www.stats.ox.ac.uk/pub/MASS3/Exegeses.pdf

?update.lm

-- Bert

On Thu, Mar 31, 2011 at 10:27 AM, Tyler Rinker  wrote:
> Bert and anyone else with info,
>
> First, Bert thank you for your quick reply.  drop1 gives the results as a
> type II anova.  Is there a way to make drop1 give you type I anova (the args
> don't appear to have a way to do so)?  Another package/function perhaps?
>
> Tyler
>
>> Date: Thu, 31 Mar 2011 09:32:02 -0700
>> Subject: Re: [R] Sequential multiple regression
>> From: gunter.ber...@gene.com
>> To: tyler_rin...@hotmail.com
>> CC: r-help@r-project.org
>>
>> ?drop1
>>
>> -- Bert
>>
>> On Thu, Mar 31, 2011 at 9:24 AM, Tyler Rinker 
>> wrote:
>> >
>> > Hello,
>> >
>> > In the past I have tended to reside more in the ANOVA camp but am trying
>> > to become more familiar with regression techniques in R.  I would like to
>> > get the F change from a model as I take away factors:
>> >
>> > SO...
>> >
>> > mod1<-lm(y~x1+x2+x3)...mod2<-lm(y~x1,x2)...mod3<-lm(y~x1)
>> >
>> >
>> > I can do this by hand by running several models in R and taking the
>> > MSr1/MSe1, MSr2/MSe2...  This is slow and I know there's a better way.
>> >
>> > In SPSS (which I no longer use) I could easily obtain these results
>> > (F-change) as documented by Professor Andy Fields:
>> > http://www.statisticshell.com/multireg.pdf
>> >
>> > You can see the F changes for his two IV model yielding 2 F changes.
>> >  Maybe it's the language I'm using (sequential multiple regression) that
>> > yields me poor results in searching the archives and Rseek.  The results
>> > tend to be around hierarchal regression (I'm not familiar with this
>> > terminology being in the ANOVA camp).  When I look at the hier.part package
>> > and run the examples it doesn't seem to give me the F change I'm looking
>> > for.  The step function in the base program reduces the model but takes 
>> > away
>> > the non sig. IV's (which is a great approach but I'm really after those F
>> > changes).  As is the usually the case I'm sure R does this simply and
>> > beautifully, I'm just not experienced with the statistical vocabulary and
>> > techniques around regression to find what I'm looking for.
>> >
>> > F-change values with R:  Any help would be appreciated.
>> >
>> > Thank you in advance,
>> > Tyler
>> >        [[alternative HTML version deleted]]
>> >
>> > __
>> > R-help@r-project.org mailing list
>> > https://stat.ethz.ch/mailman/listinfo/r-help
>> > PLEASE do read the posting guide
>> > http://www.R-project.org/posting-guide.html
>> > and provide commented, minimal, self-contained, reproducible code.
>> >
>>
>>
>>
>> --
>> "Men by nature long to get on to the ultimate truths, and will often
>> be impatient with elementary studies or fight shy of them. If it were
>> possible to reach the ultimate truths without the elementary studies
>> usually prefixed to them, these would not be preparatory studies but
>> superfluous diversions."
>>
>> -- Maimonides (1135-1204)
>>
>> Bert Gunter
>> Genentech Nonclinical Biostatistics
>



-- 
"Men by nature long to get on to the ultimate truths, and will often
be impatient with elementary studies or fight shy of them. If it were
possible to reach the ultimate truths without the elementary studies
usually prefixed to them, these would not be preparatory studies but
superfluous diversions."

-- Maimonides (1135-1204)

Bert Gunter
Genentech Nonclinical Biostatistics

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Re: [R] generate random numbers

[Greg Snow]

There are several non-normal distributions built in that you could just use 
their generationg functions.  There are packages (most with dist somewhere in 
the name) that allow for functions on standard distributions including 
combining together distributions.

If your distribution is not one of the above, but has a well defined inverse 
cumulative distribution then you can just generate random uniforms (runif) and 
put them into the inverse.

If none of the above works then there are still many tools available such as 
rejection sampling or metropolis-hastings sampling.

The logspline package can approximate a distribution from a dataset and 
generate values from the approximation.

We would need to know more about the actual distribution that you are 
interested in to be of more specific help.

-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111


> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Anna Lee
> Sent: Thursday, March 31, 2011 1:24 PM
> To: r-help@r-project.org
> Subject: [R] generate random numbers
> 
> Hey List,
> 
> does anyone know how I can generate a vector of random numbers from a
> given distribution? Something like "rnorm" just for non normal
> distributions???
> 
> Thanks a lot!
> Anna
> 
> --
> 
> 
> 
> Der Inhalt dieser E-Mail ist vertraulich. Sollte Ihnen die E-Mail
> irrtümlich zugesandt worden sein, bitte ich Sie, mich unverzüglich zu
> benachrichtigen und die E-Mail zu löschen.
> 
> This e-mail is confidential. If you have received it in error, please
> notify me immediately and delete it from your system.
> 
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-
> guide.html
> and provide commented, minimal, self-contained, reproducible code.

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Re: [R] generate random numbers

[Ted Harding]

On 31-Mar-11 19:23:33, Anna Lee wrote:
> Hey List,
> does anyone know how I can generate a vector of random numbers
> from a given distribution? Something like "rnorm" just for non
> normal distributions???
> 
> Thanks a lot!
> Anna

SUppose we give your distribution the name "Dist".

The generic approach would start by defining a function for
the inverse of its cumulative distribution. Call this qDist.
Then

  qDist(runif(1000))

would generate 1000 values from the distribution "Dist".

As a ready-made example, qnorm is the inverse of pnorm,
the cumulative distribution function of the Normal distribution.
Then

  qnorm(runif(1000))

would act just like rnorm(1000), though the sequence of values
would be different (a different algorithm) -- and also rnorm()
would be more efficient (being specially written).

Depending on what your desired distribution is, you may find
that an "rDist" has already been written for it. There are
many distributions already in R for which the family of
functions dDist, pDist, qDist and rDist are provided.

For more specific advice, please give us information about
the specific distribution you want to sample from!

Ted.


E-Mail: (Ted Harding) 
Fax-to-email: +44 (0)870 094 0861
Date: 31-Mar-11   Time: 20:50:52
-- XFMail --

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[R] fit.mult.impute() in Hmisc

[Yuelin Li]

I tried multiple imputation with aregImpute() and
fit.mult.impute() in Hmisc 3.8-3 (June 2010) and R-2.12.1.

The warning message below suggests that summary(f) of
fit.mult.impute() would only use the last imputed data set.
Thus, the whole imputation process is ignored.

  "Not using a Design fitting function; summary(fit) 
   will use standard errors, t, P from last imputation only.  
   Use vcov(fit) to get the correct covariance matrix, 
   sqrt(diag(vcov(fit))) to get s.e."

But the standard errors in summary(f) agree with the values
from sqrt(diag(vcov(f))) to the 4th decimal point.  It would
seem that summary(f) actually adjusts for multiple
imputation?

Does summary(f) in Hmisc 3.8-3 actually adjust for MI?

If it does not adjust for MI, then how do I get the
MI-adjusted coefficients and standard errors?

I can't seem to find answers in the documentations, including
rereading section 8.10 of the Harrell (2001) book  Googling
located a thread in R-help back in 2003, which seemed dated.
Many thanks in advance for the help,

Yuelin.
http://idecide.mskcc.org
---
> library(Hmisc)
Loading required package: survival
Loading required package: splines
> data(kyphosis, package = "rpart")
> kp <- lapply(kyphosis, function(x) 
+   { is.na(x) <- sample(1:length(x), size = 10); x })
> kp <- data.frame(kp)
> kp$kyp <- kp$Kyphosis == "present"
> set.seed(7)
> imp <- aregImpute( ~ kyp + Age + Start + Number, dat = kp, n.impute = 10, 
+  type = "pmm", match = "closest")
Iteration 13 
> f <- fit.mult.impute(kyp ~ Age + Start + Number, fitter=glm, xtrans=imp, 
+ family = "binomial", data = kp)

Variance Inflation Factors Due to Imputation:

(Intercept) Age   Start  Number 
   1.061.281.171.12 

Rate of Missing Information:

(Intercept) Age   Start  Number 
   0.060.220.140.10 

d.f. for t-distribution for Tests of Single Coefficients:

(Intercept) Age   Start  Number 
2533.47  193.45  435.79  830.08 

The following fit components were averaged over the 10 model fits:

  fitted.values linear.predictors 

Warning message:
In fit.mult.impute(kyp ~ Age + Start + Number, fitter = glm, xtrans = imp,  :
  Not using a Design fitting function; summary(fit) will use
standard errors, t, P from last imputation only.  Use vcov(fit) to get the
correct covariance matrix, sqrt(diag(vcov(fit))) to get s.e.


> f

Call:  fitter(formula = formula, family = "binomial", data = completed.data)

Coefficients:
(Intercept)  AgeStart   Number  
-3.6971   0.0118  -0.1979   0.6937  

Degrees of Freedom: 80 Total (i.e. Null);  77 Residual
Null Deviance:  80.5 
Residual Deviance: 58   AIC: 66 
> sqrt(diag(vcov(f)))
(Intercept) Age   Start  Number 
  1.5444782   0.0063984   0.0652068   0.2454408 
> -0.1979/0.0652068
[1] -3.0350
> summary(f)

Call:
fitter(formula = formula, family = "binomial", data = completed.data)

Deviance Residuals: 
   Min  1Q  Median  3Q Max  
-1.240  -0.618  -0.288  -0.109   2.409  

Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept)  -3.6971 1.5445   -2.39   0.0167
Age   0.0118 0.00641.85   0.0649
Start-0.1979 0.0652   -3.03   0.0024
Number0.6937 0.24542.83   0.0047

(Dispersion parameter for binomial family taken to be 1)

Null deviance: 80.508  on 80  degrees of freedom
Residual deviance: 57.965  on 77  degrees of freedom
AIC: 65.97

Number of Fisher Scoring iterations: 5

 
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Re: [R] Simple lattice question

[David Winsemius]

On Mar 31, 2011, at 8:41 AM, Peter Ehlers wrote:


On 2011-03-31 03:39, Rubén Roa wrote:


DeaR ComRades,

require(lattice)
data<- data.frame(SP=sort(rep(as.factor(c('A','B','C','D','E')),12)),
   x=rpois(60,10),
   y=rep(c(rep(0,4),rep(10,4),rep(20,4)),5),
   z=rep(1:4,15))
xyplot(x~y| 
SP 
,data 
=data,groups=z,layout=c(2,3),pch=1:4,lty=1:4,col='black',type='b')


How do I put a legend for the grouping variable in the empty upper- 
right panel?


See the help page for xyplot for an example using the iris data.
You just need to add something like

auto.key = list(x = .6, y = .7, corner = c(0, 0))

to your lattice call. See the 'key' entry on the ?xyplot page.


On my machine the default color scheme is TRUE, so the auto.key comes  
out with colored circles. This incantation gets the key to match the  
plotting symbols and lines:


xyplot(x~y|SP, data=data,groups=z, layout=c(2,3),  
par.settings=simpleTheme(pch=1:4,lty=1:4,col='black'), type="b",

auto.key = list(x = .6, y = .7, lines=TRUE, corner = c(0, 0)) )



Peter Ehlers



TIA

Rubén



Dr. Rubén Roa-Ureta
AZTI - Tecnalia / Marine Research Unit
Txatxarramendi Ugartea z/g
48395 Sukarrieta (Bizkaia)
SPAIN

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David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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[R] Effects - plot the marginal effect

[Tomii]

Hello,

I try to plot the marginal effect by using package "effects" (example of the
graph i want to get is in the attached picture).
All variables are continuous.

Here is regression function, results and error effect function gives:

> mreg01 = lm(a90$enep1 ~ a90$enpres + a90$proximity1 + (a90$enpres * 
> a90$proximity1), data=a90)> summary(mreg01)
Call:
lm(formula = a90$enep1 ~ a90$enpres + a90$proximity1 + (a90$enpres *
a90$proximity1), data = a90)

Residuals:
Min  1Q  Median  3Q Max
-2.3173 -1.3349 -0.5713  0.8938  8.1084

Coefficients:
  Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.2273 0.3090  13.683  < 2e-16 ***
a90$enpres  0.4225 0.2319   1.822 0.072250 .
a90$proximity1 -3.8797 1.0984  -3.532 0.000696 ***
a90$enpres:a90$proximity1   0.8953 0.4101   2.183 0.032025 *
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 2.029 on 78 degrees of freedom
Multiple R-squared: 0.2128, Adjusted R-squared: 0.1826
F-statistic: 7.031 on 3 and 78 DF,  p-value: 0.0003029
> plot(effect(a90$enpres:a90$proximity1, mreg01))Warning messages:1: In 
> a90$enpres:a90$proximity1 :
  numerical expression has 82 elements: only the first used2: In
a90$enpres:a90$proximity1 :
  numerical expression has 82 elements: only the first used3: In
analyze.model(term, mod, xlevels, default.levels) :
  0 does not appear in the modelError in
plot(effect(a90$enpres:a90$proximity1, mreg01)) :
  error in evaluating the argument 'x' in selecting a method for function 'plot'

>

Thanks in advance.
Tomas
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[R] nlme for exact binomial model

[Nonyane, Bareng A.S.]

Hi
I am conducting a meta-analysis on proportions and I'm interested in using the 
exact binomial likeliood for the within-study effects rather than the 
approximate approaches  and I don't want to use the quasi-likelihood approach 
in R's glmmPQL.
Does anyone know how I can do this using nlme? I can do it in SAS nlmixed but  
I would rather use R -this is my SAS code:

proc nlmixed data=work.temp   ;
parms  mup=2 vp= 08;  /*initial values*/
rawprop=1/(1+exp(-truep)); /*unkown true p logit proportion*/
model  numberofisolates~binomial(sumisolatestudy, rawprop) ;
random truep~normal(mup,vp) subject=studyid;
run;

Thanks

B. Aletta Nonyane, PhD
Assistant Scientist
Department of International Health
Johns Hopkins Bloomberg School of Public Health
615 N. Wolfe Street
Baltimore, MD, 21205


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Re: [R] VECM with UNRESTRICTED TREND

[renoir vieira]

Dear Pfaff,

Would that be possible to fit a Time varying VECM using urca?

Yours,
Renoir

On Thursday, March 31, 2011, Grzegorz Konat  wrote:
> The code you gave me works fine with Finland, but the same for my data -
> does not!
> I do:
>
> library(urca)
> data(my.data)
> dat1 <- my.data[, c("dY", "X", "dM")]
> trend <- matrix(1:nrow(dat1), ncol = 1)
> colnames(trend) <- "trd"
> yxm.vecm <- ca.jo(dat1, type = "trace", ecdet = "const", K = 2, spec =
> "longrun", dumvar = trend)
>
> and the result is again:
>
> Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), , drop =
> FALSE] :
>   non-numeric argument to binary operator
>
> I attach my dataset in xls format. If you have 5 minutes and wish to check
> it out, I'd be extremely grateful!
>
> Best,
> Greg
>
>
>
> 2011/3/31 Pfaff, Bernhard Dr. 
>
>>  Well, without further information, I do not know, but try the following
>>
>> library(urca)
>> example(ca.jo)
>> trend <- matrix(1:nrow(sjf), ncol = 1)
>> colnames(trend) <- "trd"
>> ca.jo(sjf, type = "trace", ecdet = "const", K = 2, spec = "longrun",
>> dumvar = trend)
>>
>> Best,
>> Bernhard
>>
>>
>>
>>  --
>> *Von:* Grzegorz Konat [mailto:grzegorz.ko...@ibrkk.pl]
>> *Gesendet:* Donnerstag, 31. März 2011 14:40
>>
>> *An:* Pfaff, Bernhard Dr.; r-help@r-project.org
>> *Betreff:* Re: [R] VECM with UNRESTRICTED TREND
>>
>> 'time' was a trend variable from my.data set. Equivalent to the output of
>> the command 'matrix' you just gave me.
>>
>> So now I did:
>>
>>  library(urca)
>> data(my.data)
>> names(my.data)
>> attach(my.data)
>> dat1 <- my.data[, c("dY", "X", "dM")]
>> mat1 <- matrix(seq(1:nrow(dat1)), ncol = 1)
>> args('ca.jo')
>> yxm.vecm <- ca.jo(dat1, type = "trace", ecdet = "const", K = 2, spec =
>> "longrun", dumvar=mat1)
>>
>> and the output is:
>>
>>  Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), , drop
>> = FALSE] :
>>   non-numeric argument to binary operator
>> In addition: Warning message:
>> In ca.jo(dat1, type = "trace", ecdet = "const", K = 2, spec = "longrun",
>>  :
>> No column names in 'dumvar', using prefix 'exo' instead.
>>
>> What do I do wrong?
>>
>> Best,
>> Greg
>>
>>
>> 2011/3/31 Pfaff, Bernhard Dr. 
>>
>>>
>>>
>>>
>>>  Hello Bernhard,
>>>
>>> thank You so much one again! Now I (more or less) understand the idea, but
>>> still have problem with its practical application.
>>>
>>> I do (somewhat following example 8.1 in your textbook):
>>>
>>>  library(urca)
>>> data(my.data)
>>> names(my.data)
>>> attach(my.data)
>>> dat1 <- my.data[, c("dY", "X", "dM")]
>>> dat2 <- cbind(time)
>>>
>>> What is 'time'? Just employ matrix(seq(1:nrow(dat1)), ncol = 1) for
>>> creating the trend variable.
>>>
>>> Best,
>>> Bernhard
>>>
>>>
>>>  args('ca.jo')
>>> yxm.vecm <- ca.jo(dat1, type = "trace", ecdet = "trend", K = 2, spec =
>>> "longrun", dumvar=dat2)
>>>
>>> The above code produces following output:
>>>
>>>  Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), ,
>>> drop = FALSE] :
>>>   non-numeric argument to binary operator
>>>
>>> What does that mean? Should I use cbind command to dat1 as well? And
>>> doesn't it transform the series into series of integer numbers?
>>>
>>> Thank you once again (especially for your patience).
>>>
>>> Best,
>>> Greg
>>>
>>>
>>>
>>> 2011/3/31 Pfaff, Bernhard Dr. 
>>>
  Hello Greg,

 you include your trend as a (Nx1) matrix and use this for 'dumvar'. The
 matrix 'dumvar' is just added to the VECM as deterministic regressors and
 while you are referring to case 5, this is basically what you are after, if
 I am not mistaken. But we aware that this implies a quadratic trend for the
 levels

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[R] Create Variable names dynamically

[Noah Silverman]

Hi,

I want to create variable names from within my code, but can't find any 
documentation for this.

An example is probably the best way to illustrate. I am reading data in from a 
file, doing a bunch of stuff, and want to generate variables with my output.  
(I could make a "list of lists" and name all the elements, but I really want 
separate variables.)


#
#This is just a dummy example, please excuse any shortcuts...

data <- read.table("file", )
animals <- (data[,animal])  
animals
> "cat", "dog", "horse"  # Not known what these are before I read the data file

# do a bunch of stuff

mean_cat <- abc
var_cat <- dfd
mean_dog <- 123
var_dog <- 453
etc..
##

I thought of trying to use the paste() function to create the variable name, 
but that doesn't work:
for( animal in animals){
paste("mean", animal "_") <- 123
}

Any ideas???

Thanks


--
Noah Silverman

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Re: [R] Create Variable names dynamically

[Henrique Dallazuanna]

Take a look in ?assign

On Thu, Mar 31, 2011 at 5:42 PM, Noah Silverman  wrote:
> Hi,
>
> I want to create variable names from within my code, but can't find any 
> documentation for this.
>
> An example is probably the best way to illustrate. I am reading data in from 
> a file, doing a bunch of stuff, and want to generate variables with my 
> output.  (I could make a "list of lists" and name all the elements, but I 
> really want separate variables.)
>
>
> #
> #This is just a dummy example, please excuse any shortcuts...
>
> data <- read.table("file", )
> animals <- (data[,animal])
> animals
>> "cat", "dog", "horse"  # Not known what these are before I read the data file
>
> # do a bunch of stuff
>
> mean_cat <- abc
> var_cat <- dfd
> mean_dog <- 123
> var_dog <- 453
> etc..
> ##
>
> I thought of trying to use the paste() function to create the variable name, 
> but that doesn't work:
> for( animal in animals){
>        paste("mean", animal "_") <- 123
> }
>
> Any ideas???
>
> Thanks
>
>
> --
> Noah Silverman
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

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Re: [R] read password-protected files

[Shi, Tao]

Thanks, Henrique!  I'll try that.

...Tao



- Original Message 
> From: Henrique Dallazuanna 
> To: "Shi, Tao" 
> Cc: Prof Brian Ripley ; r-help@r-project.org
> Sent: Thu, March 31, 2011 12:00:45 PM
> Subject: Re: [R] read password-protected files
> 
> Using RDCOMCliente:
> 
> library(RDCOMClient)
> eApp <-  COMCreate("Excel.Application")
> wk <-  eApp$Workbooks()$Open(Filename="your_file",Password="your_password")
> tf <-  tempfile()
> wk$Sheets(1)$SaveAs(tf, 3)
> 
> DF <-  read.table(sprintf("%s.txt", tf), header = TRUE, sep = "\t")
> 
> On Thu, Mar  31, 2011 at 3:33 PM, Shi, Tao  wrote:
> > Hi  Josh and Prof. Ripley,
> >
> > Thanks for the quick  replies!
> >
> > Sorry about the HTML email.  It's the default for my  yahoo account and 
>forgot to
> > switch.
> >
> > The question was  asked by a colleague of mine.  After double-checking, 
> > yes, 
>they
> > are  actually Excel files.  Any idea on how to approach this?
> >
> >  Thanks!
> >
> > ...Tao
> >
> >
> >
> >  
> > From: Prof Brian Ripley 
> >
> >  Cc: r-help@r-project.org
> > Sent: Thu,  March 31, 2011 11:15:35 AM
> > Subject: Re: [R] read password-protected  files
> >
> > On Thu, 31 Mar 2011, Shi, Tao wrote:
> >
> >>  Hi list,
> >>
> >> I have a bunch of .csv files that are  password-protected.  I wonder if 
>there
> > is
> >
> > There is  nothing in the CSV standard about password protection.  I
> > very much  doubt that these actually CSV files.  So please follow the
> > posting guide  [*]
> >
> >> a way to read them in in R without manually removing the  password 
>protection
> >>for
> >> each file?
> >>
> >  [[elided Yahoo spam]]
> >>
> >>
> >> ...Tao
> >>  [[alternative HTML version deleted]]
> >>
> >>  __
> >> R-help@r-project.org mailing  list
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >>  PLEASE do read the posting guide  
>http://www.R-project.org/posting-guide.html
> >> and provide commented,  minimal, self-contained, reproducible code.
> >
> > [*] as above: and as  you sent HTML you are either full of contempt for
> > the helpeRs or failed  to do your homework.
> >
> >
> > --
> > Brian D. Ripley,   rip...@stats.ox.ac.uk
> > Professor  of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
> > University of  Oxford, Tel:  +44 1865 272861 (self)
> > 1 South Parks Road,  +44 1865 272866 (PA)
> > Oxford OX1 3TG, UK Fax:  +44 1865 272595
> >
> >  __
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the  posting guide 
http://www.R-project.org/posting-guide.html
> > and  provide commented, minimal, self-contained, reproducible  code.
> >
> 
> 
> 
> -- 
> Henrique  Dallazuanna
> Curitiba-Paraná-Brasil
> 25° 25' 40" S 49° 16' 22" O
>

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Re: [R] Create Variable names dynamically

[Noah Silverman]

Perfect.

Thank You

On Mar 31, 2011, at 1:47 PM, Mark Leeds wrote:

> hi noah: assign(thing you paste, 123, envir=whatever) should work I think.
> 
> 
> On Thu, Mar 31, 2011 at 4:42 PM, Noah Silverman  
> wrote:
> Hi,
> 
> I want to create variable names from within my code, but can't find any 
> documentation for this.
> 
> An example is probably the best way to illustrate. I am reading data in from 
> a file, doing a bunch of stuff, and want to generate variables with my 
> output.  (I could make a "list of lists" and name all the elements, but I 
> really want separate variables.)
> 
> 
> #
> #This is just a dummy example, please excuse any shortcuts...
> 
> data <- read.table("file", )
> animals <- (data[,animal])
> animals
> > "cat", "dog", "horse"  # Not known what these are before I read the data 
> > file
> 
> # do a bunch of stuff
> 
> mean_cat <- abc
> var_cat <- dfd
> mean_dog <- 123
> var_dog <- 453
> etc..
> ##
> 
> I thought of trying to use the paste() function to create the variable name, 
> but that doesn't work:
> for( animal in animals){
>paste("mean", animal "_") <- 123
> }
> 
> Any ideas???
> 
> Thanks
> 
> 
> --
> Noah Silverman
> 
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
> 


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[R] Is it possible to run unit tests after a package installation?

[Juan Carlos Borrás]

Dear all,
I'm deep into Chambers' SoDA and R-exts.html but I can't find all
answers. The thing is that I would like to run my unit tests right
after a package installation.
That is while the command "R CMD check " runs all files
named /tests/*.R (so unit tests can be placed there) I
wonder if it is possible to run them on some destination/target
machine after the package has been installed.
My goal is to enforce reliable results from calls to the package
functions independently on the machine the package is being installed.
 The only way I can come up with (and is not even close to a package
post-install check) is to place them in the directory where examples
can be show using the example() command. But
package_name/tests/Examples does not look as the place it should be
(or at least it does not work for me for the time being).
One of the difficulties I foresee also is how to state that my unit
test framework of choice (be RUnit or testthat depending on my mood)
is a dependency but not something that must be loaded before my
package does. That is, it must be installed in the system but it
should not be loaded.
Any hints or comments on the above will be highly appreciated.
Thanks in advance.
jcb!

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Re: [R] generate random numbers

[David Scott]

 On 01/04/11 08:50, Ted Harding wrote:

On 31-Mar-11 19:23:33, Anna Lee wrote:

Hey List,
does anyone know how I can generate a vector of random numbers
from a given distribution? Something like "rnorm" just for non
normal distributions???

Thanks a lot!
Anna

SUppose we give your distribution the name "Dist".

The generic approach would start by defining a function for
the inverse of its cumulative distribution. Call this qDist.
Then

   qDist(runif(1000))

would generate 1000 values from the distribution "Dist".

As a ready-made example, qnorm is the inverse of pnorm,
the cumulative distribution function of the Normal distribution.
Then

   qnorm(runif(1000))

would act just like rnorm(1000), though the sequence of values
would be different (a different algorithm) -- and also rnorm()
would be more efficient (being specially written).

Depending on what your desired distribution is, you may find
that an "rDist" has already been written for it. There are
many distributions already in R for which the family of
functions dDist, pDist, qDist and rDist are provided.

For more specific advice, please give us information about
the specific distribution you want to sample from!

Ted.



I can point to one general implementation which might be helpful, and 
even the function names are the same.


In the version of DistributionUtils on R-Forge you will find functions 
pDist and qDist which should give the distribution function and quantile 
function of any continuous unimodal distribution.


Provisos: there may be problems with distributions with very heavy 
tails, and generally the routines could be slow.


David Scott

--
_
David Scott Department of Statistics
The University of Auckland, PB 92019
Auckland 1142,NEW ZEALAND
Phone: +64 9 923 5055, or +64 9 373 7599 ext 85055
Email:  d.sc...@auckland.ac.nz,  Fax: +64 9 373 7018

Director of Consulting, Department of Statistics

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[R] Linear Model with curve fitting parameter?

[stephen sefick]

I have a model Q=K*A*(R^r)*(S^s)

A, R, and S are data I have and K is a curve fitting parameter.  I
have linearized as

log(Q)=log(K)+log(A)+r*log(R)+s*log(S)

I have taken the log of the data that I have and this is the model
formula without the K part

lm(Q~offset(A)+R+S, data=x)

What is the formula that I should use?

Thanks for all of your help.  I can provide a subset of data if necessary.



-- 
Stephen Sefick

| Auburn University                                         |
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|___|
| sas0...@auburn.edu                                  |
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Let's not spend our time and resources thinking about things that are
so little or so large that all they really do for us is puff us up and
make us feel like gods.  We are mammals, and have not exhausted the
annoying little problems of being mammals.

                                -K. Mullis

"A big computer, a complex algorithm and a long time does not equal science."

                              -Robert Gentleman
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Re: [R] Create Variable names dynamically

[jim holtman]

The best thing to do is to understand how to use 'list' for this
purpose.  Much easier to handle the information.

On Thu, Mar 31, 2011 at 4:42 PM, Noah Silverman  wrote:
> Hi,
>
> I want to create variable names from within my code, but can't find any 
> documentation for this.
>
> An example is probably the best way to illustrate. I am reading data in from 
> a file, doing a bunch of stuff, and want to generate variables with my 
> output.  (I could make a "list of lists" and name all the elements, but I 
> really want separate variables.)
>
>
> #
> #This is just a dummy example, please excuse any shortcuts...
>
> data <- read.table("file", )
> animals <- (data[,animal])
> animals
>> "cat", "dog", "horse"  # Not known what these are before I read the data file
>
> # do a bunch of stuff
>
> mean_cat <- abc
> var_cat <- dfd
> mean_dog <- 123
> var_dog <- 453
> etc..
> ##
>
> I thought of trying to use the paste() function to create the variable name, 
> but that doesn't work:
> for( animal in animals){
>        paste("mean", animal "_") <- 123
> }
>
> Any ideas???
>
> Thanks
>
>
> --
> Noah Silverman
>
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?

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[R] regular expression

[array chip]

Hi, I am stuck on this: how to specify a match pattern that means not to 
include 
"abc"?

I tried:

grep("^(abc)", "hello", value=T)
should return "hello". 

while 

grep("^(abc)", "hello abcd foo", value=T)
should return character(0).

But both returned character(0).

Thanks

John

[[alternative HTML version deleted]]

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Re: [R] rank of Matrix

[Carson Farmer]

Hmm, looks like I'm 'answering' my own question here...

library(Matrix)
data(KNex)
mm <- KNex$mm
str(mmQR <- qr(mm))
# new stuff:
R <- mmQR@R
Rdiag <- diag(R)
rank <- sum(Rdiag > max(dim(mm))*.Machine$double.eps*abs(R[1,1])) #
this is the matlab default I think?
# 712

for comparison, rankMatrix from package Matrix gives the same answer,
but takes considerably more time and memory!
rankMatrix(mm)
# 712

Does the above make sense to others? Alternatively, is it possible to
derive rank from the model itself (I can't see how)?

library(MatrixModels)
trial <- data.frame(counts=c(18,17,15,20,10,20,25,13,12),
                       outcome=gl(3,1,9,labels=LETTERS[1:3]),
                       treatment=gl(3,3,labels=letters[1:3]))
glmS <- glm4(counts ~ 0+outcome + treatment, family=poisson, data=trial,
                     verbose = TRUE, sparse = TRUE)
str(glmS)


-- 
Carson J. Q. Farmer
ISSP Doctoral Fellow
National Centre for Geocomputation
National University of Ireland, Maynooth,
http://www.carsonfarmer.com/

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Re: [R] regular expression

[Bernd Weiss]

Am 31.03.2011 19:31, schrieb array chip:

Hi, I am stuck on this: how to specify a match pattern that means not
to include "abc"?

I tried:

grep("^(abc)", "hello", value=T) should return "hello".


> grep("[^(abc)]", "hello", value=T)
[1] "hello"


HTH,

Bernd

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Re: [R] regular expression

[array chip]

Thanks Bernd! I tried your approach with my real example, sometimes it worked, 
sometimes it didn't. For example

grep('[^(arg)]\\.symptom',"stomach.symptom",value=T)
[1] "stomach.symptom"

grep('[^(arg)]\\.symptom',"liver.symptom",value=T)
character(0)

I think both examples should return the text, but the 2nd example didn't.

What was wrong here?

Thanks

John





From: Bernd Weiss 

Sent: Thu, March 31, 2011 5:32:25 PM
Subject: Re: [R] regular expression

Am 31.03.2011 19:31, schrieb array chip:
> Hi, I am stuck on this: how to specify a match pattern that means not
> to include "abc"?
>
> I tried:
>
> grep("^(abc)", "hello", value=T) should return "hello".

> grep("[^(abc)]", "hello", value=T)
[1] "hello"


HTH,

Bernd

[[alternative HTML version deleted]]

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