Re: [R] Do loop

[sundar]

Gud evening sir ,I want do the cluster analysis algorithm in r software can
u guide me sir 

My mail id is :sundars...@gmail.com

And I want the brief explanation to for,do.while,if etc loops and
conditions.

Thank you sir.


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[R] (no subject)

[sundar]

Gud evening sir ,I want do the cluster analysis algorithm in r software  
can u guide me sir

My mail id is :sundars...@gmail.com
And I want the brief explanation to for,do.while,if etc loops and  
conditions.

Thank you sir.

--

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[R] Bayesian cox model: spBayesSurv package

[radhika sundar]

I am going through R's function indeptCoxph in the spBayesSurv package
which fits a bayesian Cox model. I am confused by some of the input
parameters to this function.

What is the role of the "prediction" input parameter? Should it not only
contain the predictor covariates? In the R code example, the authors have
included a vector "s" which was used to initially simulate the survival
times data in their example as well as the predictors. I'm not sure what
this "s" is.

Given that my data is just a set of survival times between 0 and 100, along
with censored (yes/no) information, how would I use this function and how
should I handle the input "s"?


Thanks for any help!

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Re: [R] Bayesian cox model: spBayesSurv package

[radhika sundar]

The example code for the indeptCoxph function  in the spBayesSurv package
has been updated(this is not cross-posted anywhere else). See code below.
The author simulates data to illustrate the Cox model -  I am stuck trying
to understand the role of the functions f0oft, S0oft, fioft and Sioft as
also Finv.

Am I right in thinking that he is only using the above mentioned functions
to simulate his data? If I wanted to run the indeptCoxph function with
different data, do I need to define those functions again?
Roughly, in the author's example, I can understand he fits a Cox model with
2 predictors x1 and x2. He simulates survival time data (but this is where
I am confused).
As for the bayesian model itself, the only prior he uses is for M, the
number of cutpoints in the baseline hazard function. (There is no function
listed as a prior for Survival times in the ideptCoxph call).

Sorry --- this is a novice question relating to understanding both the
statistical set-up and the R-code.
Thanks for any help!
Author's(updated)  code:
###
# A simulated data: Cox PH
###
rm(list=ls())
library(survival)
library(spBayesSurv)
library(coda)
library(MASS)
## True parameters
betaT = c(1,1);
n=500; npred=30; ntot=n+npred;
## Baseline Survival
f0oft = function(t) 0.5*dlnorm(t, -1, 0.5)+0.5*dlnorm(t,1,0.5);
S0oft = function(t) (0.5*plnorm(t, -1, 0.5, lower.tail=FALSE)+
   0.5*plnorm(t, 1, 0.5, lower.tail=FALSE))
## The Survival function:
Sioft = function(t,x)  exp( log(S0oft(t))*exp(sum(x*betaT)) ) ;
fioft = function(t,x) exp(sum(x*betaT))*f0oft(t)/S0oft(t)*Sioft(t,x);
Fioft = function(t,x) 1-Sioft(t,x);
## The inverse for Fioft
Finv = function(u, x) uniroot(function (t) Fioft(t,x)-u, lower=1e-100,
  upper=1e100, extendInt ="yes",
tol=1e-6)$root

## generate x
x1 = rbinom(ntot, 1, 0.5); x2 = rnorm(ntot, 0, 1); X = cbind(x1, x2);
## generate survival times
u = runif(ntot);
tT = rep(0, ntot);
for (i in 1:ntot){
  tT[i] = Finv(u[i], X[i,]);
}

## right censoring
t_obs=tT
Centime = runif(ntot, 2, 6);
delta = (tT<=Centime) +0 ;
length(which(delta==0))/ntot; # censoring rate
rcen = which(delta==0);
t_obs[rcen] = Centime[rcen]; ## observed time
## make a data frame
dtotal = data.frame(t_obs=t_obs, x1=x1, x2=x2, delta=delta,
tT=tT);
## Hold out npred=30 for prediction purpose
predindex = sample(1:ntot, npred);
dpred = dtotal[predindex,];
dtrain = dtotal[-predindex,];

# Prediction settings
xpred = cbind(dpred$x1,dpred$x2);
prediction = list(xpred=xpred);

###
# Independent Cox PH
###
# MCMC parameters
nburn=1000; nsave=1000; nskip=0;
# Note larger nburn, nsave and nskip should be used in practice.
mcmc=list(nburn=nburn, nsave=nsave, nskip=nskip, ndisplay=1000);
prior = list(M=10);
state <- NULL;
# Fit the Cox PH model
res1 = indeptCoxph( y = dtrain$t_obs, delta =dtrain$delta,
x = cbind(dtrain$x1, dtrain$x2),RandomIntervals=FALSE,
prediction=prediction,  prior=prior, mcmc=mcmc,
state=state);
save.beta = res1$beta; row.names(save.beta)=c("x1","x2")
apply(save.beta, 1, mean); # coefficient estimates
apply(save.beta, 1, sd); # standard errors
apply(save.beta, 1, function(x) quantile(x, probs=c(0.025, 0.975))) # 95% CI
## traceplot
par(mfrow = c(2,1))
traceplot(mcmc(save.beta[1,]), main="beta1")
traceplot(mcmc(save.beta[2,]), main="beta2")
res1$ratebeta; # adaptive MH acceptance rate
## LPML
LPML1 = sum(log(res1$cpo)); LPML1;
## MSPE
mean((dpred$tT-apply(res1$Tpred, 1, median))^2);

## plots
par(mfrow = c(2,1))
x1new = c(0, 0);
x2new = c(0, 1)
xpred = cbind(x1new, x2new);
nxpred = nrow(xpred);
tgrid = seq(1e-10, 4, 0.03);
ngrid = length(tgrid);
estimates = GetCurves(res1, xpred, log(tgrid), CI=c(0.05, 0.95));
fhat = estimates$fhat;
Shat = estimates$Shat;
## density in t
plot(tgrid, fioft(tgrid, xpred[1,]), "l", lwd=2,  ylim=c(0,3),
main="density")
for(i in 1:nxpred){
  lines(tgrid, fioft(tgrid, xpred[i,]), lwd=2)
  lines(tgrid, fhat[,i], lty=2, lwd=2, col=4);
}
## survival in t
plot(tgrid, Sioft(tgrid, xpred[1,]), "l", lwd=2, ylim=c(0,1),
main="survival")
for(i in 1:nxpred){
  lines(tgrid, Sioft(tgrid, xpred[i,]), lwd=2)
  lines(tgrid, Shat[,i], lty=2, lwd=2, col=4);
  lines(tgrid, estimates$Shatup[,i], lty=2, lwd=1, col=4);
  lines(tgrid, estimates$Shatlow[,i], lty=2, lwd=1, col=4);
}


On Wed, Dec 28, 2016 at 2:11 AM, David Winsemius 
wrote:

> Cross posting is deprecated on rhelp but if you do so, please at least
> post a link to the stackoverflow address for the duplicate question.
>
> Sent from my iPhone
>
> > On Dec 27, 2016, at 6:52 AM, radhika sundar 
&g

[R] need some help finding power in test about variances

[ati sundar]

Hello All

I am new to this list. I have a problem where for a single sample drawn from 
normal population, null hypothesis is that variance = k (say). Alternative 
hypothesis is variance > k. Now if we know the true variance, then I would like 
to calculate the sample size required to produce certain power (for some
significance). How do I do this ? I thought of using pwr.chisq.test, and I 
contacted the author stephane champely, but he said his package can't do this. 
Does anybody have an idea ?

Thanks
Ati

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[R] problem about finding power in test about variances

[ati sundar]

Hello All

I am new to this list. I have a problem where for a single sample drawn from 
normal population, null hypothesis is that variance = k (say). Alternative 
hypothesis is variance > k. Now if we know the true variance, then I would like 
to calculate the sample size required to produce certain power (for some
significance). How do I do this ? I thought of using pwr.chisq.test, and I 
contacted the author stephane champely, but he said his package can't do this. 
Does anybody have an idea ?

Thanks
Ati

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Re: [R] problem about finding power in test about variances

[ati sundar]

Ok sorry Uwe.

Let me put a specific problem. I have some data for which sample size
n=15 and sample standard deviation s=0.008 mm. Now there are two parts of
the question. Here the random variable is diameter of rivet hole.
a) Is there a strong evidence to indicate that the standard deviation
of hole diameter exceeds 0.01 mm. Use aplha=0.05 .
 Now here Ho: variance = (0.01)^2 = 0.0001 mm
  H1: variance > 0.0001 mm
we can use test statistic  chisq = (n-1)*s^2 / 0.0001

Chisq has degree of freedom n-1=14. So using this test  statistic
we can check the null hypothesis. I can do this part. Now lets
look at the next part.

b)If true variance (sigma) is as large as 0.0125 mm, what sample size will
be required to detect this with power of at least 0.8 

Now this is the part where I am stuck and I want to use R to solve this
I know that R has power tests for other test statistics, but how do I solve 
this ?

Ati


--- On Thu, 4/30/09, Uwe Ligges  wrote:

> From: Uwe Ligges 
> Subject: Re: [R] problem about finding power in test about variances
> To: atisun...@yahoo.com
> Cc: r-help@r-project.org
> Date: Thursday, April 30, 2009, 4:37 PM
> ati sundar wrote:
> > Hello All
> > 
> > I am new to this list. I have a problem where for a
> single sample drawn from normal population, null hypothesis
> is that variance = k (say).
> 
> Perhaps you want to tell us more precisely what you are
> going to do rather than asking the same inprecise question
> again and again (including messages to single persons on the
> list).
> Is k fixed or the variance? What are your data?  Specify a
> valid Null / Alternative combination. If you Null is
> variance = k (say) then you alternative is variance != k or
> if the alternative is variance > k, then your null is
> probably variance <= k
> 
> Uwe Ligges
> 
> 
> 
>  Alternative hypothesis is variance > k. Now if we know
> the true variance, then I would like to calculate the sample
> size required to produce certain power (for some
> > significance). How do I do this ? I thought of using
> pwr.chisq.test, and I contacted the author stephane
> champely, but he said his package can't do this. Does
> anybody have an idea ?
> > 
> > Thanks
> > Ati
> > 
> > __
> > 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] upgrade to 2.5

[Sundar Dorai-Raj]

Iasonas Lamprianou said the following on 5/2/2007 8:25 AM:
> Hi I am using R version 2.4.1. How can I upgrade to version 2.5 without 
> having to install all the packages again? 
> Thanks
> Jason
>  

You may find the following link relevant.

http://finzi.psych.upenn.edu/R/Rhelp02a/archive/75359.html

HTH,

--sundar

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Re: [R] use expression() in a loop

[Sundar Dorai-Raj]

Nanye Long said the following on 8/18/2008 3:00 PM:

Hi all,

I want to do plot() in a loop to make 10 graphs, so I have some code like

for (i in 1:10) {
   plot(... ... , xlab = expression(g[i]) )
}

I expect g_1, g_2, and so on appear on x labels, but it simply prints
g_i for each graph. Does anybody know how to get around this problem?
Thanks.

NL

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Try:

par(mfrow = c(2, 5))
for (i in 1:10) {
  plot(1, 1, xlab = bquote(g[.(i)]))
}

HTH,

--sundar

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Re: [R] Lattice graph tweaking

[Sundar Dorai-Raj]

A reproducible example would be nice.

Try grid = FALSE for the first question, though I'm unaware which
lattice plot you are using where the default is TRUE. So I can't
guarantee that will even work.

For your second question, add

par.settings = list(strip.background = list(col = "white"))

to your call (e.g. xyplot(..., par.settings = ...)).

To see what other parameters you can set in par.settings, try:

str(trellis.par.get())

HTH,

--sundar

On Tue, Aug 25, 2009 at 6:14 AM, Wallis,
David wrote:
> To: silwood-r
> Subject: Removing lattice graph gridlines and editing label box colour
>
> Hi,
>
> Is it possible to remove the background gridlines from a lattice graph (ie 
> graph made up of multiple individual graphs with annoying blue grid in the 
> backgroun)?
>
> Also, Is it possible to change the colour of the individual graph label 
> boxes? - ie the default pink boxes above the individual graphs
>
> Thanks
>
>        [[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.
>

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

Re: [R] Help with nls and error messages singular gradient

[Sundar Dorai-Raj]

Hi, Michael,

I think the SPSS answer is wrong. Your starting values are way off.
Look at this plot for verification:

con <- textConnection("time  bod
11 0.47
22 0.74
33 1.17
44 1.42
55 1.60
67 1.84
79 2.19
8   11 2.17")
mydata <- read.table(con, header = TRUE)
close(con)

beta <- c(3, -0.1) # your initial values
beta <- c(2.4979, -2.02456)  # SPSS answer
mycurve <- function(x) {
  beta[1]/(1 - exp(beta[1] * x))
}
curve(mycurve, from = 1, to = 11,
  ylim = range(mydata$bod, mycurve(mydata$time)))
points(mydata$time, mydata$bod)

You might want to look at package nls2 which allows a brute force grid
search to find some starting values. Or rethink the equation you're
trying to fit.

HTH,

--sundar

On Tue, Aug 25, 2009 at 9:10 AM, Michael Pearmain wrote:
> Hi All,
>
> I'm trying to run nls on the data from the study by Marske (Biochemical
> Oxygen Demand Interpretation Using Sum of Squares Surface. M.S. thesis,
> University of Wisconsin, Madison, 1967) and was reported in Bates and Watts
> (1988).
>
> Data is as follows, (stored as mydata)
>
>  time  bod
> 1    1 0.47
> 2    2 0.74
> 3    3 1.17
> 4    4 1.42
> 5    5 1.60
> 6    7 1.84
> 7    9 2.19
> 8   11 2.17
>
> I then run the following;
> #Plot initial curve
> plot(mydata$time, mydata$bod,xlab="Time (in days)",ylab="biochemical oxygen
> demand (mg/l) ")
>
> model <- nls(bod ~ beta1/(1 - exp(beta2*time)), data =
> mydata, start=list(beta1 = 3, beta2 = -0.1),trace=T)
>
> The start values are recommended, (I have used these values in SPSS without
> any problems, SPSS returns values of Beta1 = 2.4979 and Beta2 = -2.02 456)
>
> but return the error message,
> Error in nls(bod ~ beta1/(1 - exp(beta2 * time)), data = mydata, start =
> list(beta1 = 3,  : singular gradient
>
> Can anyone offer any advice?
>
> Thanks in advance
>
> Mike
>
>
>
>
>
>
>
>
>
> --
> Michael Pearmain
> Senior Analytics Research Specialist
>
> “Statistics are like women; mirrors of purest virtue and truth, or like
> whores to use as one pleases”
>
> Google UK Ltd
> Belgrave House
> 76 Buckingham Palace Road
> London SW1W 9TQ
> United Kingdom
> t +44 (0) 2032191684
> mpearm...@google.com
>
> If you received this communication by mistake, please don't forward it to
> anyone else (it may contain confidential or privileged information), please
> erase all copies of it, including all attachments, and please let the sender
> know it went to the wrong person. Thanks.
>
>        [[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.
>
>

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

Re: [R] Help with nls and error messages singular gradient

[Sundar Dorai-Raj]

Another alternative is to use SSlogis which is very similar to the
model you're fitting except with one additional parameter:

Asym <- 3
xmid <- 0
scal <- 10
model <- nls(bod ~ SSlogis(time, Asym, xmid, scal), data = mydata)
summary(model)
plot(bod ~ time, mydata)
newdata <- data.frame(time = seq(1, 11, length = 100))
lines(newdata$time, predict(model, newdata))

HTH,

--sundar

On Tue, Aug 25, 2009 at 10:05 AM, Sundar Dorai-Raj wrote:
> Hi, Michael,
>
> I think the SPSS answer is wrong. Your starting values are way off.
> Look at this plot for verification:
>
> con <- textConnection("time  bod
> 1    1 0.47
> 2    2 0.74
> 3    3 1.17
> 4    4 1.42
> 5    5 1.60
> 6    7 1.84
> 7    9 2.19
> 8   11 2.17")
> mydata <- read.table(con, header = TRUE)
> close(con)
>
> beta <- c(3, -0.1) # your initial values
> beta <- c(2.4979, -2.02456)  # SPSS answer
> mycurve <- function(x) {
>  beta[1]/(1 - exp(beta[1] * x))
> }
> curve(mycurve, from = 1, to = 11,
>      ylim = range(mydata$bod, mycurve(mydata$time)))
> points(mydata$time, mydata$bod)
>
> You might want to look at package nls2 which allows a brute force grid
> search to find some starting values. Or rethink the equation you're
> trying to fit.
>
> HTH,
>
> --sundar
>
> On Tue, Aug 25, 2009 at 9:10 AM, Michael Pearmain wrote:
>> Hi All,
>>
>> I'm trying to run nls on the data from the study by Marske (Biochemical
>> Oxygen Demand Interpretation Using Sum of Squares Surface. M.S. thesis,
>> University of Wisconsin, Madison, 1967) and was reported in Bates and Watts
>> (1988).
>>
>> Data is as follows, (stored as mydata)
>>
>>  time  bod
>> 1    1 0.47
>> 2    2 0.74
>> 3    3 1.17
>> 4    4 1.42
>> 5    5 1.60
>> 6    7 1.84
>> 7    9 2.19
>> 8   11 2.17
>>
>> I then run the following;
>> #Plot initial curve
>> plot(mydata$time, mydata$bod,xlab="Time (in days)",ylab="biochemical oxygen
>> demand (mg/l) ")
>>
>> model <- nls(bod ~ beta1/(1 - exp(beta2*time)), data =
>> mydata, start=list(beta1 = 3, beta2 = -0.1),trace=T)
>>
>> The start values are recommended, (I have used these values in SPSS without
>> any problems, SPSS returns values of Beta1 = 2.4979 and Beta2 = -2.02 456)
>>
>> but return the error message,
>> Error in nls(bod ~ beta1/(1 - exp(beta2 * time)), data = mydata, start =
>> list(beta1 = 3,  : singular gradient
>>
>> Can anyone offer any advice?
>>
>> Thanks in advance
>>
>> Mike
>>
>>
>>
>>
>>
>>
>>
>>
>>
>> --
>> Michael Pearmain
>> Senior Analytics Research Specialist
>>
>> “Statistics are like women; mirrors of purest virtue and truth, or like
>> whores to use as one pleases”
>>
>> Google UK Ltd
>> Belgrave House
>> 76 Buckingham Palace Road
>> London SW1W 9TQ
>> United Kingdom
>> t +44 (0) 2032191684
>> mpearm...@google.com
>>
>> If you received this communication by mistake, please don't forward it to
>> anyone else (it may contain confidential or privileged information), please
>> erase all copies of it, including all attachments, and please let the sender
>> know it went to the wrong person. Thanks.
>>
>>        [[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.
>>
>>
>

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Re: [R] xyplot: Can I identify groups but not use them for regression?

[Sundar Dorai-Raj]

I think this ought to work for you:

library(lattice)
set.seed(42)
d <- data.frame(year  = c(rep(2007,12), rep(2008,12)),
treatment = rep(LETTERS[1:3], each = 4, times = 2))
d$cover <- rnorm(nrow(d))
d$variable <- rnorm(nrow(d))

xyplot(variable ~ cover | year, d,
   panel = function(x, y, ...) {
 panel.superpose(x, y, ...)
 panel.lmline(x, y, ...)
   },
   groups = treatment)

HTH,

--sundar

On Fri, Sep 18, 2009 at 3:42 PM, Seth W Bigelow  wrote:
> I wish to identify groups representing different treatments, but to plot
> them and do a regression using a continuous variable ("cover")
> ignoring the groupings.
>
> d$year <- NA
> d$year <-c(rep(2007,12), rep(2008,12))
> d$treatment <- c(rep("A",4),rep("B",4),rep("C",4), rep("A",4), rep("B",4),
> rep("C",4))
> d$cover <- rnorm(24)
> d$variable <- rnorm(24)
>
> xyplot(variable ~ cover | year, d,
>        type=c("p","r"),
>        groups=treatment
>        )
>
> As it stands, a different regression line is plotted for each treatment.
> Oh, and how do I display the actual numeric value of year (e.g., "2007")
> in the strip, rather than the word "year"?
>
> --Seth
>
>
>
> Dr. Seth  W. Bigelow
> Biologist, USDA-FS Pacific Southwest Research Station
> 1731 Research Park Drive, Davis California
>        [[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.
>

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Re: [R] Skipping missing files when importing data

[Sundar Dorai-Raj]

Try ?file.exists.

if (file.exists(fxxx)) {
  read.table(fxxx)
} else {
  cat("\"", fxxx, "\" is missing\n", sep = "")
}

HTH,

--sundar

On Sun, Sep 20, 2009 at 9:28 PM, jiangrm  wrote:
> Trying to import a bunch of data files named like f001, f002, f999. Some 
> of the files may be
> missing and the missing files vary from time to time.
>
> Used for loop and read.table. When it reaches the missing file (say f100), it 
> shows:
>
> Error in file(file, "r") : cannot open the connection
> In addition: Warning message:
> In file(file, "r") :
>  cannot open file 'f100': No such file or directory
>
> and the program stops.
>
> How can I skip the missing ones and keep the program running? Guess either 
> checking the validity of
> filenames, or ignore the error message may work. Which functions should be 
> used or any better ideas?
>
>
> -RJ
>
> __
> 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] capturing errors in Sweave

[Sundar Dorai-Raj]

Hi,

I'm writing a manual using Sweave and I want to be able to print errors from
bad code. Here's an example:

>=
MySqrt <- function(x) {
  if (missing(x)) {
stop("'x' is missing with no default")
  }
  if (!is.numeric(x)) {
stop("'x' should only be numeric")
  }
  if (x < 0) {
stop("'x' should be non-negative")
  }
  return(sqrt(x))
}
MySqrt()
MySqrt("a")
MySqrt(-2)
MySqrt(4)
@

And I would like the output to be:

MySqrt <- function(x) {
...
}
> MySqrt()
Error in MySqrt() : 'x' is missing with no default
> MySqrt("a")
Error in MySqrt("a") : 'x' should only be numeric
> MySqrt(-2)
Error in MySqrt(-2) : 'x' should be non-negative
> MySqrt(2)
[1] 1.414214

I.e. I want the Error statements to print in the tex file and not just make
Sweave to bomb.

Thanks,

--sundar

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Re: [R] capturing errors in Sweave

[Sundar Dorai-Raj]

Thanks for the input, but I don't want "try" in the Sweave output. I want
the output to look just like it does in the console, as if an uncaptured
error really did occur.

--sundar

On Mon, Mar 1, 2010 at 11:42 PM, Sharpie  wrote:

>
>
> Sundar Dorai-Raj-2 wrote:
> >
> > Hi,
> >
> > I'm writing a manual using Sweave and I want to be able to print errors
> > from
> > bad code. Here's an example:
> >
> > >=
> > MySqrt <- function(x) {
> >   if (missing(x)) {
> > stop("'x' is missing with no default")
> >   }
> >   if (!is.numeric(x)) {
> > stop("'x' should only be numeric")
> >   }
> >   if (x < 0) {
> > stop("'x' should be non-negative")
> >   }
> >   return(sqrt(x))
> > }
> > MySqrt()
> > MySqrt("a")
> > MySqrt(-2)
> > MySqrt(4)
> > @
> >
> > And I would like the output to be:
> >
> > MySqrt <- function(x) {
> > ...
> > }
> >> MySqrt()
> > Error in MySqrt() : 'x' is missing with no default
> >> MySqrt("a")
> > Error in MySqrt("a") : 'x' should only be numeric
> >> MySqrt(-2)
> > Error in MySqrt(-2) : 'x' should be non-negative
> >> MySqrt(2)
> > [1] 1.414214
> >
> > I.e. I want the Error statements to print in the tex file and not just
> > make
> > Sweave to bomb.
> >
> > Thanks,
> >
> > --sundar
> >
>
> You can catch errors in R using the try() function:
>
>  foo <- try( log("A"), silent = TRUE )
>
>  if( class( foo ) == 'try-error' ){
>
>print( unclass( foo ) )
>
>  }
>
> [1] "Error in log(\"A\") : Non-numeric argument to mathematical function\n"
>
>
> There might be a more clever way to override the default error handling
> that
> would save you from wrapping everything in try()-- but this method should
> work.
>
> Hope it helps!
>
> -Charlie
>
> -
> Charlie Sharpsteen
> Undergraduate-- Environmental Resources Engineering
> Humboldt State University
> --
> View this message in context:
> http://n4.nabble.com/capturing-errors-in-Sweave-tp1574642p1574686.html
> Sent from the R help mailing list archive at Nabble.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.
>

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Re: [R] capturing errors in Sweave

[Sundar Dorai-Raj]

Thanks, Berwin. That works just great!

--sundar

On Tue, Mar 2, 2010 at 12:57 AM, Berwin A Turlach
wrote:

> G'day Sundar,
>
> On Mon, 1 Mar 2010 23:46:55 -0800
> Sundar Dorai-Raj  wrote:
>
> > Thanks for the input, but I don't want "try" in the Sweave output. I
> > want the output to look just like it does in the console, as if an
> > uncaptured error really did occur.
>
> I don't think that you will get around using "try"; and you will have
> to work moderately hard to make the output appear as it does on the
> console.  Probably somewhere along the lines:
>
>  Sweave code start ++
> <>=
> MySqrt <- function(x) {
>  if (missing(x)) {
>stop("'x' is missing with no default")
>  }
>  if (!is.numeric(x)) {
>stop("'x' should only be numeric")
>  }
>  if (x < 0) {
>stop("'x' should be non-negative")
>  }
>  return(sqrt(x))
> }
> @
>
> <>=
> tmp <- try(MySqrt())
> @
> <>=
> MySqrt()
> @
> <>=
>  cat(tmp[1])
> @
>
> <>=
> tmp <- try(MySqrt("a"))
> @
> <>=
> MySqrt("a")
> @
> <>=
>  cat(tmp[1])
> @
>
> <>=
> tmp <- try(MySqrt(-2))
> @
> <>=
> MySqrt(-2)
> @
> <>=
>  cat(tmp[1])
> @
>
> <<>>=
> MySqrt(4)
> @
> +++ Sweave code end ++
>
> Now what I would like to know is how to include easily warning messages
> in my Sweave output without having to try whether Jean Lobry's [1] hack
> still works. :)
>
> HTH.
>
> Cheers,
>
>Berwin
>
> [1]
> https://www.stat.math.ethz.ch/pipermail/r-help/2006-December/121975.html
>
> == Full address 
> Berwin A Turlach  Tel.: +61 (8) 6488 3338 (secr)
> School of Maths and Stats (M019)+61 (8) 6488 3383 (self)
> The University of Western Australia   FAX : +61 (8) 6488 1028
> 35 Stirling Highway
> Crawley WA 6009e-mail: ber...@maths.uwa.edu.au
> Australiahttp://www.maths.uwa.edu.au/~berwin
>

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Re: [R] capturing errors in Sweave

[Sundar Dorai-Raj]

What I ended up using was:

cat(unclass(tmp))

--sundar

On Tue, Mar 2, 2010 at 8:58 AM, Berwin A Turlach wrote:

> G'day Sundar,
>
> On Tue, 2 Mar 2010 01:03:54 -0800
> Sundar Dorai-Raj  wrote:
>
> > Thanks, Berwin. That works just great!
>
> You are welcome.
>
> I noticed by now that "cat(tmp)" is sufficient; the "tmp[1]" in
> "cat(tmp[1])" was a left over from earlier attempts to get the output
> to look correct.
>
> Cheers,
>
>Berwin
>

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Re: [R] Source code for the t-distribution

[Sundar Dorai-Raj]

Here it is.

https://svn.r-project.org/R/trunk/src/nmath/pt.c

--sundar

On Tue, Mar 9, 2010 at 4:24 AM, Ravi Kulkarni  wrote:

>
> I have tried looking for the source code for the pt() function in
>
> https://svn.r-project.org/R/trunk/src/library/stats/
>
> and am unable to find it there. Can someone please tell me where to find
> it?
>
> Thanks,
>  Ravi Kulkarni
> --
> View this message in context:
> http://n4.nabble.com/Source-code-for-the-t-distribution-tp1585875p1585875.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
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>

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Re: [R] sapply, lattice functions

[Sundar Dorai-Raj]

Or perhaps more clearly,

histogram(~a1 + b1 + c1, data = aa, outer = TRUE)

--sundar

On Fri, Mar 19, 2010 at 3:50 PM, Gabor Grothendieck  wrote:

> Try this:
>
> histogram(~ values | ind, stack(aa))
>
>
> On Fri, Mar 19, 2010 at 5:44 PM, Santosh  wrote:
> > Dear R-gurus
> >
> > aa <- data.frame(a1=rnorm(20),b1=rnorm(20,0.8),c1=rnorm(20,0.5))
> > sapply(aa,function(x) histogram(x,breaks=NULL))
> >
> > or px <- sapply(aa,function(x) histogram(x,breaks=NULL))
> > print(px,split=c(1,1,1,1),more=F)
> >
> > The above code does not seem to work. am I missing something?
> >
> > Thanks,
> > Santosh
> >
> >[[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.
> >
>
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Re: [R] sapply, lattice functions

[Sundar Dorai-Raj]

You're right. It's necessary for xyplot though to prevent grouping.

On Mar 20, 2010 10:43 AM, "Dieter Menne" 
wrote:



Sundar Dorai-Raj-2 wrote:
>
> Or perhaps more clearly,
>
> histogram(~a1 + b1 + c1, data = aa, o...
Why outer=TRUE? Looks same for me without:
Dieter

library(lattice)

aa <- data.frame(a1=rnorm(20),b1=rnorm(20,0.8),c1=rnorm(20,0.5))
histogram(~a1 + b1 + c1, data = aa)


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Re: [R] how to check if ... is empty

[Sundar Dorai-Raj]

Try

dots <- list(...)
if (length(dots) == 0) {
  ## do something
}

On Thu, Jul 16, 2009 at 6:46 AM, Thomas Roth (geb.
Kaliwe) wrote:
> Hi,
>
> I was wondering what would be the best way to check if the three dots
> argument contains any arguments (i.e. does ... contain any arguments or not?
> )
>
> #Example
>
> test = function(x,y, ...)
> {
>   #Wanted R-Code
>   # if(empty(...))
>   #    do some calculation
>
>
>   plot(x,y,...)
>
> }
>
> Thanks
>
> Thomas Roth
>
> __
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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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Re: [R] How to speed up R with version 2.9.2?

[Sundar Dorai-Raj]

Another possibility is a very large .RData file in the directory where
you're starting R. You can try

Rgui --no-restore

(I don't have windows, so I'm not sure if this an option with RGui,
though I know it is with R.)

--sundar

On Fri, Oct 2, 2009 at 8:50 AM, Gabor Grothendieck
 wrote:
> Its under 5 seconds on my Vista laptop.  Do you have any startup files?  If
>  Rgui --vanilla
> is much faster then your startup files are the problem.
>
> On Fri, Oct 2, 2009 at 11:45 AM, FMH  wrote:
>> Thank you for your answer. I'm using Win XP with 2GB RAM in memory.
>>
>> Cheers
>> Fir
>>
>>
>>
>> - Original Message 
>> From: stephen sefick 
>> To: FMH 
>> Cc: r-help@r-project.org
>> Sent: Fri, October 2, 2009 4:38:10 PM
>> Subject: Re: [R] How to speed up R with version 2.9.2?
>>
>> You're fine, but please do read the posting guide.  What OS etc.
>> Where you doing anything else on the computer?  Is this a RAM
>> limitation?  I have 2.9.2 running on two flavours of linux, mac os x
>> and windows all 2.9.2 and there doesn't seem to be a problem.
>> regards,
>>
>> Stephen
>>
>> On Fri, Oct 2, 2009 at 10:35 AM, FMH  wrote:
>>> Dear All,
>>>
>>> I'm sorry if my question does not suit with this R group.
>>>
>>> I have recently installed R software with version 2.9.2, but i found the 
>>> program took almost 1 minute as soon as it was opened, before it can be 
>>> used. However, the previous version 2.9.1 only take few seconds after the 
>>> menu bar was clicked. This circumstance has caused me to wait for couple of 
>>> minutes as several R windows were opened simultaneously.
>>>
>>> Are there any ways to speed up on this 2.9.2 version? Could someone give 
>>> some hints, please?
>>>
>>> Thank you
>>> Fir
>>>
>>>
>>>
>>>
>>> __
>>> 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.
>>>
>>
>>
>>
>> --
>> Stephen Sefick
>>
>> 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
>>
>>
>>
>>
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
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>> and provide commented, minimal, self-contained, reproducible code.
>>
>
> __
> 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] Error in family$family : $ operator is invalid for atomic vectors

[Sundar Dorai-Raj]

Check to see if you have an old workspace being loaded. You might have an
object called 'family' which you might need to remove.

--sundar

On Oct 11, 2009 12:15 PM, "romunov"  wrote:

Thank you Jorge and Barry for your input.

I've fiddled around a bit and as a result, am even more confused. If I start
R console via Notepad++ (I use Npp2R) and execute the model1, it goes
through just fine. Here is the sessionInfo() for this "working" session:

> sessionInfo()
R version 2.9.2 (2009-08-24)
i386-pc-mingw32

locale:
LC_COLLATE=Slovenian_Slovenia.1250;LC_CTYPE=Slovenian_Slovenia.1250;LC_MONETARY=Slovenian_Slovenia.1250;LC_NUMERIC=C;LC_TIME=Slovenian_Slovenia.1250

attached base packages: [1] stats graphics grDevices utils datasets methods
base

loaded via a namespace (and not attached):
[1] tools_2.9.2

And if I run R "normally", via an icon from the desktop (Rgui.exe) it gives
the aforementioned error. Here is the sessionInfo() for "non-working"
session. Is it possible that grid, reshape, plyr, ggplot2 and proto could be
causing this? If so, how can I prevent them from loading automatically or
unloading from a live session?

> sessionInfo()
R version 2.9.2 (2009-08-24)
i386-pc-mingw32

locale:
LC_COLLATE=Slovenian_Slovenia.1250;LC_CTYPE=Slovenian_Slovenia.1250;LC_MONETARY=Slovenian_Slovenia.1250;LC_NUMERIC=C;LC_TIME=Slovenian_Slovenia.1250

attached base packages:
[1] stats graphics  grDevices utils datasets  grid  methods
[8] base

other attached packages:
[1] reshape_0.8.3 plyr_0.1.9proto_0.3-8

loaded via a namespace (and not attached):
[1] ggplot2_0.8.3


Cheers,
Roman



On Sun, Oct 11, 2009 at 6:32 PM, Jorge Ivan Velez
wrote:

> Hi Romain, > It works for me: > > model1 <- glm(as.vector(x)
~dept*sex*admit,poisson) > model1 > ...

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Re: [R] List mappings and variable creation

[Sundar Dorai-Raj]

Hi, Michael,

Seems like all you need is aggregate and rbind:

x <- aggregate(saw.aggr.data["value"],
   saw.aggr.data[c("conversion.type", "filteredID", "metric")],
   sum)
x$bucketID <- "combined"
y <- rbind(saw.aggr.data, x)

Is this what you need?

--sundar

On Mon, Oct 12, 2009 at 6:09 AM, Michael Pearmain  wrote:
> Hi All,
>
> I have a questions about associative list mappings in R, and if they are
> possible?
>
> I have data in the form show below, and want to make a new 'bucket' variable
> called combined. Which is the sum of the control and the exposed metric
> values
> This combined variable is a many to many matching as values only appear in
> the file if they have a value > 0.
>
> conversion.type   filteredID        bucketID      Metric       Value
>    counter            true              control           a              1
>    counter            true              control           b              1
>    counter            true              control           c              2
>    counter            true              control           d              3
>
>    counter            true              exposed         a             4
>    counter            true              exposed         e             1
>
> ASIDE:
>
> At the minute i read the data into my file and and then create all the
> 'missing' row values
> (in this case,
>    counter            true              control             e             0
>    counter            true              exposed           b              0
>    counter            true              exposed           c              0
>    counter            true              exposed           d              0)
>
>
> and then run  a sort on the data, and count the number of times control
> appears, and then use this as an index matcher.
>
> saw.aggr.data <- [order(saw.aggr.data$bucketID, saw.aggr.data$metric), ]
> no.of.metrics <- length(saw.aggr.data$bucketID[grep("control",
> saw.aggr.data$bucketID)])
>
> for (i in (1:no.of.metrics)) {
>  assign(paste("combined", as.character(saw.aggr.data$metric[i])),
> (saw.aggr.data$value[i] + saw.aggr.data$value[i + no.of.metrics]))
> }
>
> This does what i want it to but is very very weak and could be open to large
> errors, ( error handling currently via grepping the names of the metric[i]
> == name of metric [i + no.of.metrics])
>
> Is there a more powerful way of doing this using some kind of list mapping?
> I've looked at the older threads in this area and it looks like something
> that should be possible but i can't figure out how to do this?
> Ideally i'd like a final dataset  / list that is of the following form.
>
> conversion.type   filteredID        bucketID      Metric       Value
>    counter            true              control           a              1
>    counter            true              control           b              1
>    counter            true              control           c              2
>    counter            true              control           d              3
>
>    counter            true              exposed         a             4
>    counter            true              exposed         e             1
>    counter            true              combined        a             5
>    counter            true              combined        b             1
>    counter            true              combined        c             2
>    counter            true              combined        d             3
>    counter            true              combined        e             1
>
> So i dont have to create the dummy variables.
>
> does this make sense?
>
> Many thanks in advance
>
> Mike
>
>
>
> --
> Michael Pearmain
> "I abhor averages.  I like the individual case.  A man may have six meals
> one day and none the next, making an average of three meals per day, but
> that is not a good way to live.  ~Louis D. Brandeis"
>
> f you received this communication by mistake, please don't forward it to
> anyone else (it may contain confidential or privileged information), please
> erase all copies of it, including all attachments, and please let the sender
> know it went to the wrong person. Thanks.
>
>        [[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.
>

__
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Re: [R] How can I run a function to a piece of text?

[Sundar Dorai-Raj]

Based solely on what you told us, this can be done using eval(parse(text=...))

cmd <- sprintf("mean(%s)", script)
eval(parse(text = cmd))

However, with more context, there may be a better solution. See, for example,

install.packages("fortunes")
library(fortunes)
fortune("parse()")

HTH,

--sundar

On Fri, Oct 16, 2009 at 11:39 AM, Javier PB
 wrote:
>
> Dear users,
>
> I got really stuck trying to apply a function to a piece of code that I
> created using different string functions.
>
> To make things really easy, this is a wee example:
>
> x<-c(1:10)
> script<-"x, trim = 0, na.rm = FALSE"        ##script created by a number of
> paste() and rep() steps
> mean(script)                                       ##function that I want to
> apply: doesn't work
>
> Is there any way to convert the "script" class so that the function mean()
> can read it as if it were text typed in the console?
>
> Thanks and have a superb day
>
> Javier
>
>
>
> --
> View this message in context: 
> http://www.nabble.com/How-can-I-run-a-function-to-a-piece-of-text--tp25930315p25930315.html
> Sent from the R help mailing list archive at Nabble.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.
>

__
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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] tapply function

[Sundar Dorai-Raj]

you must have missing values in "data". Try

tapply(data, group, mean, na.rm = TRUE)

If that's not the case, read the bottom of this email about the posting guide.

HTH,

--sundar

On Tue, Nov 3, 2009 at 5:28 AM, FMH  wrote:
> Hi,
>
> I tried to use tapply function to find the mean of the data in each group as 
> the following command, but the result are NA, as there are several missing 
> values in each group.
>
> tapply(data,group,mean)
>
> Could someone please advice me the way to  ignore the missing data in order 
> for the fucntion to run successfully?
>
> Thanks
>
> Fir
>
>
>
>
> __
> 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
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Issue with %in% - not matching identical rows in data frames

[Sundar Dorai-Raj]

?"%in%" says "x" and "table" must be vectors. You supplied
data.frames. So %in% is coercing your today.sequence to a vector using

as.character(today.sequence)

Perhaps you should paste the columns together first:

x <- do.call("paste", c(sequence, sep = "::"))
table <- do.call("paste", c(today.sequence, sep = "::"))
x[7] %in% table

I'm not sure if this is what you want/need, but it does match your example.

HTH,

--sundar

On Tue, Nov 3, 2009 at 7:53 AM, Kaushik Krishnan
 wrote:
> Hi folks
>
> I have two data frames.  I know that the nth (let's say the 7th) row
> in the first data frame (sequence) is there in the second
> (today.sequence).  When I try to check that by doing 'sequence[7,]
> %in% today.sequence', I get all FALSE when it should be all TRUE.
>
> I'm certain I'm making some trivial mistake.  Any solutions?
>
> The code to recreate the data frames and see for yourself is:
> 
> sequence <- structure(list(DATE = structure(c(14549, 14549, 14553, 14550,
> 14557, 14550, 14551, 14550), class = "Date"), DATASET = c(1L,
> 2L, 1L, 2L, 2L, 3L, 3L, 4L), REP = c(1L, 0L, 2L, 2L, 3L, 0L,
> 1L, 0L), WRONGS_ABS = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), WRONGS_RATIO = c(0L,
> 0L, 0L, 0L, 0L, 0L, 0L, 0L), DONE = c(1L, 1L, 0L, 1L, 0L, 1L,
> 0L, 0L)), .Names = c("DATE", "DATASET", "REP", "WRONGS_ABS",
> "WRONGS_RATIO", "DONE"), class = "data.frame", row.names = c(NA,
> -8L))
>
> today.sequence <- structure(list(DATE = structure(c(14551, 14550),
> class = "Date"),
>    DATASET = 3:4, REP = c(1L, 0L), WRONGS_ABS = c(0L, 0L),
> WRONGS_RATIO = c(0L,
>    0L), DONE = c(0L, 0L)), .Names = c("DATE", "DATASET", "REP",
> "WRONGS_ABS", "WRONGS_RATIO", "DONE"), row.names = 7:8, class = "data.frame")
>
> sequence[7,] #You should see '2009-11-03       3   1          0
>    0    0'
>
> today.sequence #You can clearly see that sequence [7,] is the first
> row in today.sequence
>
> sequence[7,] %in% today.sequence #This should show 'TRUE TRUE TRUE
> TRUE TRUE TRUE'.  Instead
> # it shows 'FALSE FALSE FALSE FALSE FALSE FALSE'
> 
>
> Thanks
>
> --
> Kaushik Krishnan
> (kaushik.s.krish...@gmail.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.
>

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

Re: [R] Trellis settings get lost when printing to pdf

[Sundar Dorai-Raj]

Did you make the changes before or after starting the device:

library(lattice)
## before doesn't change the settings on the device:
trellis.par.set(plot.symbol = list(col = "red"))
trellis.device(pdf, file = "tmp.pdf")
xyplot(1 ~ 1)
dev.off()

## after does
trellis.device(pdf, file = "tmp.pdf")
trellis.par.set(plot.symbol = list(col = "red"))
xyplot(1 ~ 1)
dev.off()

I never do things like this, though. I would suggest creating a theme
instead and supplying it to xyplot (or whatever plot you're using)
using par.settings:

my.theme <- list(plot.symbol = list(col = "red"))
trellis.device(pdf, file = "tmp.pdf")
xyplot(1 ~ 1, par.settings = my.theme)
dev.off()

HTH,

--sundar


2009/11/13 Joel Fürstenberg-Hägg :
>
> Hi all,
>
>
>
> I've got some problems when changing the trellis settings for the lattice 
> plots. The plots look exactly as I want them to when calling show.settings() 
> as well as when plotting them in the graphical window. But when printing to a 
> pdf file, none of the settings are used!? Does anyone know what might have 
> happened? Because the when changing the trellis settings, these should remain 
> in the new state until you close R right..?
>
>
>
> # Change settings for the boxplot appearance
> new.dot=trellis.par.get("box.dot")
> new.rectangle=trellis.par.get("box.rectangle")
> new.umbrella=trellis.par.get("box.umbrella")
> new.symbol=trellis.par.get("plot.symbol")
> new.strip.background=trellis.par.get("strip.background")
> new.strip.shingle=trellis.par.get("strip.shingle")
> new.dot$pch="|"
> new.dot$col="black"
> new.rectangle$col="black"
> new.rectangle$fill="grey65"
> new.umbrella$col="black"
> new.umbrella$lty=1 # Continous line, not dotted
> new.symbol$col="black"
> new.strip.background$col="grey87" # Background colour in the upper label
> new.strip.shingle$col="black" # Border colour around the upper label
> trellis.par.set(box.dot=new.dot, box.rectangle=new.rectangle, 
> box.umbrella=new.umbrella, plot.symbol=new.symbol, 
> strip.background=new.strip.background, strip.shingle=new.strip.shingle)
>
>
>
> Best regards,
>
>
>
> Joel
>
> _
> Nya Windows 7 - Hitta en dator som passar dig! Mer information.
> http://windows.microsoft.com/shop
>        [[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.
>

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

Re: [R] lapply() not converting columns to factors (no error message)

[Sundar Dorai-Raj]

Works for me:

x <- 
read.csv(url("http://dc170.4shared.com/download/153147281/a5c78386/Testvcomp10.csv?tsid=20091116-075223-c3093ab0";))
names(x)
x[2:13] <- lapply(x[2:13], factor)

> levels(x$P1L55)
[1] "0" "1"
> is.factor(x$P1L96)
[1] TRUE

> sessionInfo()
R version 2.10.0 (2009-10-26)
i386-apple-darwin9.8.0

locale:
[1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8

attached base packages:
[1] stats graphics  grDevices utils datasets  methods   base

other attached packages:
[1] lattice_0.17-26

loaded via a namespace (and not attached):
[1] grid_2.10.0  tools_2.10.0

On Mon, Nov 16, 2009 at 4:50 AM, A Singh  wrote:
> Sorry, my file is at:
>
>
> http://www.4shared.com/file/153147281/a5c78386/Testvcomp10.html
>
>
> --
> A Singh
> aditi.si...@bristol.ac.uk
> School of Biological Sciences
> University of Bristol
>
> __
> 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] lapply() not converting columns to factors (no error message)

[Sundar Dorai-Raj]

Could it be you have "factor" redefined in your workspace? Have you
tried it in a clean directory? I.e. a directory where no .RData
exists?

On Mon, Nov 16, 2009 at 5:07 AM, A Singh  wrote:
> Oh, strange!
>
> I thought it might be a problem with the 'base' package installation,
> because the same thing's worked for me too before but won't do now.
>
> I tried to reinstall it (base), but R says its there already which I
> expected it to be anyway.
>
> I don't quite know where the issue is. Very odd.
>
>
> --On 16 November 2009 04:59 -0800 Sundar Dorai-Raj 
> wrote:
>
>> Works for me:
>>
>> x <-
>> read.csv(url("http://dc170.4shared.com/download/153147281/a5c78386/Testvc
>> omp10.csv?tsid=20091116-075223-c3093ab0")) names(x)
>> x[2:13] <- lapply(x[2:13], factor)
>>
>>> levels(x$P1L55)
>>
>> [1] "0" "1"
>>>
>>> is.factor(x$P1L96)
>>
>> [1] TRUE
>>
>>> sessionInfo()
>>
>> R version 2.10.0 (2009-10-26)
>> i386-apple-darwin9.8.0
>>
>> locale:
>> [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
>>
>> attached base packages:
>> [1] stats     graphics  grDevices utils     datasets  methods   base
>>
>> other attached packages:
>> [1] lattice_0.17-26
>>
>> loaded via a namespace (and not attached):
>> [1] grid_2.10.0  tools_2.10.0
>>
>> On Mon, Nov 16, 2009 at 4:50 AM, A Singh 
>> wrote:
>>>
>>> Sorry, my file is at:
>>>
>>>
>>> http://www.4shared.com/file/153147281/a5c78386/Testvcomp10.html
>>>
>>>
>>> --
>>> A Singh
>>> aditi.si...@bristol.ac.uk
>>> School of Biological Sciences
>>> University of Bristol
>>>
>>> __
>>> 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.
>>>
>
>
>
> --
> A Singh
> aditi.si...@bristol.ac.uk
> School of Biological Sciences
> University of Bristol
>
>
>
>
>

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] denoting max value in ylim

[Sundar Dorai-Raj]

ylim = c(0, max(log10(D10$Part.P)))

Make sure you remove any 0s or NAs before computing the max though.

--sundar

On Fri, Nov 20, 2009 at 6:12 AM, helene frigstad
 wrote:
>
> Hi,
>
> is there any way to set the ylim range from zero to whatever is the max
> value in that dataset? I am plotting many similar plots to the one below,
> and would like to avoid having to find the max value each time.
>
>
> plot (D10$Part.P ~ D10$Klorofyll,pch=16,log = "xy", xlab = ("Chla"), ylab =
> ("POP"), ylim = c (0, ???))
>        abline(m3, untf=TRUE, lty=1,col="blue")
>        text(5, 0.05,paste(round(glm(D10$Part.P ~ D10$Klorofyll, data = D10, 
> family
> = Gamma(link =      "identity"))$coef, 2),collapse = " "))
>
> thank you very much for your time and help.
>
> Best regards,
> Helene Frigstad
> --
> View this message in context: 
> http://old.nabble.com/denoting-max-value-in-ylim-tp26441590p26441590.html
> Sent from the R help mailing list archive at Nabble.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.
>

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Re: [R] Why F value and Pr are not show in summary() of an aov() result?

[Sundar Dorai-Raj]

It's hard to read your code, so I won't comment on your specific
example. So when all else fails read the documentation for
?summary.aov:

They have columns ‘"Df"’, ‘"Sum Sq"’, ‘"Mean
 Sq"’, as well as ‘"F value"’ and ‘"Pr(>F)"’ if there are non-zero
 residual degrees of freedom.

So if you do df.residual(afit), is it >0?

--sundar

On Sun, Nov 22, 2009 at 7:19 AM, Peng Yu  wrote:
> I have the following code. I'm wondering why summary() doesn't show F
> value and Pr?
>
> Rscript multi_factor.R
>> a=3
>> b=4
>> c=5
>> d=6
>> e=7
>>
>> A=1:a
>> B=1:b
>> C=1:c
>> D=1:d
>> E=1:e
>>
>> X=matrix(nr=a*b*c*d*e,nc=5)
>> colnames(X)=LETTERS[1:5]
>>
>> for(i_a in 1:a-1) {
> +   for(i_b in 1:b-1) {
> +     for(i_c in 1:c-1) {
> +       for(i_d in 1:d-1) {
> +         for(i_e in 1:e-1) {
> +           X[(((i_a * b + i_b) * c + i_c) * d + i_d) * e + i_e + 1, ]
> = c(i_a+1, i_b+1, i_c+1, i_d+1, i_e+1)
> +         }
> +       }
> +     }
> +   }
> + }
>>
>> Y=matrix(nr=a*b*c*d*e,nc=1)
>> for(i in 1:(a*b*c*d*e)) {
> +   fa=X[i,'A']
> +   fb=X[i,'B']
> +   fc=X[i,'C']
> +   fd=X[i,'D']
> +   fe=X[i,'E']
> +
> +   Y[i,1]= fa +fb +fc +fe +fa*fb +fa*fc +fb*fc +fa*fe +fc*fe
> +fa*fb*fc +fa*fc*fe + rnorm(1)
> + }
>>
>> aframe = data.frame(
> +     A=as.factor(X[,'A'])
> +     , B=as.factor(X[,'B'])
> +     , C=as.factor(X[,'C'])
> +     , D=as.factor(X[,'D'])
> +     , E=as.factor(X[,'E'])
> +     ,Y)
>>
>> afit=aov(Y ~ A * B * C * D * E, aframe)
>>
>> summary(afit)
>             Df  Sum Sq Mean Sq
> A             2 1512240  756120
> B             3  453324  151108
> C             4 2549895  637474
> D             5       2  0.3693
> E             6 1451057  241843
> A:B           6   33875    5646
> A:C           8  189839   23730
> B:C          12   56024    4669
> A:D          10       7       1
> B:D          15      25       2
> C:D          20      18       1
> A:E          12  107574    8964
> B:E          18      21       1
> C:E          24  180413    7517
> D:E          30      16       1
> A:B:C        24    4167     174
> A:B:D        30      37       1
> A:C:D        40      42       1
> B:C:D        60      63       1
> A:B:E        36      30       1
> A:C:E        48   13298     277
> B:C:E        72      62       1
> A:D:E        60      79       1
> B:D:E        90      87       1
> C:D:E       120     122       1
> A:B:C:D     120     140       1
> A:B:C:E     144     131       1
> A:B:D:E     180     145       1
> A:C:D:E     240     225       1
> B:C:D:E     360     398       1
> A:B:C:D:E   720     713       1
>
> __
> 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] makefile for sweave

[Sundar Dorai-Raj]

Is texi2dvi in your PATH? What happens if you open a CMD window and
type texi2dvi at the prompt?

--sundar

On Thu, Nov 26, 2009 at 6:14 AM, Wolfgang Raffelsberger  wrote:
> Dear all,
>
> I can't get texi2dvi working right. Basically I'd like to convert a .lex to
> .pdf without having to fiddle with the issue Sweave.sty not being in my
> current directory (as this was sugested in other posts on this list).
>
> When I'm in the R-Gui I can get the help via
> ?texi2dvi
> (So I conclude its installed.)
>
> However, when I try to use it to concert a .tex to .pdf I get trouble ...
>
> For example :
> A)
> The file "test02.r" contains :
> Sweave("Sweave_test01.rnw")
> library(tools)
> texi2dvi("Sweave_test01.tex", pdf =T)
>
> Now, when I run on the linux command line :
> R --vanilla -q < test02.r
>
> I get :
>> Sweave("Sweave_test01.rnw")
> Writing to file Sweave_test01.tex
> Processing code chunks ...
> 1 : term hide (label=chunk_ini)
> 2 : term verbatim eps pdf (label=Fig01)
> 3 : term tex (label=packageVersionInfo)
> Loading required package: xtable
>
> You can now run LaTeX on 'Sweave_test01.tex'
>> library(tools)
>> texi2dvi("Sweave_test01.tex", pdf =T)
> Error in texi2dvi("Sweave_test01.tex", pdf = T) :
>  Running 'texi2dvi' on 'Sweave_test01.tex' failed.
> Messages:
> sh: texi2dvi: command not found
> Execution halted
>
>
> B)
> In a previous message on this list I found the following command line(s)
> suggested, but I my case it won't work
>
> star5_R_test_> R CMD texi2dvi --help
> /usr/local/lib64/R/bin/Rcmd: line 62: exec: texi2dvi: not found
>
> similarly, when execute (as sugested) I get the same error message
>
> star5_R_test_> R CMD texi2dvi -p Sweave_test01.tex
> /usr/local/lib64/R/bin/Rcmd: line 62: exec: texi2dvi: not found
>
>
> I don't understand how can a command can be present (= installed) and still
> not being found as the error messages suggest ?
>
> For completeness :
>> sessionInfo()
> R version 2.10.0 (2009-10-26)
> x86_64-unknown-linux-gnu
>
> locale:
> [1] C
>
> attached base packages:
> [1] stats     graphics  grDevices utils     datasets  methods   base
> other attached packages:
> [1] xtable_1.5-6         mouse4302probe_2.5.0 AnnotationDbi_1.8.1
> [4] mouse4302cdf_2.5.0   MASS_7.3-3           fdrtool_1.2.5      [7]
> limma_3.2.1          affyPLM_1.22.0       preprocessCore_1.8.0
> [10] gcrma_2.18.0         affy_1.24.2          Biobase_2.6.0
> loaded via a namespace (and not attached):
> [1] Biostrings_2.14.3 DBI_0.2-4         IRanges_1.4.4     RSQLite_0.7-3
> [5] affyio_1.14.0     splines_2.10.0    tools_2.10.0
>
>
>
> Thank's in advance,
> Wolfgang
>
>
> Charles C. Berry a écrit :
>>
>> On Tue, 8 Sep 2009, Welma Pereira wrote:
>>
>>> Hello, I have the following makefile. The problem is that the
>>> bibliography
>>> doesn t work. Any help would be appreciated! I really don t don t what to
>>> do..:-(
>>>
>>>
>>>
>>> # The sources of the report (tex, Rnw and other files (e.g. bib, idx))
>>> TEX_CMPS = Report problem
>>> RNW_CMPS = prop1 prop2 ExeExps
>>> OTHER =  Report.bib
>>>
>>> # The name of the report to produce
>>> all: Report.pdf
>>>
>>> code: $(RNW_CMPS:=.R)
>>>
>>> clean:
>>>   rm -f  *.log *.dvi *~
>>>
>>> # On what does the report depends?
>>> Report.pdf: $(TEX_CMPS:=.tex) $(RNW_CMPS:=.tex) ${OTHER} makefile
>>>   TEXINPUTS=${TPUTS} pdflatex $<
>>>   TEXINPUTS=${TPUTS} pdflatex $<
>>
>> IIRC
>>
>>    R CMD texi2dvi -p 
>>
>> takes care of finding sweave.sty and running latex thru all the iterations
>> needed to build cross-references and a usable pdf.
>>
>> Try
>>
>>    R CMD texi2dvi --help
>>
>> at the shell prompt.
>>
>> HTH,
>>
>> Chuck
>>
>>>
>>>   rm *.log
>>> #    mv *.aux  $(dir $<)
>>>
>>> # How to build the tex files from the Rnw (Sweave) files
>>> %.tex: %.Rnw
>>>   echo "library(utils); options(width=60);  Sweave('$<')" | ${R_PRG}
>>> --no-save --vanilla
>>>   mv $(notdir $*.tex)  $(dir $<)
>>>
>>>
>>> # How to build the R code files from the Rnw (Sweave) files
>>> %.R: %.Rnw
>>>   echo "library(utils); Stangle(

Re: [R] convert large integers to hex

[Sundar Dorai-Raj]

Thanks for both answers. In the end I decided to use Gabor's bc package.

Thanks,

--sundar

On Thu, May 7, 2009 at 5:10 AM, Gabor Grothendieck
 wrote:
> There is an interface between R and bc -- not on CRAN but available
> from its home page here:
> http://r-bc.googlecode.com
>
>> source("http://r-bc.googlecode.com/svn/trunk/R/bc.R";)
>> bc("obase = 16; 123456789123456789", retclass = "character")
> [1] "1B69B4BACD05F15"
>
>
> On Wed, May 6, 2009 at 9:59 PM, jim holtman  wrote:
>> You can use the 'bc' command (use Cygwin if on Windows);
>>
>> /cygdrive/c: bc
>> bc 1.06
>> Copyright 1991-1994, 1997, 1998, 2000 Free Software Foundation, Inc.
>> This is free software with ABSOLUTELY NO WARRANTY.
>> For details type `warranty'.
>> x=6595137340052185552
>> obase=16
>> x
>> 5B86A277DEB9A1D0
>>
>> You can call this from R.
>>
>> On Wed, May 6, 2009 at 3:26 PM, Sundar Dorai-Raj wrote:
>>
>>> Hi,
>>>
>>> I'm wondering if someone has solved the problem of converting very
>>> large integers to hex. I know about format.hexmode and as.hexmode, but
>>> these rely on integers. The numbers I'm working with are overflowing
>>> and losing precision. Here's an example:
>>>
>>> x <- "6595137340052185552" # stored as character
>>> as.integer(x) # warning about inaccurate conversion
>>> format.hexmode(as.numeric(x)) # warnings about loss of precision
>>> as.hexmode(x) # more warnings and does not do what I expected
>>>
>>> I'm planning on writing a function that will do this, but would like
>>> to know if anybody already has a solution. Basically, I would like the
>>> functionality of format.hexmode on arbitrarily large integers.
>>>
>>> Thanks,
>>>
>>> --sundar
>>>
>>> __
>>> 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<http://www.r-project.org/posting-guide.html>
>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>
>>
>>
>> --
>> Jim Holtman
>> Cincinnati, OH
>> +1 513 646 9390
>>
>> What is the problem that you are trying to solve?
>>
>>        [[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.
>>
>

__
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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] what is wrong with this code?

[Sundar Dorai-Raj]

You're missing a ")" off end of the first line. You should consider
using an editor (e.g. ESS/Emacs) that does parentheses matching. I
found this in less than 5 sec (less time than I'm taking to write you
a note) by cut and pasting in Emacs.

--sundar

On Tue, May 19, 2009 at 12:52 PM, deanj2k  wrote:
>
> dlogl <- -(n/theta)-sum((y/(theta)^2)*((1-exp(y/theta))/(1+exp(y/theta)))
>
> d2logl <- (n/theta^2) - sum((-2y/theta^3)*(1-exp(y/theta))/(1+exp(y/theta)))
> - sum(((2*y/theta^4)*exp(y/theta))/((1+exp(y/theta))^2))
>
> returns the error message:
> Error: unexpected symbol in:
> "dlogl <- -(n/theta)-sum((y/(theta)^2)*((1-exp(y/theta))/(1+exp(y/theta)))
> d2logl"
>
> do you know what i have done wrong
> --
> View this message in context: 
> http://www.nabble.com/what-is-wrong-with-this-code--tp23623227p23623227.html
> Sent from the R help mailing list archive at Nabble.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.
>

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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] error message re: max(i), but code and output seen O.K.

[Sundar Dorai-Raj]

This error is thrown if the argument to max is either NULL or length zero:

[~] Rscript -e "max(NULL)"
[1] -Inf
Warning message:
In max(NULL) : no non-missing arguments to max; returning -Inf
[~] Rscript -e "max(numeric(0))"
[1] -Inf
Warning message:
In max(numeric(0)) : no non-missing arguments to max; returning -Inf

HTH,

--sundar

On Wed, May 20, 2009 at 11:23 AM, Kirsten Miles  wrote:
> I have a researcher who is consistently get the warning message:
>
> In max(i) : no non-missing arguments to max; returning -Inf
>
> Best as I can tell the code is working properly and the output is as
> expected. I would like some help in understanding why he is getting this
> error message and what its implications are.  I have his code.
>
> Sincerely,
> Kirsten Miles
> Support Specialist
> Research Computing Lab
> Charles L. Brown Science and Engineering Library
>
> kd...@virginia.edu
>
>        [[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.
>

__
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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] find a sequence of characters in a vector

[Sundar Dorai-Raj]

use gregexpr and paste

> aze <- paste(c("a", "z", "e"), collapse = "")
> sequence <- paste(c("a","z","e","r","t","a","z","a","z","e","c"), collapse = 
> "")
> gregexpr(aze, sequence, fixed = TRUE)
[[1]]
[1] 1 8
attr(,"match.length")
[1] 3 3

HTH,

--sundar

On Fri, Jun 5, 2009 at 6:22 AM, Ptit_Bleu wrote:
>
> Hello,
>
> I'm just looking for an easy way to find the positions of a complete
> sequence in a bigger vector.
> For example :
> c("a","z","e") in c("a","z","e","r","t","a","z","a","z","e","c")
> and the result should be
> 1 8
> that is the positions of the beginning of the complete sequence.
>
> I tried with %in%, match, is.element but all I get is, for example
> which(c("a","z","e") in c("a","z","e","r","t","a","z","a","z","e","c"))
> 1 2 3
> meaning that each character is in the bigger vector.
>
> It must be easy, except for me. Sorry.
>
> If you have a solution, thanks in advance to share it.
> Have a good week-end,
> Ptit Bleu.
>
> --
> View this message in context: 
> http://www.nabble.com/find-a-sequence-of-characters-in-a-vector-tp23888063p23888063.html
> Sent from the R help mailing list archive at Nabble.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.
>

__
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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] reading in multiple files

[Sundar Dorai-Raj]

You could try:

do.call("rbind", lapply(list.files("path/to/files", full = TRUE), read.csv))

And add more arguments to lapply if the files are not csv, have no header, etc.

--sundar

On Tue, Jun 9, 2009 at 11:18 AM, Erin Hodgess wrote:
> Dear R People:
>
> I have about 6000 files to be read in that I'd like to go to one
> matrix.  There are two columns, 1 line in each file.
>
> Is there a way to bring them in to produce one matrix or data frame, please?
>
> Thanks,
> Erin
>
>
> --
> Erin Hodgess
> Associate Professor
> Department of Computer and Mathematical Sciences
> University of Houston - Downtown
> mailto: erinm.hodg...@gmail.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.
>

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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] Lattice group colors?

[Sundar Dorai-Raj]

Look at show.settings() and str(trellis.par.get()). This will show you
what the default settings are. The group colors are set by the
superpose.* elements (e.g. superpose.line is for group lines). To set
them, I usually create a list and pass it to par.settings. For
example,

my.theme <- list(superpose.line = list(col = c("darkred", "darkblue"),
lty = 1:2))
xyplot(y ~ x, data = mydata, groups = g, par.settings = my.theme,
  auto.key = list(points = FALSE, lines = TRUE))

HTH,

--sundar

On Mon, Jun 22, 2009 at 6:30 AM, Fredrik Karlsson wrote:
> Dear list,
>
> I have been struggling to find how I would go about changing the
> bakground colors of groups in a lattice barchart in a way so that the
> auto.key generated also does the right thing and pick it up for the
> key.
> I have used the "col" argument (which I guess is sent to par()) in a
> way so that the colors are like I would want them, but now I guess I
> need to know which part part of the theme I need to change in order to
> make this change permanent for all the plots, and all the keys.
>
> I am of course thankful for all the help I can get.
>
> (Also, how does one know these things about the theme variables? Is
> there some documentation somewhere?)
>
> /Fredrik
>
> --
> "Life is like a trumpet - if you don't put anything into it, you don't
> get anything out of it."
>
> __
> 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
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Requesting help with lattice again

[Sundar Dorai-Raj]

For the first question, add a groups argument. E.g.

barchart(HSI ~ Scenario | Region, Wbirdsm, groups = HydroState)

Also note that using Wbirdsm$HSI makes your call less readable, so I
added the data argument.

For your second question, setting the key does not set the color
theme. You want to set the colors using par.setting. E.g.

dotplot(HSI ~ Scenario | Region, Wbirdsm,
groups = HydroState,
par.settings = list(superpose.symbol =
  list(col = c("red", "green", "blue"))),
auto.key = list(space = "right"))

Then use auto.key instead of key.

HTH,

--sundar

On Wed, Mar 25, 2009 at 7:02 AM,   wrote:
>
> Hello, this is a request for assistance that I submitted earlier, this time
> with the dataset. My mistake for taking up bandwidth.  I've also rephrased
> the question to address an additional concern.
>
> I'm working on a windows XP machine with R 2.8.1
>
> 1).  I'd like a barchart (or other lattice type display) HSI ~ of the three
> factors (Region, Scenario and HydroState).
>      However barchart complains
>
>> library(reshape)
>> library(lattice)
>
>>  barchart(Wbirdsm$HSI ~ WBirdsm$Scenario | WBridsm$Region)
>
>>  # This works, but only marginally.  I'd like specify the interaction
> between the variables ( Scenario and HydroState)
>>    What is the best way to combine these factors so they can be treated
> as a single variable and rewrite the barchart call?
>
>>  barchart(Wbirdsm$HSI ~ WBirdsm$"interaction between Scenario and
> HydroState") ~ Wbirdsm$Region)
>
>
>
> The second question refers back to my original posting.
> 2)  #  using dotplot instead of barchart I use the following code.
>
>> key.variable <- list(space = "right", text =
> list(levels(wbirdm$variable)), points = list(pch = 1:3,
> col=c("red","green", "blue")))
>> dotplot(wbirdm$value  ~ wbirdm$variable | wbirdm$Region, col=c(1:9), pch=
> rep(c(1:3), key = key.variable,  groups=wbirdm$variable,     ylab= "Mean
> HSI"))
>
>> # However the key legend doesn't appear in the plot with this sequence
> and the labels are too along the panels.  Is there a way to address this?
>
>
> thank you very much for the assistance.
>
> Steve
>
>
> This is part of the larger data base, after I passed it thru melt.
>
>
>
>                                Region Species Scenario HydroState
> HSI
> 1                    Eastern Panhandle  WBLong      NSM        Ave
> 0.165945170
> 2                    Eastern Panhandle  WBLong      NSM        Dry
> 0.056244263
> 3                    Eastern Panhandle  WBLong      NSM        Wet
> 0.290692607
> 4                    Eastern Panhandle  WBLong      ECB        Ave
> 0.165945170
> 5                    Eastern Panhandle  WBLong     ECB         Dry
> 0.056244263
> 6                    Eastern Panhandle  WBLong     ECB         Wet
> 0.290692607
> 7                    Eastern Panhandle  WBLong     CERP        Ave
> 0.165945170
> 8                    Eastern Panhandle  WBLong     CERP        Dry
> 0.056244263
> 9                    Eastern Panhandle  WBLong     CERP        Wet
> 0.290692607
> 10 Long Pine Key / South Taylor Slough  WBLong      NSM        Ave
> 0.151159734
> 11 Long Pine Key / South Taylor Slough  WBLong      NSM        Dry
> 0.067348863
> 12 Long Pine Key / South Taylor Slough  WBLong      NSM        Wet
> 0.20738
> 13 Long Pine Key / South Taylor Slough  WBLong      ECB        Ave
> 0.151159734
> 14 Long Pine Key / South Taylor Slough  WBLong     ECB         Dry
> 0.067348863
> 15 Long Pine Key / South Taylor Slough  WBLong     ECB         Wet
> 0.20738
> 16 Long Pine Key / South Taylor Slough  WBLong     CERP        Ave
> 0.151159734
> 17 Long Pine Key / South Taylor Slough  WBLong     CERP        Dry
> 0.067348863
> 18 Long Pine Key / South Taylor Slough  WBLong     CERP        Wet
> 0.20738
> 19              Northern Taylor Slough  WBLong      NSM        Ave
> 0.115503291
> 20              Northern Taylor Slough  WBLong      NSM        Dry
> 0.005617136
> 21              Northern Taylor Slough  WBLong      NSM        Wet
> 0.252428530
> 22              Northern Taylor Slough  WBLong      ECB        Ave
> 0.115503291
> 23              Northern Taylor Slough  WBLong     ECB         Dry
> 0.005617136
> 24              Northern Taylor Slough  WBLong     ECB         Wet
> 0.252428530
> 25              Northern Taylor Slough  WBLong     CERP        Ave
> 0.115503291
> 26              Northern Taylor Slough  WBLong     CERP        Dry
> 0.005617136
> 27              Northern Taylor S

Re: [R] use R Group SFBA April meeting reminder; video of Feb k

[Sundar Dorai-Raj]

Could be that you have some sort of ad filter in your browser that's
blocking the video? It appears just fine for me in Firefox 3.

On Mon, Mar 30, 2009 at 3:55 PM, Ted Harding
 wrote:
> On 30-Mar-09 22:13:04, Jim Porzak wrote:
>> Next week Wednesday evening, April 8th, Mike Driscoll will be talking
>> about "Building Web Dashboards using R"
>> see: http://www.meetup.com/R-Users/calendar/9718968/ for details & to
>> RSVP.
>>
>> Also of interest, our member Ron Fredericks has just posted a well
>> edited video of the February kickoff panel discussion at Predictive
>> Analytics World "The R and Science of Predictive Analytics: Four Case
>> Studies in R" with
>>     * Bo Cowgill, Google
>>     * Itamar Rosenn, Facebook
>>     * David Smith, Revolution Computing
>>     * Jim Porzak, The Generations Network
>> and chaired by Michael Driscoll, Dataspora LLC
>>
>> see: http://www.lecturemaker.com/2009/02/r-kickoff-video/
>>
>> Best,
>> Jim Porzak
>
> It could be very interesting to watch that video! However, I have
> had a close look at the web page you cite:
>
>  http://www.lecturemaker.com/2009/02/r-kickoff-video/
>
> and cannot find a link to a video. Lots of links to non-video
> things, but none that I could see to a video.
>
> There is a link on that page at:
>  How Google and Facebook are using R
>  by Michael E. Driscoll | February 19, 2009
>  
>
> Following that link leads to a page, on which the first link, in:
>
>  <(March 26th Update: Video now available)>
>  Last night, I moderated our Bay Area R Users Group kick-off
>  event with a panel discussion entitled "The R and Science of
>  Predictive Analytics", co-located with the Predictive Analytics
>  World conference here in SF.
>
> leads you back to where you came from, and likewise the link at
> the bottom of the page:
>
>   is now available courtesy of Ron Fredericks
>  and LectureMaker.
>
> Could you help by describing where on that web page it can be found?
> With thanks,
> Ted.
>
> 
> E-Mail: (Ted Harding) 
> Fax-to-email: +44 (0)870 094 0861
> Date: 30-Mar-09                                       Time: 23:55:07
> -- XFMail --
>
> __
> 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
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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] labeling panels in lattice plots

[Sundar Dorai-Raj]

Try converting year to a factor

xyplot(min + max + ave ~ month | factor(year), data = rain.stats, ...)

Also, notice the inclusion of the "data" argument.

HTH,

--sundar

On Tue, Mar 31, 2009 at 6:28 AM,   wrote:
>
> I am using windows XP with R 2.8.1
>
>
> I am generating a lattice plot of annual rain patterns using the following
> function:
>
>
>>  xyplot(rain.stats$min+ rain.stats$max + rain.stats$ave ~
> rain.stats$month |rain.stats$year,
>            lty = 1,  data = rain.stats,  type = c("l","l", "l"), col =
> c("red", "blue", "green"), distribute.type = TRUE,
>            main = "Annual Monthly Minimum, Average and Maximum PPT")
>
>
> Each panel is labels with "rain.stats$year"  instead of the actual value
> referenced by this variable.
>
> the data.frame I'm using has the following form (year values range from
> 1965 to 2000 and month from 1 to 12).
>
>> dim(rain.stats)
>> 432 6
>
>>rain.stats[1:10,]
>   year month X2    min    max       ave
> 1  1965     1  1 0. 1.9196 0.3650112
> 2  1966     1  1 2.1483 4.9615 3.5247034
> 3  1967     1  1 0.4038 3.9145 1.7133045
> 4  1968     1  1 0.2033 3.2119 1.1844769
> 5  1969     1  1 1.2533 5.6226 2.9505545
> 6  1970     1  1 0.9142 3.8861 2.6248453
> 7  1971     1  1 0.1191 1.6109 0.6570289
> 8  1972     1  1 0.2309 2.9380 0.9259674
> 9  1973     1  1 0.9471 3.6342 1.9019848
> 10 1974     1  1 0.1739 9.0225 1.0672980
>
>
> The plot illustrates exactly what I'm after with the exception of the panel
> labels.  I'm following an example in the Lattice Book by Deepayan Sarkar
> (which I find very informative  - thanks), but I'm not getting the results
> as per the example.
>
> Can anyone offer a solution?
>
> Thanks in advance.
>
> Steve
>
>
>
>
>
>
> Steve Friedman Ph. D.
> Spatial Statistical Analyst
> Everglades and Dry Tortugas National Park
> 950 N Krome Ave (3rd Floor)
> Homestead, Florida 33034
>
> steve_fried...@nps.gov
> Office (305) 224 - 4282
> Fax     (305) 224 - 4147
>
> __
> 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] Can not get a prediction interval from Predict

[Sundar Dorai-Raj]

?predict.glm has no "interval" argument. Perhaps you're thinking of
?predict.lm, which is different.

To get intervals in glm, I've used:

example(predict.glm)
pr <- predict(budworm.lg, se.fit = TRUE)
family <- family(budworm.lg)
lower <- family$linkinv(pr$fit - qnorm(0.95) * pr$se.fit)
upper <- family$linkinv(pr$fit + qnorm(0.95) * pr$se.fit)

Note that these are "confidence" limits and not "prediction" limits.
The latter would require more thought.

You could also try RSiteSearch("glm interval").

HTH,

--sundar

On Tue, Mar 31, 2009 at 9:58 AM, Taylor Davis
 wrote:
> I am trying to get a prediction interval from a glm regression.
>
> With newdat being my set of values to be fitted, and glmreg the name of my
> regression, I am using the following code.
>
> predict(glmreg, newdat, se.fit = TRUE, interval = "confidence", level =
> 0.90)
>
> The problem is that I am only getting the standard error and the fitted
> value, not a prediction interval.
>
> Any help would be great! Thanks so much.
>
> ___
> TAYLOR DAVIS    |
> KLAS
> taylor.da...@klasresearch.com    |    801.734.6279
>
> __
> 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] Changing the y-axis units of a lattice histogram

[Sundar Dorai-Raj]

Try:

library(lattice)
histogram( ~ height | voice.part,
  data = singer, type = "c",
  scales = list(y =
list(at = seq(0, 20, 5),
 labels = seq(0, 200, 50

HTH,

--sundar

On Fri, Apr 3, 2009 at 2:01 PM, Judith Flores  wrote:
>
> Hello,
>
>   I need to multiply the number of counts in the y-axis of a lattice 
> histogram by a constant factor, such that the plot would represent a 
> different type of variable plotted in the y-axis, not counts. Can this be 
> done?
>
> Thank you,
>
> Judith
>
> __
> 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] question on using lattice panel plots

[Sundar Dorai-Raj]

Try:

z <- cbind(rep(c("BIC", "hist"), each = 150), rep(rep(c(5, 10, 30),
each = 50),2))

z <- as.data.frame(z)

z <- cbind(z, runif(300))
names(z) <- c("Method", "sigma", "Error")
z$sigma <- factor(z$sigma, c("5", "10", "30"))
library(lattice)

sigma <- as.numeric(levels(z$sigma))
sigmaExprList <- lapply(sigma, function(s) bquote(sigma == .(s)))
sigmaExpr <- as.expression(sigmaExprList)
bwplot(Error~Method | sigma, data = z,
   horiz = F, xlab = "Method",
   strip = strip.custom(var.name = sigmaExpr,
 strip.levels = FALSE, strip.names = TRUE),
   layout = c(3,1))

HTH,

--sundar

On Thu, Apr 16, 2009 at 12:43 PM, Ranjan Maitra  wrote:
> Hi,
>
> I think this question is best explained using the following
> self-contained toy example:
>
>
>
>
> ## cut code here and paste to R window
>
> z <- cbind(rep(c("BIC", "hist"), each = 150), rep(rep(c(5, 10, 30),
> each = 50),2))
>
> z <- as.data.frame(z)
>
> z <- cbind(z, runif(300))
>
> names(z) <- c("Method", "sigma", "Error")
>
> library(lattice)
>
> bwplot(Error~Method | sigma, data = z, horiz = F, xlab = "Method",
> layout = c(3,1))
>
>
> ## end code
>
>
>
>
> Now the question:
>
> I would like the panels to be in the order of sigma, i. e. 5, 10, 30
> and not 10, 30 and 5 as is currently the case. Is this possible?
>
> Not to seek too much indulgence, but to ask anyway, I wonder if it is
> possible to have a Greek sigma = 5, a Greek sigma = 10 and a Greek
> sigma = 30. (Sort of what we would get using expression(sigma == 5),
> expression(sigma == 10), expression(sigma == 10) on "base" R figures).
>
> Please let me know if my question is not clear.
>
> Many thanks for any suggestions and help and best wishes!
> Ranjan
>
> __
> 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] question on using lattice panel plots

[Sundar Dorai-Raj]

Sorry, that should be:

sigma <- as.numeric(levels(z$sigma))
sigmaExprList <- lapply(sigma, function(s) bquote(sigma == .(s)))
sigmaExpr <- as.expression(sigmaExprList)
bwplot(Error~Method | sigma, data = z,
   horiz = F, xlab = "Method",
   strip = function(which.given, which.panel, var.name,
strip.levels = FALSE,
strip.names = TRUE, ...) {
 strip.default(which.given, which.panel,
   var.name = sigmaExpr[which.panel],
   strip.levels = FALSE,
   strip.names = TRUE, ...)
   },
   layout = c(3,1))

Not sure how to do this with strip.custom.

--sundar

On Thu, Apr 16, 2009 at 1:20 PM, Sundar Dorai-Raj  wrote:
> Try:
>
> z <- cbind(rep(c("BIC", "hist"), each = 150), rep(rep(c(5, 10, 30),
> each = 50),2))
>
> z <- as.data.frame(z)
>
> z <- cbind(z, runif(300))
> names(z) <- c("Method", "sigma", "Error")
> z$sigma <- factor(z$sigma, c("5", "10", "30"))
> library(lattice)
>
> sigma <- as.numeric(levels(z$sigma))
> sigmaExprList <- lapply(sigma, function(s) bquote(sigma == .(s)))
> sigmaExpr <- as.expression(sigmaExprList)
> bwplot(Error~Method | sigma, data = z,
>       horiz = F, xlab = "Method",
>       strip = strip.custom(var.name = sigmaExpr,
>         strip.levels = FALSE, strip.names = TRUE),
>       layout = c(3,1))
>
> HTH,
>
> --sundar
>
> On Thu, Apr 16, 2009 at 12:43 PM, Ranjan Maitra  wrote:
>> Hi,
>>
>> I think this question is best explained using the following
>> self-contained toy example:
>>
>>
>>
>>
>> ## cut code here and paste to R window
>>
>> z <- cbind(rep(c("BIC", "hist"), each = 150), rep(rep(c(5, 10, 30),
>> each = 50),2))
>>
>> z <- as.data.frame(z)
>>
>> z <- cbind(z, runif(300))
>>
>> names(z) <- c("Method", "sigma", "Error")
>>
>> library(lattice)
>>
>> bwplot(Error~Method | sigma, data = z, horiz = F, xlab = "Method",
>> layout = c(3,1))
>>
>>
>> ## end code
>>
>>
>>
>>
>> Now the question:
>>
>> I would like the panels to be in the order of sigma, i. e. 5, 10, 30
>> and not 10, 30 and 5 as is currently the case. Is this possible?
>>
>> Not to seek too much indulgence, but to ask anyway, I wonder if it is
>> possible to have a Greek sigma = 5, a Greek sigma = 10 and a Greek
>> sigma = 30. (Sort of what we would get using expression(sigma == 5),
>> expression(sigma == 10), expression(sigma == 10) on "base" R figures).
>>
>> Please let me know if my question is not clear.
>>
>> Many thanks for any suggestions and help and best wishes!
>> Ranjan
>>
>> __
>> 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] Setting lattice par parameters

[Sundar Dorai-Raj]

Because you're not calling trellis.par.set correctly. It should be:

trellis.par.set(par.ylab.text = list(cex = 0.65), par.xlab.text =
list(cex = 0.65))

However, I usually do things like this:

my.theme <- list(par.ylab.text = list(cex = 0.65), par.xlab.text =
list(cex = 0.65))
barchart(..., par.settings = my.theme)

so the settings are only changed for the current plot and not globally.

HTH,

--sundar

On Thu, Apr 23, 2009 at 6:25 AM,   wrote:
>
> Hello
>
> I'm plotting a large suite of barcharts and need to modify the size of the
> text for both the yaxis and xaxis labels.
>
> I've tried using the following:
>
>> trellis.par.set(list(par.ylab.text = list(cex = 0.65)),
> trellis.par.set(list = par.xlab.text = list(cex = 0.65
>
> On inspection,  however after I invoke this line,
>
>>  trellis.par.get("par.ylab.text")
>>  trellis.par.get("par.xlab.text")
>
> It looks like the parameters have not been modified.
>
> Do I need to do something different than this approach ?
>
> Thank you very much in advance.
>
> Steve
>
> Steve Friedman Ph. D.
> Spatial Statistical Analyst
> Everglades and Dry Tortugas National Park
> 950 N Krome Ave (3rd Floor)
> Homestead, Florida 33034
>
> steve_fried...@nps.gov
> Office (305) 224 - 4282
> Fax     (305) 224 - 4147
>
> __
> 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] Flipping axes of qqnorm

[Sundar Dorai-Raj]

Try (re)reading ?qqnorm. Use datax = TRUE.

--sundar

On Sun, Apr 26, 2009 at 4:37 PM, Chris_d  wrote:
>
> Hi all,
>       I have just started using R to produce qqnorm plots. I am trying to
> switch the x and y axes so that the theoretical values are plotted on the y
> axis and my data on the x axis. Can anyone help me with this?
>
> Thanks
> --
> View this message in context: 
> http://www.nabble.com/Flipping-axes-of-qqnorm-tp23248007p23248007.html
> Sent from the R help mailing list archive at Nabble.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.
>

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Re: [R] Cannot clean infinite values

[Sundar Dorai-Raj]

Use ?is.infinite

inf <- is.infinite(data)
data[inf] <- 0.3 * sign(data[inf])

On Sun, Apr 26, 2009 at 5:44 PM, Nigel Birney  wrote:
>
> Hello all,
>
> I have to import numeric data from file but found it contains Infinite
> values which need to be eliminated. I tried to replace them in this way:
>
>        data[which(data=="-Inf")] <- -0.3
>        data[which(data=="+Inf")] <-  0.3
>
> But, somehow, the Infinite values stayed there. Any suggestions?
>
> regards,
>
> N.
> --
> View this message in context: 
> http://www.nabble.com/Cannot-clean-infinite-values-tp23248409p23248409.html
> Sent from the R help mailing list archive at Nabble.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.
>

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Re: [R] evaluate a expression

[Sundar Dorai-Raj]

Hi, Ning,

Try: eval(parse(text = expr))

HTH,

--sundar

On Sat, May 2, 2009 at 5:39 AM, Ning Ma  wrote:
> Hi,
>
> I am new to R. Can anyone tell me how to evaluate an expression stored
> in a string?
> such as:
>> expr <- "3*5"
> I want to get the result 15.
>
> Thanks in advance.
>
> __
> 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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Re: [R] setting trellis auto.key color values

[Sundar Dorai-Raj]

Set the colors in graph.sets and not auto.key.

graph.sets <- list(axis.text = list(cex = 0.65),
  par.ylab.text = list(cex = 1.25),
  par.xlab.text = list(cex = 1.25),
  superpose.polygon = list(col = 3:5))

Then remove the "col = 3:5" from auto.key and barchart.

Also, you can simplify your code by removing "gator_IR$" and including
"data = gator_IR". I.e.

 barchart(MEAN ~ Hydro | as.factor(IR_ID),
 data = gator_IR, layout = c(4, 1),
 groups = Rain, ylim = c(0, 1), ...)

HTH,

--sundar

On Tue, May 5, 2009 at 8:32 AM,   wrote:
>
>
> I'm working with Lattice graphics and I would like very much to color code
> the auto.key fill color with the same corresponding colors that I use in
> the panels.  I've looked on the web for clues, and on the CRAN-R help sites
> searching on "trellis auto.key color" and variations, unfortunately the
> responses there are not very specific.
>
>  Would someone please explain   I have the Book, if you can point me to the
> pages that explain this I'd appreciate that too
>
>
>
>> graph.sets <- list(axis.text = list(cex = 0.65),
>              par.ylab.text = list(cex = 1.25),
>              par.xlab.text = list(cex = 1.25))
>
>
>>   barchart(gator_IR$MEAN ~ gator_IR$Hydro |
> as.factor(gator_IR$IR_ID),layout=c(4,1),col = c(3:5),
>                 groups=gator_IR$Rain, ylim=c(0,1), par.settings =
> graph.sets,
>                 main = "Alligator Nesting Performance", ylab= "Mean HSI",
>                 auto.key = list("top", columns=3, col=c(3:5))
>
>
> this script creates the graph nicely, with the qualification that the color
> for the key titles are the correct, but the filling colors are default
> pastels.  Where do I change them ?
>
>
> Windows XP  with R 2.8.1
>
> Thank you.
>
> Steve
>
> Steve Friedman Ph. D.
> Spatial Statistical Analyst
> Everglades and Dry Tortugas National Park
> 950 N Krome Ave (3rd Floor)
> Homestead, Florida 33034
>
> steve_fried...@nps.gov
> Office (305) 224 - 4282
> Fax     (305) 224 - 4147
>
> __
> 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] convert large integers to hex

[Sundar Dorai-Raj]

Hi,

I'm wondering if someone has solved the problem of converting very
large integers to hex. I know about format.hexmode and as.hexmode, but
these rely on integers. The numbers I'm working with are overflowing
and losing precision. Here's an example:

x <- "6595137340052185552" # stored as character
as.integer(x) # warning about inaccurate conversion
format.hexmode(as.numeric(x)) # warnings about loss of precision
as.hexmode(x) # more warnings and does not do what I expected

I'm planning on writing a function that will do this, but would like
to know if anybody already has a solution. Basically, I would like the
functionality of format.hexmode on arbitrarily large integers.

Thanks,

--sundar

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Re: [R] Dynamic arguments in "rbind" function

[Sundar Dorai-Raj]

Use a list instead of assign then do.call("rbind", thelist).

import.files <- c("a.txt", "b.txt",  "c.txt",  "d.txt",  "e.txt")
imp <- vector("list", length(import.files))
for (i in 1:length(import.files)) {
  imp[[i]] <- read.delim(import.files[i], sep = "", header = TRUE)
}

combined <- do.call("rbind", imp)

HTH,

--sundar

On Mon, Jan 4, 2010 at 4:31 PM, Steven Kang wrote:

> Hi, all
>
> Basically, I have unknown number of data that need to be imported and
> collapsed row-wisely.
>
> The code below works fine, however the "rbind" function may require 50
> arguments if there are 50 data files...
>
> Thus, I would like to explore whether there are any methods in using
> dynamic
> objects (i.e from the resulting objects in the for loop) as an argument in
> the *"rbind"* function.
>
>
>
> setwd(.)
>
> import.files <- c("a.txt", "b.txt",  "c.txt",  "d.txt",  "e.txt")
> for (i in 1:length(import.files)) {
>  assign(paste("imp", i, sep = "."), read.delim(eval(paste(".\\",
> import.files[i], sep = "")), header = TRUE))
> }
>
> combined <- rbind(*imp.1, imp.2, imp.3, imp.4, imp.5, imp.6*)
>
>
>
> Your expertise in resolving this issue would be greatly appreciated.
>
>
>
> Steve
>
>[[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.
>

[[alternative HTML version deleted]]

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Re: [R] How do I juxtapose two lattice graphs with common X axes such that the X axes line up?

[Sundar Dorai-Raj]

Try googling "latticeExtra x.same" for some examples. Here's one:

http://www.mail-archive.com/r-help@r-project.org/msg39048.html

On Wed, Jan 20, 2010 at 9:44 AM, George Chen  wrote:

> Hello,
>
> I would like to juxtapose two lattice graphs with common X axes such that
> the X axes line up.  I am using plot right now but the edges are not neat
> and it would be nice if I could just draw 1 X axis and not both of them.
>
> Here is my code:
>
>
>
> upper<-bwplot(SignalUsed~as.factor(AllNormalHitsNamesCount),data=NmlOverviewArray2,
>xlab="",
>ylab="Intensity of Individual Antibody Responses",
>main="Intensity, Frequency, Distribution, & Quantity of Normal
> Antibody Responses",
>box.ratio=1,
>panel = function (AllNormalHitsNamesCount,...) {
>panel.bwplot(...)
>}
>)
>
> lower<-barchart(as.vector(table(NmlOverviewArray2$AllNormalHitsNamesCount))
>
>  
> ~as.factor(as.numeric(names(table(NmlOverviewArray2$AllNormalHitsNamesCount,
>data=NmlOverviewArray2,
>ylab="Number of Individual Antibody Responses",
>xlab="Occurrence of Individual Antibody Responses
> (Out of 45 Normals)",
>box.ratio=1)
>
> plot (upper, newpage=TRUE, more=TRUE, position = c(0,.15,1,1))
> plot (lower, newpage=FALSE, more=TRUE, position = c(0,0,1,.3))
>
>
> George
>
> __
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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.
>

[[alternative HTML version deleted]]

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Re: [R] as.numeric with tclvalue redux

[Sundar Dorai-Raj]

Erin Hodgess said the following on 3/24/2008 10:39 AM:
> Hi again R People:
> 
> This works fine:
>> library(tcltk)
>> a <- tclVar("4.5")
>> as.numeric(tclvalue(a))
> [1] 4.5
>> #But if you have:
>> b <- tclVar("pi")
>> as.numeric(tclvalue(b))
> [1] NA
> Warning message:
> NAs introduced by coercion
> 
> Is anyone aware of a way around this, please?
> 
> thanks,
> Erin
> 
> 

Does this help?

eval.tclvalue <- function(x, ...) {
   x <- type.convert(tclvalue(x), as.is = TRUE)
   if(is.character(x) && exists(x, ...)) {
 get(x)
   } else {
 x
   }
}

a <- tclVar("4.5")
b <- tclVar("pi")
c <- tclVar("abcd")

eval.tclvalue(a)
eval.tclvalue(b)
eval.tclvalue(c)

HTH,

--sundar

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Re: [R] Accessing items in a list of lists

[Sundar Dorai-Raj]

[EMAIL PROTECTED] said the following on 5/14/2008 
12:40 PM:

Using R 2.6.2, say I have the following list of lists, "comb":

data1 <- list(a = 1, b = 2, c = 3)
data2 <- list(a = 4, b = 5, c = 6)
data3 <- list(a = 3, b = 6, c = 9)
comb <- list(data1 = data1, data2 = data2, data3 = data3)

So that all names for the lowest level list are common.  How can I most
efficiently access all of the sublist items "a" indexed by the outer
list names?  For example, I can loop through comb[[i]], unlisting as I
go, and then look up the field "a", as below, but there has got to be a
cleaner way.

finaldata <- double(0)
for(i in 1:length(names(comb))) {
test <- unlist(comb[[i]])
finaldata <- c(finaldata, test[which(names(test) == "a")])
}
data.frame(names(comb), finaldata)

Gives what I want:
  names.comb. finaldata
1   data1 1
2   data2 4
3   data3 3

Any help you can give would be greatly appreciated.  Thanks.


Try

data.frame(names.comb = names(comb),
   finaldata = sapply(comb, "[[", "a"))

HTH,

--sundar

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Re: [R] Using plotmath expressions in lattice key text

[Sundar Dorai-Raj]

Bert Gunter said the following on 11/15/2007 1:12 PM:
> Folks:
> 
> delta <- 1:5
> 
> I would like to put 5 separate lines of text of the form "10 %+-% delta[i]"
> into a lattice key legend, where ""%+-%" is the plotmath plus/minus symbol
> and delta[i] is the ith value of delta.
> 
> The construct:
> 
> lapply(delta,function(d)bquote(10%+-%.(d)))
> 
> appears to produce a list of expressions of the correct form, and, indeed,
> if I assign the above to (a list!) test,
> 
> plot(0:1,0:1)
> text(.5,.5,test[[1]])
> 
> produces the correctly formatted plotmath expression. However, note that I
> have to use test[[1]] to extract the expression; test[1] doesn't work (it is
> a list containing an expression, not an expression) -- and therein may lie
> the problem. For if I try to use the above expression as the lab component
> of the text component in key, e.g. by
> 
> xyplot(,
> key = list( text = list(lab =
> lapply(delta,function(d)bquote(10%+-%.(d))),...),...)
> 
> 
> I get an error:
> 
> Error in fun(key = list(text = list(lab = list(10 %+-% 1, 10 %+-% 2 : 
>   first component of text has to be vector of labels
> 
> 
> So how should I do this?? I suspect it's simple, but I just can't figure it
> out.
> 
> Note: I'd be happy to supply reproducible code if needed. Just complain and
> I'll do so.
> 
> Thanks.
> 
> 
> Bert Gunter
> Genentech Nonclinical Statistics

Hi, Bert,

This is how I've done it in the past. Perhaps there's a better way:

library(lattice)

delta <- 1:5
expr <- paste("10%+-%", delta, sep = "", collapse = ",")
expr <- paste("expression(", expr, ")", sep = "")
expr <- eval(parse(text = expr))

xyplot(0 ~ 0 | factor(delta),
key = list(text = list(expr)),
## bonus: adding it to the strip
strip = function(factor.levels, ...) {
  strip.default(factor.levels = expr, ...)
})


Thanks,

--sundar

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Re: [R] How can R be used to solve algebra equations?

[Sundar Dorai-Raj]

francogrex said the following on 11/29/2007 1:00 PM:
> suppose I have this equation:
> (x^2+y^2+3z^3)/(5*z^2*x^3)=0
> 
> and I want to find x in relation to the other variables which actually is:
> 
> x=sqrt(-3*z^3-y^2) or x=-sqrt(-3*z^3-y^2)
> 
> Can R give me this expression solution? I know there is uniroot, but here I
> want the "expression" not the value, because I do  not have values for y and
> z.
> 
> Thanks

Try:

library(Ryacas)
yacas("Solve((x^2+y^2+3*z^3)/(5*z^2*x^3)==0, x)")

expression(list(x == root(-(4 * (y^2 + 3 * z^3)), 2)/2, x ==
     -root(-(4 * (y^2 + 3 * z^3)), 2)/2))

HTH,

--sundar

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Re: [R] Accesing the value

[Sundar Dorai-Raj]

get(x)

This is a FAQ:

http://cran.cnr.berkeley.edu/doc/FAQ/R-FAQ.html#How-can-I-turn-a-string-into-a-variable_003f

--sundar

Shubha Vishwanath Karanth said the following on 12/28/2007 8:38 AM:
> Hi R,
> 
>  
> 
> x="A"
> 
> A=5
> 
>  
> 
> I need to get the value of A using x only. How do I do this?
> 
>  
> 
> Thanks in advance, Shubha
> 
> This e-mail may contain confidential and/or privileged i...{{dropped:13}}
> 
> __
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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] mfrow for levelplot?

[Sundar Dorai-Raj]

marcg said the following on 1/5/2008 3:48 AM:
> hello
> 
> could anyone tell my, why I do not suceed with mfrow?
> 
> par(mfrow=c(4,4))
> 
> for (i in 5:17){
> levelplot(maxwater[,i]~maxwater$V1*maxwater$V2, col.regions=whiteblue(5), 
> xlab="", cuts=4)
> }
> 
> Thanks
> 
> Marc
> --
> 

Because "par" settings have little to no effect on lattice. From ?Lattice:

  Lattice plots are highly customizable via user-modifiable
  settings. However, these are completely unrelated to base graphics
  settings; in particular, changing 'par()' settings usually have no
  effect on lattice plots.

To do what you're doing, you need to understand how Lattice works with 
panels. Try:

## Since you didn't supply the data.frame 'maxwater'
## here's a fake dataset to demonstrate
library(lattice)
set.seed(1)
z <- expand.grid(V1 = 1:10, V2 = 1:10)
z <- cbind(z, matrix(rnorm(100 * 16), 100, 16))
names(z) <- sprintf("V%d", 1:ncol(z))

## now create a formula
left <- paste(names(z)[3:18], collapse = "+")
right <- paste(names(z)[1:2], collapse = "*")
form <- formula(sprintf("%s~%s", left, right))

## not sure how you defined 'whiteblue', but here's my version
whiteblue <- colorRampPalette(c("white", "blue"))

## now call levelplot using a formula
levelplot(form, z, col.regions = whiteblue(5),
   xlab = "", cuts = 4, as.table = TRUE)

In the future, please read the posting guide about providing "commented, 
minimal, self-contained, reproducible code."

HTH,

--sundar

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Re: [R] side by side plots

[Sundar Dorai-Raj]

[EMAIL PROTECTED] said the following on 1/7/2008 2:59 PM:
> Hello everyone,
> 
> I have an overlay plot it's nice but you can't see all the data. I would
> like to know if there is a way to get a plot that gives a side by side
> plot so that  each plot would be next to each other. The two plots have
> the same data are of different species. At the moment this is the code I'm
> using:
> 
> exp<-cbind(abs(round(rnorm(10),2)*10), seq(100, 200, by=10))
> ref<-cbind(abs(round(rnorm(10),2)*10), seq(100, 200, by=10))
> 
> plot(ref, ylab="Intensity", xlab="wavelength", type="h")
> points(exp, type="h", col="red")
> 
> This is working in a script and I would like to have a single pdf/png file
> for the user with this plot, rather than asking the user to manually
> compare them.
> 
> Any ideas on how I would do this?
> 
> Thanks
> 
> Paul
> 
> __
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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.

Hi, Paul,

Simple enough in lattice:

library(lattice)
z <- rbind(cbind(as.data.frame(exp), type = "exp"),
cbind(as.data.frame(ref), type = "ref"))
z$type <- factor(z$type, levels = c("ref", "exp"))
xyplot(V2 ~ V1 | type, data = z, type = "h")

HTH,

--sundar

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Re: [R] lattice color problem with symbols: bug?

[Sundar Dorai-Raj]

Stefan Grosse said the following on 1/11/2008 10:04 AM:
> Dear useR's,
> 
> I have a problem with the lattice plotting of some symbols:
> 
> library(lattice)
> 
> test<-data.frame(x=c(2,3,1,5),u=c(rep(1,2),rep(2,2)),g=c(rep(c(1,2),2)))
> 
> xyplot(x~u,groups=g,
>  data=test,
>  par.settings=list(
> superpose.symbol=list(pch=c(22, 23),cex=c(1.7,1.6),col="black")
> ),
>  key=list(
> text=list(c("t1","t2")),
> space = "bottom",pch=c(23, 22),
> points=F,
> cex=1.0,
> col="black"
> ),
> )
> 
> As you see the symbols which have been plotted into the plotting area appear 
> to have some filling color while in the legend there is no filling color 
> although the specification of the symbols is the same.
> 
> If I use a normal plot command, the symbols are also not filled:
> plot(c(1,2,3),c(1,2,3),pch=c(22,23,24),cex=1.5)
> 
> That problem must have occured during a recent lattice update since the color 
> was not there when I was plotting a year ago with R2.5.x and some older 
> lattice. It occurs on my linux as well as on my windows machine both with R 
> 2.6.1 and latest lattice from CRAN.
> 
> So here my questions: How do I get rid of the color? (or is it a bug?) If it 
> is a feature, so how do I determine the color in both the symbols in the 
> legend and in the plot itself? (Actually I was overlaying two plots e.g. one 
> colored diamond and one empty diamond to achieve that effect but if there is 
> a more efficient way to draw bordered symbols with customized color that 
> would be preferable...)
> 
> Stefan
> 

Hi, Stefan

You need to use "fill = 'transparent'" in your par.settings call. If 
you're not using a fill color, you probably should use pch = c(0, 5) 
instead. This is explained under pch on the ?points help page.

Here's a modified version of your example:

library(lattice)
test <- data.frame(x = c(2, 3, 1, 5),
u = rep(1:2, each = 2),
g = paste("t", rep(1:2, 2), sep = ""))

xyplot(x ~ u, data = test, groups = g,
    par.settings = list(
  superpose.symbol = list(
pch = c(22, 23),
cex = c(1.7, 1.6),
fill = c("red", "blue"),
col = "black")),
auto.key = list(space = "bottom"))

HTH,

--sundar

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[R] using bquote to construct function

[Sundar Dorai-Raj]

Hi, R-help,

(sessionInfo at the end)

I'm trying to construct a function using bquote and running into a 
strange error message. As an example, what I would like to do is this:


z <- 2
eval(bquote(function(x, y) { x^.(z) + y }))(2, 3)

However, I get the following:

Error in eval(expr, envir, enclos) :
  invalid formal argument list for "function"

However, if I change the command to following, it works:

z <- 2
eval(bquote(function(x) { x^.(z) }))(2)
# [1] 4

In other words, I remove the second argument. Is there a workaround for 
this without using eval(parse(text = ))?


Thanks,

--sundar

> sessionInfo()
R version 2.7.2 (2008-08-25)
i386-pc-mingw32

locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United 
States.1252;LC_MONETARY=English_United 
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252


attached base packages:
[1] stats graphics  grDevices utils datasets  methods   base

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Re: [R] using bquote to construct function

[Sundar Dorai-Raj]

Thanks, Gabor! This workaround fixes my immediate needs.

--sundar

Gabor Grothendieck said the following on 10/1/2008 10:42 PM:

That may be a bug in R but I think there is another problem on top of that
as I don't think bquote descends into function bodies:


z <- 2
bquote(function(x) {x^.(z)})

function(x) {x^.(z)}


bquote(function(x, y) { x^.(z) + y})

function(x, y) { x^.(z) + y}


R.version.string # Vista

[1] "R version 2.7.2 (2008-08-25)"

Try it this way:

z <- 2
f <- function(x, y) {}
body(f) <- bquote({ x^.(z) + y })
eval(f)(2, 3)

On Thu, Oct 2, 2008 at 1:14 AM, Sundar Dorai-Raj
<[EMAIL PROTECTED]> wrote:

Hi, R-help,

(sessionInfo at the end)

I'm trying to construct a function using bquote and running into a strange
error message. As an example, what I would like to do is this:

z <- 2
eval(bquote(function(x, y) { x^.(z) + y }))(2, 3)

However, I get the following:

Error in eval(expr, envir, enclos) :
 invalid formal argument list for "function"

However, if I change the command to following, it works:

z <- 2
eval(bquote(function(x) { x^.(z) }))(2)
# [1] 4

In other words, I remove the second argument. Is there a workaround for this
without using eval(parse(text = ))?

Thanks,

--sundar


sessionInfo()

R version 2.7.2 (2008-08-25)
i386-pc-mingw32

locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252

attached base packages:
[1] stats graphics  grDevices utils datasets  methods   base

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Re: [R] xyplot with lowess curves

[Sundar Dorai-Raj]

You'll need a custom panel function. It would also help if you
provided a reproducible example:

xyplot (
  SnowLineElevation ~ Year | Model,
  data = data,
  panel = function(x, y, col, ...) {
col <- ifelse(panel.number() == 1, "red", "green")
panel.xyplot(x, y, col = "blue", ...)
panel.loess(x, y, col = col)
  },
  ylim = c(0,1800),
  pch = 21,
  xlab = 'Year',
  ylab = 'Snowline Elevation [m]'
)

Alternatively, you can use the group argument in conjunction with the panels:

xyplot(SnowLineElevation ~ Year | Model, data, groups = Model, type =
c("p", "smooth"))

if you want the points and the lines to be the same color.

--sundar

On Mon, Feb 2, 2009 at 10:20 AM, Hutchinson,David [PYR]
 wrote:
> I am trying to change the attributes of the lowess lines fit to an
> xyplot command, but have been unsuccessful in my search of the online
> help. Right now, both the points and lowess line come out in the same
> color (blue). I am unsure how I can change the properties of the lowess
> line separately.
>
> xyplot (
>  SnowLineElevation ~ Year | Model,
>  data = data,
>  ylim = c(0,1800),
>  type = c('p','smooth'),
>  col = 'blue',
>  pch = 21,
>  xlab = 'Year',
>  ylab = 'Snowline Elevation [m]'
> )
>
> Any help would be much appreciated,
>
> Dave
>
>[[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.
>

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Re: [R] xyplot with lowess curves

[Sundar Dorai-Raj]

Does this do what you want? The "panel" argument has the custom pane
function I referred to before.

Col <- c("red", "green", "blue", "purple")
xyplot (
 SnowLineElevation ~ Year | Model,
 data = d,
 panel = function(x, y, col, ...) {
   Col <- Col[panel.number()]
   panel.xyplot(x, y, col = "blue", ...)
   panel.loess(x, y, col = Col)
 },
 ylim = c(0,100),
 type = c('p','smooth'),
 col = 'blue',
 pch = 21,
 xlab = 'Year',
 ylab = 'Snowline Elevation [m]'
)

On Mon, Feb 2, 2009 at 12:18 PM, Hutchinson,David [PYR]
 wrote:
> I haven't had much luck with a custom panel function; mainly because I
> don't truly understand how to embedd the functionality into the xyplot
> command.
>
> Here's a reproducible example if you can help out.
>
> Thanks,
> Dave
>
> library (lattice)
>
> d <- NULL
> models <- c('A','B','C','D')
> n = 100
> for (i in seq(along = models)){
>  d <- rbind(
>d, data.frame (
>  Model = models[i],
>  Year = seq(1960, length.out=n, by = 1),
>  SnowLineElevation = runif(n, 0, 100)
>)
>  )
> }
>
> xyplot (
>  SnowLineElevation ~ Year | Model,
>  data = d,
>  ylim = c(0,100),
>  type = c('p','smooth'),
>  col = 'blue',
>  pch = 21,
>  xlab = 'Year',
>  ylab = 'Snowline Elevation [m]'
> )
>
> -Original Message-
> From: Sundar Dorai-Raj [mailto:sdorai...@gmail.com]
> Sent: Monday, February 02, 2009 11:43 AM
> To: Hutchinson,David [PYR]
> Cc: r-help@r-project.org
> Subject: Re: [R] xyplot with lowess curves
>
> You'll need a custom panel function. It would also help if you provided
> a reproducible example:
>
> xyplot (
>  SnowLineElevation ~ Year | Model,
>  data = data,
>  panel = function(x, y, col, ...) {
>col <- ifelse(panel.number() == 1, "red", "green")
>panel.xyplot(x, y, col = "blue", ...)
>panel.loess(x, y, col = col)
>  },
>  ylim = c(0,1800),
>  pch = 21,
>  xlab = 'Year',
>  ylab = 'Snowline Elevation [m]'
> )
>
> Alternatively, you can use the group argument in conjunction with the
> panels:
>
> xyplot(SnowLineElevation ~ Year | Model, data, groups = Model, type =
> c("p", "smooth"))
>
> if you want the points and the lines to be the same color.
>
> --sundar
>
> On Mon, Feb 2, 2009 at 10:20 AM, Hutchinson,David [PYR]
>  wrote:
>> I am trying to change the attributes of the lowess lines fit to an
>> xyplot command, but have been unsuccessful in my search of the online
>> help. Right now, both the points and lowess line come out in the same
>> color (blue). I am unsure how I can change the properties of the
>> lowess line separately.
>>
>> xyplot (
>>  SnowLineElevation ~ Year | Model,
>>  data = data,
>>  ylim = c(0,1800),
>>  type = c('p','smooth'),
>>  col = 'blue',
>>  pch = 21,
>>  xlab = 'Year',
>>  ylab = 'Snowline Elevation [m]'
>> )
>>
>> Any help would be much appreciated,
>>
>> Dave
>>
>>[[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.
>>
>

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Re: [R] Assigning colnames in loop

[Sundar Dorai-Raj]

It's always best to do this with list operations (e.g. lapply) rather
than a loop:

DF1 <- split(DF, DF$male)
DF2 <- lapply(DF1, function(x) {
  x2 <- t(as.matrix(x[3:5], dimnames = list(levels(x$age), NULL)))
  as.data.frame(x2)
})

Then DF2[["0"]] and DF2[["1"]] are the data.frames you want.

HTH,

--sundar

On Mon, Feb 2, 2009 at 2:30 PM, Peter Jepsen  wrote:
> Dear R-listers,
>
> I am trying to assign colnames to a data frame within a loop, but I keep
> getting a "target of assignment expands to non-language object"-error. I
> need to split up a large dataset into about 20 smaller ones, and I would
> like to assign colnames within the loop, so I won't have to type the
> same thing 20 times over.
>
> I have concocted this really goofy example which constructs two
> datasets:
>
> -
> male <- rep(0:1, each=5)
> age <- factor(c(10:14,10:14))
> DF <- data.frame(male, age, res1=rnorm(10), res2=rnorm(10),
> res3=rnorm(10))
>
> for(n in 0:1) {
>assign(paste("test",n, sep="."), as.data.frame(t(subset(DF,
> male==n, select=c(res1, res2, res3)
>colnames(get(paste("test",n, sep="."))) <-
> paste("age",levels(age), "m", n, sep="") # This line gives an error.
>assign(colnames(paste("test",n, sep="."))) <-
> paste("age",levels(age), "m", n, sep="") # This line gives the same
> error.
> }
> ---
> The following command assigns the right colnames to the 'test.0' data
> frame, but I want this line inside the loop so I won't have to type it
> 20 times over.
> colnames(test.0) <- paste("age",levels(age), "m", 0, sep="")
>
>
> Thank you in advance for any assistance.
> Peter.
>
>> sessionInfo()
> R version 2.8.1 (2008-12-22)
> i386-pc-mingw32
>
> locale:
> LC_COLLATE=Danish_Denmark.1252;LC_CTYPE=Danish_Denmark.1252;LC_MONETARY=
> Danish_Denmark.1252;LC_NUMERIC=C;LC_TIME=Danish_Denmark.1252
>
> attached base packages:
> [1] tools stats graphics  grDevices utils datasets  methods
> base
>
> other attached packages:
> [1] epiR_0.9-14 maptools_0.7-18 sp_0.9-29   foreign_0.8-30
> chron_2.3-28
>
> loaded via a namespace (and not attached):
> [1] grid_2.8.1  lattice_0.17-20
>
> __
> 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
and provide commented, minimal, self-contained, reproducible code.

Re: [R] color and fontfamily in lattice

[Sundar Dorai-Raj]

Try this:

dados <- data.frame(varsep = factor(rep(1:2,10)),
i = runif(20))

library(lattice)
font.settings <- list(
 font = 2,
 cex = 2,
 fontfamily = "serif")
my.theme <- list(
 box.umbrella = list(col = "red"),
 box.rectangle = list(col = "purple"),
 box.dot = list(col = "blue", pch = 15),
 par.xlab.text = font.settings,
 par.ylab.text = font.settings,
 axis.text = font.settings)
bwplot(varsep ~ i, dados,
   xlab = names(dados)[1],
   ylab = names(dados)[2],
   panel = function(...) {
 panel.grid(v = -1, h = 0)
 panel.bwplot(...)
   },
   par.settings = my.theme)

Type "trellis.par.get()" at the R command line to see other parameters
you can change. To change the settings on all plots, you can remove
the "par.settings" from the call to "bwplot" and simply use:

trellis.par.set(theme = my.theme)

HTH,

--sundar


2009/2/3 Leandro Marino :
>
>
>
>
> Hi,
> I am having some problems using bwplot(lattice) in my data. I want change 
> some parameters:
> 1) Fontfamily to serif
> 2) The size of the font
> 3) Put it in a bold face
> 4) Change de color of the lines
>
> How can I do that?! Now, I am using this to plot my boxplot.
> dados <- data.frame(varsep=as.factor(rep(1:2,10)),i=runif(20))
> bwplot(dados[,'varsep']~dados[,'i'],xlab=names(dados)[2],ylab=names(dados)[1],panel
>  =function(...){panel.grid(v = -1, h = 
> 0);panel.bwplot(...)},font=2,fontfamily='serif')
>
> Thanks for any help on advance and sorry about my English.
>
>
>
>
>
>
>
>
>
>
> Atenciosamente,
> Leandro Lins Marino
> Centro de Avaliação
> Fundação CESGRANRIO
> Rua Santa Alexandrina, 1011 - 2º andar
> Rio de Janeiro, RJ - CEP: 20261-903
> R (21) 2103-9600 R.:236
> 0 (21) 8777-7907
> ( lean...@cesgranrio.org.br
>
> "Aquele que suporta o peso da sociedade
>é precisamente aquele que obtém
>  as menores vantagens". (SMITH, Adam)
>
>   Antes de imprimir pense em sua responsabilidade e compromisso com o MEIO 
> AMBIENTE
>
> __
> 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] Foreign function call

[Sundar Dorai-Raj]

You're missing that "R_TSConv" is an R object. You can use
stats:::R_TSConv to see the value. Not sure how this helps you though.

On Wed, Feb 4, 2009 at 10:08 AM,   wrote:
> Let me get more specific. I think it this can be answered then I can 
> translate the information to other calls. In the arima 'R' code there is a 
> reference to
>
> .Call(R_TSconv, a, b)
>
> If from the console I type:
>
>> .Call(R_TSConv, c(1,-1), c(1,-1))
>
> I get:
>
> Error: object "R_TSConv" not found
>
> If I do
>
>> getNativeSymbolInfo("R_TSConv")
>
> I get:
>
> Error in FUN("R_TSConv"[[1L]], ...) : no such symbol R_TSConv
>
> What am I missing?
>
> Thank you.
>
> Kevin
>
> __
> 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] sapply

[Sundar Dorai-Raj]

I'm not sure what you really want, so perhaps a simple example would
help (i.e. what a sample of the input looks like and what the output
you need looks like). My guess would be

sapply(df, diff)

but again, I'm not sure.

--sundar

On Sun, Feb 8, 2009 at 4:24 PM, glenn  wrote:
> Newbie question sorry (have tried the help pages I promise)
>
>
>
> I have a dataframe (date,stockprice) say and looking how I might get the
> return of: dataframe (difference in days, change in stock price) using
> sapply - I require a very simple function and don't really want to go down
> the zoo and quant mod route
>
>
>
> Regards
>
>
>
> glenn
>
>
>[[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.
>

__
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] subsets problem

[Sundar Dorai-Raj]

you can try

lapply(lapply(uniques, function(x) subset(df, date == x)), myfun)

or possibly more accurate (subset may be finicky due to scoping):

lapply(lapply(uniques, function(x) df[df$date == x, ]), myfun)

or use ?split

lapply(split(df, df$date), myfun)

HTH,

--sundar

On Sun, Feb 8, 2009 at 5:00 PM, glenn  wrote:
> Help with this much appreciated
>
>
>
> I have a large dataframe that I would like to subset where the constraint
>
>
>
> Test1 <- subset(df, date == uniques[[1]]), where uniques is a list of dates
> that must be matched to create Test1.
>
>
>
> I would like to perform an operation on Test1 that results in a single
> column of data. So far so good.
>
>
>
> How do loop through all values in the uniques list (say there is 50),
> perform an operationon Test1Test50, and then bolt all the lists together
> in a single list please ?
>
>
>
> Regards
>
>
>
>
>
> Glenn
>
>
>
>
>[[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.
>

__
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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] changing settings on a barchart (lattice)

[Sundar Dorai-Raj]

Add the "adj" argument to panel.text to left (adj = 0) or right(adj =
1) justify the text. Add the "font" argument to change the font. See
?text.

--sundar

On Wed, Feb 11, 2009 at 1:25 PM, Dimitri Liakhovitski  wrote:
> Thanks a lot, Sundar. I experimented somewhat and here is the code
> that works well - it allows me to modify most of the stuff I want to
> modify:
>
> p<-as.vector(c(0.1, 0.2, 0.3, 0.4))
> names(p)<-c("A","BB","","")
> barchart(~sort(p), main=list("Chart Title",cex=1),xlab=list("X axis
> title",cex=1),xlim=c(0,0.42),
>layout = c(1,1),
>stack = TRUE,
>auto.key = list(points = FALSE, rectangles = TRUE, space = 
> "top"),
>panel = function(y,x,...){
>panel.grid(h = 0, v = -1, col = "gray60", lty 
> ="dotted")
>panel.barchart(x,y,col="brown")
>panel.text(x,y,label = round(x,2),cex=1)
>}
> )
>
> One last question: How can I modify the way the value labels (those
> that are at the end of the bars) appear? Can I make them bold? Make
> them appear a bit to the right or to the left of where they currently
> are?
> Thanks a lot!
>
> Dimitri
>
> On Wed, Feb 11, 2009 at 12:46 PM, Sundar Dorai-Raj  
> wrote:
>> Pass a list to xlab and main for the font sizes:
>>
>> barchart(..., xlab = list("x-axis", cex = 2), main = list("title", cex = 2))
>>
>> For value labels and a grid you'll need a custom panel function:
>>
>> barchart(..., panel = function(x, y, ...) {
>>  panel.barchart(x, y, ...)
>>  panel.text(x, y, format(y), cex = 1.2)
>>  panel.grid(h = -1, v = -1)
>> })
>>
>> This is untested, but I think it should get you started.
>>
>> --sundar
>>
>> On Wed, Feb 11, 2009 at 9:10 AM, Dimitri Liakhovitski  
>> wrote:
>>> Hello!
>>>
>>> I apologize - I never used lattice before, so my question is probably
>>> very basic - but I just can't find the answer in the archive nor in
>>> the documentation:
>>>
>>> I have a named numeric vector p of 6 numbers (of the type 6 numbers
>>> with people's names to whom those numbers belong). I want a simple bar
>>> chart.
>>>
>>> I am doing:
>>>
>>> library(lattice)
>>> trellis.par.set(fontsize=list(text=12))  # Changes only axes font
>>> barchart(~sort(p),main="Text for main",xlab="Text for X axis",
>>> col=PrimaryColors[3])
>>>
>>> It works just fine.
>>> Question: Where how can I change such things as font size for X axis
>>> label (below the numbers), font size for Title, value labels (to label
>>> bars with actual numbers), add grids, etc.?
>>>
>>> Thanks a lot!
>>>
>>> --
>>> Dimitri Liakhovitski
>>> MarketTools, Inc.
>>> dimitri.liakhovit...@markettools.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.
>>>
>>
>
>
>
> --
> Dimitri Liakhovitski
> MarketTools, Inc.
> dimitri.liakhovit...@markettools.com
>

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Re: [R] Efficent way to create an nxn upper triangular matrix of one's

[Sundar Dorai-Raj]

Try

x <- diag(n)
x[upper.tri(x)] <- 1

On Wed, Feb 11, 2009 at 1:22 PM, Dale Steele  wrote:
> The code below create an nxn upper triangular matrix of one's.  I'm
> stuck on finding a more efficient vectorized way - Thanks.  --Dale
>
> n <- 9
> data <- matrix(data=NA, nrow=n, ncol=n)
> data
> for (i in 1:n) {
>data[,i] <- c(rep(1,i), rep(0,n-i))
> }
> 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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Adding abline in Lattice graph

[Sundar Dorai-Raj]

Try:

coplot(lbxglu~lbxgh|eth, data = reg.dat.5,
   panel= function(...) {
panel.smooth(...)
panel.abline(h = 126, col = "red")
panel.abline(v = 6.5, col = "blue")
  },
  xlab="ABC", ylab="FBG")

Also note that you removed your "with" call and give coplot a data argument.

HTH,

--sundar

On Thu, Feb 12, 2009 at 3:41 PM, Veerappa Chetty  wrote:
> Hi,I would like add a horizontal line at 126 (col=red) and a vertical line
> at 6.5 ( col= blue) in each panel .How should I use the panel.abline
> function in the following code I am using:
> --
> library(lattice)
> with(reg.dat.5,coplot(lbxglu~lbxgh|eth,panel=panel.smooth,xlab="ABC",
> ylab="FBG"))
> 
> Thanks a lot.
>
> Professor of Family Medicine
> Boston University
> Tel: 617-414-6221, Fax:617-414-3345
> emails: chett...@gmail.com,vche...@bu.edu
>
>[[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.
>

__
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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] Adding abline in Lattice graph

[Sundar Dorai-Raj]

Sorry, that was my lack of understanding on how coplot works. Try the following:

coplot(lbxglu~lbxgh|eth, data = reg.dat.5,
   panel= function(...) {
 panel.smooth(...)
 abline(h = 126, col = "red")
 abline(v = 6.5, col = "blue")
  },
  xlab="ABC", ylab="FBG")

On Thu, Feb 12, 2009 at 3:49 PM, Sundar Dorai-Raj  wrote:
> Try:
>
> coplot(lbxglu~lbxgh|eth, data = reg.dat.5,
>   panel= function(...) {
>panel.smooth(...)
>panel.abline(h = 126, col = "red")
>panel.abline(v = 6.5, col = "blue")
>  },
>  xlab="ABC", ylab="FBG")
>
> Also note that you removed your "with" call and give coplot a data argument.
>
> HTH,
>
> --sundar
>
> On Thu, Feb 12, 2009 at 3:41 PM, Veerappa Chetty  wrote:
>> Hi,I would like add a horizontal line at 126 (col=red) and a vertical line
>> at 6.5 ( col= blue) in each panel .How should I use the panel.abline
>> function in the following code I am using:
>> --
>> library(lattice)
>> with(reg.dat.5,coplot(lbxglu~lbxgh|eth,panel=panel.smooth,xlab="ABC",
>> ylab="FBG"))
>> 
>> Thanks a lot.
>>
>> Professor of Family Medicine
>> Boston University
>> Tel: 617-414-6221, Fax:617-414-3345
>> emails: chett...@gmail.com,vche...@bu.edu
>>
>>[[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.
>>
>

__
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Re: [R] Uninstall question

[Sundar Dorai-Raj]

This is on the Mac FAQ:

http://cran.cnr.berkeley.edu/bin/macosx/RMacOSX-FAQ.html#How-can-R-for-Mac-OS-X-be-uninstalled_003f

HTH,

--sundar

On Tue, Feb 17, 2009 at 7:17 AM, ANJAN PURKAYASTHA
 wrote:
> I need to uninstall R 2.7.1 from my Mac. What is the best way to uninstall
> it? Simply delete the R icon in the Applications folder?
> Or is it more involved?
> TIA,
> Anjan
>
> --
> =
> anjan purkayastha, phd
> bioinformatics analyst
> whitehead institute for biomedical research
> nine cambridge center
> cambridge, ma 02142
>
> purkayas [at] wi [dot] mit [dot] edu
> 703.740.6939
>
>[[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.
>

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Re: [R] difference between assignment syntax <- vs =

[Sundar Dorai-Raj]

Read the help page as to their differences: ?"<-"

On Sat, Feb 21, 2009 at 7:30 AM, Thomas Mang  wrote:
> Hi,
>
> Both operators <- and = can be used to make an assignment. My question is:
> Is there a semantic difference between these two? Some time ago, I remember
> I have read that because of some reason, one should be given preference over
> the other - but I cannot remember the source, nor the argument, nor which
> operator the preferred was.
>
> What is the present state ?
> Is still one version better than the other, or is it only a matter of taste
> what to use ?
>
> thanks
> Thomas
>
> __
> 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] lattice contourplot line types

[Sundar Dorai-Raj]

The only way I can figure out to do this is to use two calls to
panel.contourplot:

library(lattice)
x <- seq(-2, 2, length = 20)
y <- seq(-2, 2, length = 20)
grid <- expand.grid(x=x, y=y)
grid$z <- dnorm(grid$x) * dnorm(grid$y)
contourplot(z ~ x * y, grid,
panel = function(at, lty, col, ...) {
  at.o <- at[seq(1, length(at), 2)]
  at.e <- at[seq(2, length(at), 2)]
  panel.contourplot(at = at.o, lty = 1, col = "blue", ...)
  panel.contourplot(at = at.e, lty = 4, col = "red", ...)
},
at = pretty(grid$z, 10))

HTH,

--sundar

On Sun, Feb 22, 2009 at 12:45 PM, Andrew Beckerman
 wrote:
> Dear all -
>
> I would like to adjust the line type of specific contours in contourplot
> from the lattice package, but it seems like lty does not take a list in the
> call.
>
> Here is my call to contourplot:
>
> contourplot(preds~size+trt|Size.Name,
>data=pred.dat,layout=c(2,4),
>at=c(0.025,0.5,0.975),
>par.strip.text=list(cex=1.2),
>scales=list(cex=0.5),
>xlab=list("Size",cex=1.2),
>ylab=list("Treatment",cex=1.2),
>panel=function(x,y,z,...){
>panel.contourplot(x,y,z,lwd=1,...)
>panel.grid(h=-1,v=-1,col="grey",...)})
>
> I would like to specify lty=c(2,1,2) corresponding to the
> at=c(0.025,0.5,0.975), and have tried this in both the core part of the
> call, and in panel.countourplot.  However, it only recognises the first
> type.
>
> If there is no straightforward answer, I can provide the data.
>
> Best wishes,
> Andrew
>
> R version 2.8.1 (2008-12-22)
> i386-apple-darwin8.11.1
>
> locale:
> en_GB.UTF-8/en_GB.UTF-8/C/C/en_GB.UTF-8/en_GB.UTF-8
>
> attached base packages:
> [1] stats graphics  grDevices utils datasets  methods   base
>
> other attached packages:
> [1] MASS_7.2-45 lattice_0.17-17
>
> loaded via a namespace (and not attached):
> [1] grid_2.8.1
>
> __
> 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] levelplot help needed

[Sundar Dorai-Raj]

To reorder the y-labels, simply reorder the factor levels:

df <- data.frame(x_label = factor(x_label),
 y_label = factor(y_label, rev(y_label)),
 values = as.vector(my.data))

Not sure about putting the strips at the bottom. A quick scan of
?xyplot and ?strip.default suggests that this is not possible, but I'm
sure Deepayan will correct me if I'm wrong (he often does).

--sundar

On Fri, Feb 27, 2009 at 5:51 AM, Antje  wrote:
> Hi there,
>
> I'm looking for someone who can give me some hints how to make a nice
> levelplot. As an example, I have the following code:
>
> # create some example data
> # --
> xl <- 4
> yl <- 10
>
> my.data <- sapply(1:xl, FUN = function(x) { rnorm( yl, mean = x) })
>
> x_label <- rep(c("X Label 1", "X Label 2", "X Label 3", "X Label 4"), each =
> yl)
> y_label <- rep(paste("Y Label ", 1:yl, sep=""), xl)
>
> df <- data.frame(x_label = factor(x_label),y_label = factor(y_label), values
> = as.vector(my.data))
>
> df1 <- data.frame(df, group = rep("Group 1", xl*yl))
> df2 <- data.frame(df, group = rep("Group 2", xl*yl))
> df3 <- data.frame(df, group = rep("Group 3", xl*yl))
>
> mdf <- rbind(df1,df2,df3)
>
> # plot
> # --
>
> graph <- levelplot(mdf$values ~ mdf$x_label * mdf$y_label | mdf$group,
>                                aspect = "xy", layout = c(3,1),
>                                scales = list(x = list(labels =
> substr(levels(factor(mdf$x_label)),0,5), rot = 45)))
>            print(graph)
>
> # --
>
>
> (I need to put this strange x-labels, because in my real data the values of
> the x-labels are too long and I just want to display the first 10 characters
> as label)
>
> My questions:
>
> * I'd like to start with "Y Label 1" in the upper row (that's a more general
> issue, how can I have influence on the order of x,y, and groups?)
> * I'd like to put the groups at the bottom
>
> Can anybody give me some help?
>
> Antje
>
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Re: [R] lattice: remove box around a wireframe

[Sundar Dorai-Raj]

(Sorry for the repeat. Forgot to copy R-help)

Try,

test = data.frame(expand.grid(c(1:10), c(1:10)))
z = test[,1] + test[,2]
test = cbind(test, z)
names(test) = c("x", "y", "z")
require(lattice)
wireframe(z ~ x*y, data = test,
 par.settings = list(axis.line = list(col = "transparent")),
 par.box = c(col = "transparent") )

--sundar

On Wed, Mar 4, 2009 at 8:17 AM, Thomas Roth (geb. Kaliwe)
 wrote:
> #Hi,
> #
> #somebody knows how to  remove the outer box around a wireframe and reduce
> the height
> #
> #
>
> test = data.frame(expand.grid(c(1:10), c(1:10)))
> z = test[,1] + test[,2]
> test = cbind(test, z)
> names(test) = c("x", "y", "z")
> require(lattice)
> wireframe(z ~ x*y, data = test, par.box = c(col = "transparent") )  #not
> this one but the remaining outer box.
>
> Thanks in advance
>
> Thomas Roth
>
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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>

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Re: [R] Date conversion

[Sundar Dorai-Raj]

Hi,

There are possibly several ways to do this. My approach would be:

dates <- strptime(as.character(DATE), "%d%b%Y")
year <- dates$year + 1900
week <- floor(dates$yday/365 * 52)

HTH,

--sundar

On Thu, Mar 5, 2009 at 8:58 AM, Pele  wrote:
>
> Hi R users,
>
> I have a factor variable called date as shown below:  Can anyone share the
> best / most efficient way to extract year and week (e.g.  year = 2006, week
> = 52 for first record, etc..)?  My data set has 1 million records.
>
> DATE
> 11DEC2006
> 11SEP2006
> 01APR2007
> 02DEC2007
>
>
> Thanks in advance for any help!
> --
> View this message in context: 
> http://www.nabble.com/Date-conversion-tp22355788p22355788.html
> Sent from the R help mailing list archive at Nabble.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.
>

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Re: [R] Lattice: Customizing point-sizes with groups

[Sundar Dorai-Raj]

Try this:

xyplot(y ~ x, temp, groups = groups,
   par.settings = list(
 superpose.symbol = list(
   cex = c(1, 3),
   pch = 19,
   col = c("blue", "red"

See:

str(trellis.par.get())

for other settings you might want to change.

Also, you should drop the ";" from all your scripts.

HTH,

--sundar

On Mon, Mar 9, 2009 at 6:49 PM, Paul Boutros  wrote:
> Hello,
>
> I am creating a scatter-plot in lattice, and I would like to customize the
> size of each point so that some points are larger and others smaller.
>  Here's a toy example:
>
> library(lattice);
>
> temp <- data.frame(
>        x = 1:10,
>        y = 1:10,
>        cex = rep( c(1,3), 5),
>        groups = c( rep("A", 5), rep("B", 5) )
>        );
>
> xyplot(y ~ x, temp, cex = temp$cex, pch = 19);
>
> This works just fine if I create a straight xy-plot, without groups.
>  However when I introduce groupings the cex argument specifies the
> point-size for the entire group.  For example:
>
> xyplot(y ~ x, temp, cex = temp$cex, pch = 19, group = groups);
>
> Is it possible to combine per-spot sizing with groups in some way?  One
> work-around is to manually specify all graphical parameters, but I thought
> there might be a better way than this:
>
> temp$col <- rep("blue", 10);
> temp$col[temp$groups == "B"] <- "red";
> xyplot(y ~ x, temp, cex = temp$cex, pch = 19, col = temp$col);
>
> Any suggestions/advice is much appreciated!
> 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] Lattice: Customizing point-sizes with groups

[Sundar Dorai-Raj]

Sorry, I missed your point the first time. Why not create a group for
each subset then?

xyplot(y ~ x, temp, groups = interaction(cex, groups),
   par.settings = list(
 superpose.symbol = list(
   cex = c(1, 2, 3, 4),
   pch = 19,
   col = c("blue", "red", "green", "purple"


On Tue, Mar 10, 2009 at 8:11 AM, Paul C. Boutros
 wrote:
> Hi Sundar,
>
> Thanks for your help!  Unfortunately your code seems to give the same
> result.  Compare this:
>
> temp <- data.frame(
>       x = 1:10,
>       y = 1:10,
>       cex = rep( c(1,3), 5),
>       col = c( rep("blue", 5), rep("red", 5) ),
>       groups = c( rep("A", 5), rep("B", 5) )
>       );
>
> xyplot(y ~ x, temp, groups = groups,
>       par.settings = list(
>         superpose.symbol = list(
>           cex = c(1, 3),
>           pch = 19,
>           col = c("blue", "red"
>
> And this:
> xyplot(y ~ x, temp, cex = temp$cex, col = temp$col, pch = 19);
>
> Once I introduce groups, I lose the ability to customize individual
> data-points and seem only to be able to customize entire groups.
>
> Paul
>
> -Original Message-
> From: Sundar Dorai-Raj [mailto:sdorai...@gmail.com]
> Sent: Tuesday, March 10, 2009 5:49 AM
> To: paul.bout...@utoronto.ca
> Cc: r-help@r-project.org
> Subject: Re: [R] Lattice: Customizing point-sizes with groups
>
> Try this:
>
> xyplot(y ~ x, temp, groups = groups,
>       par.settings = list(
>         superpose.symbol = list(
>           cex = c(1, 3),
>           pch = 19,
>           col = c("blue", "red"
>
> See:
>
> str(trellis.par.get())
>
> for other settings you might want to change.
>
> Also, you should drop the ";" from all your scripts.
>
> HTH,
>
> --sundar
>
> On Mon, Mar 9, 2009 at 6:49 PM, Paul Boutros 
> wrote:
>> Hello,
>>
>> I am creating a scatter-plot in lattice, and I would like to customize the
>> size of each point so that some points are larger and others smaller.
>>  Here's a toy example:
>>
>> library(lattice);
>>
>> temp <- data.frame(
>>        x = 1:10,
>>        y = 1:10,
>>        cex = rep( c(1,3), 5),
>>        groups = c( rep("A", 5), rep("B", 5) )
>>        );
>>
>> xyplot(y ~ x, temp, cex = temp$cex, pch = 19);
>>
>> This works just fine if I create a straight xy-plot, without groups.
>>  However when I introduce groupings the cex argument specifies the
>> point-size for the entire group.  For example:
>>
>> xyplot(y ~ x, temp, cex = temp$cex, pch = 19, group = groups);
>>
>> Is it possible to combine per-spot sizing with groups in some way?  One
>> work-around is to manually specify all graphical parameters, but I thought
>> there might be a better way than this:
>>
>> temp$col <- rep("blue", 10);
>> temp$col[temp$groups == "B"] <- "red";
>> xyplot(y ~ x, temp, cex = temp$cex, pch = 19, col = temp$col);
>>
>> Any suggestions/advice is much appreciated!
>> 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] 2 Simple Lattice Plot Questions

[Sundar Dorai-Raj]

Convert Year to a factor and both problems will be solved.

--sundar

On Tue, Mar 10, 2009 at 3:48 PM, jimdare  wrote:
>
> Hi,
>
> I have created the plot below and have a few questions about changes.
>
> 1) How do I change the "Year" title of each plot so it reads from the top
> "2006","2007","2008","2009".
> 2) How do I get rid of those vertical grey bars in the title bar of each
> plot?
>
> I apologise for my ignorance... one of those days :(
>
> James
>
>
> http://www.nabble.com/file/p22445242/PLOT.jpg
> --
> View this message in context: 
> http://www.nabble.com/2-Simple-Lattice-Plot-Questions-tp22445242p22445242.html
> Sent from the R help mailing list archive at Nabble.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.
>

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Re: [R] 2 Simple Lattice Plot Questions

[Sundar Dorai-Raj]

I don't believe Elena's suggestion will work. However, the following will:

xyplot(..., scales = list(y = list(at = seq(5, 25, 5

though you may need to extend the limits a little as well:

xyplot(..., ylim = lattice:::extend.limits(c(0, 30)))

and add the scales argument from the first example to place explicit
tick marks rather than let xyplot do so.

HTH,

--sundar

On Tue, Mar 10, 2009 at 7:04 PM, Elena Wilson  wrote:
> Have you tried specifying the levels of y's you want to display, e.g. 
> ylim=c(0,5,10,15,20,30)?
>
>  -Original Message-
> From:   r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]  
> On Behalf Of jimdare
> Sent:   Wednesday, 11 March 2009 12:50 PM
> To:     r-help@r-project.org
> Subject:        Re: [R] 2 Simple Lattice Plot Questions
>
>
> Thanks very much.    One other thing... see how the Y axis begins before 0.
> At first glance it appears that the 0's are actually worth something.  When
> I use ylim=c(0,30) I get the graph I want, but the tick marks show only
> 5,10,15,20,25.  I want them to show 0,5,10,15,20,30.  Does anyone know how
> to change this setting?
>
> Cheers,
>
>
>
>
>
>
> jimdare wrote:
>>
>> Hi,
>>
>> I have created the plot below and have a few questions about changes.
>>
>> 1) How do I change the "Year" title of each plot so it reads from the top
>> "2006","2007","2008","2009".
>> 2) How do I get rid of those vertical grey bars in the title bar of each
>> plot?
>>
>> I apologise for my ignorance... one of those days :(
>>
>> James
>>
>>
>>  http://www.nabble.com/file/p22445242/PLOT.jpg
>>
>
> --
> View this message in context: 
> http://www.nabble.com/2-Simple-Lattice-Plot-Questions-tp22445242p22447415.html
> Sent from the R help mailing list archive at Nabble.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.
> --
> Message  protected by MailGuard: e-mail anti-virus, anti-spam and content 
> filtering.
> http://www.mailguard.com.au/mg
>
>
> https://login.mailguard.com.au/report/1x1TEaABIw/4soNUVnBh04s1coMEHC4LA/7.998
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Re: [R] Matrix Construction; Subdiagonal

[Sundar Dorai-Raj]

Does this help?

A <- matrix(0, 6, 6)
vec <- 1:5
A[row(A) == col(A) + 1] <- vec

--sundar

On Wed, Mar 11, 2009 at 4:42 PM, Stu Field  wrote:
> I'm trying to enter a vector into the subdiagonal of a matrix but
> cannot find a command in R which corresponds to the MatLab version of
> diag(vec, k), where vec = the vector of interest, and k = the diagonal
> (k=0 for the diagonal; k=-1 for the subdiagonal; k=1 for
> superdiagonal, etc.)
> Is there an equivalent command in R?
>
> I'm looking for something like this:
> vec = seq(1, 5, 1)        # vector of interest
>
> A = xyz(vec,-1)           # creates a 6x6 matrix with vec on the
> subdiagonal
> where xyz is some function similar to diag, but with differing
> arguments.
>
> I can't believe there is not a simple way to do this...
> Thanks for your help,
>
> ~~
> Stu Field, PhD
> Postdoctoral Fellow
> Department of Biology
> Colorado State University
> 1878 Campus Delivery
> Fort Collins, CO 80523-1878
> Office: E208 Anatomy/Zoology
> Phone: (970) 491-5744
> ~~
>
>
>
>
>
>        [[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.
>

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Re: [R] Matrix Construction; Subdiagonal

[Sundar Dorai-Raj]

You can always write your own function:

myDiag <- function(x, vec, k) {
  x[row(x) == col(x) - k] <- vec
  x
}

myDiag(A, vec, -1)

Of course, you should probably do some input checking too.

--sundar

On Wed, Mar 11, 2009 at 4:57 PM, Stu Field  wrote:
> Sure, that'll work fine, thanks.
> But I guess I was looking for something more similar to MatLab, I'm really
> surprised R doesn't have a preset command for this (?)
> Thanks again,
> Stu
> On 11 • Mar • 2009, at 5:49 PM, Sundar Dorai-Raj wrote:
>
> Does this help?
>
> A <- matrix(0, 6, 6)
> vec <- 1:5
> A[row(A) == col(A) + 1] <- vec
>
> --sundar
>
> On Wed, Mar 11, 2009 at 4:42 PM, Stu Field  wrote:
>
> I'm trying to enter a vector into the subdiagonal of a matrix but
>
> cannot find a command in R which corresponds to the MatLab version of
>
> diag(vec, k), where vec = the vector of interest, and k = the diagonal
>
> (k=0 for the diagonal; k=-1 for the subdiagonal; k=1 for
>
> superdiagonal, etc.)
>
> Is there an equivalent command in R?
>
> I'm looking for something like this:
>
> vec = seq(1, 5, 1)        # vector of interest
>
> A = xyz(vec,-1)           # creates a 6x6 matrix with vec on the
>
> subdiagonal
>
> where xyz is some function similar to diag, but with differing
>
> arguments.
>
> I can't believe there is not a simple way to do this...
>
> Thanks for your help,
>
> ~~
>
> Stu Field, PhD
>
> Postdoctoral Fellow
>
> Department of Biology
>
> Colorado State University
>
> 1878 Campus Delivery
>
> Fort Collins, CO 80523-1878
>
> Office: E208 Anatomy/Zoology
>
> Phone: (970) 491-5744
>
> ~~
>
>
>
>
>
>        [[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.
>
>
> ~~
> Stu Field, PhD
> Postdoctoral Fellow
> Department of Biology
> Colorado State University
> 1878 Campus Delivery
> Fort Collins, CO 80523-1878
> Office: E208 Anatomy/Zoology
> Phone: (970) 491-5744
> ~~
>
>
>
>

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Re: [R] XYplot simple question

[Sundar Dorai-Raj]

Convert "Plot" to factor:

xyplot(AbvBioAnnProd ~ Year | Plot, type = c("b", "r"), pch = 16)

Also note that using the "type" argument with multiple values prevents
the necessity of a custom panel function.

HTH,

--sundar

On Mon, Mar 16, 2009 at 5:54 AM, AllenL  wrote:
>
> Hello R friends,
> Simple question today: I am desiring to do an xyplot with the below code,
> which graphs time series across different experimental Plots-
>
> xyplot(AbvBioAnnProd~Year|Plot)        ### Plots each monoculture biomass vs
> time
> xyplot(AbvBioAnnProd~Year|Plot,panel=function(x,y){
> panel.xyplot(x,y,type="b",pch=16)
> panel.abline(lm(y~x))
> })
>
> What I want to add are unique labels for each panel, where instead of all
> labeled "plot" with the "slide-bar visual" (although that is okay) I want
> the Plot number to appear (i.e. the value of "Plot" for that panel).
>
> This should be easy, right?
> -Al
> --
> View this message in context: 
> http://www.nabble.com/XYplot-simple-question-tp22537350p22537350.html
> Sent from the R help mailing list archive at Nabble.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.
>

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Re: [R] XYplot simple question

[Sundar Dorai-Raj]

Sorry, I should have

xyplot(AbvBioAnnProd ~ Year | factor(Plot), type = c("b", "r"), pch = 16)

On Mon, Mar 16, 2009 at 6:17 AM, Sundar Dorai-Raj  wrote:
> Convert "Plot" to factor:
>
> xyplot(AbvBioAnnProd ~ Year | Plot, type = c("b", "r"), pch = 16)
>
> Also note that using the "type" argument with multiple values prevents
> the necessity of a custom panel function.
>
> HTH,
>
> --sundar
>
> On Mon, Mar 16, 2009 at 5:54 AM, AllenL  wrote:
>>
>> Hello R friends,
>> Simple question today: I am desiring to do an xyplot with the below code,
>> which graphs time series across different experimental Plots-
>>
>> xyplot(AbvBioAnnProd~Year|Plot)        ### Plots each monoculture biomass vs
>> time
>> xyplot(AbvBioAnnProd~Year|Plot,panel=function(x,y){
>> panel.xyplot(x,y,type="b",pch=16)
>> panel.abline(lm(y~x))
>> })
>>
>> What I want to add are unique labels for each panel, where instead of all
>> labeled "plot" with the "slide-bar visual" (although that is okay) I want
>> the Plot number to appear (i.e. the value of "Plot" for that panel).
>>
>> This should be easy, right?
>> -Al
>> --
>> View this message in context: 
>> http://www.nabble.com/XYplot-simple-question-tp22537350p22537350.html
>> Sent from the R help mailing list archive at Nabble.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.
>>
>

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Re: [R] Get user system name

[Sundar Dorai-Raj]

Assuming "USER" is defined on your system then

Sys.getenv("USER")

ought to work.

--sundar

On Mon, Mar 16, 2009 at 3:04 PM, Etienne Bellemare Racine
 wrote:
> I would like to get the name of the user form the system. Is it possible ?
> Something like
>  >system.user()
> returning something like
> [1] "etber12"
>
> Thanks,
> Etienne
>
>        [[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.
>

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Re: [R] Get user system name

[Sundar Dorai-Raj]

If that's the case, it's best to post the output from "version" to
this thread. Type "version" from the R command prompt and paste the
results here.

On Mon, Mar 16, 2009 at 3:14 PM, Etienne Bellemare Racine
 wrote:
> It seems like it is not defined. I get an empty string. Is there a
> workaround or another solution ?
>
> --
> Etienne
>
> Sundar Dorai-Raj a écrit :
>
> Assuming "USER" is defined on your system then
>
> Sys.getenv("USER")
>
> ought to work.
>
> --sundar
>
> On Mon, Mar 16, 2009 at 3:04 PM, Etienne Bellemare Racine
>  wrote:
>
>
> I would like to get the name of the user form the system. Is it possible ?
> Something like
>  >system.user()
> returning something like
> [1] "etber12"
>
> Thanks,
> Etienne
>
>        [[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.
>
>

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Re: [R] maps and lattice

[Sundar Dorai-Raj]

Jon Loehrke said the following on 2/5/2008 2:29 PM:
> Is it possible to place maps onto lattice plots?
> 
> With basic plotting you can add a map to a plot
> 
> library(lattice)
> long<-c(-69.2, -69.5, -70.1, -70.3)
> lat<-c(41, 41.5, 43.2, 42.8)
> plot(long, lat)
> map('state', c("massachusetts"),add=TRUE)
> 
> but is it possible with lattice?
> 
> 
> library(lattice)
> factor<-c(1,1,2,2)
> xyplot(lat~long|fact)
> ...now what?
> 
> I have looked at panel and found through google an extravagant shape  
> file export/import to R.
> Is there a simpler fix?
> 
> Thank you very much for your help.
> 
> Jon
> 
> School for Marine Science and Technology
> UMASS-Dartmouth
> 


Try this:

library(lattice)
library(maps)
long <- c(-69.2, -69.5, -70.1, -70.3)
lat <- c(41, 41.5, 43.2, 42.8)
fact <- c(1, 1, 2, 2)
xyplot(lat ~ long | fact,
panel = function(...) {
  panel.xyplot(...)
  mp <- map("state", "massachusetts", plot = FALSE)
  lpolygon(mp$x, mp$y)
})

HTH,

--sundar

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Re: [R] fun.aggregate=mean in reshape

[Sundar Dorai-Raj]

[Ricardo Rodriguez] Your XEN ICT Team said the following on 2/12/2008 
12:23 AM:
> Hi all,
> 
> We are facing a problem while introducing ourselves to Reshape package 
> use. Melt seems to work fine, but cast fails when we use mean as 
> fun.aggregate. As you see here, length and sum work fine, but mean 
> throws this same error whatever dataset we use.
> 
>  > cast(aqm, month ~ variable, length)
>   month ozone solar.r wind temp
> 1 526  27   31   31
> 2 6 9  30   30   30
> 3 726  31   31   31
> 4 826  28   31   31
> 5 929  30   30   30
>  > cast(aqm, month ~ variable, sum)
>   month ozone solar.r  wind temp
> 1 5   6144895 360.3 2032
> 2 6   2655705 308.0 2373
> 3 7  15376711 277.2 2601
> 4 8  15594812 272.6 2603
> 5 9   9125023 305.4 2307
>  > cast(aqm, month ~ variable, mean)
> Error in get(as.character(FUN), mode = "function", envir = envir) :
>   variable "fun" of mode "function" was not found
>  >
> 
> 
> Our environment:
> 
>  > version
>_  
> platform   i386-apple-darwin8.10.1
> arch   i386   
> os darwin8.10.1   
> system i386, darwin8.10.1 
> status
> major  2  
> minor  6.2
> year   2008   
> month  02 
> day08 
> svn rev44383  
> language   R  
> version.string R version 2.6.2 (2008-02-08)
> 
> 
>  > installed.packages()
> 
> reshape"reshape"
> "/Library/Frameworks/R.framework/Resources/library" "0.8.0"   
> NANA
> 
> 
> Please, could you help use to work out this issue? Thanks!
> 

Do you have an object called 'mean' that's masking the base::mean 
function? I can replicate your error using the following:

 > library(reshape)
 > names(airquality) <- tolower(names(airquality))
 > aqm <- melt(airquality, id=c("month", "day"), na.rm=TRUE)
 > mean <- 1
 > cast(aqm, month ~ variable, mean)
Error in get(as.character(FUN), mode = "function", envir = envir) :
   variable "fun" of mode "function" was not found
 > cast(aqm, month ~ variable, base::mean)
   monthozone  solar.r  wind temp
1 5 23.61538 181.2963 11.622581 65.54839
2 6 29.4 190.1667 10.27 79.1
3 7 59.11538 216.4839  8.941935 83.90323
4 8 59.96154 171.8571  8.793548 83.96774
5 9 31.44828 167.4333 10.18 76.9
 > find("mean")
[1] ".GlobalEnv"   "package:base"


HTH,

--sundar

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Re: [R] wire.frame tick labels from matrix

[Sundar Dorai-Raj]

Marlin Keith Cox said the following on 2/15/2008 11:39 AM:
> Dear R Users, close to the end of this I used wireframe to create a 3D plot
> from a matrix.  The x and y axis tick labels (1-6) for each were created
> from the matrix being a 6X6 matrix. I need the axis tick labels to be the
> row and column headings which you can see in the output (mat.x).  I have
> tried several work arounds, but they have not been successful.
> 
> Thanks in advance. keith
> 
> 
> rm(list=ls())
> sen<-read.csv("brooktroutsort.csv")
> 
> #Resistance
> R=c(660, 548, 676, 763, 768, 692, 657, 630, 748, 680, 786, 645, 710, 677,
> 692, 732, 737, 651, 396,
> 601, 640, 448, 464, 472, 434, 487, 495, 426, 429, 456)
> 
> #Detector length
> Lend=c(37.0,  39.0,  39.0,  39.0,  40.0,  41.5,  44.0,  45.0,  46.0,  47.0,
> 47.0,  48.0,
> 48.5,  49.0,  51.0,  53.0, 53.0,  60.0,  89.0, 103.0, 108.5, 118.0, 118.0,
> 123.0,
> 126.0, 138.0, 139.0, 141.0, 141.0, 151.0)
> 
> #Errors to be multiplied by Restistance
> x=c(0,.05,.10,.15,.20,.25)
> 
> #Errors to be multiplied by Detector length
> y=c(0,.01,.02,.03,.04,.05)
> 
> #equation to predict water weight in grams
> a=3.453*((Lend^2)/R)+1.994
> 
> X=(R%o%x+R)
> 
> Y=((Lend%o%y+Lend)^2)
> 
> num.x.col <- length(X[1,])
> num.y.col <- length(Y[1,])
> num.rows <- length(X[,1])
> 
> Z <- matrix(nrow=num.rows, ncol=num.x.col*num.y.col)
> 
> for( i in 1:num.rows)  {
>Z[i,] <- as.vector( Y[i,] %*% t(X[i,])^-1 )
> }
> 
> pred.est <- 3.453*Z+1.994
> z=(pred.est-a)/a
> 
> colnames(z)<- rep(c("X1","X2","X3","X4","X5","X6"),6)
> 
> meanz=colMeans(z)
> mat.x <- matrix(meanz, nrow=6, ncol=6, byrow=TRUE)
> 
> colnames(mat.x)<- c(0,1,2,3,4,5)
> rownames(mat.x)<-c(0,5,10,15,20,25)
> mat.x
> 
> 
> library(lattice)
> wireframe(mat.x,drape=TRUE,zlab=list("Proportion Error of Estimate",
> rot=90), xlab="Resistance Error (%) ",ylab="Length Error
> (%)",scale=list(arrows=FALSE))
> detach(z)
> detach(sen)
> 
> 

Try:

mat.df <- data.frame(z = as.vector(mat.x))
mat.df$x <- rep(c(0,5,10,15,20,25), each = 6)
mat.df$y <- rep(c(0,1,2,3,4,5), times = 6)

library(lattice)
wireframe(z ~ x * y, mat.df,
   drape = TRUE,
   zlab = list("Proportion Error of Estimate", rot=90),
   xlab = "Resistance Error (%) ",
   ylab = "Length Error (%)",
   scales = list(arrows = FALSE))

HTH,

--sundar

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Re: [R] LaTeX in R

[Sundar Dorai-Raj]

Or my personal favorite if the length of mySigma is variable:

mySigma <- 2:3
plot(1:10, dnorm(1:10, sd = mySigma[1]), type = 'l')
lines(dnorm(1:10,sd = mySigma[2]),lty = 2)
leg <- as.expression(lapply(mySigma, function(x) bquote(sigma == .(x
legend(x = "topright", lty = c(1,2),legend = leg)

Thanks,

--sundar

Uwe Ligges said the following on 3/7/2008 8:15 AM:
> You might want to read
>   Ligges, U. (2002): R Help Desk: Automation of Mathematical Annotation 
> in Plots. R News 2 (3), 32-34.
> with an example at the end that meets your requirements:
> 
> (please note that I removed those ugly ";"
> 
> 
> mySigma[1] <- 2
> mySigma[2] <- 3
> plot(1:10, dnorm(1:10, sd = mySigma[1]), type = 'l')
> lines(dnorm(1:10, sd = mySigma[2]), lty = 2)
> legend1 <- substitute(sigma == myS, list(myS = mySigma[1]))
> legend2 <- substitute(sigma == myS, list(myS = mySigma[2]))
> legend(x = "topright", lty = c(1,2),
> legend = do.call(expression, list(legend1, legend2)))
> 
> 
> Uwe Ligges
> 
> 
> 
> Mario Maiworm wrote:
>> Finally, this should work for an array of sigmas. I just realized that the
>> substitute()-command is not evaluated within a c()-environment :(
>>
>> mySigma[1] <- 2; mySigma[2] <- 3;
>> plot(1:10, dnorm(1:10, sd = mySigma[1]), type = 'l') ;
>> lines(dnorm(1:10,sd = mySigma[2]),lty = 2);
>> legend(x = "topright", lty = c(1,2),legend = c(substitute(sigma == myS,
>> list(myS = mySigma[1])),substitute(sigma == myS, list(myS = mySigma[2]
>>
>> Mario.
>>
>>  
>> __
>>
>> Mario Maiworm
>> Biological Psychology and Neuropsychology
>> University of Hamburg
>> Von-Melle-Park 11
>> D-20146 Hamburg
>>
>> Tel.: +49 40 42838 3515
>> Fax.: +49 40 42838 6591
>>
>> http://bpn.uni-hamburg.de/Maiworm_e.html
>> http://cinacs.org
>> __
>>
>>>>> -Ursprüngliche Nachricht-
>>>>> Von: Uwe Ligges [mailto:[EMAIL PROTECTED]
>>>>> Gesendet: Freitag, 7. März 2008 16:30
>>>>> An: Mario Maiworm
>>>>> Cc: r-help@r-project.org
>>>>> Betreff: Re: AW: [R] LaTeX in R
>>>>>
>>>>>
>>>>>
>>>>> Mario Maiworm wrote:
>>>>>> Thank you, uwe and jeremy. I was actually looking exactly for that!
>>>>> But
>>>>>> something still doesn't work:
>>>>>> I want to plot a symbol in a legend of a plot, lets say "\sigma = 2".
>>>>> 2
>>>>>> should be the value of a variable. So, when I try
>>>>>>
>>>>>> mySigma=2;plot(1:10,dnorm(1:10,sd=mySigma),type='l')
>>>>>> legend(x="topright",legend=paste(expression(sigma)," =
>>>>> ",mySigma),lty=1)
>>>>>> , the sigma is not plotted as a symbol. This version:
>>>>>>
>>>>>> mySigma=2;plot(1:10,dnorm(1:10,sd=mySigma),type='l')
>>>>>> legend(x="topright",legend=expression(paste(sigma," =
>>>>> ",mySigma)),lty=1)
>>>>>> gives me a 'real' sigma but the mySigma variable is not evaluated. Any
>>>>>> ideas?
>>>>> Yes:
>>>>>
>>>>>
>>>>> mySigma <- 2
>>>>> plot(1:10, dnorm(1:10, sd = mySigma), type='l')
>>>>> legend(x = "topright", lty = 1,
>>>>> legend = substitute(sigma == myS, list(myS = mySigma)))
>>>>>
>>>>>
>>>>> Uwe Ligges
>>>>>
>>>>>
>>>>>> Mario.
>>>>>>
>>>>>> __
>>>>>>
>>>>>> Mario Maiworm
>>>>>> Biological Psychology and Neuropsychology
>>>>>> University of Hamburg
>>>>>> Von-Melle-Park 11
>>>>>> D-20146 Hamburg
>>>>>>
>>>>>> Tel.: +49 40 42838 3515
>>>>>> Fax.: +49 40 42838 6591
>>>>>>
>>>>>> http://bpn.uni-hamburg.de/Maiworm_e.html
>>>>>> http://cinacs.org
>>>>>> __
>>>>>>
>>>>>&g

Re: [R] Distances between two datasets of x and y co-ordinates

[Sundar Dorai-Raj]

Andrew McFadden said the following on 3/12/2008 1:47 PM:
> Hi all
> 
> I am trying to determine the distances between  two datasets of x and y
> points. The number of points in dataset One is very small i.e. perhaps
> 5-10. The number of points in dataset Two is likely to be very large
> i.e. 20,000-30,000. My initial approach was to append the first dataset
> to the second and then carry out the calculation:
> 
> dists <- as.matrix(dist(gis data from 2 * datasets)) 
> 
> However, the memory of the computer is not sufficient. A lot of
> calculations carried out in this situation are unnecessary as I only
> want approx 5 * 20,000 calculations versus 20,000 *20,000. 
> 
> x <- c(2660156,2663703,2658165,2659303,2661531,2660914)
> y <- c(6476767,6475013,6475487,6479659,6477004,6476388)
> data2<-cbind(x,y)
> 
> x <- c(266500,261)
> y <- c(6478767,6485013)
> data1<-cbind(x,y)
> 
> Any suggestions on how to do this would be appreciated.
> 
> Regards
> 
> Andrew

If you're trying to find only the closest point in data1 to data2, then 
use knn (or knn1) in the 'class' package:

library(class)
nn <- knn1(data2, data1, 1:nrow(data2))

which gives you the rows in data1 closest to each row in data2. Then 
compute the distance:

rowSums((data2[nn, ] - data1)^2)^0.5

HTH,

--sundar

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Re: [R] Lme does not work without a random effect (UNCLASSIFIED)

[Sundar Dorai-Raj]

Park, Kyong H Mr ECBC said the following on 3/14/2008 12:25 PM:
> Classification:  UNCLASSIFIED 
> Caveats: NONE
> 
> Dear R users,
> 
> I'm interested in finding a random effect of the Block in the data shown
> below, but 'lme' does not work without the random effect. I'm not sure how
> to group the data without continuous value which is shown in the error
> message at the bottom line. If I use 'aov' with Error(Block), is there a
> test method comparing between with and without the Block random effect. I'm
> using R 2.4.1.
> 
> Appreciate your help.
> 
> Kyong  
> 
>  LCU ST1 SURF Block
> 1  6.71   AN 1
> 2  6.97   AY 1
> 3  6.77   BN 1
> 4  6.90   BY 1
> 5  6.63   CN 1
> 6  6.94   CY 1
> 7  6.79   DN 1
> 8  6.93   DY 1
> 9  6.23   AN 2
> 10 6.83   AY 2
> 11 6.61   BN 2
> 12 6.86   BY 2
> 13 6.51   CN 2
> 14 6.90   CY 2
> 15 5.90   DN 2
> 16 6.97   DY 2
> 
> A result with the random effect:
> 
> Anal1<-lme(LCU~ST1*SURF,random=~1|Block,data=data1)
>> summary(Anal1)
> Linear mixed-effects model fit by REML
>  Data: data1 
>AIC  BIClogLik
>   25.38958 26.18399 -2.694789
> 
> Random effects:
>  Formula: ~1 | Block
> (Intercept) Residual
> StdDev:   0.1421141 0.218483
> 
> Fixed effects: LCU ~ ST1 * SURF 
>  Value Std.Error DF  t-value p-value
> (Intercept)  6.470 0.1842977  7 35.10625  0.
> ST1B 0.220 0.2184830  7  1.00694  0.3475
> ST1C 0.100 0.2184830  7  0.45770  0.6610
> ST1D-0.125 0.2184830  7 -0.57213  0.5851
> SURFY0.430 0.2184830  7  1.96812  0.0897
> ST1B:SURFY  -0.240 0.3089816  7 -0.77675  0.4627
> ST1C:SURFY  -0.080 0.3089816  7 -0.25892  0.8031
> ST1D:SURFY   0.175 0.3089816  7  0.56638  0.5888
> 
> Without the random effect:
> 
> Anal2<-lme(LCU~ST1*SURF,data=data1)
> Error in getGroups.data.frame(dataMix, groups) : 
> Invalid formula for groups
> Classification:  UNCLASSIFIED 
> Caveats: NONE
> 
> 

Use "lm" to fit the model without random effect and use anova to compare:

z <- read.table(con <- textConnection(" LCU ST1 SURF Block
1  6.71   AN 1
2  6.97   AY 1
3  6.77   BN 1
4  6.90   BY 1
5  6.63   CN 1
6  6.94   CY 1
7  6.79   DN 1
8  6.93   DY 1
9  6.23   AN 2
10 6.83   AY 2
11 6.61   BN     2
12 6.86   BY 2
13 6.51   CN 2
14 6.90   CY 2
15 5.90   DN 2
16 6.97   DY 2"), header = TRUE)
close(con)

library(nlme)
fit <- lme(LCU~ST1*SURF,random=~1|Block,data=z)
fit0 <- lm(LCU~ST1*SURF,data=z)
anova(fit, fit0)

HTH,

--sundar

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[R] ttkcombobox

[Sundar Dorai-Raj]

Hi, all,

(sessionInfo at the end)

I've been struggling with the tcltk package and can't seem to get the 
ttkcombobox to work. Here's an example:


library(tcltk)
p <- tktoplevel()
l <- tclVar()
## I don't know if I'm even calling it correctly
cb <- ttkcombobox(p, values = letters[1:4], textvariable = l)
tkpack(cb)

1. How do I know when the value of the combobox has been changed from, 
say, "a" to "b"?
2. How can I get the value of the box? E.g. if I switch the box to "b", 
how can I access this value?
3. How can I set the default value to something of my choice? I've tried 
the "set" argument as described in the "Tcl/Tk 8.5 Manual" but I 
received an error:

cb <- ttkcombobox(p, values = letters[1:4], textvariable = l, set = "b")
Error in structure(.External("dotTclObjv", objv, PACKAGE = "tcltk"), 
class = "tclObj") :

  [tcl] unknown option "-set".

Thanks,

--sundar

> sessionInfo()
R version 2.7.2 (2008-08-25)
i386-pc-mingw32

locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United 
States.1252;LC_MONETARY=English_United 
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252


attached base packages:
[1] tcltk stats graphics  grDevices utils datasets  methods
[8] base

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Re: [R] ttkcombobox

[Sundar Dorai-Raj]

Hi, Greg,

Yes, that helps immensely!

Thanks,

--sundar

Greg Snow said the following on 10/27/2008 7:14 PM:

Here is one example of using ttkcombobox, hopefully it helps:

library(tcltk)

have.ttk <- function () { # from Prof. Ripley
as.character(tcl("info", "tclversion")) >= "8.5"
}


testfunc <- function() {
tt <- tktoplevel()
if( !have.ttk() ) stop( "Need version 8.5 or greater of tcl" )

tkpack(tklabel(tt, text='Choose a package'), side='top')

tmp <- tclVar()
tclvalue(tmp) <- 'TeachingDemos'

tkpack(ttkcombobox( tt, values=c('blockrand',
'ObsSens','TeachingDemos','sudoku'), textvariable=tmp ),
side='left' )

tkpack(tkbutton(tt,text='print',command=function() 
print(tclvalue(tmp
tkpack(tkbutton(tt,text='exit', command=function() 
tkdestroy(tt)),side='right')
}


testfunc()

If you want something to happen when you change the combobox (rather than just 
getting the value when you press a button or something else), then you will 
probably have to use tkbind to bind an event to it, or use tkafter to check it 
on a regular basis.  I usually do as above and just have a button or something 
else run the code when I want and it just gets the current value.

Hope this helps,


--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
801.408.8111



-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
project.org] On Behalf Of Sundar Dorai-Raj
Sent: Monday, October 27, 2008 3:02 PM
To: r-help@r-project.org
Subject: [R] ttkcombobox

Hi, all,

(sessionInfo at the end)

I've been struggling with the tcltk package and can't seem to get the
ttkcombobox to work. Here's an example:

library(tcltk)
p <- tktoplevel()
l <- tclVar()
## I don't know if I'm even calling it correctly
cb <- ttkcombobox(p, values = letters[1:4], textvariable = l)
tkpack(cb)

1. How do I know when the value of the combobox has been changed from,
say, "a" to "b"?
2. How can I get the value of the box? E.g. if I switch the box to "b",
how can I access this value?
3. How can I set the default value to something of my choice? I've
tried
the "set" argument as described in the "Tcl/Tk 8.5 Manual" but I
received an error:
cb <- ttkcombobox(p, values = letters[1:4], textvariable = l, set =
"b")
Error in structure(.External("dotTclObjv", objv, PACKAGE = "tcltk"),
class = "tclObj") :
   [tcl] unknown option "-set".

Thanks,

--sundar

 > sessionInfo()
R version 2.7.2 (2008-08-25)
i386-pc-mingw32

locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252

attached base packages:
[1] tcltk stats graphics  grDevices utils datasets  methods
[8] base

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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.