Re: [R] system() command in R

[rasanpreet kaur suri]

Thank you all for the suggestions. I've been trying different things to make
this work.

Mike I did try what you suggested . It would work normally But i guess here
the problem is my local server itself. It by default takes time to start and
nothing I do is helping . So I came up with the idea to communicate with the
server through R-java connection.
I am halfway through that , I hope it works:)

Thank you all for the response.

On Tue, Apr 5, 2011 at 4:36 PM, Jeff Newmiller wrote:

> The ampersand is a good idea, but nohup is best avoided in scripts. Rather
> the server itself should handle the transition to daemon status to avoid
> ending up with many duplicate server processes running.
> ---
> Jeff Newmiller The . . Go Live...
> DCN: Basics: ##.#. ##.#. Live Go...
> Live: OO#.. Dead: OO#.. Playing
> Research Engineer (Solar/Batteries O.O#. #.O#. with
> /Software/Embedded Controllers) .OO#. .OO#. rocks...1k
> ---
>
> Sent from my phone. Please excuse my brevity.
>
> Mike Marchywka  wrote:
>>
>> --
>> > Date: Tue, 5 Apr 2011 13:37:12 +0530
>> > From: nandan.a...@gmail.com
>> > To: rasanpreet.k...@gmail.com
>> > CC: r-help@r-project.org
>> > Subject: Re: [R] system() command in R
>> >
>> > On 4 April 2011 16:54, rasanpreet kaur suri wrote:
>> >
>> > > Hi all,
>> > > I have a local server insalled on my system and have to start that from
>> > > within my R function.
>> > >
>> > > here is how I start it:
>> > >
>> > > cmd<-"sh start-server.sh"
>> > >
>> > > system(cmd, wait=FALSE)
>> > >
>> > > My function has to start the server and proceed with further steps. The
>> > > server starts but the further steps of the program are not executed.The
>> > > cursor keeps waiting after the server is started.
>> > >
>> > > How r u executing further steps after starting server, meant for server
>> > from R ??
>> >
>> >
>> > i tried removing the wait=FALSE, but it still keeps waiting.
>> > >
>> > > I also tried putting the start-server in a separate function and my 
>> > > further
>> > > script in a separate function and then run them together, but it still
>> > > waits. The transition from the start of server to next step is not
>> > > happening.
>> > >
>> > > Please help. I have been stuck on this for quite some time now.
>> > >
>> > > --
>>
>> I hadn't done this in R but expect to do so soon.
>> I just got done with some java code to do something similar and you can 
>> expect in any implementation these things will be  system dependent. It 
>> often helps to have simple test cases to isolate the problem. Here I made a 
>> tst script called "foo"
>> that takes a minute or so to exevute and generates some output.
>>
>> If I type
>> system("./foo",wait=F)
>>  the prompt comes back right away but stdout seems to still go to my console 
>> and maybe stdin is not redicrected either and it could eat your input ( no 
>> idea, but this is probably not what you want).
>>
>> I did try this that could fix your problem, on debian anyway it seems to 
>> work,
>>
>> system("nohup ./foo &")
>>  you can "man nohup" for details.
>>
>> > > Rasanpreet Kaur
>> > >
>> > > [[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.
>> > >
>> >
>> >
>> >
>> > --
>> > Amar Kumar Nandan
>> > Karnataka, India, 560100
>> > http://aknandan.co.nr
>> >
>> > [[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 listhttps://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.
>>
>>


-- 
Rasanpreet Kaur

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[R] MLE where loglikelihood function is a function of numerical solutions

[Kristian Lind]

Hi there,

I'm trying to solve a ML problem where the likelihood function is a function
of two numerical procedures and I'm having some problems figuring out how to
do this.

The log-likelihood function is of the form L(c,psi) = 1/T sum [log (f(c,
psi)) - log(g(c,psi))], where c is a 2xT matrix of data and psi is the
parameter vector. f(c, psi) is the transition density which can be
approximated. The problem is that in order to approximate this we need to
first numerically solve 3 ODEs. Second, numerically solve 2 non-linear
equations in two unknowns wrt the data. The g(c,psi) function is known, but
dependent on the numerical solutions.
I have solved the ODEs using the deSolve package and the 2 non-linear
equations using the BB package, but the results are dependent on the
parameters.

How can I write a program that will maximise this log-likelihood function,
taking into account that the numerical procedures needs to be updated for
each iteration in the maximization procedure?

Any help will be much appreciated.


Kristian

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[R] maxLik package.

[hello bye]

Dear Sir/ Madam,
 
I have some enquiry in R about maxLik package where, in this package we have 
the 
usage 
maxLik(logLik, grad, hess, start, method, iterlim, print.level)   when I used 
this with print.level equals to 3 I could have estimates of parameters at each 
iteration but I do not know how can I call the information in the level. Is 
there any way can help me to call the information within the print. level 
argument?. Also when I changed the number of  iterlim argument to one or zero 
for example I got the same estimates of parameters at print. level but with 
message that  "Iteration limit exceeded". So what is the meaning of this 
message 
and how may it effect my results? and if I change the number of iterlim to get 
the estimates of parameters that I need  at this number is it reasonable?.

Thank you.
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[R] Duncan's multiple range test

[mustafabinar]

Hello;
 
I don't find quantile function of duncan's multiple range
test. 
 
Example 
 
quantile function of Tukey test is qtukey(...). What is
quantile function for Duncan's multiple range test?
If you help me, i am very happy. 
 
Thanks.


Facebook ve Twitter hesaplarýný tek yerden güncelle, anýnda 
paylaþ!
Hemen týkla!


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[R] Multinomial Logit Model with lots of Dummy Variables

[ghpow1]

Hi All,

I am attempting to build a Multinomial Logit model with dummy variables of
the following form:

Dependent Variable : 0-8 Discrete Choices

Dummy Variable 1: 965 dummy varsgh...@student.monash.edu.augh@gp1.com
Dummy Variable 2: 805 dummy vars

The data set I am using has the dummy columns pre-created, so it's a table
of 72,381 rows and 1770 columns.

The first 965 columns represent the dummy columns for Variable 1
The next 805 columns represent the dummy columns for Variable 2

My code to build the mlogit model looks like the following. I want to
know...is there a better way of doing this without these huge equations? (I
probably also need a more powerful PC to do all of this).

I'll also want to perform a joint test of significance on the first 805
coefficients...

Is this possible?

Thanks

GP

[code]

#install MLOGIT
library(mlogit)

#load mydata
mydata = 0
mydata<-read.csv(file="G:\\data.csv",head=TRUE)
my_data=0

num.rows=length(mydata[,1])
num.cols=965+805+1


my_data=matrix(0,nr=num.rows,nc=num.cols)

for(i in 1:num.rows) {

nb=mydata[i,2]
np=mydata[i,3]

my_data[i,nb]=1
my_data[i,965+np]=1
my_data[i,1+1770]=mydata[i,1]


}

#convert matrix to data.frame
# convert to data frame
my_data_frame<-as.data.frame(my_data)

#check data frame headers
head(my_data_frame)

#load dataframe into mldata with choice variable
mldata<-mlogit.data(my_data_frame, varying=NULL, choice="V1771",
shape="wide")

#V1771 = dependent var
#V1-V965 = variable 1 dummies
#V966-V1700 = variable 2 dummies

#regress V1771 against all 1700 variables...
mlogit.model<-mlogit(V1771~0|V1+V2+V3...+V1700,data=mldata, reflevel="0")


[/code]



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Re: [R] Adding text labels to lattice plots with multiple panels

[Jeff Stevens]

Hi Peter,

Thanks for the solutions.  The only problem now is that I have
duplicate values in my labels that are removed by the duplicated
function.  Perhaps a better example of the labels vector would be:
lab <- c(1,2,4,4,6,6)

Your first version would maintain the second 4 (because they are in
different panels) but leave a blank for the second 6.
1 2 4 4 6
The second version removes the second 4 and 6.
1 2 4   6

The following works, but it is not very efficient for my full data set
with six levels of f1, two levels of f2, and 64 replicates of each of
these combinations.

bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
 panel = function(x, y, ..., subscripts) {
   lab <<- df$lab[subscripts]
   lab[c(2:3,5:6,8:9)] <<- ""
   panel.bwplot(x, y,  ...)
   panel.text(x, 0.55, labels = lab)
 }
)

Thanks,
Jeff


On Sat, Apr 9, 2011 at 7:52 PM, Peter Ehlers  wrote:
> On 2011-04-09 06:21, Dennis Murphy wrote:
>>
>> Hi:
>>
>> One hopes that there is a more elegant solution than this bit of
>> ad-hockery.
>>>
>>> From your posted example:
>>
>> f1<- c(rep(c(rep("a", 3), rep("b", 3), rep("c", 3)), 2))
>> f2<- c(rep("A", 9), rep("B", 9))
>> dv<- c(0.9, 0.8, 0.85, 0.6, 0.65, 0.7, 0.8, 0.85, 0.8, 0.95, 0.85,
>>  0.9, 0.65, 0.7, 0.75, 0.85, 0.9, 0.85)
>> df<- data.frame(f1, f2, dv)
>> df$lab<- rep(1:6, each = 3)
>>
>> df$lab2<- ''
>> df$lab2[seq(1, 16, by = 3)]<- 1:6        # adapt to your situation -
>> seq(1,
>> nrow(df) - 63, by = 64), perhaps
>>
>> bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
>>  panel = function(x, y, ..., subscripts) {
>>    lab<- df$lab2[subscripts]
>>    panel.bwplot(x, y,  ...)
>>    panel.text(x, 0.55, labels = lab)
>>  }
>> )
>>
>> Alternatively, panel.text() takes an alpha = argument; for example,
>>
>> bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
>>  panel = function(x, y, ..., subscripts) {
>>    lab<- df$lab[subscripts]
>>    panel.bwplot(x, y,  ...)
>>    panel.text(x, 0.55, labels = lab, alpha = 0.5)
>>  }
>> )
>>
>> You could toy with the value of alpha until something acceptable emerges.
>> But as I said, there is probably a better solution and I'm happy to be
>> educated if there is.
>
> Here's a slight variation on your first solution which doesn't
> require the data to be appropriately sorted, using your df:
>
>  bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
>  panel = function(x, y, ..., subscripts) {
>    lab <- df$lab[subscripts]
>    lab[duplicated(lab)] <- ""
>    panel.bwplot(x, y,  ...)
>    panel.text(x, 0.55, labels = lab)
>  }
>  )
>
> and another variation which sets the text positions to NA for
> all but the first pass through the panel.text() function:
>
>  bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
>  panel = function(x, y, ..., subscripts) {
>    at.y <- rep(0.55, nrow(df))
>    is.na(at.y) <- which(duplicated(df$lab))
>    panel.bwplot(x, y,  ...)
>    panel.text(x, at.y[subscripts], labels = df$lab[subscripts])
>  }
>  )
>
> I think that the alpha argument is too one-off, i.e. dependent on
> how many levels in the boxplot.
>
> Peter Ehlers
>
>>
>> HTH,
>> Dennis
>>
>> On Sat, Apr 9, 2011 at 4:56 AM, Jeff
>> Stevenswrote:
>>
>>> Hi,
>>>
>>> Thanks for the work around, Dennis.  My actual data set has 64
>>> replicates for each factor level combination (rather than the 3 in the
>>> example), so the overplotting is quite messy.  Any ideas on how to
>>> avoid the overplotting?
>>>
>>> Jeff
>>>
>>>
>>> On Fri, Apr 8, 2011 at 7:32 PM, Dennis Murphy  wrote:

 Hi:

 After a number of false starts, I finally consulted Deepayan's book and
>>>
>>> the

 example on p. 73, suitably adapted, yielded a solution. Add a variable
>>>
>>> for

 the labels and then...

 df$lab<- rep(1:6, each = 3)
 bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
  panel = function(x, y, ..., subscripts) {
    lab<- df$lab[subscripts]
    panel.bwplot(x, y,  ...)
    panel.text(x, 0.55, labels = lab)
  }
 )

 If you look closely, you'll see that each label is overplotted three
>>>
>>> times.

 A similar plot in ggplot2  would be

 library(ggplot2)
 ggplot(df, aes(x = f1, y = dv)) + geom_boxplot() +
     geom_text(aes(x = as.numeric(f1), lab = lab), y = 0.55, alpha = 0.5)
>>>
>>> +

     facet_wrap( ~ f2) + ylim(0.5, 1)

 The alpha argument in geom_text() is designed to mitigate the
>>>
>>> overplotting

 effect somewhat.

 HTH,
 Dennis


 On Fri, Apr 8, 2011 at 5:32 AM, Jeff Stevens
 wrote:
>
> Hi,
>
> I am trying to add text to the bottom of a lattice bwplot with
> multiple panels.  I would like to add a label below each boxplot, but
> the labels do not come from the data.  I've tried the following, code:
>
> f1<- c(rep(c(rep("a", 3), rep("b", 3), rep("c", 3)), 2))
> f2<- c(rep("A", 9), rep("B", 9))
> dv<- c(0.9, 0.8, 0.85, 0.6, 0.65, 0.7, 0.8, 0.85, 0.8, 0.95, 0.85,
> 0.9, 0.65, 0.7, 0

[R] Fwd: open.exe Virus W32.ATRAPS

[Costas Vorlow]

Hello,

I am not sure if this is the correct mailing list to send this email...

I was trying to install on windows XP latest R-2.12.2 and my unvirus program
(Immunet 3.0) reports a virus at open.exe while  this is unpacked:

W32.ATRAPS

Has anyone come across the same problem?

Best,
Costas

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[R] Help in cluster MPI and R

[Ronaldo Reis Junior]

Hi,

ImountedanMPIclusterwithdebian.I 
installedthepackagesnecessarytorunusingRinthiscluster,butunsuccessfully, 
thefunctionsofexampleswere runbutonlyin theserver processorin the 
cluster.I thinkitlacks an essentialconfigurationofR packagesfor him 
tocommunicatewith thecluster.AnyoneknowanymorebasictutorialforusingRin 
thiscluster?Theclusteralreadyexistsand is functioning.

Anyhelp iswelcome.

Thanks
Ronaldo

-- 
2ª lei - Na pesquisa, o que importa é o que está correto,
  e não quem está correto.

   --Herman, I. P. 2007. Following the law. NATURE, Vol 445, p. 228.

>  Prof. Ronaldo Reis Júnior
|  .''`. UNIMONTES/DBG/Lab. Ecologia Comportamental e Computacional
| : :'  : Campus Universitário Prof. Darcy Ribeiro, Vila Mauricéia
| `. `'` CP: 126, CEP: 39401-089, Montes Claros - MG - Brasil
|   `- Fone: (38) 3229-8192 | ronaldo.r...@unimontes.br
| http://www.ppgcb.unimontes.br/lecc | LinuxUser#: 205366


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Re: [R] Multinomial Logit Model with lots of Dummy Variables

[Jeremy Hetzel]

If you are just looking to collapse the dummy variables into two factor 
variables, the following will work.

## Generate some example data
set.seed(1234)
n <- 100
# Generate outcome
outcome <- rbinom(n, 3, 0.5)
colnames(exposures) <- paste("V", seq(1:10), sep = "")

#Generate dummy variables for A and B
A <- t(apply(matrix(nrow = 100, ncol = 5), 1, function(x)
{
sample(c(1, 0, 0, 0, 0))
}))
B <- t(apply(matrix(nrow = 100, ncol = 5), 1, function(x)
{
sample(c(1, 0, 0, 0, 0))
}))

# Combine into data frame
dat <- data.frame(outcome, A, B)
names(dat) <- c('outcome', paste("A", seq(1:5), sep = ""), paste("B", 
seq(1:5), sep = ""))
head(dat)


## Collapse dummies to factor variable
A <- apply(dat, 1, function(x)
{
A <- x[2:6]
A.names <- names(x[2:6])
A.value <- A.names[A==1]
return(A.value)
})

B <- apply(dat, 1, function(x)
{
B <- x[7:11]
B.names <- names(x[7:11])
B.names
B.value <- B.names[B==1]
return(B.value)
})

# Combine into new data frame
dat.new <- data.frame(dat$outcome, A, B)

head(dat.new)



Jeremy

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Re: [R] set locale information in R

[raji sankaran]

Hi Uwe,

  Thank you very much.. That worked :)

Regards,
Raji

2011/4/9 Uwe Ligges 

> Try
>
> Sys.setenv("LANGUAGE"="Fr")
>
> Uwe Ligges
>
>
> On 07.04.2011 10:29, Raji wrote:
>
>> Hi R-helpers,
>>
>>   Is it possible to localise the error messages/warnings that comes from
>> R.My application takes in a locale information.I used the following
>> command
>> to set the locale in R.But in RGui, i still get the error messages in
>> English only.Can you please help me out with this?
>>
>> Sys.setlocale("LC_ALL", "French")
>>
>> Regards,
>> Raji
>>
>> --
>> View this message in context:
>> http://r.789695.n4.nabble.com/set-locale-information-in-R-tp3432760p3432760.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] Adding text labels to lattice plots with multiple panels

[Peter Ehlers]

On 2011-04-10 04:50, Jeff Stevens wrote:

Hi Peter,

Thanks for the solutions.  The only problem now is that I have
duplicate values in my labels that are removed by the duplicated
function.  Perhaps a better example of the labels vector would be:
lab<- c(1,2,4,4,6,6)

Your first version would maintain the second 4 (because they are in
different panels) but leave a blank for the second 6.
1 2 4 4 6
The second version removes the second 4 and 6.
1 2 4   6

The following works, but it is not very efficient for my full data set
with six levels of f1, two levels of f2, and 64 replicates of each of
these combinations.

bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
  panel = function(x, y, ..., subscripts) {
lab<<- df$lab[subscripts]
lab[c(2:3,5:6,8:9)]<<- ""
panel.bwplot(x, y,  ...)
panel.text(x, 0.55, labels = lab)
  }
)


Good point; I hadn't thought of that. Now I realize that a
solution that fiddles with the labels inside the panel
function is likely to be less efficient anyway. So let's
adjust the labels to be printed before we do the bwplot()
call. What's needed is to set to blanks the labels that
are duplicates because they correspond to duplicated
(f1,f2) combinations.

## add an adjusted labels variable to the data;
## lab is the current vector of labels;
 names(df)
 #[1] "f1"   "f2"   "dv"   "lab"

 df <- transform(df,
lab2 = ifelse(duplicated(df[, c("f1","f2")]), "", lab))

## now use lab2 in bwplot()
 bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
  panel = function(x, y, ..., subscripts) {
lab <- df$lab2[subscripts]   # note the lab2
panel.bwplot(x, y,  ...)
panel.text(x, 0.55, labels = lab)
  }
 )

Peter Ehlers



Thanks,
Jeff


On Sat, Apr 9, 2011 at 7:52 PM, Peter Ehlers  wrote:

On 2011-04-09 06:21, Dennis Murphy wrote:


Hi:

One hopes that there is a more elegant solution than this bit of
ad-hockery.


 From your posted example:


f1<- c(rep(c(rep("a", 3), rep("b", 3), rep("c", 3)), 2))
f2<- c(rep("A", 9), rep("B", 9))
dv<- c(0.9, 0.8, 0.85, 0.6, 0.65, 0.7, 0.8, 0.85, 0.8, 0.95, 0.85,
  0.9, 0.65, 0.7, 0.75, 0.85, 0.9, 0.85)
df<- data.frame(f1, f2, dv)
df$lab<- rep(1:6, each = 3)

df$lab2<- ''
df$lab2[seq(1, 16, by = 3)]<- 1:6# adapt to your situation -
seq(1,
nrow(df) - 63, by = 64), perhaps

bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
  panel = function(x, y, ..., subscripts) {
lab<- df$lab2[subscripts]
panel.bwplot(x, y,  ...)
panel.text(x, 0.55, labels = lab)
  }
)

Alternatively, panel.text() takes an alpha = argument; for example,

bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
  panel = function(x, y, ..., subscripts) {
lab<- df$lab[subscripts]
panel.bwplot(x, y,  ...)
panel.text(x, 0.55, labels = lab, alpha = 0.5)
  }
)

You could toy with the value of alpha until something acceptable emerges.
But as I said, there is probably a better solution and I'm happy to be
educated if there is.


Here's a slight variation on your first solution which doesn't
require the data to be appropriately sorted, using your df:

  bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
  panel = function(x, y, ..., subscripts) {
lab<- df$lab[subscripts]
lab[duplicated(lab)]<- ""
panel.bwplot(x, y,  ...)
panel.text(x, 0.55, labels = lab)
  }
  )

and another variation which sets the text positions to NA for
all but the first pass through the panel.text() function:

  bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
  panel = function(x, y, ..., subscripts) {
at.y<- rep(0.55, nrow(df))
is.na(at.y)<- which(duplicated(df$lab))
panel.bwplot(x, y,  ...)
panel.text(x, at.y[subscripts], labels = df$lab[subscripts])
  }
  )

I think that the alpha argument is too one-off, i.e. dependent on
how many levels in the boxplot.

Peter Ehlers



HTH,
Dennis

On Sat, Apr 9, 2011 at 4:56 AM, Jeff
Stevenswrote:


Hi,

Thanks for the work around, Dennis.  My actual data set has 64
replicates for each factor level combination (rather than the 3 in the
example), so the overplotting is quite messy.  Any ideas on how to
avoid the overplotting?

Jeff


On Fri, Apr 8, 2011 at 7:32 PM, Dennis Murphywrote:


Hi:

After a number of false starts, I finally consulted Deepayan's book and


the


example on p. 73, suitably adapted, yielded a solution. Add a variable


for


the labels and then...

df$lab<- rep(1:6, each = 3)
bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
  panel = function(x, y, ..., subscripts) {
lab<- df$lab[subscripts]
panel.bwplot(x, y,  ...)
panel.text(x, 0.55, labels = lab)
  }
)

If you look closely, you'll see that each label is overplotted three


times.


A similar plot in ggplot2  would be

library(ggplot2)
ggplot(df, aes(x = f1, y = dv)) + geom_boxplot() +
 geom_text(aes(x = as.numeric(f1), lab = lab), y = 0.55, alpha = 0.5)


+


 facet_wrap( ~ f2) + ylim(0.5, 1)

The alpha argument in geom_text() is designed to mitigate the


overplotting


effect somewhat.

HTH,
Dennis

[R] howto calculate column means in data frame

[Maas James Dr (MED)]

Long story short, I have a big iterative procedure that produces a long list of 
data.frames such as the one called
"results" here.  Is there an easy way to produce a similar list of data.frames 
comprised of the mean of each of the
columns in results, such that it ends up like the one I've shown in 
"resultsmean" below?

I've tried apply and lapply, still not got the correct arguments.  As usual, 
TIA.

Jim

> results
[[1]]
   nameLOR23   BIA23   MSE23  H0R23
1 0.222   -1.012228   -0.0959370.0356501.00
2 0.222   -0.8363000.0799910.0423220.75
3 0.222   -0.5186310.3976590.2145930.50


[[2]]
   nameLOR23   BIA23   MSE23  H0R23
1 0.2211122   -0.7246300.1916600.051308   1
2 0.2211122   -0.7818120.1344780.033872   1
3 0.2211122   -0.5221090.3941810.1646280.75


would like

> resultsmean
[[1]]
   nameLOR23   BIA23   MSE23  H0R23
1 0.222   -0.783330.12734 0.0971600.75

[[2]]
   nameLOR23   BIA23   MSE23  H0R23
1 0.2211122   -0.675660.2400  0.08266 0.916

===
Dr. Jim Maas
University of East Anglia

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Re: [R] howto calculate column means in data frame

[Joshua Wiley]

Hi Jim,

Using ?lapply with ?colMeans should do the trick.  Here is a little sample:

eg <- list(mtcars, mtcars) # mtcars data frame twice in a list
resultsmean <- lapply(eg, colMeans) # calculate column means for each
element of "eg"
resultsmean # show the results

Hope this helps,

Josh

On Sun, Apr 10, 2011 at 8:27 AM, Maas James Dr (MED)  wrote:
> Long story short, I have a big iterative procedure that produces a long list 
> of data.frames such as the one called
> "results" here.  Is there an easy way to produce a similar list of 
> data.frames comprised of the mean of each of the
> columns in results, such that it ends up like the one I've shown in 
> "resultsmean" below?
>
> I've tried apply and lapply, still not got the correct arguments.  As usual, 
> TIA.
>
> Jim
>
>> results
> [[1]]
>       name    LOR23       BIA23       MSE23          H0R23
> 1 0.222   -1.012228   -0.095937    0.035650        1.00
> 2 0.222   -0.836300    0.079991    0.042322        0.75
> 3 0.222   -0.518631    0.397659    0.214593        0.50
>
>
> [[2]]
>       name    LOR23       BIA23       MSE23          H0R23
> 1 0.2211122   -0.724630    0.191660    0.051308           1
> 2 0.2211122   -0.781812    0.134478    0.033872           1
> 3 0.2211122   -0.522109    0.394181    0.164628        0.75
>
>
> would like
>
>> resultsmean
> [[1]]
>       name    LOR23       BIA23       MSE23          H0R23
> 1 0.222   -0.78333    0.12734     0.097160        0.75
>
> [[2]]
>       name    LOR23       BIA23       MSE23          H0R23
> 1 0.2211122   -0.67566    0.2400      0.08266         0.916
>
> ===
> Dr. Jim Maas
> University of East Anglia
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



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

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Re: [R] howto calculate column means in data frame

[Berend Hasselman]

jamaas wrote:
> 
> Long story short, I have a big iterative procedure that produces a long
> list of data.frames such as the one called
> "results" here.  Is there an easy way to produce a similar list of
> data.frames comprised of the mean of each of the
> columns in results, such that it ends up like the one I've shown in
> "resultsmean" below?
> 
> I've tried apply and lapply, still not got the correct arguments.  As
> usual, TIA.
> 
> Jim
> 
>> results
> [[1]]
>nameLOR23   BIA23   MSE23  H0R23
> 1 0.222   -1.012228   -0.0959370.0356501.00
> 2 0.222   -0.8363000.0799910.0423220.75
> 3 0.222   -0.5186310.3976590.2145930.50
> 
> 
> [[2]]
>nameLOR23   BIA23   MSE23  H0R23
> 1 0.2211122   -0.7246300.1916600.051308   1
> 2 0.2211122   -0.7818120.1344780.033872   1
> 3 0.2211122   -0.5221090.3941810.1646280.75
> 
> 
> would like
> 
>> resultsmean
> [[1]]
>nameLOR23   BIA23   MSE23  H0R23
> 1 0.222   -0.783330.12734 0.0971600.75
> 
> [[2]]
>nameLOR23   BIA23   MSE23  H0R23
> 1 0.2211122   -0.675660.2400  0.08266 0.916
> 

lapply(results, FUN=colMeans)

Berend

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Re: [R] Orthoblique rotation on eigenvectors (SAS VARCLUS)

[Frank Harrell]

See the varclus function in the Hmisc package.
Frank


Axel Urbiz wrote:
> 
> Hi All,
> 
> I'd like to build a package for the community that replicates the output
> produced by SAS "proc varclus". According to the SAS documentation, the
> first few steps are:
> 
> 1. Find the first two principal components.
> 2. Perform an orthoblique rotation (quartimax rotation) on eigenvectors.
> 3. Assign each variable to the rotated component with which it has the
> higher
>squared correlation.
> The cartoon example below attempt to do this, but I found my results
> differ
> from SAS in "pc3" (i.e,  the standardize component scores).
> 
> I'd appreciate your help in whether you see anything wrong in "pc2" or
> "pc3"?
> 
> set.seed(1)
> x1=rnorm(200); x2=0.9*x1; x3=0.7*x1; x4=x1*x1; x5=x1*x1*x1;
> x6=rnorm(200); x7=0.9*x6; x8=0.7*x6; x9=x6*x6; x10=x6*x6*x6;
> x <- cbind(x1,x2,x3,x4,x5,x6,x7,x8,x9,x10)
> 
> require(GPArotation)
> pc1 <- princomp(x, cor = TRUE, scores = TRUE)
> pc2 <- quartimax(pc1$loadings[,1:2],normalize=TRUE)$loadings
> pc3 <- scale(x%*% pc2)
> pc4 <- apply(x, 2, function(x) cor(x, pc3)^2)
> 
> Thanks in advance for any help!
> 
> Axel.
> 
>   [[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.
> 


-
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
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Re: [R] MLE where loglikelihood function is a function of numerical solutions

[Albyn Jones]

Hi Kristian

The obvious approach is to treat it like any other MLE problem:  evaluation
of the log-likelihood is done as often as necessary for the optimizer  
you are using: eg a call to optim(psi,LL,...)  where LL(psi) evaluates  
the log likelihood at psi.  There may be computational shortcuts that  
would work if you knew that LL(psi+eps) were well approximated by  
LL(psi), for the values of eps used to evaluate numerical derivatives  
of LL.  Of course, then you might need to write your own custom  
optimizer.


albyn

Quoting Kristian Lind :


Hi there,

I'm trying to solve a ML problem where the likelihood function is a function
of two numerical procedures and I'm having some problems figuring out how to
do this.

The log-likelihood function is of the form L(c,psi) = 1/T sum [log (f(c,
psi)) - log(g(c,psi))], where c is a 2xT matrix of data and psi is the
parameter vector. f(c, psi) is the transition density which can be
approximated. The problem is that in order to approximate this we need to
first numerically solve 3 ODEs. Second, numerically solve 2 non-linear
equations in two unknowns wrt the data. The g(c,psi) function is known, but
dependent on the numerical solutions.
I have solved the ODEs using the deSolve package and the 2 non-linear
equations using the BB package, but the results are dependent on the
parameters.

How can I write a program that will maximise this log-likelihood function,
taking into account that the numerical procedures needs to be updated for
each iteration in the maximization procedure?

Any help will be much appreciated.


Kristian

[[alternative HTML version deleted]]

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Re: [R] MLE where loglikelihood function is a function of numerical solutions

[Albyn Jones]

to clarify: by  "if you knew that LL(psi+eps) were well approximated  
by LL(psi), for the values of eps used to evaluate numerical  
derivatives of LL. "

I mean the derivatives of LL(psi+eps) are close to the derivatives of LL(psi),
and perhaps you would want the hessian to be close as well.

albyn

Quoting Albyn Jones :


Hi Kristian

The obvious approach is to treat it like any other MLE problem:  evaluation
of the log-likelihood is done as often as necessary for the  
optimizer you are using: eg a call to optim(psi,LL,...)  where  
LL(psi) evaluates the log likelihood at psi.  There may be  
computational shortcuts that would work if you knew that LL(psi+eps)  
were well approximated by LL(psi), for the values of eps used to  
evaluate numerical derivatives of LL.  Of course, then you might  
need to write your own custom optimizer.


albyn

Quoting Kristian Lind :


Hi there,

I'm trying to solve a ML problem where the likelihood function is a function
of two numerical procedures and I'm having some problems figuring out how to
do this.

The log-likelihood function is of the form L(c,psi) = 1/T sum [log (f(c,
psi)) - log(g(c,psi))], where c is a 2xT matrix of data and psi is the
parameter vector. f(c, psi) is the transition density which can be
approximated. The problem is that in order to approximate this we need to
first numerically solve 3 ODEs. Second, numerically solve 2 non-linear
equations in two unknowns wrt the data. The g(c,psi) function is known, but
dependent on the numerical solutions.
I have solved the ODEs using the deSolve package and the 2 non-linear
equations using the BB package, but the results are dependent on the
parameters.

How can I write a program that will maximise this log-likelihood function,
taking into account that the numerical procedures needs to be updated for
each iteration in the maximization procedure?

Any help will be much appreciated.


Kristian

[[alternative HTML version deleted]]

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[R] list to data frame

[Franklin Tamborello II]

I need to make a data frame out of the data that I currently have in a list. 
This works, but is ugly:
ineffData<-rbind(ineffFilesList[[1]], ineffFilesList[[2]], ineffFilesList[[3]], 
ineffFilesList[[4]], ineffFilesList[[5]], ineffFilesList[[6]], 
ineffFilesList[[7]], ineffFilesList[[8]], ineffFilesList[[9]], 
ineffFilesList[[10]], ineffFilesList[[11]], ineffFilesList[[12]], 
ineffFilesList[[13]], ineffFilesList[[14]], ineffFilesList[[15]], 
ineffFilesList[[16]], ineffFilesList[[17]], ineffFilesList[[18]], 
ineffFilesList[[19]], ineffFilesList[[20]], ineffFilesList[[21]], 
ineffFilesList[[22]], ineffFilesList[[23]], ineffFilesList[[24]], 
ineffFilesList[[25]], ineffFilesList[[26]], ineffFilesList[[27]])


What's an efficient way of doing this such that the computer will do the work 
of recurring through the list of elements of ineffFilesList?

Much appreciation,




Frank Tamborello, PhD
W. M. Keck Postdoctoral Fellow
School of Biomedical Informatics
University of Texas Health Science Center at Houston

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Re: [R] list to data frame

[Jorge Ivan Velez]

Hi Franklin,

Try

do.call(rbind, ineffFilesList)

See ?do.call for more details.

HTH,
Jorge


On Sun, Apr 10, 2011 at 2:01 PM, Franklin Tamborello II <> wrote:

> I need to make a data frame out of the data that I currently have in a
> list. This works, but is ugly:
> ineffData<-rbind(ineffFilesList[[1]], ineffFilesList[[2]],
> ineffFilesList[[3]], ineffFilesList[[4]], ineffFilesList[[5]],
> ineffFilesList[[6]], ineffFilesList[[7]], ineffFilesList[[8]],
> ineffFilesList[[9]], ineffFilesList[[10]], ineffFilesList[[11]],
> ineffFilesList[[12]], ineffFilesList[[13]], ineffFilesList[[14]],
> ineffFilesList[[15]], ineffFilesList[[16]], ineffFilesList[[17]],
> ineffFilesList[[18]], ineffFilesList[[19]], ineffFilesList[[20]],
> ineffFilesList[[21]], ineffFilesList[[22]], ineffFilesList[[23]],
> ineffFilesList[[24]], ineffFilesList[[25]], ineffFilesList[[26]],
> ineffFilesList[[27]])
>
>
> What's an efficient way of doing this such that the computer will do the
> work of recurring through the list of elements of ineffFilesList?
>
> Much appreciation,
>
>
>
>
> Frank Tamborello, PhD
> W. M. Keck Postdoctoral Fellow
> School of Biomedical Informatics
> University of Texas Health Science Center at Houston
>
> __
> 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.
>

[[alternative HTML version deleted]]

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Re: [R] Fwd: open.exe Virus W32.ATRAPS

[peter dalgaard]

On Apr 10, 2011, at 15:10 , Costas Vorlow wrote:

> Hello,
> 
> I am not sure if this is the correct mailing list to send this email...
> 
> I was trying to install on windows XP latest R-2.12.2 and my unvirus program
> (Immunet 3.0) reports a virus at open.exe while  this is unpacked:
> 
> W32.ATRAPS
> 
> Has anyone come across the same problem?

Yes, with Avira and 2.12.0. It was a false positive. See

http://finzi.psych.upenn.edu/Rhelp10/2010-October/256518.html

-- 
Peter Dalgaard
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk  Priv: pda...@gmail.com

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Re: [R] By function

[Raoni Rodrigues]

Liviu,

Thanks for your attention and help!

Sorry for attached csv file, I didn't read R-Help Posting guide
careful. Another R-helper thaught me use de dput() function, and this
output are in email's end.

Your correction in my code works pretty well!! Now I can understand a
little more how the R "think"! Thanks a lot!

Also thanks for the suggetion to use sessionInfo(). I will do that on
next questions to list.

All best, and thanks very much again,

Raoni

The dput() of data is:

structure(list(Data = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L,
4L, 4L, 4L, 4L, 5L, 5L, 5L), .Label = c("2011-02-04", "2011-02-05",
"2011-02-07", "2011-02-08", "2011-02-09", "2011-02-10", "2011-02-11",
"2011-02-14", "2011-02-15", "2011-02-16", "2011-02-17", "2011-02-18",
"2011-02-21", "2011-02-22", "2011-02-23", "2011-02-24", "2011-02-25",
"2011-02-28", "2011-03-01", "2011-03-02", "2011-03-03", "2011-03-04",
"2011-03-10", "2011-03-11", "2011-03-14", "2011-03-15", "2011-03-16",
"2011-03-17", "2011-03-28", "2011-03-29", "2011-03-30", "2011-03-31",
"2011-04-01", "2011-04-04", "2011-04-05", "2011-04-06", "2011-04-07"
), class = "factor"), Fisherman = structure(c(5L, 1L, 2L,
5L, 1L, 2L, 1L, 2L, 5L, 5L, 2L, 1L, 2L, 5L, 1L, 2L, 5L, 1L, 5L,
5L, 2L, 1L, 5L, 1L, 2L, 5L, 1L, 2L, 5L, 1L), .Label = c("Yellow",
"Blue", "White", "Green", "Red"), class = "factor"), N = c(0L,
1L, 1L, 0L, 0L, 0L, 0L, 0L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 1L, 1L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L)), .Names = c("Date",
"Fisherman", "N"), row.names = c(NA, 30L), class = "data.frame")


2011/4/9 Liviu Andronic :
> Hello
>
> On Fri, Apr 8, 2011 at 12:38 AM, Raoni Rodrigues
>  wrote:
>> Hello all!
>>
>> I have a data frame with nine variables and 293 cases. (attached goes
>> the csv file).
>>
> The CSV didn't get through so it's difficult to replicate your
> example. Please post the output of:
> str(cpue)
>
>
>> I need to calculate a index using the sum of one variable (N) divided
>> by the length of other variable (Fisherman), but for each day (Date).
>>
>> I tried to use that codes:
>>
>> by (cpue, cpue$Date, function (x) sum (cpue$N)/length(cpue$Fisherman))
>>
> [..]
>
>> In both codes, as result, I received the same value for all Date
>> value. And, oddly, this result is the value of calculation for all
>> data frame, as I used just:
>>
>> sum (cpue$N)/length(cpue$Fishermans)
>>
> From the code above, this is expected since in the function()
> statement you use the entire df (cpue) and not the subset (x). Try
> this:
> by (cpue, cpue$Date, function (x) sum (x$N)/length(x$Fisherman))
>
>
>> Sorry for the basic question, but I worked on this simple code all day
>> and can't do it work out! And have no idea why...
>>
>> I'm using R 2.11 in Windows XP.
>>
> It is often a good a idea to post the output of
> sessionInfo()
>
> Regards
> Liviu
>
>
>
>> Since now, thanks for the attention and all help,
>>
>> Best,
>>
>> Raoni
>>
>> __
>> 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.
>>
>>
>
>
>
> --
> Do you know how to read?
> http://www.alienetworks.com/srtest.cfm
> http://goodies.xfce.org/projects/applications/xfce4-dict#speed-reader
> Do you know how to write?
> http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail
>

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Re: [R] Adding text labels to lattice plots with multiple panels

[Peter Ehlers]

On 2011-04-10 12:19, Jeff Stevens wrote:

Many thanks, Peter.  This works brilliantly, and I prefer to have the
labels assigned outside of panel function as well.

Cheers,
Jeff


Small add-on:
It's probably best to ensure that the labels are of
type character, not factor.

Peter

[...snip...]

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Re: [R] list to data frame

[Philipp Pagel]

On Sun, Apr 10, 2011 at 06:01:39PM +, Franklin Tamborello II wrote:
> I need to make a data frame out of the data that I currently have in
> a list. This works, but is ugly:
> ineffData<-rbind(ineffFilesList[[1]], ineffFilesList[[2]],
> ineffFilesList[[3]], ineffFilesList[[4]], ineffFilesList[[5]],
> ineffFilesList[[6]], ineffFilesList[[7]], ineffFilesList[[8]],
> ineffFilesList[[9]], ineffFilesList[[10]], ineffFilesList[[11]],
> ineffFilesList[[12]], ineffFilesList[[13]], ineffFilesList[[14]],
> ineffFilesList[[15]], ineffFilesList[[16]], ineffFilesList[[17]],
> ineffFilesList[[18]], ineffFilesList[[19]], ineffFilesList[[20]],
> ineffFilesList[[21]], ineffFilesList[[22]], ineffFilesList[[23]],
> ineffFilesList[[24]], ineffFilesList[[25]], ineffFilesList[[26]],
> ineffFilesList[[27]])
> 
> 
> What's an efficient way of doing this such that the computer will do
> the work of recurring through the list of elements of
> ineffFilesList?

as.data.frame(ineffFilesList)


cu
Philipp

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

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Re: [R] Converting edgelist to symmetric matrix

[Mike Marchywka]

> Date: Sat, 9 Apr 2011 14:34:28 -0700
> From: kmshafi...@yahoo.com
> To: r-help@r-project.org
> Subject: [R] Converting edgelist to symmetric matrix
>
> Hi,
> I have network data in the form of a couple of edgelists containing weights in
> the format "x,y,weight" whereby x represents row header and y represents 
> column
> header. All edgelists are based on links among 634 nodes and I need to convert
> them into a 634*634 weighted matrix.
>
>

not find
> a clue how to do this in R. Any help will be appreciated.

I'm trying to do something related and found

?read.graph 

will format=ncol do what you need? This apparently creates a graph object that 
likely
has capacilities you need.  Again, I haven't actually used any of this
just found while trying to solve a different problem. 

'It is a simple text file
with one edge per line. An edge is defined by two symbolic vertex
names separated by whitespace. (The symbolic vertex names
themselves cannot contain whitespace. They might followed by an
optional number, this will be the weight of the edge; the number
can be negative and can be in scientific notation. If there is no
weight specified to an edge it is assumed to be zero. 
'

>
> Best regards,
> Shafique
>

  
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[R] Fitting controlled released data

[EmaDaCuz]

Hi,

I am new to the forum/mailing list. I have been using R for a while and I
find it incredible.
I was just wondering whether someone has ever written a library to calculate
the best fit of experimental data to some controlled release models, having
only the release cumulative drug release at given time points. For example,
there is an extension for SigmaPlot

http://www.sigmaplot.co.uk/products/sigmaplot/productuses/prod-uses15.php

which allows rapid fitting of 5 standard model. I prefer to use free
software and therefore I rather use R than Sigma Plot. 
Is there anyone who can help?

Thank you very much 

Marco

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[R] Question about levels/as.numeric

[Thibault Vatter]

Hi,

I am still new to R and this is my first post on this mailing-list.

I have two .csv (each one being a column of real numbers) coming from the
same database (the first one is just longer than the second) and I read them
in R the following way:

returns  <- read.csv("test.csv", header = FALSE)
returns2  <- read.csv("test2.csv", header = FALSE)

However, the two objects clearly don't seem to be equivalent:

> returns[2528:2537,1]
 [1] -0.002206 0.115696  -0.015192 0.008719  -0.004654 -0.010688 0.009453
0.002676  0.001334  -0.011326
7470 Levels: -0.78 -0.85 -0.86 -0.0001 -0.000112 -0.000115
-0.000152 -0.000154 -0.000157 -0.00016 -0.000171 -0.000185 -0.000212
-0.000238 -0.000256 -0.000259 -0.000263 -0.000273 ... C

> returns2[1:10,1]
 [1] -0.002206  0.115696 -0.015192  0.008719 -0.004654 -0.010688  0.009453
0.002676  0.001334 -0.011326

> as.numeric(returns[2528:2537,1])
 [1]  341 7444 2244 5149  787 1717 5251 4122 3878 1811

> as.numeric(returns2[1:10,1])
 [1] -0.002206  0.115696 -0.015192  0.008719 -0.004654 -0.010688  0.009453
0.002676  0.001334 -0.011326

I would like to understand what's happening and how to handle the longer
one. This problem may seem stupid, but I've been trying to figure it out for
a while and nothing seems to work. I checked in excel and both seems to be
completely normal lists of real numbers).

What am I missing here? What are those "levels" and why the as.numeric
doesn't work the same with the longer one?

My final goal  is to extract small parts of those columns the following way:

> cbind(returns[which(names == id)[2528:2537],1])
  [,1]
 [1,]  341
 [2,] 7444
 [3,] 2244
 [4,] 5149
 [5,]  787
 [6,] 1717
 [7,] 5251
 [8,] 4122
 [9,] 3878
[10,] 1811

Wich should be equivalent to:

> cbind(returns2[which(names == id)[1:10],1])
   [,1]
 [1,] -0.002206
 [2,]  0.115696
 [3,] -0.015192
 [4,]  0.008719
 [5,] -0.004654
 [6,] -0.010688
 [7,]  0.009453
 [8,]  0.002676
 [9,]  0.001334
[10,] -0.011326

Thanks a lot,
Thibault

-
*Thibault Vatter*
 EPFL- Master, 1ère année
Laboratory of Statistical Biophysics 

Tel: +41 78 820 18 64
 @: thibault.vat...@epfl.ch
Web: http://personnes.epfl.ch/thibault.vatter

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[R] Password-protect R script files

[Vijayan Padmanabhan]

There was a question in R forum very long time back.. on how to protect R
Script files from inadvertent editing by users.
There is a way to do this  from within R, atleast in Windows XP I have tried
this and it certainly works , The method  is very different from the OS
based folder protection route,  however making available such a method in
the open forum would only kill the very spirit of R.
But if someone is able to convince me the genuineness of his reasons to
achieve such a purpose, I might decide to provide selective service to
achieve the same.

Regards
Vijayan Padmanabhan

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Re: [R] Adding text labels to lattice plots with multiple panels

[Jeff Stevens]

Many thanks, Peter.  This works brilliantly, and I prefer to have the
labels assigned outside of panel function as well.

Cheers,
Jeff


On Sun, Apr 10, 2011 at 5:18 PM, Peter Ehlers  wrote:
> On 2011-04-10 04:50, Jeff Stevens wrote:
>>
>> Hi Peter,
>>
>> Thanks for the solutions.  The only problem now is that I have
>> duplicate values in my labels that are removed by the duplicated
>> function.  Perhaps a better example of the labels vector would be:
>> lab<- c(1,2,4,4,6,6)
>>
>> Your first version would maintain the second 4 (because they are in
>> different panels) but leave a blank for the second 6.
>> 1 2 4 4 6
>> The second version removes the second 4 and 6.
>> 1 2 4   6
>>
>> The following works, but it is not very efficient for my full data set
>> with six levels of f1, two levels of f2, and 64 replicates of each of
>> these combinations.
>>
>> bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
>>  panel = function(x, y, ..., subscripts) {
>>    lab<<- df$lab[subscripts]
>>    lab[c(2:3,5:6,8:9)]<<- ""
>>    panel.bwplot(x, y,  ...)
>>    panel.text(x, 0.55, labels = lab)
>>  }
>> )
>
> Good point; I hadn't thought of that. Now I realize that a
> solution that fiddles with the labels inside the panel
> function is likely to be less efficient anyway. So let's
> adjust the labels to be printed before we do the bwplot()
> call. What's needed is to set to blanks the labels that
> are duplicates because they correspond to duplicated
> (f1,f2) combinations.
>
> ## add an adjusted labels variable to the data;
> ## lab is the current vector of labels;
>  names(df)
>  #[1] "f1"   "f2"   "dv"   "lab"
>
>  df <- transform(df,
>        lab2 = ifelse(duplicated(df[, c("f1","f2")]), "", lab))
>
> ## now use lab2 in bwplot()
>  bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
>  panel = function(x, y, ..., subscripts) {
>    lab <- df$lab2[subscripts]   # note the lab2
>    panel.bwplot(x, y,  ...)
>    panel.text(x, 0.55, labels = lab)
>  }
>  )
>
> Peter Ehlers
>
>>
>> Thanks,
>> Jeff
>>
>>
>> On Sat, Apr 9, 2011 at 7:52 PM, Peter Ehlers  wrote:
>>>
>>> On 2011-04-09 06:21, Dennis Murphy wrote:

 Hi:

 One hopes that there is a more elegant solution than this bit of
 ad-hockery.
>
>  From your posted example:

 f1<- c(rep(c(rep("a", 3), rep("b", 3), rep("c", 3)), 2))
 f2<- c(rep("A", 9), rep("B", 9))
 dv<- c(0.9, 0.8, 0.85, 0.6, 0.65, 0.7, 0.8, 0.85, 0.8, 0.95, 0.85,
  0.9, 0.65, 0.7, 0.75, 0.85, 0.9, 0.85)
 df<- data.frame(f1, f2, dv)
 df$lab<- rep(1:6, each = 3)

 df$lab2<- ''
 df$lab2[seq(1, 16, by = 3)]<- 1:6        # adapt to your situation -
 seq(1,
 nrow(df) - 63, by = 64), perhaps

 bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
  panel = function(x, y, ..., subscripts) {
    lab<- df$lab2[subscripts]
    panel.bwplot(x, y,  ...)
    panel.text(x, 0.55, labels = lab)
  }
 )

 Alternatively, panel.text() takes an alpha = argument; for example,

 bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
  panel = function(x, y, ..., subscripts) {
    lab<- df$lab[subscripts]
    panel.bwplot(x, y,  ...)
    panel.text(x, 0.55, labels = lab, alpha = 0.5)
  }
 )

 You could toy with the value of alpha until something acceptable
 emerges.
 But as I said, there is probably a better solution and I'm happy to be
 educated if there is.
>>>
>>> Here's a slight variation on your first solution which doesn't
>>> require the data to be appropriately sorted, using your df:
>>>
>>>  bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
>>>  panel = function(x, y, ..., subscripts) {
>>>    lab<- df$lab[subscripts]
>>>    lab[duplicated(lab)]<- ""
>>>    panel.bwplot(x, y,  ...)
>>>    panel.text(x, 0.55, labels = lab)
>>>  }
>>>  )
>>>
>>> and another variation which sets the text positions to NA for
>>> all but the first pass through the panel.text() function:
>>>
>>>  bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
>>>  panel = function(x, y, ..., subscripts) {
>>>    at.y<- rep(0.55, nrow(df))
>>>    is.na(at.y)<- which(duplicated(df$lab))
>>>    panel.bwplot(x, y,  ...)
>>>    panel.text(x, at.y[subscripts], labels = df$lab[subscripts])
>>>  }
>>>  )
>>>
>>> I think that the alpha argument is too one-off, i.e. dependent on
>>> how many levels in the boxplot.
>>>
>>> Peter Ehlers
>>>

 HTH,
 Dennis

 On Sat, Apr 9, 2011 at 4:56 AM, Jeff
 Stevenswrote:

> Hi,
>
> Thanks for the work around, Dennis.  My actual data set has 64
> replicates for each factor level combination (rather than the 3 in the
> example), so the overplotting is quite messy.  Any ideas on how to
> avoid the overplotting?
>
> Jeff
>
>
> On Fri, Apr 8, 2011 at 7:32 PM, Dennis Murphy
>  wrote:
>>
>> Hi:
>>
>> After a number of false starts, I finally consulted Deepayan's book
>> and
>
> th

[R] About Tinn-R

[Marcos Prunello]

Hi everybody!!

I don`t know if it is correct to make a question about R and Tinn-R in this 
mailing-list, but I need some help with a problem for which I don't find a 
solution anywhere.

I have Tinn-R 2.3.5.2 in my notebook and I always used it with no problems, 
until a few days ago when this signs appears everytime I start the programme:

"Key Violation.
C:/Documents and Settings/Marcos/Datos de programa/Tinn-R
Serious problem reading ini files!
Tinn-R can not be initiated.
Please, try to rename (or remove) the folder above and restart the program!"

I don't know what that means, I don`t have such folder in my notebook, and I 
always used it normally and suddenly it doesn't work anymore.

I unistalled it and installed again several times, and I tried with the newest 
version of R and TinnR too. I can`t find help in internet, and I also cleaned 
the registry of the computer.

Thanks in advance!!

Marcos from Argentina


  
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Re: [R] Question about levels/as.numeric

[Petr Savicky]

On Sun, Apr 10, 2011 at 05:47:59PM +0200, Thibault Vatter wrote:
> Hi,
> 
> I am still new to R and this is my first post on this mailing-list.
> 
> I have two .csv (each one being a column of real numbers) coming from the
> same database (the first one is just longer than the second) and I read them
> in R the following way:
> 
> returns  <- read.csv("test.csv", header = FALSE)
> returns2  <- read.csv("test2.csv", header = FALSE)
> 
> However, the two objects clearly don't seem to be equivalent:
> 
> > returns[2528:2537,1]
>  [1] -0.002206 0.115696  -0.015192 0.008719  -0.004654 -0.010688 0.009453
> 0.002676  0.001334  -0.011326
> 7470 Levels: -0.78 -0.85 -0.86 -0.0001 -0.000112 -0.000115
> -0.000152 -0.000154 -0.000157 -0.00016 -0.000171 -0.000185 -0.000212
> -0.000238 -0.000256 -0.000259 -0.000263 -0.000273 ... C

Hi.

It seems that the first file contains a non-numeric row. It may contain "C",
which is the last of the levels. In this case, the whole column is
considered as a character vector and is converted to a factor.

> > returns2[1:10,1]
>  [1] -0.002206  0.115696 -0.015192  0.008719 -0.004654 -0.010688  0.009453
> 0.002676  0.001334 -0.011326
> 
> > as.numeric(returns[2528:2537,1])
>  [1]  341 7444 2244 5149  787 1717 5251 4122 3878 1811

These are indices to the levels of the factor.

Petr Savicky.

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[R] count number of TRUEs in each row

[Wendy]

Hi all, 

I have a huge matrix of TRUE/FALSE table like following, and I want to count
the number of TRUEs in each row. Instead of looping through each row and do
length(Z[Z==TRUE]), I am wondering if there is an easier way of doing this. 

   [,1]   [,2][,3]
[1,]TRUE FALSE FALSE
[2,]FALSE TRUE TRUE

Thank you in advance.
Wendy

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Re: [R] count number of TRUEs in each row

[Peter Langfelder]

On Sun, Apr 10, 2011 at 2:24 PM, Wendy  wrote:
> Hi all,
>
> I have a huge matrix of TRUE/FALSE table like following, and I want to count
> the number of TRUEs in each row. Instead of looping through each row and do
> length(Z[Z==TRUE]), I am wondering if there is an easier way of doing this.
>
>       [,1]   [,2]    [,3]
> [1,]TRUE FALSE FALSE
> [2,]FALSE TRUE TRUE
>

rowSums(Z)

Peter

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Re: [R] count number of TRUEs in each row

[Tóth Dénes]

Hi Wendy,

try this:
Z <- matrix(runif(1000)>0.5,10,100)
rowSums(Z)

HTH,
  Denes


> Hi all,
>
> I have a huge matrix of TRUE/FALSE table like following, and I want to
> count
> the number of TRUEs in each row. Instead of looping through each row and
> do
> length(Z[Z==TRUE]), I am wondering if there is an easier way of doing
> this.
>
>[,1]   [,2][,3]
> [1,]TRUE FALSE FALSE
> [2,]FALSE TRUE TRUE
>
> Thank you in advance.
> Wendy
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/count-number-of-TRUEs-in-each-row-tp3440486p3440486.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
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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] Question about levels/as.numeric

[Peter Ehlers]

Thibault,

Your questions indicate that you would benefit enormously
from reading 'An Introduction to R'.

A very useful function is str().
Understanding the concept of "factors" is crucial in R.

"Checking" anything with Excel is never much use.

Peter Ehlers

On 2011-04-10 08:47, Thibault Vatter wrote:

Hi,

I am still new to R and this is my first post on this mailing-list.

I have two .csv (each one being a column of real numbers) coming from the
same database (the first one is just longer than the second) and I read them
in R the following way:

returns<- read.csv("test.csv", header = FALSE)
returns2<- read.csv("test2.csv", header = FALSE)

However, the two objects clearly don't seem to be equivalent:


returns[2528:2537,1]

  [1] -0.002206 0.115696  -0.015192 0.008719  -0.004654 -0.010688 0.009453
0.002676  0.001334  -0.011326
7470 Levels: -0.78 -0.85 -0.86 -0.0001 -0.000112 -0.000115
-0.000152 -0.000154 -0.000157 -0.00016 -0.000171 -0.000185 -0.000212
-0.000238 -0.000256 -0.000259 -0.000263 -0.000273 ... C


returns2[1:10,1]

  [1] -0.002206  0.115696 -0.015192  0.008719 -0.004654 -0.010688  0.009453
0.002676  0.001334 -0.011326


as.numeric(returns[2528:2537,1])

  [1]  341 7444 2244 5149  787 1717 5251 4122 3878 1811


as.numeric(returns2[1:10,1])

  [1] -0.002206  0.115696 -0.015192  0.008719 -0.004654 -0.010688  0.009453
0.002676  0.001334 -0.011326

I would like to understand what's happening and how to handle the longer
one. This problem may seem stupid, but I've been trying to figure it out for
a while and nothing seems to work. I checked in excel and both seems to be
completely normal lists of real numbers).

What am I missing here? What are those "levels" and why the as.numeric
doesn't work the same with the longer one?

My final goal  is to extract small parts of those columns the following way:


cbind(returns[which(names == id)[2528:2537],1])

   [,1]
  [1,]  341
  [2,] 7444
  [3,] 2244
  [4,] 5149
  [5,]  787
  [6,] 1717
  [7,] 5251
  [8,] 4122
  [9,] 3878
[10,] 1811

Wich should be equivalent to:


cbind(returns2[which(names == id)[1:10],1])

[,1]
  [1,] -0.002206
  [2,]  0.115696
  [3,] -0.015192
  [4,]  0.008719
  [5,] -0.004654
  [6,] -0.010688
  [7,]  0.009453
  [8,]  0.002676
  [9,]  0.001334
[10,] -0.011326

Thanks a lot,
Thibault

-
*Thibault Vatter*
  EPFL- Master, 1ère année
Laboratory of Statistical Biophysics

Tel: +41 78 820 18 64
  @: thibault.vat...@epfl.ch
Web: http://personnes.epfl.ch/thibault.vatter

*Please consider the environment before printing this email.*

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Re: [R] Adding margin text to lattice graphics

[baptiste auguie]

Yes, very sorry about this -- I had subconsciously ignored the
hypothetical possibility that anyone wouldn't have ggplot2 loaded in
their .Rprofile ;)

Replacing mpg with beaver1 (datasets) should be more reproducible.

That being said, I was told off-list that this is not answering at all
the question, whatever it was.

Best,

baptiste

On 11 April 2011 09:54, Mark Leeds  wrote:
> hi  baptiste: thanks for that but how do I get mpg ? I got an error that R
> couldn't find it. thanks again.
>
>
> On Sat, Apr 9, 2011 at 9:45 PM, baptiste auguie
>  wrote:
>>
>> Hi,
>>
>> You may want to read about ?viewport in the grid package. They allow
>> you to position graphical elements wherever you want on a page, such
>> as lattice plots and text (grid.text). For a high-level interface, you
>> could try the following,
>>
>>
>> library(gridExtra)
>> library(lattice)
>>
>> p1 = xyplot(1~1)
>> p2 = levelplot(volcano)
>> p3 = tableGrob(head(mpg[, 1:3]))
>> p4 = textGrob("some text")
>>
>> grid.arrange(p1, p2, p3, p4, main="global page title",
>>                    sub=p4, left="page y-label")
>>
>> HTH,
>> baptiste
>>
>> On 10 April 2011 13:33, Dennis Fisher  wrote:
>> > Colleagues
>> >
>> > I am learning lattice graphics (R 2.12.2; OS X).  Several days ago, I
>> > inquired about adding margin text to lattice graphics.  Jim Price offered a
>> > useful reply, suggesting that I add:
>> >        page = function(page) grid.text('words', x = 0.5, y = 0.01)
>> > to my call to the function.  The entire function that he suggested was;
>> >        xyplot(1 ~ 1,
>> >         par.settings = list(layout.heights = list(bottom.padding = 10)),
>> >         page = function(page) grid.text('words', x = 0.5, y = 0.01))
>> > That worked initially and I also had success with panel.text.
>> >
>> > However, I am now working with more complicated objects in which more
>> > than one image is displayed on a page.  In this instance, the text added by
>> > the command above appears with each image.  I would like it to appear only
>> > once, scaled across the entire page, not relative to a single panel.
>> >
>> > Is there a different command that accomplishes my goal?  Or a different
>> > implementation of this same command?  Any help would be greatly 
>> > appreciated.
>> > Also, because of my naivete with lattice graphics, I may be asking the
>> > question in entirely the wrong way -- please feel free to redirect me.
>> >
>> > Dennis
>> >
>> > Dennis Fisher MD
>> > P < (The "P Less Than" Company)
>> > Phone: 1-866-PLessThan (1-866-753-7784)
>> > Fax: 1-866-PLessThan (1-866-753-7784)
>> > www.PLessThan.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.
>
>

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Re: [R] Help with basic loop

[Daniel Malter]

The loop is correct, you just need to make sure that your result is computed
and stored as the n-th element that is returned by the loop. Pick up any
manual of R, and looping will be explained there. Also, I would recommend
that you draw a random number for every iteration of the loop. Defining the
random vectors outside the loop make sense to me only if they are the same
length as n.

prob<-numeric(1000)

for (n in 1:1000) { 
task1 <- runif(1, min=0.8, max= 0.9) 
task2 <- runif(1, min=0.75, max= 0.85) 
task3 <- runif(1, min=0.81, max= 0.89)
prob[n]<-task1*task2*task3
} 

If you wanted to store the individual probabilities (task1..3), you would
proceed accordingly by defining them outside the loop and storing the value
in the loop as the n-th element of that vector just like for prob.

HTH,
Daniel

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Re: [R] For->lapply->parallel apply

[Steve Lianoglou]

Hi,

On Sat, Apr 9, 2011 at 5:03 AM, Alaios  wrote:
> Dear all,
> I would like to ask your help understand the subsequent steps for making my 
> program faster.
>
> The following code:
> Gauslist<-array(data=NA,dim=c(dimx,dimy,dimz))
> for (i in c(1:dimz)){
>    print(sprintf('Creating the %d map',i));
>    Gauslist[,,i]<-f <- GaussRF(x=x, y=y, model=model, 
> grid=TRUE,param=c(mean,variance,nugget,scale,Whit.alpha))
> }
>
>
> creates 100 GaussMaps (each map is of 256*256 dim) and stores them in a 
> matrix called Gauslist.
>
> This process takes too long, so I was thinking if you can help me understand 
> what should I do to make it run in parallel (in work there is a system with 
> 16 cores).
>
> There is mclapply (parralel version of lapply) . If I make run my code run 
> with lapply then I will be able to run it with mclapply also (they have same 
> syntax).
> If I understand it correct the sequence for doing that is to understand the 
> following:
>
> for..loop->lapply->mcapply
>
> Can you please help me understand if my for loop can be converted to lapply 
> or not?

Your loop can be converted quite easily.

The lapply function simply takes an object to iterate over as its
first argument (this can be a list of things, a vector of things,
etc.) and a function to apply to each element in the iteration.
`lapply` will build a list of results that your function returns for
each element.

A simple example is to iterate over the words in a character vector
and return how many characters are in each word.

R> words <- c('cat', 'dog's, 'people')
R> sizes <- lapply(words, function(x) nchar(x))
R> sizes
[[1]]
[1] 3

[[2]]
[1] 4

[[3]]
[1] 6

So in your example:

> for (i in c(1:dimz)){
>    print(sprintf('Creating the %d map',i));
>    Gauslist[,,i]<-f <- GaussRF(x=x, y=y, model=model,
>  grid=TRUE,param=c(mean,variance,nugget,scale,Whit.alpha))
> }

Could be something like:

gauslist <- lapply(1:dimz, function(i) {
  GaussRF(x=x, y=y, model=model, ... WHATEVER ELSE)
})

using mclapply would be exactly the same, except replace lapply with mclapply.

Actually, is it correct that you aren't doing anything different in
the iterations of the for loop -- I mean, nothing in your code really
depends on your value for `i`, right?

-- 
Steve Lianoglou
Graduate Student: Computational Systems Biology
 | Memorial Sloan-Kettering Cancer Center
 | Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact

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Re: [R] About Tinn-R

[Steve Lianoglou]

Hi,

On Sun, Apr 10, 2011 at 5:13 PM, Marcos Prunello  wrote:
> Hi everybody!!
>
> I don`t know if it is correct to make a question about R and Tinn-R in this 
> mailing-list, but I need some help with a problem for which I don't find a 
> solution anywhere.

If you go to Tinn-R's "Support" section at sourceforge:

http://sourceforge.net/projects/tinn-r/support

There is a sentence there that says:

Tinn-R says the best way to get help with its software is by using its
forum Open Discussion.

And provides a link to:

http://sourceforge.net/projects/tinn-r/forums/forum/481900

So I guess you'll have better luck with an answer if you post your
question there.

Also, rstudio is meant to be a nice IDE (in development) that you
might want to try if Tinn-R keeps giving you problems:

http://www.rstudio.org/

-steve

>
> I have Tinn-R 2.3.5.2 in my notebook and I always used it with no problems, 
> until a few days ago when this signs appears everytime I start the programme:
>
> "Key Violation.
> C:/Documents and Settings/Marcos/Datos de programa/Tinn-R
> Serious problem reading ini files!
> Tinn-R can not be initiated.
> Please, try to rename (or remove) the folder above and restart the program!"
>
> I don't know what that means, I don`t have such folder in my notebook, and I 
> always used it normally and suddenly it doesn't work anymore.
>
> I unistalled it and installed again several times, and I tried with the 
> newest version of R and TinnR too. I can`t find help in internet, and I also 
> cleaned the registry of the computer.
>
> Thanks in advance!!
>
> Marcos from Argentina
>
>
>
> __
> 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.
>



-- 
Steve Lianoglou
Graduate Student: Computational Systems Biology
 | Memorial Sloan-Kettering Cancer Center
 | Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact

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Re: [R] About Tinn-R

[Kenneth Roy Cabrera Torres]

Dear Marcos:

Did you try this version (version 2.3.7.1)?
http://sourceforge.net/projects/tinn-r/

Does is show the same messages?

What R version are you using?

It seems more a problem with the OS that with R or Tinn-R.

When you uninstall Tinn-R, do you check any file left in the folder:
C:/Documents and Settings/Marcos/Datos de programa/

If there are some, erase them manualy.

If the problem persist, try to rename them with different extension
like ".XXX" and reinstall the program.

Maybe is a problem with a bad sector in the hard disk.

Check the hard disk to see if it is the case.

Hope this helps.

Kenneth

El dom, 10-04-2011 a las 14:13 -0700, Marcos Prunello escribió:
> Hi everybody!!
> 
> I don`t know if it is correct to make a question about R and Tinn-R in this 
> mailing-list, but I need some help with a problem for which I don't find a 
> solution anywhere.
> 
> I have Tinn-R 2.3.5.2 in my notebook and I always used it with no problems, 
> until a few days ago when this signs appears everytime I start the programme:
> 
> "Key Violation.
> C:/Documents and Settings/Marcos/Datos de programa/Tinn-R
> Serious problem reading ini files!
> Tinn-R can not be initiated.
> Please, try to rename (or remove) the folder above and restart the program!"
> 
> I don't know what that means, I don`t have such folder in my notebook, and I 
> always used it normally and suddenly it doesn't work anymore.
> 
> I unistalled it and installed again several times, and I tried with the 
> newest version of R and TinnR too. I can`t find help in internet, and I also 
> cleaned the registry of the computer.
> 
> Thanks in advance!!
> 
> Marcos from Argentina
> 
> 
>   
> __
> 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] rtmvt

[statfan]

I have been using the rtmvt function in the {tmvtnorm} package i'm getting
the warning:

"Acceptance rate is very low and rejection sampling becomes inefficient.
Consider using Gibbs sampling."

but i AM specifying the gibbs algorithm!!:

rtmvt(M, mean=q[,,i,j], sigma=((u[i,j] + nu[i])/(p+nu[i]))*delta[,,i],
df=ceiling(nu[i]+p), lower=c(0,0), algorithm="gibbs")

Any ideas why I am getting this warning and how I can fix it?  

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Re: [R] count number of TRUEs in each row

[Hasan Diwan]

Ms Qiao,

On 10 April 2011 14:24, Wendy  wrote:

> I have a huge matrix of TRUE/FALSE table like following, and I want to
> count
> the number of TRUEs in each row. Instead of looping through each row and do
> length(Z[Z==TRUE]), I am wondering if there is an easier way of doing this.
>
>   [,1]   [,2][,3]
> [1,]TRUE FALSE FALSE
> [2,]FALSE TRUE TRUE
>

Does this help...
Z[Z==T]
?

-- 
Sent from my mobile device
Envoyait de mon telephone mobil

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Re: [R] Question about levels/as.numeric

[Rolf Turner]

On 11/04/11 10:08, Peter Ehlers wrote:



"Checking" anything with Excel is never much use.




Fortune?

cheers,

Rolf Turner

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[R] Specifying the color of points and the plot symbol to be used on a graph

[John Sorkin]

R 2.12.2
Windows 7

I would like to know how to specify the color used in plotting points on a 
scatter plot, generated using plot(x,y) and how I can specify the particular 
symbol used. I would like, for example, to plot some points as blue circles and 
other points a red squares. If someone will tell me how to specify the symbol 
and color used when plotting points, I can overlay graphs [par(new=TRUE] to get 
the two colors I desire.

Thanks,
John



John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)

Confidentiality Statement:
This email message, including any attachments, is for th...{{dropped:6}}

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Re: [R] Specifying the color of points and the plot symbol to be used on a graph

[Peter Alspach]

Tena koe John

If I understand you correctly I think it is easier than that:

plot(x, y, col=yourColourVariate, pch=yourPchVariate)

where yourColourVariate and yourPchVariate are numeric variates of the same 
lengths as x and y (recycling occurs).  Actually the former can be a character 
variate of legal names.

HTH 

Peter Alspach

> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of John Sorkin
> Sent: Monday, 11 April 2011 2:29 p.m.
> To: r-help@r-project.org
> Subject: [R] Specifying the color of points and the plot symbol to be
> used on a graph
> 
> R 2.12.2
> Windows 7
> 
> I would like to know how to specify the color used in plotting points
> on a scatter plot, generated using plot(x,y) and how I can specify the
> particular symbol used. I would like, for example, to plot some points
> as blue circles and other points a red squares. If someone will tell me
> how to specify the symbol and color used when plotting points, I can
> overlay graphs [par(new=TRUE] to get the two colors I desire.
> 
> Thanks,
> John
> 
> 
> 
> John David Sorkin M.D., Ph.D.
> Chief, Biostatistics and Informatics
> University of Maryland School of Medicine Division of Gerontology
> Baltimore VA Medical Center
> 10 North Greene Street
> GRECC (BT/18/GR)
> Baltimore, MD 21201-1524
> (Phone) 410-605-7119
> (Fax) 410-605-7913 (Please call phone number above prior to faxing)
> 
> Confidentiality Statement:
> This email message, including any attachments, is for
> th...{{dropped:6}}
> 
> __
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> guide.html
> and provide commented, minimal, self-contained, reproducible code.

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Re: [R] In svm(), how to connect quantitative prediction result to categorical result?

[Steve Lianoglou]

Hi Yunfei,

On Fri, Apr 8, 2011 at 8:35 PM, Li, Yunfei  wrote:
> Hi,
>
> I am studying using SVM functions of e1071 package to do prediction, and I 
> found during the training data are "factor" type, then svm.predict() can 
> predict data directly by categories; but if response variables are 
> "numerical", the predicted value from svm will be continuous quantitative 
> numbers, then how can I connect these quantitative numbers to categories? 
> (for example:in an example data set, the response variables are numerical and 
> have two categories: 0 and 1, and the predicted value are continuous 
> quantitative numbers from 0 to 1.3, how can I know which of them represent 
> category 0 and which represent 1?)

You have to figure out if you want the svm to do classification or regression.

If I remember correctly, a "vanilla" call to SVM will try to pick one
or the other in a "smart way" by looking at the types (and values) of
your labels (y vector).

You can be more explicit and tell the SVM what you want by specifying
a value for the `type` argument in your original `svm` call.

See ?svm for more info.

I'm not sure if I'm answering your question or not(?). If I didn't
understand what you wanted, perhaps you can rephrase your question, or
maybe explain how my answer is not what you were after ... otherwise
hopefully someone else can provide a better answer.

-steve

-- 
Steve Lianoglou
Graduate Student: Computational Systems Biology
 | Memorial Sloan-Kettering Cancer Center
 | Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact

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[R] predict ordered regresssion

[Joe P King]

Is there a way to get confidence intervals around an ordered regression like
polr() in the MASS package?

 

---

Joe King, M.A.

Ph.D. Student 

University of Washington - Seattle

206-913-2912  

  j...@joepking.com

---

Ad astra per aspera  - "Through hardships to the stars"

 


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Re: [R] Specifying the color of points and the plot symbol to be used on a graph

[John Sorkin]

Thank you! An elegant solution to my problem!
John




John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)>>> Peter 
Alspach  4/10/2011 10:59 PM >>>
Tena koe John

If I understand you correctly I think it is easier than that:

plot(x, y, col=yourColourVariate, pch=yourPchVariate)

where yourColourVariate and yourPchVariate are numeric variates of the same 
lengths as x and y (recycling occurs).  Actually the former can be a character 
variate of legal names.

HTH 

Peter Alspach

> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of John Sorkin
> Sent: Monday, 11 April 2011 2:29 p.m.
> To: r-help@r-project.org 
> Subject: [R] Specifying the color of points and the plot symbol to be
> used on a graph
> 
> R 2.12.2
> Windows 7
> 
> I would like to know how to specify the color used in plotting points
> on a scatter plot, generated using plot(x,y) and how I can specify the
> particular symbol used. I would like, for example, to plot some points
> as blue circles and other points a red squares. If someone will tell me
> how to specify the symbol and color used when plotting points, I can
> overlay graphs [par(new=TRUE] to get the two colors I desire.
> 
> Thanks,
> John
> 
> 
> 
> John David Sorkin M.D., Ph.D.
> Chief, Biostatistics and Informatics
> University of Maryland School of Medicine Division of Gerontology
> Baltimore VA Medical Center
> 10 North Greene Street
> GRECC (BT/18/GR)
> Baltimore, MD 21201-1524
> (Phone) 410-605-7119
> (Fax) 410-605-7913 (Please call phone number above prior to faxing)
> 
> Confidentiality Statement:
> This email message, including any attachments, is for\...{{dropped:31}}

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Re: [R] predict ordered regresssion

[Prof Brian Ripley]

On Sun, 10 Apr 2011, Joe P King wrote:


Is there a way to get confidence intervals around an ordered regression like
polr() in the MASS package?


Please don't both ask the author directly *and* post to a forum.

The answer is the same as the one I sent you directly.

- For what do you want CIs, and why?  The fits/predictions are 
probabilities, and CIs for probabilities are rarely useful.


- You can do this by simulation, and simulate() has a polr method.


---

Joe King, M.A.

Ph.D. Student

University of Washington - Seattle

206-913-2912

 j...@joepking.com

---

Ad astra per aspera  - "Through hardships to the stars"

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--
Brian D. Ripley,  rip...@stats.ox.ac.uk
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University of Oxford, Tel:  +44 1865 272861 (self)
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Re: [R] Question about levels/as.numeric

[Petr Savicky]

On Sun, Apr 10, 2011 at 05:47:59PM +0200, Thibault Vatter wrote:
> Hi,
> 
> I am still new to R and this is my first post on this mailing-list.
> 
> I have two .csv (each one being a column of real numbers) coming from the
> same database (the first one is just longer than the second) and I read them
> in R the following way:
> 
> returns  <- read.csv("test.csv", header = FALSE)
> returns2  <- read.csv("test2.csv", header = FALSE)
> 
> However, the two objects clearly don't seem to be equivalent:
> 
> > returns[2528:2537,1]
>  [1] -0.002206 0.115696  -0.015192 0.008719  -0.004654 -0.010688 0.009453
> 0.002676  0.001334  -0.011326
> 7470 Levels: -0.78 -0.85 -0.86 -0.0001 -0.000112 -0.000115
> -0.000152 -0.000154 -0.000157 -0.00016 -0.000171 -0.000185 -0.000212
> -0.000238 -0.000256 -0.000259 -0.000263 -0.000273 ... C

There is probably a non-numeric row in the data. In order to locate this
row, try the following

  which(is.na(as.numeric(as.character(returns[, 1]

This will show the indices of the rows, which cannot be converted
to numeric type.

Petr Savicky.

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Re: [R] random sampling with levels and with replacement

[taby gathoni]

Andreas,

Thanks alot. I  combined below and other suggestions given on r-help and it 
worked.







--- On Fri, 4/8/11, Andreas Borg  wrote:

From: Andreas Borg 
Subject: Re: [R] random sampling with levels and with replacement
To: tab...@yahoo.com
Cc: "R help" 
Date: Friday, April 8, 2011, 11:13 AM

Hi,

I am not perfectly sure what you want to do, but here is what I would do 
to maintain good/bad ratio in the sample (as Daniel posted, split the 
data and sample from the groups):

df <- data.frame(V1 = 1:400, V2 = c(rep("good",360), rep("bad",40)))
isGood <- which(df$V2=="good")
isBad <- which(df$V2=="bad")
sampleGood <- df[sample(isGood, replace=TRUE),]
sampleBad <- df[sample(isBad, replace=TRUE),]
summary(rbind(sampleGood, sampleBad))

Please include a more specific example with test data (for "final" in 
this case) next time.

Best regards,

Andreas


taby gathoni schrieb:
> Dear all,
> i have a dataset of about 400 records , with a variable that has  two levels 
> 40 bad and 360 good among other variables,how do i come up  with10 random 
> samples that have the composition of as the main sample  but maintaining the 
> 40 bad 360 good with replacement, i recently discovered that my random 
> samples generated dont maintain the ratio. My code is as  :
>
> mysample <- final[sample(1:nrow(final), 400,replace=TRUE),] 
>
> does not give me the ratio of 40 bad and 360 good can anyone give me some 
> pointers please?
>
>
>
> Thanks,
> Taby
>
>
>
>
>     [[alternative HTML version deleted]]
>
>   
> 
>
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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>   


-- 
Andreas Borg
Medizinische Informatik

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Re: [R] help question

[Petr PIKAL]

Hi

r-help-boun...@r-project.org napsal dne 08.04.2011 18:24:37:

> On 08/04/2011 9:20 AM, DEBERGH Patrick wrote:
> > hello
> >
> > I am at the very beginneing of using the R program
> >
> > I just don't understand how one can save a programfile
> >
> > For exemple, if I type in R  23+456 and want to save this file under a 

> ceratin name to reload it later, i just don't get the way to do it
> 
> R is not Microsoft Word.  When you type 23+456, you're asking R to do 
> something for you, you're not writing a document.  So you can save a 
> record of what you asked for, but you can't reload it later in R to 
> continue on.

If you want such functionality you shall look to some editor which is 
capable to cooperate with R like TINN-R or maybe ESS.

Regards
Petr


> 
> Duncan Murdoch
> > I can save it with the save function;I acheive to see that I have a 
file 
> with the name, but no way to understand how to reload or re-suse it
> >
> > in the official r manuel this is very badly described for first 
users..
> >
> > thanks helping me saving R programs
> >
> > Best regards
> >
> >
> >
> > Patrick Debergh
> > VP Product Development
> >
> > Colibrys (Suisse) Ltd - Maladière 83 - CH-2000 Neuchâtel, Switzerland
> > Phone: +41 32 720  5696 / Fax: +41 32 720 57 84
> > mailto:patrick.debe...@colibrys.com
> > http://www.colibrys.com
> > _
> > This message may contain confidential and proprietary material for the 
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contact
> > the sender and delete all copies
> > _
> >
> >
> >
> >
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> >
> >
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> > and provide commented, minimal, self-contained, reproducible code.
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Re: [R] Question about levels/as.numeric

[Thibault Vatter]

Thanks Kent and Petr, the problem was indeed the "C"/missing value that I
had to convert!

Thanks Peter too, the "factor" explanation will also be quite usefull for
further work.

Best regards,
Thibault

On 11 April 2011 03:48, Rolf Turner  wrote:

> On 11/04/11 10:08, Peter Ehlers wrote:
>
> 
>
>  "Checking" anything with Excel is never much use.
>>
>
> 
>
> Fortune?
>
>cheers,
>
>Rolf Turner
>
>
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>

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