date:20130117

I was unable to run your code; 'fac' is missing and npk$fac in teh 
interaction.plot returns NA.

 data(npk, package=MASS)
 fit - by(npk, npk$block, function(bydata) fitted.values(lm(yield ~ N,
 data=bydata)))
 fit - unlist(fit)
 interaction.plot(npk$N, npk$block, fit, xlab=N, 
 ylab=yield) # fake factor, numeric fac - 
 c(rep(1,12),rep(2,12)) # plots everything in black 
 interaction.plot(npk$N, npk$block, fit, xlab=N, 
 ylab=yield,col=fac)
 cbind(npk$block,fac)
 # plots everything transparent
 fac - as.factor(fac)
 interaction.plot(npk$N, npk$block, fit, xlab=N,
 ylab=yield,col=as.numeric(npk$fac))

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[R] Making an aggregated factor variable from an existing one

2013-01-17 Thread Antti Simola


Hi,

This is quite simple data manipulation task and I need help for it. I 
want to make new factor variable that is an aggregation of an existing 
factor.


This works as I intended:

X[Y == original label] -  new label

How to make following work then (to make coding more convenient):

original.labels - c(ol.1,ol.2,ol.3)

X[Y == original.labels] - new label

Cheers,

Antti

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Re: [R] Making an aggregated factor variable from an existing one

2013-01-17 Thread Gerrit Eichner


Hi, Antti,

you should look at

?levels

(and particular its Examples section) to find out how to use

levels( X) - c( new1, ..., newk)

to achieve what you want.

 Regards  --  Gerrit


On Thu, 17 Jan 2013, Antti Simola wrote:


Hi,

This is quite simple data manipulation task and I need help for it. I want to 
make new factor variable that is an aggregation of an existing factor.


This works as I intended:

X[Y == original label] -  new label

How to make following work then (to make coding more convenient):

original.labels - c(ol.1,ol.2,ol.3)

X[Y == original.labels] - new label

Cheers,

Antti

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Re: [R] How to change R file in stats?

 

 -Original Message-
 I have changed some code in R file inside the stats package 
 (dendrogram.R). 
That was brave. Others have already commented on its wisdom...

 Now I wan to test and run the stats package 
 with the new updated code, what should I do in detail?

1. Read and follow the instructions in 'Writing R extensions' to build an 
installable .tar.gz or (for windows) .zip binary of the stats package. 
2. Back up your copy of R.
3. Install the newly built stats package in the normal way.
4. Test.
5. Revert to the backup.

Safer still to create a different installation of  R locally and apply the new 
package install to that.

S

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Re: [R] How to change R file in stats?


On 13-01-17 8:31 AM, S Ellison wrote:




-Original Message-
I have changed some code in R file inside the stats package
(dendrogram.R).

That was brave. Others have already commented on its wisdom...


Now I wan to test and run the stats package
with the new updated code, what should I do in detail?


1. Read and follow the instructions in 'Writing R extensions' to build an 
installable .tar.gz or (for windows) .zip binary of the stats package.
2. Back up your copy of R.
3. Install the newly built stats package in the normal way.


Stats is a base package, so that won't work.  Base packages are built 
and installed differently from other packages.


He could modify the sources and build a custom copy of R for himself, 
but that seems kind of extreme.


A better way would be to copy the source for the function into his own 
package.  If he intends to release it, he must retain copyright notices 
on it and release his own package with a GPL-compatible license.


Duncan Murdoch



4. Test.
5. Revert to the backup.

Safer still to create a different installation of  R locally and apply the new 
package install to that.

S

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Re: [R] How to change R file in stats?

 Stats is a base package, so that won't work.  Base packages 
 are built and installed differently from other packages.
Dang! Of course it is. Scratch previous response. 

The irony is that my first thought was indeed Read 'R Installation and 
configuration' with special attention to 'Building from source'. As you say, 
rather extreme ... and having recently done that myself I recalled the 'finish 
PhD thesis' bit of those instructions.)

Agree entirely, though, that modifying base package source is not at all a good 
way for an ordinary user to get a bettter dendrogram.  Really something one 
would only consider if intending to suggest a patch to R-core.


S

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Re: [R] Issue while installing Hmisc package

2013-01-17 Thread R. Michael Weylandt

On Thu, Jan 17, 2013 at 12:57 PM, vaseem shaikh vsma...@gmail.com wrote:
 Hello Michael,

 I have tried with different CRAN but still i am getting the same error.



Most likely you've got something blocking your network. Can you
download the source / binary from the CRAN pages (using your
web-browser) and then install from the command line? Hmisc doesn't
require any special tools to build so that shouldn't be too hard.


 But still, i am also trying to install the package by locally giving
 absolute path with repos= Null and type = Source, will CRAN have any role
 to play here???

You can't escape CRAN entirely in this case (unless you know where
the source is, which I don't) -- it's the distribution network.

MW



 BR-
 Vaseem Shaikh
 On Thu, Jan 17, 2013 at 3:03 PM, Michael Weylandt
 michael.weyla...@gmail.com wrote:

 Try a different CRAN mirror.

 MW

 On Jan 17, 2013, at 7:42 AM, vaseem shaikh vsma...@gmail.com wrote:

  Hi,
 
  Please find the snap shot attached for the error reported wile
  installing
  Hmisc Package.
 
  Is there any thing you can help me with.
 
  Currently using R i386 2.15.2 Version of R on windows 7 platform.
 
  Ragards
  Vaseem Shaikh
  Instalation_Issue_HMISC.png
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[R] how to use ...

2013-01-17 Thread Ivan Calandra


Dear users,

I'm trying to learn how to use the 

I have written a function (simplified here) that uses doBy::summaryBy():
# 'dat' is a data.frame from which the aggregation is computed
# 'vec_cat' is a integer vector defining which columns of the data.frame 
should be use on the right side of the formula

# 'stat_fun' is the function that will be run to aggregate
stat.group - function(dat, vec_cat, stat_fun){
require(doBy)
df - 
summaryBy(as.formula(paste0(.~,paste0(names(dat)[vec_cat],collapse=+))), 
data=dat, FUN=stat_fun)

return(df)
}

Example, works fine:
my_data - structure(list(cat = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L,
2L, 2L, 2L), .Label = c(A, B), class = factor), varnum = 
c(-0.754816565434373,
-1.94101630973709, -0.102461836059522, -0.519952759645808, 
-1.73772800855664,

-1.13939178585609, 0.522356715260142, -0.701428514907824, 1.45197576541159,
0.0844567413828095)), .Names = c(cat, varnum), row.names = c(NA,
-10L), class = data.frame)
stat.group(dat=my_data, vec_cat=1, stat_fun=mean)


Now summaryBy() has an ... argument and I would like to use it.
For example, I would like to be able to add the trim argument to my call 
like this:

stat.group(dat=my_data, vec_cat=1, stat_fun=mean, trim=0.2)


I know I can do it using this ... but I have no idea how to do it. 
I've tried to search for it, but a search with ... doesn't yield 
interesting results!



Thank you in advance for your help!
Ivan

--
Ivan CALANDRA
Université de Bourgogne
UMR CNRS/uB 6282 Biogéosciences
6 Boulevard Gabriel
21000 Dijon, FRANCE
+33(0)3.80.39.63.06
ivan.calan...@u-bourgogne.fr
http://biogeosciences.u-bourgogne.fr/calandra

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[R] coxph with smooth survival

2013-01-17 Thread Bond, Stephen

Hello users,

I would like to obtain a survival curve from a Cox model that is smooth and 
does not have zero differences due to no events for those particular days.
I have:
 sum((diff(surv))==0)
[1] 18

So you can see 18 days where the survival curve did not drop due to no events. 

Is there a way to ask survfit to fit a nice spline for the survival??

Note: I tried survreg and it did not work, but maybe I did not do it properly??

Thank you very much.

Stephen B

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Re: [R] how to use ...

2013-01-17 Thread R. Michael Weylandt

On Thu, Jan 17, 2013 at 2:36 PM, Ivan Calandra
ivan.calan...@u-bourgogne.fr wrote:
 Dear users,

 I'm trying to learn how to use the 

 I have written a function (simplified here) that uses doBy::summaryBy():
 # 'dat' is a data.frame from which the aggregation is computed
 # 'vec_cat' is a integer vector defining which columns of the data.frame
 should be use on the right side of the formula
 # 'stat_fun' is the function that will be run to aggregate
 stat.group - function(dat, vec_cat, stat_fun){
 require(doBy)
 df -
 summaryBy(as.formula(paste0(.~,paste0(names(dat)[vec_cat],collapse=+))),
 data=dat, FUN=stat_fun)
 return(df)
 }
 [SNIP EXAMPLE -- THANK YOU FOR IT]

 Now summaryBy() has an ... argument and I would like to use it.
 For example, I would like to be able to add the trim argument to my call
 like this:
 stat.group(dat=my_data, vec_cat=1, stat_fun=mean, trim=0.2)



Thanks for the great working examples!

It's actually not too hard here --  just pass ... down as if it were
an argument and let summaryBy() do the hard work of actually handling
the dots:

stat.group - function(dat, vec_cat, stat_fun, ...){
require(doBy)
df - 
summaryBy(as.formula(paste0(.~,paste0(names(dat)[vec_cat],collapse=+))),
data=dat, FUN=stat_fun, ...)
return(df)
}

Also, note that as a matter of style, you can actually clean this up a
little bit: R follows the trend of many functional languages in
automatically returning the value of the last expression evaluated:

stat.group - function(dat, vec_cat, stat_fun, ...){
require(doBy)
summaryBy(as.formula(paste0(.~,paste0(names(dat)[vec_cat],collapse=+))),
data=dat, FUN=stat_fun, ...)
}

Cheers,
Michael

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Re: [R] how to use ...

2013-01-17 Thread Berend Hasselman


On 17-01-2013, at 15:36, Ivan Calandra ivan.calan...@u-bourgogne.fr wrote:

 Dear users,
 
 I'm trying to learn how to use the 
 
 I have written a function (simplified here) that uses doBy::summaryBy():
 # 'dat' is a data.frame from which the aggregation is computed
 # 'vec_cat' is a integer vector defining which columns of the data.frame 
 should be use on the right side of the formula
 # 'stat_fun' is the function that will be run to aggregate
 stat.group - function(dat, vec_cat, stat_fun){
require(doBy)
df - 
 summaryBy(as.formula(paste0(.~,paste0(names(dat)[vec_cat],collapse=+))), 
 data=dat, FUN=stat_fun)
return(df)
 }
 
 Example, works fine:
 my_data - structure(list(cat = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L,
 2L, 2L, 2L), .Label = c(A, B), class = factor), varnum = 
 c(-0.754816565434373,
 -1.94101630973709, -0.102461836059522, -0.519952759645808, -1.73772800855664,
 -1.13939178585609, 0.522356715260142, -0.701428514907824, 1.45197576541159,
 0.0844567413828095)), .Names = c(cat, varnum), row.names = c(NA,
 -10L), class = data.frame)
 stat.group(dat=my_data, vec_cat=1, stat_fun=mean)
 
 
 Now summaryBy() has an ... argument and I would like to use it.
 For example, I would like to be able to add the trim argument to my call like 
 this:
 stat.group(dat=my_data, vec_cat=1, stat_fun=mean, trim=0.2)
 
 
 I know I can do it using this ... but I have no idea how to do it. I've 
 tried to search for it, but a search with ... doesn't yield interesting 
 results!
 


This seems to work:

stat.group - function(dat, vec_cat, stat_fun,...){
   require(doBy) 
   df - 
summaryBy(as.formula(paste0(.~,paste0(names(dat)[vec_cat],collapse=+))), 
data=dat, FUN=stat_fun,...)
   return(df)
}

and

stat.group(dat=my_data, vec_cat=1, stat_fun=mean, trim=0.2)

as the example for sumfun in  the help for summaryBy shows.

Berend

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Re: [R] R and Windows 8

2013-01-17 Thread Gabor Grothendieck

On Wed, Jan 16, 2013 at 12:39 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
 On Wed, Jan 16, 2013 at 11:06 AM, Claire Oswald
 claire.j.osw...@gmail.com wrote:
 Hello:

 I'd like to know if R will run under Windows 8?


 I am running R on Windows 8 with no apparent problems.

Actually I found one tiny problem.   If you ask which Windows version
you are using when running under Windows 8 it tells you that you are
using Windows 7 :

 R.version.string
[1] R version 2.15.2 Patched (2013-01-13 r61650)
 win.version() ## Should be Windows 8
[1] Windows 7 x64 (build 9200)




--
Statistics  Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com

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Re: [R] coxph with smooth survival

2013-01-17 Thread Bond, Stephen

I also tried fitting a spline to the resulting survival curve and the result 
was horrible.   
maybe spline won't work or knots need special handling. 

overall, I must have the final point of the smooth survival to be same as the 
final point of the raw Cox survival and no flat days, the drops should be 
spread around. 

Thanks again everybody.

Stephen 

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Re: [R] equivalent code that doesn't return same results?

2013-01-17 Thread cwladis

William Dunlap wrote
 eval(parse(text=paste(dataset,IVcat[k],sep=$)))-relevel(eval(parse(text=paste(dataset,IVcat[k],sep=$))),ref=online)
This code returns the following error:
Error in eval(parse(text = paste(dataset, IVcat[k], sep = $))) -
relevel(eval(parse(text = paste(dataset,  : 
  target of assignment expands to non-language object
 
 Replace
 eval(parse(text=paste(dataset, IVcat[k], sep=$)))
 with
 dataset[[ IVcat[k] ]]
 everywhere.

Thanks so much for the response, Bill - this was very helpful.  I feel silly
for not thinking of this myself - I had originally started with this
approach, but abandoned it because for some inexplicable reason it seemed to
call the wrong dataset column (which it did again when I substituted your
code for mine).  However, I figured out a modification that did work:
apparently I needed to convert the output of IVcat[k] to a string, like
this, and then the code ran fine:

dataset[[ toString(IVcat[k]) ]]

Thanks again!



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[R] importing a SAS syntax-files (value labels)

2013-01-17 Thread David Studer

Hello everybody,

I imported an SAS data-file into R. open.sas7bdat() did not work,
so I had to convert it to csv first. Now I would like to recode the
value values into factors. Unfortunately I only have a SAS
syntax file, having this form:

proc format;
value $resstatus
'B'= 'Jahresaufenthalter'
'C' = 'Niedergelassene'
'I' = 'Dipl./int. Funkt. und AngehÃ¶rige'
;
run;

Does anyone know if there is a possibility to change the numeric
value labels into factor levels acording to the SAS syntax-file?

I cannot do this manually as there are several hundred labels...

Thank you!

[[alternative HTML version deleted]]

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Re: [R] how to use ...

2013-01-17 Thread Ivan Calandra

Ok, it is that simple... Actually I had tried it but messed up so that 
it didn't work.

Do you know where I can find some documentation about it?

Regarding return(), I know that it's not necessary, but when the 
function gets more complicated, I like to have it because it becomes 
clearer to me.


Thanks all!
Ivan

--
Ivan CALANDRA
Université de Bourgogne
UMR CNRS/uB 6282 Biogéosciences
6 Boulevard Gabriel
21000 Dijon, FRANCE
+33(0)3.80.39.63.06
ivan.calan...@u-bourgogne.fr
http://biogeosciences.u-bourgogne.fr/calandra

Le 17/01/13 15:55, R. Michael Weylandt a écrit :

On Thu, Jan 17, 2013 at 2:36 PM, Ivan Calandra
ivan.calan...@u-bourgogne.fr wrote:

Dear users,

I'm trying to learn how to use the 

I have written a function (simplified here) that uses doBy::summaryBy():
# 'dat' is a data.frame from which the aggregation is computed
# 'vec_cat' is a integer vector defining which columns of the data.frame
should be use on the right side of the formula
# 'stat_fun' is the function that will be run to aggregate
stat.group - function(dat, vec_cat, stat_fun){
 require(doBy)
 df -
summaryBy(as.formula(paste0(.~,paste0(names(dat)[vec_cat],collapse=+))),
data=dat, FUN=stat_fun)
 return(df)
}
[SNIP EXAMPLE -- THANK YOU FOR IT]

Now summaryBy() has an ... argument and I would like to use it.
For example, I would like to be able to add the trim argument to my call
like this:
stat.group(dat=my_data, vec_cat=1, stat_fun=mean, trim=0.2)



Thanks for the great working examples!

It's actually not too hard here --  just pass ... down as if it were
an argument and let summaryBy() do the hard work of actually handling
the dots:

stat.group - function(dat, vec_cat, stat_fun, ...){
 require(doBy)
 df - 
summaryBy(as.formula(paste0(.~,paste0(names(dat)[vec_cat],collapse=+))),
data=dat, FUN=stat_fun, ...)
 return(df)
}

Also, note that as a matter of style, you can actually clean this up a
little bit: R follows the trend of many functional languages in
automatically returning the value of the last expression evaluated:

stat.group - function(dat, vec_cat, stat_fun, ...){
 require(doBy)
 
summaryBy(as.formula(paste0(.~,paste0(names(dat)[vec_cat],collapse=+))),
data=dat, FUN=stat_fun, ...)
}

Cheers,
Michael



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Re: [R] R and Windows 8


On 17/01/2013 9:59 AM, Gabor Grothendieck wrote:

On Wed, Jan 16, 2013 at 12:39 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
 On Wed, Jan 16, 2013 at 11:06 AM, Claire Oswald
 claire.j.osw...@gmail.com wrote:
 Hello:

 I'd like to know if R will run under Windows 8?


 I am running R on Windows 8 with no apparent problems.

Actually I found one tiny problem.   If you ask which Windows version
you are using when running under Windows 8 it tells you that you are
using Windows 7 :

 R.version.string
[1] R version 2.15.2 Patched (2013-01-13 r61650)
 win.version() ## Should be Windows 8
[1] Windows 7 x64 (build 9200)


I'll fix this if someone else doesn't get there first.

Duncan Murdoch

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Re: [R] how to use ...

Well..

On Thu, Jan 17, 2013 at 7:42 AM, Ivan Calandra
ivan.calan...@u-bourgogne.fr wrote:
 Ok, it is that simple... Actually I had tried it but messed up so that it
 didn't work.
 Do you know where I can find some documentation about it?

The R language definition manual would be the logical place to look,
no? And sure enough, it's there!

The Introduction to R tutorial also contains info in the writing
functions section.

A very good, though now dated, exposition can be found in VR's S
Programming book. Although a little care is needed due to age, it's
still my favorite.

-- Bert


 Regarding return(), I know that it's not necessary, but when the function
 gets more complicated, I like to have it because it becomes clearer to me.

 Thanks all!
 Ivan

 --
 Ivan CALANDRA
 Université de Bourgogne
 UMR CNRS/uB 6282 Biogéosciences
 6 Boulevard Gabriel
 21000 Dijon, FRANCE
 +33(0)3.80.39.63.06
 ivan.calan...@u-bourgogne.fr
 http://biogeosciences.u-bourgogne.fr/calandra

 Le 17/01/13 15:55, R. Michael Weylandt a écrit :

 On Thu, Jan 17, 2013 at 2:36 PM, Ivan Calandra
 ivan.calan...@u-bourgogne.fr wrote:

 Dear users,

 I'm trying to learn how to use the 

 I have written a function (simplified here) that uses doBy::summaryBy():
 # 'dat' is a data.frame from which the aggregation is computed
 # 'vec_cat' is a integer vector defining which columns of the data.frame
 should be use on the right side of the formula
 # 'stat_fun' is the function that will be run to aggregate
 stat.group - function(dat, vec_cat, stat_fun){
  require(doBy)
  df -

 summaryBy(as.formula(paste0(.~,paste0(names(dat)[vec_cat],collapse=+))),
 data=dat, FUN=stat_fun)
  return(df)
 }
 [SNIP EXAMPLE -- THANK YOU FOR IT]

 Now summaryBy() has an ... argument and I would like to use it.
 For example, I would like to be able to add the trim argument to my call
 like this:
 stat.group(dat=my_data, vec_cat=1, stat_fun=mean, trim=0.2)


 Thanks for the great working examples!

 It's actually not too hard here --  just pass ... down as if it were
 an argument and let summaryBy() do the hard work of actually handling
 the dots:

 stat.group - function(dat, vec_cat, stat_fun, ...){
  require(doBy)
  df -
 summaryBy(as.formula(paste0(.~,paste0(names(dat)[vec_cat],collapse=+))),
 data=dat, FUN=stat_fun, ...)
  return(df)
 }

 Also, note that as a matter of style, you can actually clean this up a
 little bit: R follows the trend of many functional languages in
 automatically returning the value of the last expression evaluated:

 stat.group - function(dat, vec_cat, stat_fun, ...){
  require(doBy)

 summaryBy(as.formula(paste0(.~,paste0(names(dat)[vec_cat],collapse=+))),
 data=dat, FUN=stat_fun, ...)
 }

 Cheers,
 Michael


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-- 

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
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[R] help with error: DV converted to a factor

2013-01-17 Thread cwladis

I've spent several days compiling the following code (I apologize in advance
- this code is very inelegant, and I'm sure could be written much more
efficiently, but I've stuck with whatever method I could get to work -
sometimes the more efficient code I just couldn't get to work without an
error, because of my R inexperience).  

My main motivation for writing the code is that I'd like to be able to
assess interaction effects, and so I want to be able to rotate through the
reference values so that I can assess different interaction effects than
simply the default ones reported by the single set of reference values given
in a single analysis.  The code below generates a lot more than that, but
since I am such an R newbie (and therefore struggling to learn so many
different things at once) I thought that if I could just generate one bit
matrix with all the pairwise comparisons, I could then pick out the stuff I
actually want later (probably just manually at first, but later hopefully by
automating my code to return a matrix with just the comparisons I want).

Here is the code:
DV-factors[,1]
IV-factors[,2]
int-factors[,3]
IV-IV[!is.na(IV)]
int-int[!is.na(int)]
#Limit our modification of reference values to categorical variables only by
redefining IV vector as IVcat
IVcat-IV
for(p in 1:length(IV)){

if((class(dataset[[toString(IV[p])]])==character)|(class(dataset[[toString(IV[p])]])==factor)){
IVcat[p]-IV[p]
}
else{IVcat[p]-}
}
IVcat-IVcat[!is.na(IVcat)]
#Create vectors (IVvalslist[n] for nth factor) for each IVcat containing
each possible value for that IV
IVvalslist-vector('list', length(IVcat))
for(i in 1:length(IVcat)){

assign(paste(IVvalues,i,sep=),unique(dataset[[toString(IVcat[i])]])[!is.na(unique(dataset[[toString(IVcat[i])]]))])
IVvalslist[i]-list(get(paste(IVvalues,i,sep=)))
}
#Create a data frame (refM) with every combination of values for each IVcat
refM-expand.grid(IVvalslist)
#Loop through all possible reference values, and then run the model, 
#and then compile the model summary output into a single matrix
#Go through each row of the matrix of possible reference value 
combinations 
for(j in 1:nrow(refM)){
#Go through each reference value for each factor in that row, 
and assign
#that reference value for that factor
for(k in 1:length(IVcat)){
dataset[[ toString(IVcat[k]) ]]-relevel(dataset[[ 
toString(IVcat[k])
]],ref=toString(refM[j,k]))
}
#Run model with new reference values from row j

model-paste(paste(DV[1],~1,sep=),paste(IV,collapse=+),paste(int,collapse=+),sep=+)

modeloutput-glm(model,family=binomial(logit),data=dataset)
#assigning all output from every possible combo of ref 
values to a single
matrix named coeffM
if(j==1){coeffM-coef(summary(modeloutput))}

if(j1){coeffM-rbind(coeffM,coef(summary(modeloutput)))}
}


*I tested each step of the code individually, and each individual step runs
fine.  However, when I run the whole set of code at once, I get the
following error message after the last line of R input above:

Warning in model.matrix.default(mt, mf, contrasts) :
  variable 'retention' converted to a factor
Error in weights * y : non-numeric argument to binary operator

And if I ask for R to return coeffM or modeloutput, it tells me that no such
objects exist.  *

The object factors is as follows:
 factors
   var1   var2 var3
1 retention method method*level
2  NA  level NA
3  NAgpa NA

And I can't give the actual dataset (named dataset above) here for human
subject reasons, but here is a made-up sample of what it looks like:

id  instructor   method success retention  levelcareer
1 1001  NAME1   online 1 1   LL 
  
career
2 1002  NAME2   face-to-face   1 1  UL  
lib. arts

  STEM required ethnicity   
   
gender age finaid  gpa
1 non-STEM  electiveAsian or Pacific Islander  M
 
28   1.97
2 non-STEM  maj. req.  Asian or Pacific Islander  F 

21   none3.01

  experiencecreditsyrsenrolled  course
1 no online exp.  22   1   NAME1_MAT100
2 no online exp.  33   2   NAME2_ENG100

If anyone can help me figure out what is going wrong here, I'd be incredibly
grateful!  I've tried searching repeatedly for this error, and the other
instances that I've found

Re: [R] Colors in interaction plots

Iuri:

Your code as emailed reads:

##
data(npk, package=MASS)
fit - by(npk, npk$block, function(bydata) fitted.values(lm(yield ~ N, 
data=bydata)))
fit - unlist(fit)
interaction.plot(npk$N, npk$block, fit, xlab=N, ylab=yield) # fake factor, 
numeric fac - c(rep(1,12),rep(2,12)) # plots everything in black 
interaction.plot(npk$N, npk$block, fit, xlab=N, ylab=yield,col=fac)
cbind(npk$block,fac)
# plots everything black as well
fac - as.factor(fac)
interaction.plot(npk$N, npk$block, fit, xlab=N,
ylab=yield,col=as.numeric(fac))
#

Close inspection shows that the definition of fac is buried in a comment on the 
fifth line of the code, so it was not running.

Having disentangled that, did you inspect fac?

fac is a length 24 vector with the first 12 values equal to 1 - corresponding 
to black. You are plotting only six lines  all of which plot correctly as 
black when I run the code because the first 6 values in the colour vector you 
supply are all black.

If you want to plot 6 different colour lines, try, for example, col=1:6.

Hope that helps you clear things up!

S Ellison

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Re: [R] importing a SAS syntax-files (value labels)

2013-01-17 Thread Anthony Damico

it is easy to parse through yourself.  if you don't care about the labels
and just want to import fixed-width file data, you can use the SAScii
package.  if you do, run this code to get 'em  :)


# load the stringr package to trim strings quickly
library(stringr)

# example proc format block--

# store the proc format text.. you can use ?readLines to import your file
instead
proc.format - proc format;
value $resstatus
'B'= 'Jahresaufenthalter'
'C' = 'Niedergelassene'
'I' = 'Dipl./int. Funkt. und AngehÃ¶rige'
;
run;


# separate all strings by return characters
x - strsplit( proc.format , \n )[[1]]

# then break the strings apart by the equals signs
y - strsplit( x , = )


# throw out the lines you don't need by hand
# only the third through fifth rows contain anything of value.
z - y[ 3:5 ]


# extract the first and second elements of the list `z`
value - sapply( z , [[ , 1 )
label - sapply( z , [[ , 2 )


# trim outside whitespace
value - str_trim( value )
label - str_trim( label )

# remove first and last ' character
value - substr( value , 2 , nchar( value ) - 1 )
label - substr( label , 2 , nchar( label ) - 1 )

# print the results to the screen
value
label





On Thu, Jan 17, 2013 at 10:39 AM, David Studer stude...@gmail.com wrote:

 Hello everybody,

 I imported an SAS data-file into R. open.sas7bdat() did not work,
 so I had to convert it to csv first. Now I would like to recode the
 value values into factors. Unfortunately I only have a SAS
 syntax file, having this form:

 proc format;
 value $resstatus
 'B'= 'Jahresaufenthalter'
 'C' = 'Niedergelassene'
 'I' = 'Dipl./int. Funkt. und Angehörige'
 ;
 run;

 Does anyone know if there is a possibility to change the numeric
 value labels into factor levels acording to the SAS syntax-file?

 I cannot do this manually as there are several hundred labels...

 Thank you!

 [[alternative HTML version deleted]]


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Re: [R] create block diagonal with each rows

2013-01-17 Thread Martin Maechler

 arun  smartpink...@yahoo.com
 on Wed, 16 Jan 2013 19:20:46 -0800 writes:

 Hi,
 May be this helps:
 library(Matrix)
 res1-lapply(split(x,1:nrow(x)),function(y) 
sparseMatrix(i=rep(1:4,each=5),j=1:(4*5),x=y))
  do.call(rbind,lapply(seq_along(res1),function(i) res1[[i]][i,]))
 # [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] 
[,13] [,14]
 #[1,]    1    5    9   13   17    0    0    0    0 0 0 0 
0 0
 #[2,]    0    0    0    0    0    2    6   10   14    18 0 0 
0 0
 #[3,]    0    0    0    0    0    0    0    0    0 0 3 7    
11    15
 #[4,]    0    0    0    0    0    0    0    0    0 0 0 0 
0 0
 # [,15] [,16] [,17] [,18] [,19] [,20]
 #[1,] 0 0 0 0 0 0
 #[2,] 0 0 0 0 0 0
 #[3,]    19 0 0 0 0 0
 #[4,] 0 4 8    12    16    20

Thank you for thinking of  Matrix  (the package, not the movie)
here.

If you do,

  x - matrix(1:20, 4,5)
  require(Matrix)

the following is  a tiny bit  nicer, using the  bdiag()
function which has been written to create block diagonal matrices.

 t(bdiag(split(x,1:nrow(x

4 x 20 sparse Matrix of class dgCMatrix
   
[1,] 1 5 9 13 17 . .  .  .  . . .  .  .  . . .  .  .  .
[2,] . . .  .  . 2 6 10 14 18 . .  .  .  . . .  .  .  .
[3,] . . .  .  . . .  .  .  . 3 7 11 15 19 . .  .  .  .
[4,] . . .  .  . . .  .  .  . . .  .  .  . 4 8 12 16 20
 

and if you don't want a sparse matrix for some reason,
(and think twice: it may be more efficient to keep it !)
you wrap the result with a 

as.matrix(  result  )

---
Martin Maechler, ETH Zurich





 - Original Message -
 From: Kathryn Lord kathryn.lord2...@gmail.com
 To: r-help r-help@r-project.org
 Cc: 
 Sent: Wednesday, January 16, 2013 9:11 PM
 Subject: [R] create block diagonal with each rows

 Dear R users,

 I'd like to create a block diagonal matrix with each rows in a matrix.

 Here is a simple example. (In fact, the matrix is big)


 x - matrix(1:20, 4,5)

 x
      [,1] [,2] [,3] [,4] [,5]
 [1,]    1    5    9   13   17
 [2,]    2    6   10   14   18
 [3,]    3    7   11   15   19
 [4,]    4    8   12   16   20


 With each rows in matrix x, I'd like to make the matrix below.


      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9][,10]
 [,11][,12][,13][,14][,15][,16][,17][,18][,19][,20]
 [1,]    1    5    9   13   17   0    0    0    0    0     0    0    0
 0    0     0    0    0    0    0
 [2,]    0    0    0    0    0    2    6   10   14   18   0    0    0
 0    0     0    0    0    0    0
 [3,]    0    0    0    0    0    0    0    0    0    0     3    7   11
 15   19   0    0    0    0    0
 [4,]    0    0    0    0    0    0    0    0    0    0     0    0    0
 0    0     4    8   12   16   20


 Any suggestion will be greatly appreciated.

 Best,

 Kathryn Lord

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[R] How do I get Garamond font in R?

2013-01-17 Thread Keith Weintraub

Folks,
  I run R on a early 2009 MacBook Pro running Mountain Lion.

I have a bunch of fonts in my user Library one of which is Garamond.

I have tried the ttf_import function to no avail. I played with this for a 
couple of hours at least and I have gotten nowhere.

Here is a bit of one of my sessions (note that I used file.choose to get down 
to the Font level)

 require(extrafont)
Loading required package: extrafont
Registering fonts with R
Warning messages:
1: In loadfonts(pdf, quiet = TRUE) :
  More than one version of regular/bold/italic found for Apple Braille. 
Skipping setup for this font.
2: In loadfonts(pdf, quiet = TRUE) :
  No regular (non-bold, non-italic) version of Brush Script MT. Skipping setup 
for this font.
3: In loadfonts(postscript, quiet = TRUE) :
  More than one version of regular/bold/italic found for Apple Braille. 
Skipping setup for this font.
4: In loadfonts(postscript, quiet = TRUE) :
  No regular (non-bold, non-italic) version of Brush Script MT. Skipping setup 
for this font.

 fonts()
 [1] .Keyboard Andale Mono   Apple Braille 
AppleMyungjo 
 [5] Arial Black   Arial Arial Narrow  
Arial Rounded MT Bold
 [9] Arial Unicode MS  BatangBrush Script MT   
Calibri  
[13] Cambria   Candara   Comic Sans MS 
Consolas 
[17] ConstantiaCorbelCourier New   
Georgia  
[21] Gulim Humor SansImpact
Jazz 
[25] JazzCord  JazzPerc  JazzText  
JazzTextExtended 
[29] Khmer Sangam MN   Krini Krinitsky 
Lao Sangam MN
[33] Microsoft Sans Serif  MS Gothic MS Mincho 
MS PGothic   
[37] MS PMinchoMusicalSymbolsMyanmar Sangam MN 
PMingLiU 
[41] SimSunSinfonia  Tahoma
Times New Roman  
[45] Trebuchet MS  Verdana   Webdings  
Wingdings
[49] Wingdings 2   Wingdings 3  

 ttf_import(file.choose())
Scanning ttf files in /Users/kw/Library/Fonts/Garamond ...
Extracting .afm files from .ttf files...
Error in data.frame(fontfile = ttfiles, FontName = , stringsAsFactors = 
FALSE) : 
  arguments imply differing number of rows: 0, 1

Here is what Font Book has to say about the font:

Garamond
Garamond

PostScript name Garamond
Full name   Garamond
Family  Garamond
Style   Regular
KindTrueType
LanguageAfrikaans, Albanian, Basque, Cornish, Danish, Dutch, 
English, Faroese, French, Galician, German, Icelandic, Indonesian, Irish, 
Italian, Malay, Manx, Norwegian Bokmål, Norwegian Nynorsk, Oromo, Portuguese, 
Somali, Spanish, Swahili, Swedish, Swiss German, Zulu
Script  Latin
Version Version 2.35
Location/Users/kw/Library/Fonts/Garamond
Unique name Monotype - Garamond Regular
ManufacturerMonotype Typography, Inc.
DesignerClaude Garamond
Copyright   Digitized data copyright Monotype Typography, Ltd 
1991-1995. All rights reserved. Monotype Garamond® is a trademark of Monotype 
Typography, Ltd which may be registered in certain jurisdictions.
Trademark   Monotype Garamond® is a trademark of Monotype 
Typography, Ltd which may be registered in certain jurisdictions.
Description Monotype Drawing Office 1922. This typeface is based on 
roman types cut by Jean Jannon in 1615. Jannon followed the designs of Claude 
Garamond which had been cut in the previous century. Garamond's types were, in 
turn, based on those used by Aldus Manutius in 1495 and cut by Francesco 
Griffo. The italic is based on types cut in France circa 1557 by Robert 
Granjon. Garamond is a beautiful typeface with an air of informality which 
looks good in a wide range of applications. It works particularly well in books 
and lengthy text settings.
License NOTIFICATION OF LICENSE AGREEMENT

This typeface is the property of Monotype Typography and its use by you is 
covered under the terms of a license agreement. You have obtained this typeface 
software either directly from Monotype or together with software distributed by 
one of Monotypes licensees.

This software is a valuable asset of Monotype. Unless you have entered into a 
specific license agreement granting you additional rights, your use of this 
software is limited to your workstation for your own publishing use. You may 
not copy or distribute this software.

If you have any question concerning your rights you should review the license 
agreement you received with the software or contact Monotype for a copy of the 
license agreement.

Monotype can be contacted at:

USA 847-718-0400  UK 44(0)1737 765959

For license

[R] How to calculate monthly average from daily files in R?

2013-01-17 Thread Jonsson

I have 365 binary files:  
https://echange-fichiers.inra.fr/get?k=oy3CN1yV1Um7ouRWm2U   ,I want to
calculate the monthly average. So from the 365 files, I will get 12 files.I
would like also to tell R not to take into account the no-data value
(-32765).for example, for the first month, there are 31 records: 3 of these
records has the value -32765,I want R to take the average of the rest
records(28  records) and so on with all months.

This code will take the average of every 30 files(any idea on how to make it
according to number of days in a month?and not to take into account the
no-data values)

   files- list.files(C:\\New folder (4)\\New folder,
*.bin,full.names=TRUE)
   # assume that we want to take the average of every 30 files
files.group- split(files , rep(seq_along(files), each = 30,
length =length(files)))
  results- list()
 for (.files in files.group){
   # read in the 30 files as a vector of numbers that you take
the average of
 x- do.call(rbind,(lapply(.files, readBin  , double() , size =
4 ,n =360 * 720 , signed =T)))
  ## take the means across the 30 files
 results[[length(results) + 1L]]- colMeans(x)}
  close(x)
 for (i in seq_along(results)){
fileName - sprintf(C:/New folder/glo_%d.flt, i)
writeBin(as.double(results[[i]]), fileName, size = 4)}



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[R] Naming an object after another object...can it be done?

Hello R-helpers,

I have run the following line of code:

x-dat$col

and now I would like to assign names(x) to be dat$col (e.g., a character
string equal to the column name that I assigned to x).

What I am trying to do is to assign columns in my dataframe to new objects
called x and y. Then I will use x and y within a new function to make plots
with informative axis labels (e.g., dat$col instead of x. So, for
example, I would like to plot (y~x,xlab=names(x)) and have dat$col
printed in the x-axis label. I can do this all manually, by typing

names(x)- dat$col)

but I'd like to do it with non-specific code within my function so I don't
have to type the variable names manually each time.

Many thanks,

Mark Na

[[alternative HTML version deleted]]

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

Re: [R] Naming an object after another object...can it be done?

2013-01-17 Thread Wensui Liu

are you looking for assign()?
On Jan 17, 2013 1:56 PM, mtb...@gmail.com wrote:

 Hello R-helpers,

 I have run the following line of code:

 x-dat$col

 and now I would like to assign names(x) to be dat$col (e.g., a character
 string equal to the column name that I assigned to x).

 What I am trying to do is to assign columns in my dataframe to new objects
 called x and y. Then I will use x and y within a new function to make plots
 with informative axis labels (e.g., dat$col instead of x. So, for
 example, I would like to plot (y~x,xlab=names(x)) and have dat$col
 printed in the x-axis label. I can do this all manually, by typing

 names(x)- dat$col)

 but I'd like to do it with non-specific code within my function so I don't
 have to type the variable names manually each time.

 Many thanks,

 Mark Na

 [[alternative HTML version deleted]]

 __
 R-help@r-project.org mailing list
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 PLEASE do read the posting guide
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[[alternative HTML version deleted]]

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Re: [R] Naming an object after another object...can it be done?

Hi Liu - I have been trying with assign() but it's not working. I don't
think that's what I'm looking forany other ideas? Many thanks, Mark

On Thu, Jan 17, 2013 at 1:11 PM, Wensui Liu liuwen...@gmail.com wrote:

 are you looking for assign()?
 On Jan 17, 2013 1:56 PM, mtb...@gmail.com wrote:

 Hello R-helpers,

 I have run the following line of code:

 x-dat$col

 and now I would like to assign names(x) to be dat$col (e.g., a character
 string equal to the column name that I assigned to x).

 What I am trying to do is to assign columns in my dataframe to new objects
 called x and y. Then I will use x and y within a new function to make
 plots
 with informative axis labels (e.g., dat$col instead of x. So, for
 example, I would like to plot (y~x,xlab=names(x)) and have dat$col
 printed in the x-axis label. I can do this all manually, by typing

 names(x)- dat$col)

 but I'd like to do it with non-specific code within my function so I don't
 have to type the variable names manually each time.

 Many thanks,

 Mark Na

 [[alternative HTML version deleted]]

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[[alternative HTML version deleted]]

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[R] How can i keep only latest entries?

2013-01-17 Thread Mat

Hello togehter,

i have a data.frame like this one:

No. Date last change
1  1 2012-10-04 change settings
2  1 2012-10-20 bug fix
3  1 2012-11-05 final
4  2 2013-01-15new task
5  2 2013-01-16Bug fix
6  2 2013-01-17final

now i want a new data.frame, who show me only the newest entries according
to the No.
The solution look like this one:

   No. Datelast change
1  1 2012-11-05   final
2  2 2013-01-17   final

can anyone help me?

thanks.

Mat



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Re: [R] Issue while installing Hmisc package

2013-01-17 Thread vaseem shaikh

Hello Michael,

I have tried with different CRAN but still i am getting the same error.


But still, i am also trying to install the package by locally giving
absolute path with repos= Null and type = Source, will CRAN have any role
to play here???


BR-
Vaseem Shaikh
On Thu, Jan 17, 2013 at 3:03 PM, Michael Weylandt 
michael.weyla...@gmail.com wrote:

 Try a different CRAN mirror.

 MW

 On Jan 17, 2013, at 7:42 AM, vaseem shaikh vsma...@gmail.com wrote:

  Hi,
 
  Please find the snap shot attached for the error reported wile installing
  Hmisc Package.
 
  Is there any thing you can help me with.
 
  Currently using R i386 2.15.2 Version of R on windows 7 platform.
 
  Ragards
  Vaseem Shaikh
  Instalation_Issue_HMISC.png
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[R] Explore patterns with GAM

2013-01-17 Thread spyros

Dear all,
I new to r and I would like your help. 
I want to explore the patterns (unimodal, monotonically increased/decreased)
of species richness~altitude using GAM in R. Although I run the gam function
in mgcv package I do not know how to manually define knots and degrees of
freedom.
Any help would be greatly appreciated.
Spyros  



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

[R] Line breaks in documentation \useage

2013-01-17 Thread Markku Karhunen




Hi again, R community.

I wonder how you do line breaks in \useage{} section in .Rd files. I  
am sure there's some tutorial for this somewhere, but I just haven't  
found it.


I have tried \\, \cr, \br and \newline, admittedly arbitrarily, but  
all of these produce warnings or errors.


br, Markku

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Re: [R] create block diagonal with each rows

2013-01-17 Thread arun

Hi,

I tried with kronecker()
do.call(rbind,lapply(1:4,function(i) t(kronecker(diag(4), x[i,]))[i,]))
  # [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
[1,]    1    5    9   13   17    0    0    0    0 0 0 0 0 0
#[2,]    0    0    0    0    0    2    6   10   14    18 0 0 0 0
#[3,]    0    0    0    0    0    0    0    0    0 0 3 7    11    15
#[4,]    0    0    0    0    0    0    0    0    0 0 0 0 0 0
  #   [,15] [,16] [,17] [,18] [,19] [,20]
#[1,] 0 0 0 0 0 0
#[2,] 0 0 0 0 0 0
#[3,]    19 0 0 0 0 0
#[4,] 0 4 8    12    16    20


Not sure if there are any shortcuts with kronecker()
A.K.





- Original Message -
From: Martin Maechler maech...@stat.math.ethz.ch
To: arun smartpink...@yahoo.com
Cc: Kathryn Lord kathryn.lord2...@gmail.com; R help r-help@r-project.org
Sent: Thursday, January 17, 2013 12:18 PM
Subject: Re: [R] create block diagonal with each rows

 arun  smartpink...@yahoo.com
     on Wed, 16 Jan 2013 19:20:46 -0800 writes:

     Hi,
     May be this helps:
     library(Matrix)
     res1-lapply(split(x,1:nrow(x)),function(y) 
sparseMatrix(i=rep(1:4,each=5),j=1:(4*5),x=y))
      do.call(rbind,lapply(seq_along(res1),function(i) res1[[i]][i,]))
     # [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] 
[,13] [,14]
     #[1,]    1    5    9   13   17    0    0    0    0 0 0 0 
0 0
     #[2,]    0    0    0    0    0    2    6   10   14    18 0 0 
0 0
     #[3,]    0    0    0    0    0    0    0    0    0 0 3 7    
11    15
     #[4,]    0    0    0    0    0    0    0    0    0 0 0 0 
0 0
     # [,15] [,16] [,17] [,18] [,19] [,20]
     #[1,] 0 0 0 0 0 0
     #[2,] 0 0 0 0 0 0
     #[3,]    19 0 0 0 0 0
     #[4,] 0 4 8    12    16    20

Thank you for thinking of  Matrix  (the package, not the movie)
here.

If you do,

  x - matrix(1:20, 4,5)
  require(Matrix)

the following is  a tiny bit  nicer, using the  bdiag()
function which has been written to create block diagonal matrices.

 t(bdiag(split(x,1:nrow(x

4 x 20 sparse Matrix of class dgCMatrix
                                                      
[1,] 1 5 9 13 17 . .  .  .  . . .  .  .  . . .  .  .  .
[2,] . . .  .  . 2 6 10 14 18 . .  .  .  . . .  .  .  .
[3,] . . .  .  . . .  .  .  . 3 7 11 15 19 . .  .  .  .
[4,] . . .  .  . . .  .  .  . . .  .  .  . 4 8 12 16 20
 

and if you don't want a sparse matrix for some reason,
    (and think twice: it may be more efficient to keep it !)
you wrap the result with a 

    as.matrix(  result  )

---
Martin Maechler, ETH Zurich
    




     - Original Message -
     From: Kathryn Lord kathryn.lord2...@gmail.com
     To: r-help r-help@r-project.org
     Cc: 
     Sent: Wednesday, January 16, 2013 9:11 PM
     Subject: [R] create block diagonal with each rows

     Dear R users,

     I'd like to create a block diagonal matrix with each rows in a matrix.

     Here is a simple example. (In fact, the matrix is big)


     x - matrix(1:20, 4,5)

     x
          [,1] [,2] [,3] [,4] [,5]
     [1,]    1    5    9   13   17
     [2,]    2    6   10   14   18
     [3,]    3    7   11   15   19
     [4,]    4    8   12   16   20


     With each rows in matrix x, I'd like to make the matrix below.


          [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9][,10]
     [,11][,12][,13][,14][,15][,16][,17][,18][,19][,20]
     [1,]    1    5    9   13   17   0    0    0    0    0     0    0    0
     0    0     0    0    0    0    0
     [2,]    0    0    0    0    0    2    6   10   14   18   0    0    0
     0    0     0    0    0    0    0
     [3,]    0    0    0    0    0    0    0    0    0    0     3    7   11
     15   19   0    0    0    0    0
     [4,]    0    0    0    0    0    0    0    0    0    0     0    0    0
     0    0     4    8   12   16   20


     Any suggestion will be greatly appreciated.

     Best,

     Kathryn Lord

         [[alternative HTML version deleted]]

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


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[R] crontab to Run R Code

2013-01-17 Thread Pinaki

Could not figure out where I am going wrong. Following is my code in
crontab -e:-

MAILTO: users

MAILTO= users

# m h  dom mon dow   command
3 19 * * * $HOME/users/REPORT/MAIL; time R --slave  report.R

[[alternative HTML version deleted]]

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

[R] Help with interpolation

2013-01-17 Thread beanbandit

hi guys

I need to interpolate values for the zero coupon yield curve. Following data
is given





 
datedays  rate

1996 01 02  155.74590
1996 01 02  505.67332
1996 01 02  785.60888
1996 01 02 1695.47376
1996 01 02 2605.35267
1996 01 02 3515.27619

1996 01 03  145.74740
1996 01 03  495.67226
1996 01 03  775.60371
1996 01 03 1685.47058
1996 01 03 2595.34662
1996 01 03 3505.26630
 
For every day i have to interpolate 10 values, for example for maturities of
30,60 or 90 days. I have interpolate data for a one year period, 10
interpolation values a day, so that equals 3600 values. 

what's the easiest way to implement this in R?

please hlep!




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Re: [R] Naming an object after another object...can it be done?

2013-01-17 Thread Patrick Burns


You are thinking that 'names' does something different
than it does.  What you seem to be after is the
deparse-substitute idiom:

dat - data.frame(Col1=1:10, Col2=rnorm(10))
myPlotFun - function(x, y) {
   plot(y ~ x, xlab=deparse(substitute(x)), ylab=deparse(substitute(y)))
}
myPlotFun(dat$Col1, dat$Col2)


Pat

On 17/01/2013 18:53, mtb...@gmail.com wrote:

Hello R-helpers,

I have run the following line of code:

x-dat$col

and now I would like to assign names(x) to be dat$col (e.g., a character
string equal to the column name that I assigned to x).

What I am trying to do is to assign columns in my dataframe to new objects
called x and y. Then I will use x and y within a new function to make plots
with informative axis labels (e.g., dat$col instead of x. So, for
example, I would like to plot (y~x,xlab=names(x)) and have dat$col
printed in the x-axis label. I can do this all manually, by typing

names(x)- dat$col)

but I'd like to do it with non-specific code within my function so I don't
have to type the variable names manually each time.

Many thanks,

Mark Na

[[alternative HTML version deleted]]

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--
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pbu...@pburns.seanet.com
twitter: @portfolioprobe
http://www.portfolioprobe.com/blog
http://www.burns-stat.com
(home of 'Some hints for the R beginner'
and 'The R Inferno')

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[R] plotting from dataframes

2013-01-17 Thread condor

thanks to your guys help I am closer to solving my problem but I have some
small problem. So let's say I start with

data
number day  hour 
1   17  10  
2   17  11
3   17  6
4   18  4
5   18  10
6   19  8
7   19  8

I want to split to odd days, which I am able to do, I call this object
frames, which looks like:

 frames
$`1`
  c1 day1 hour1
1  1   1710
2  2   1711
3  3   17 6

$`2`
  c1 day1 hour1
4  6   19 8
5  7   19 8

Now I want to make plot of the hours of both days, but not by hand. I need
some sort of loop for this. How is this done?

So
par(mfrow=c(1,2))
for(…) plot(…hours…)

thanks for the help



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Re: [R] Getting discrete colors on plot

2013-01-17 Thread Mary

Hi,

This is my first post; I'm new to R but am a senior statistical programmer.  I 
have done a lot of graphs using SAS Graph but now am trying to transition to 
using graphs in R.

I'm trying to produce a graph where the colors have three categories- ideally I 
would like them to be Green for good, Yellow for Questionable, and Red for bad.

So I tried to do this in GGPLOT; here is my code:

id -   c(1,2,3,4,5)
x1 -   c(5,2,3,5,1)
x2 -   c(4,1,3,5,1)
x3 -   c(5,2,3,5,1)
x4 -   c(4,3,3,5,1)
x5 -   c(3,1,3,5,1)
colorvar - c(3,1,2,3,1)

mydata - data.frame(id,x1,x2,x3,x4,x5,colorvar)
head(mydata)

# convert to long format
require(reshape)
mydata_long - melt(mydata, id=c(id, colorvar))
head(mydata_long)

require(ggplot2)
p - ggplot(data=mydata_long,
   aes(x=variable, y=value, 
group=id, colour = colorvar)) +
geom_line()
p
  

This works, but I get more colors on the graph than my colorvar has.  I have 3 
colors on my colorvar, but 5 colors show up on the graph, including 1.5 and 
2.5.   How do I tell ggplot only to use the 3 colors and not give me a gradient 
of colors?  Also how would I specify the colors that I want, such as the RGB 
equivalents of green, yellow, and red?  My real data will have many more 
records.

-Mary

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Re: [R] how to use ...

2013-01-17 Thread Steve Taylor

The ellipsis object is not listed in the base help pages!

help(`+`) # this works - help on arithmetic operators
help(+) # also works
help(`...`) # fails with Error: '...' used in an incorrect context
help(...) # fails also with No documentation for '...' in specified packages 
and libraries: you could try '??...'


-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of Ivan Calandra
Sent: Friday, 18 January 2013 4:43a
To: R. Michael Weylandt
Cc: R list
Subject: Re: [R] how to use ...

Do you know where I can find some documentation about it?

__
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Re: [R] Explore patterns with GAM

2013-01-17 Thread Andrew Robinson

Hi Spyros,

I suggest that you borrow / buy the book that was written by the author of
that package, and study it. It's Generalized Additive Models: An
Introduction with R. There's a lot of stuff going on in GAM fitting that
it would be worth paying close attention to.

I hope that this helps,

Andrew


On Fri, Jan 18, 2013 at 12:45 AM, spyros stsif...@bio.auth.gr wrote:

 Dear all,
 I new to r and I would like your help.
 I want to explore the patterns (unimodal, monotonically
 increased/decreased)
 of species richness~altitude using GAM in R. Although I run the gam
 function
 in mgcv package I do not know how to manually define knots and degrees of
 freedom.
 Any help would be greatly appreciated.
 Spyros



 --
 View this message in context:
 http://r.789695.n4.nabble.com/Explore-patterns-with-GAM-tp4655838.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.




-- 
Andrew Robinson
Director (A/g), ACERA
Senior Lecturer in Applied Statistics  Tel:
+61-3-8344-6410
Department of Mathematics and StatisticsFax: +61-3-8344 4599
University of Melbourne, VIC 3010 Australia
Email: a.robin...@ms.unimelb.edu.auWebsite: http://www.ms.unimelb.edu.au

FAwR: http://www.ms.unimelb.edu.au/~andrewpr/FAwR/
SPuR: http://www.ms.unimelb.edu.au/spuRs/

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

Re: [R] crontab to Run R Code

2013-01-17 Thread Gergely Daróczi

Not sure if it would solve all your problems, but try to specify absolute
path first e.g. for R, as *your* PATH is not available for the cronjob.

Best,
Gergely

On Thu, Jan 17, 2013 at 2:53 PM, Pinaki pinakimah...@gmail.com wrote:

 Could not figure out where I am going wrong. Following is my code in
 crontab -e:-

 MAILTO: users

 MAILTO= users

 # m h  dom mon dow   command
 3 19 * * * $HOME/users/REPORT/MAIL; time R --slave  report.R

 [[alternative HTML version deleted]]

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


[[alternative HTML version deleted]]

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

Re: [R] Naming an object after another object...can it be done?

2013-01-17 Thread Patrick Burns


If you want the column names but not
the data frame name, then you could do:

with(dat, myPlotFun(Col1, Col2))

Pat

On 17/01/2013 20:07, Patrick Burns wrote:

You are thinking that 'names' does something different
than it does.  What you seem to be after is the
deparse-substitute idiom:

dat - data.frame(Col1=1:10, Col2=rnorm(10))
myPlotFun - function(x, y) {
plot(y ~ x, xlab=deparse(substitute(x)), ylab=deparse(substitute(y)))
}
myPlotFun(dat$Col1, dat$Col2)


Pat

On 17/01/2013 18:53, mtb...@gmail.com wrote:

Hello R-helpers,

I have run the following line of code:

x-dat$col

and now I would like to assign names(x) to be dat$col (e.g., a
character
string equal to the column name that I assigned to x).

What I am trying to do is to assign columns in my dataframe to new
objects
called x and y. Then I will use x and y within a new function to make
plots
with informative axis labels (e.g., dat$col instead of x. So, for
example, I would like to plot (y~x,xlab=names(x)) and have dat$col
printed in the x-axis label. I can do this all manually, by typing

names(x)- dat$col)

but I'd like to do it with non-specific code within my function so I
don't
have to type the variable names manually each time.

Many thanks,

Mark Na

[[alternative HTML version deleted]]

__
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.





--
Patrick Burns
pbu...@pburns.seanet.com
twitter: @portfolioprobe
http://www.portfolioprobe.com/blog
http://www.burns-stat.com
(home of 'Some hints for the R beginner'
and 'The R Inferno')

__
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] plotting from dataframes

2013-01-17 Thread Andrew Robinson

It's not really clear to me what you mean when you say that you want to
plot the hours, so it's hard to help.  Regardless, take a look at looping
and plotting in any of the free documentation on CRAN.

http://cran.r-project.org/other-docs.html

I hope that this helps,

Andrew


On Fri, Jan 18, 2013 at 2:21 AM, condor radonniko...@hotmail.nl wrote:

 thanks to your guys help I am closer to solving my problem but I have some
 small problem. So let's say I start with

 data
 number day  hour
 1   17  10
 2   17  11
 3   17  6
 4   18  4
 5   18  10
 6   19  8
 7   19  8

 I want to split to odd days, which I am able to do, I call this object
 frames, which looks like:

  frames
 $`1`
   c1 day1 hour1
 1  1   1710
 2  2   1711
 3  3   17 6

 $`2`
   c1 day1 hour1
 4  6   19 8
 5  7   19 8

 Now I want to make plot of the hours of both days, but not by hand. I need
 some sort of loop for this. How is this done?

 So
 par(mfrow=c(1,2))
 for() plot(hours)

 thanks for the help



 --
 View this message in context:
 http://r.789695.n4.nabble.com/plotting-from-dataframes-tp4655851.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.




-- 
Andrew Robinson
Director (A/g), ACERA
Senior Lecturer in Applied Statistics  Tel:
+61-3-8344-6410
Department of Mathematics and StatisticsFax: +61-3-8344 4599
University of Melbourne, VIC 3010 Australia
Email: a.robin...@ms.unimelb.edu.auWebsite: http://www.ms.unimelb.edu.au

FAwR: http://www.ms.unimelb.edu.au/~andrewpr/FAwR/
SPuR: http://www.ms.unimelb.edu.au/spuRs/

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Re: [R] how to use ...

2013-01-17 Thread Steve Taylor

Here's a link (on my local CRAN)...

http://cran.stat.auckland.ac.nz/doc/manuals/r-release/R-intro.html#The-three-dots-argument

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of Bert Gunter
Sent: Friday, 18 January 2013 4:54a
To: ivan.calan...@u-bourgogne.fr
Cc: R list
Subject: Re: [R] how to use ...

Well..

On Thu, Jan 17, 2013 at 7:42 AM, Ivan Calandra
ivan.calan...@u-bourgogne.fr wrote:
 Ok, it is that simple... Actually I had tried it but messed up so that it
 didn't work.
 Do you know where I can find some documentation about it?

The R language definition manual would be the logical place to look,
no? And sure enough, it's there!

The Introduction to R tutorial also contains info in the writing
functions section.

A very good, though now dated, exposition can be found in VR's S
Programming book. Although a little care is needed due to age, it's
still my favorite.

-- Bert

 Regarding return(), I know that it's not necessary, but when the function
 gets more complicated, I like to have it because it becomes clearer to me.

 Thanks all!
 Ivan

 --
 Ivan CALANDRA
 Université de Bourgogne
 UMR CNRS/uB 6282 Biogéosciences
 6 Boulevard Gabriel
 21000 Dijon, FRANCE
 +33(0)3.80.39.63.06
 ivan.calan...@u-bourgogne.fr
 http://biogeosciences.u-bourgogne.fr/calandra

 Le 17/01/13 15:55, R. Michael Weylandt a écrit :

 On Thu, Jan 17, 2013 at 2:36 PM, Ivan Calandra
 ivan.calan...@u-bourgogne.fr wrote:

 Dear users,

 I'm trying to learn how to use the 

 I have written a function (simplified here) that uses doBy::summaryBy():
 # 'dat' is a data.frame from which the aggregation is computed
 # 'vec_cat' is a integer vector defining which columns of the data.frame
 should be use on the right side of the formula
 # 'stat_fun' is the function that will be run to aggregate
 stat.group - function(dat, vec_cat, stat_fun){
  require(doBy)
  df -

 summaryBy(as.formula(paste0(.~,paste0(names(dat)[vec_cat],collapse=+))),
 data=dat, FUN=stat_fun)
  return(df)
 }
 [SNIP EXAMPLE -- THANK YOU FOR IT]

 Now summaryBy() has an ... argument and I would like to use it.
 For example, I would like to be able to add the trim argument to my call
 like this:
 stat.group(dat=my_data, vec_cat=1, stat_fun=mean, trim=0.2)

 Thanks for the great working examples!

 It's actually not too hard here --  just pass ... down as if it were
 an argument and let summaryBy() do the hard work of actually handling
 the dots:

 stat.group - function(dat, vec_cat, stat_fun, ...){
  require(doBy)
  df -
 summaryBy(as.formula(paste0(.~,paste0(names(dat)[vec_cat],collapse=+))),
 data=dat, FUN=stat_fun, ...)
  return(df)
 }

 Also, note that as a matter of style, you can actually clean this up a
 little bit: R follows the trend of many functional languages in
 automatically returning the value of the last expression evaluated:

 stat.group - function(dat, vec_cat, stat_fun, ...){
  require(doBy)

 summaryBy(as.formula(paste0(.~,paste0(names(dat)[vec_cat],collapse=+))),
 data=dat, FUN=stat_fun, ...)
 }

 Cheers,
 Michael

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

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

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Re: [R] Naming an object after another object...can it be done?

?
But Pat...

The canonical way to do this is:

myPlotFin(Col2 ~ Col1, data = dat)

I have no idea what the OP wants, but my guess is that the right
answer is: Don't do that.

Cheers,
Bert

On Thu, Jan 17, 2013 at 12:25 PM, Patrick Burns
pbu...@pburns.seanet.com wrote:
 If you want the column names but not
 the data frame name, then you could do:

 with(dat, myPlotFun(Col1, Col2))

 Pat

 On 17/01/2013 20:07, Patrick Burns wrote:

 You are thinking that 'names' does something different
 than it does.  What you seem to be after is the
 deparse-substitute idiom:

 dat - data.frame(Col1=1:10, Col2=rnorm(10))
 myPlotFun - function(x, y) {
 plot(y ~ x, xlab=deparse(substitute(x)), ylab=deparse(substitute(y)))
 }
 myPlotFun(dat$Col1, dat$Col2)


 Pat

 On 17/01/2013 18:53, mtb...@gmail.com wrote:

 Hello R-helpers,

 I have run the following line of code:

 x-dat$col

 and now I would like to assign names(x) to be dat$col (e.g., a
 character
 string equal to the column name that I assigned to x).

 What I am trying to do is to assign columns in my dataframe to new
 objects
 called x and y. Then I will use x and y within a new function to make
 plots
 with informative axis labels (e.g., dat$col instead of x. So, for
 example, I would like to plot (y~x,xlab=names(x)) and have dat$col
 printed in the x-axis label. I can do this all manually, by typing

 names(x)- dat$col)

 but I'd like to do it with non-specific code within my function so I
 don't
 have to type the variable names manually each time.

 Many thanks,

 Mark Na

 [[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.



 --
 Patrick Burns
 pbu...@pburns.seanet.com
 twitter: @portfolioprobe
 http://www.portfolioprobe.com/blog
 http://www.burns-stat.com
 (home of 'Some hints for the R beginner'
 and 'The R Inferno')

 __
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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.



-- 

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

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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] Naming an object after another object...can it be done?

Hi everyone, and thanks for your replies.

Let me make this a little simpler. Please forget the plotting, that's not
the issue.

I have run the following line of code:

x-dat.col

Now, is there a function (or combination of functions) that will let me
assign the character string dat.col to a new object (called y) without
actually typing the characters dat$col, i.e. just by referring to x?

Many thanks,

Mark Na




On Thu, Jan 17, 2013 at 3:20 PM, Bert Gunter gunter.ber...@gene.com wrote:

 ?
 But Pat...

 The canonical way to do this is:

 myPlotFin(Col2 ~ Col1, data = dat)

 I have no idea what the OP wants, but my guess is that the right
 answer is: Don't do that.

 Cheers,
 Bert

 On Thu, Jan 17, 2013 at 12:25 PM, Patrick Burns
 pbu...@pburns.seanet.com wrote:
  If you want the column names but not
  the data frame name, then you could do:
 
  with(dat, myPlotFun(Col1, Col2))
 
  Pat
 
  On 17/01/2013 20:07, Patrick Burns wrote:
 
  You are thinking that 'names' does something different
  than it does.  What you seem to be after is the
  deparse-substitute idiom:
 
  dat - data.frame(Col1=1:10, Col2=rnorm(10))
  myPlotFun - function(x, y) {
  plot(y ~ x, xlab=deparse(substitute(x)),
 ylab=deparse(substitute(y)))
  }
  myPlotFun(dat$Col1, dat$Col2)
 
 
  Pat
 
  On 17/01/2013 18:53, mtb...@gmail.com wrote:
 
  Hello R-helpers,
 
  I have run the following line of code:
 
  x-dat$col
 
  and now I would like to assign names(x) to be dat$col (e.g., a
  character
  string equal to the column name that I assigned to x).
 
  What I am trying to do is to assign columns in my dataframe to new
  objects
  called x and y. Then I will use x and y within a new function to make
  plots
  with informative axis labels (e.g., dat$col instead of x. So, for
  example, I would like to plot (y~x,xlab=names(x)) and have dat$col
  printed in the x-axis label. I can do this all manually, by typing
 
  names(x)- dat$col)
 
  but I'd like to do it with non-specific code within my function so I
  don't
  have to type the variable names manually each time.
 
  Many thanks,
 
  Mark Na
 
  [[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.
 
 
 
  --
  Patrick Burns
  pbu...@pburns.seanet.com
  twitter: @portfolioprobe
  http://www.portfolioprobe.com/blog
  http://www.burns-stat.com
  (home of 'Some hints for the R beginner'
  and 'The R Inferno')
 
  __
  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.



 --

 Bert Gunter
 Genentech Nonclinical Biostatistics

 Internal Contact Info:
 Phone: 467-7374
 Website:

 http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm


[[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.

Re: [R] How can i keep only latest entries?

2013-01-17 Thread Rui Barradas


Hello,

Try the following.
(I've named your data.frame 'dat')

do.call(rbind, lapply(split(dat, dat$`No.`), tail, 1))


Hope this helps,

Rui Barradas

Em 17-01-2013 10:50, Mat escreveu:

Hello togehter,

i have a data.frame like this one:

 No. Date last change
1  1 2012-10-04 change settings
2  1 2012-10-20 bug fix
3  1 2012-11-05 final
4  2 2013-01-15new task
5  2 2013-01-16Bug fix
6  2 2013-01-17final

now i want a new data.frame, who show me only the newest entries according
to the No.
The solution look like this one:

No. Datelast change
1  1 2012-11-05   final
2  2 2013-01-17   final

can anyone help me?

thanks.

Mat



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[R] How to convert a string to the column it represents in a dataframe, with a reproducible example

Hello R-helpers,

I have run the following lines of code:

x-cars$dist
y-noquote(x)


Now y is a string containing the characters cars$dist

My questionis there an R function (or combination of functions) that I
can apply to y that will cause y to contain the numbers in cars$dist? Even
better, can I do it without using noquote()?

Many thanks,

Mark Na

[[alternative HTML version deleted]]

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

Re: [R] Naming an object after another object...can it be done?

On Thu, Jan 17, 2013 at 1:29 PM,  mtb...@gmail.com wrote:
 Hi everyone, and thanks for your replies.

 Let me make this a little simpler. Please forget the plotting, that's not
 the issue.

 I have run the following line of code:

 x-dat.col

 Now, is there a function (or combination of functions) that will let me
 assign the character string dat.col to a new object (called y) without
 actually typing the characters dat$col, i.e. just by referring to x?

No.  x contains only the values. As I said, don't do this.

-- Bert


 Many thanks,

 Mark Na




 On Thu, Jan 17, 2013 at 3:20 PM, Bert Gunter gunter.ber...@gene.com wrote:

 ?
 But Pat...

 The canonical way to do this is:

 myPlotFin(Col2 ~ Col1, data = dat)

 I have no idea what the OP wants, but my guess is that the right
 answer is: Don't do that.

 Cheers,
 Bert

 On Thu, Jan 17, 2013 at 12:25 PM, Patrick Burns
 pbu...@pburns.seanet.com wrote:
  If you want the column names but not
  the data frame name, then you could do:
 
  with(dat, myPlotFun(Col1, Col2))
 
  Pat
 
  On 17/01/2013 20:07, Patrick Burns wrote:
 
  You are thinking that 'names' does something different
  than it does.  What you seem to be after is the
  deparse-substitute idiom:
 
  dat - data.frame(Col1=1:10, Col2=rnorm(10))
  myPlotFun - function(x, y) {
  plot(y ~ x, xlab=deparse(substitute(x)),
  ylab=deparse(substitute(y)))
  }
  myPlotFun(dat$Col1, dat$Col2)
 
 
  Pat
 
  On 17/01/2013 18:53, mtb...@gmail.com wrote:
 
  Hello R-helpers,
 
  I have run the following line of code:
 
  x-dat$col
 
  and now I would like to assign names(x) to be dat$col (e.g., a
  character
  string equal to the column name that I assigned to x).
 
  What I am trying to do is to assign columns in my dataframe to new
  objects
  called x and y. Then I will use x and y within a new function to make
  plots
  with informative axis labels (e.g., dat$col instead of x. So, for
  example, I would like to plot (y~x,xlab=names(x)) and have dat$col
  printed in the x-axis label. I can do this all manually, by typing
 
  names(x)- dat$col)
 
  but I'd like to do it with non-specific code within my function so I
  don't
  have to type the variable names manually each time.
 
  Many thanks,
 
  Mark Na
 
  [[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.
 
 
 
  --
  Patrick Burns
  pbu...@pburns.seanet.com
  twitter: @portfolioprobe
  http://www.portfolioprobe.com/blog
  http://www.burns-stat.com
  (home of 'Some hints for the R beginner'
  and 'The R Inferno')
 
  __
  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.



 --

 Bert Gunter
 Genentech Nonclinical Biostatistics

 Internal Contact Info:
 Phone: 467-7374
 Website:

 http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm





-- 

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

__
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.

[R] How to delete repeated values in MCMC sampling and get index of unique values?

2013-01-17 Thread C W

Dear list,
How do you delete repeated samples?  In MCMC, when your candidate value has
been reject, so you remain on the same point, so you keep that value.

Say I have this toy example,

 c(1,6,6,6,3,5,4,4,2,3,5)

The 6 and 4 are repeated, I only want the index of the non-repeated values.

I thought of using which() and unique(), but that does not give you the
index of the unique values.

Thanks in advance.

Mike

[[alternative HTML version deleted]]

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

Re: [R] How to delete repeated values in MCMC sampling and get index of unique values?


On 13-01-17 4:50 PM, C W wrote:

Dear list,
How do you delete repeated samples?  In MCMC, when your candidate value has
been reject, so you remain on the same point, so you keep that value.

Say I have this toy example,


c(1,6,6,6,3,5,4,4,2,3,5)


The 6 and 4 are repeated, I only want the index of the non-repeated values.

I thought of using which() and unique(), but that does not give you the
index of the unique values.


You could use x[!duplicated(x)] or rle(x)$values, depending on your 
definition of repeated.  I hope you're aware that you can't use either 
for things like quantiles and moments of the limiting distribution.


 x - c(1,6,6,6,3,5,4,4,2,3,5)
 x[!duplicated(x)]
[1] 1 6 3 5 4 2
 rle(x)$values
[1] 1 6 3 5 4 2 3 5

Duncan Murdoch

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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.

[R] fastICA package: C, R codes provide vastly different results

2013-01-17 Thread Ranjan Maitra

Dear friends,

I have been trying out the C and the R codes in the fastICA package.
However, it turns out that these often give vastly different results,
especially when row.norm is set to T. This happens even though I have
initialized the input matrix to be exactly the same for both of them.

Here is an example:

### cut code here

xx - read.table(file =
http://maitra.public.iastate.edu/SharadUtshob05.dat;)


win - matrix(c(-0.434590,  1.288207, -0.597128, -0.501621,  1.434678,
0.812076, -1.281559, -1.798178,  0.409151), ncol = 3)

library(fastICA)

adeflate.C - fastICA(X = xx, alg.typ=deflation, row.norm=T,
w.init=win, n.comp=3, fun = logcosh, method = C)

adeflate.R - fastICA(X = xx, alg.typ=deflation, row.norm=T,
w.init=win, n.comp=3, fun = logcosh, method = R)


### end code here.

The answers follow: note that the final un-mixing matrix estimate W is
not even close to each other. 

adeflate.R$W
[,1]   [,2]   [,3]
[1,] -0.85074291 -0.2385149  0.4683451
[2,]  0.02907471 -0.9110909 -0.4111789
[3,] -0.52477726  0.3361905 -0.7820388


adeflate.C$W
 [,1]  [,2] [,3]
[1,] -0.001317682 -0.9904552102 -0.137828767
[2,] -0.006132130  0.1378342956 -0.990436316
[3,]  0.80330 -0.0004598964 -0.006255222

Can anyone throw light on this matter?

On the other hand, using row.norm = F provides the following results:

adeflate.R$W
[,1]   [,2]   [,3]
[1,] -0.86309623 -0.2103872  0.4591319
[2,] -0.01433631 -0.8985285 -0.4386811
[3,] -0.50483598  0.3852062 -0.7725003

which is not as much different (but still quite different) from 

adeflate.C$W
[,1]   [,2]   [,3]
[1,] -0.83111268  0.2247869  0.5086478
[2,] -0.01217191  0.9070906 -0.4207594
[3,] -0.55597085 -0.3558897 -0.7511584

What is the problem? Am I assuming something here that i should not in
my inputs?

Many thanks again for any help!

Best wishes,
Ranjan






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Re: [R] Naming an object after another object...can it be done?


On Jan 17, 2013, at 1:29 PM, mtb...@gmail.com wrote:

 Hi everyone, and thanks for your replies.
 
 Let me make this a little simpler. Please forget the plotting, that's not
 the issue.
 
 I have run the following line of code:
 
 x-dat.col
 
 Now, is there a function (or combination of functions) that will let me
 assign the character string dat.col to a new object (called y) without
 actually typing the characters dat$col, i.e. just by referring to x?

It is unclear what you intend. It is possible to rename the x-object and it 
is possible to name individual elements of a multi-element object named x. It 
is furthermore possble to construct an attribute that will remain with x. So, 
provide a complete example and it may become clearer. 

Also specify whether you are attempting to do this on an object which has been 
attached, and if so, then the right answer will be don't use attach().

-- 
David.
 
 Many thanks,
 
 Mark Na
 
 
 
 
 On Thu, Jan 17, 2013 at 3:20 PM, Bert Gunter gunter.ber...@gene.com wrote:
 
 ?
 But Pat...
 
 The canonical way to do this is:
 
 myPlotFin(Col2 ~ Col1, data = dat)
 
 I have no idea what the OP wants, but my guess is that the right
 answer is: Don't do that.
 
 Cheers,
 Bert
 
 On Thu, Jan 17, 2013 at 12:25 PM, Patrick Burns
 pbu...@pburns.seanet.com wrote:
 If you want the column names but not
 the data frame name, then you could do:
 
 with(dat, myPlotFun(Col1, Col2))
 
 Pat
 
 On 17/01/2013 20:07, Patrick Burns wrote:
 
 You are thinking that 'names' does something different
 than it does.  What you seem to be after is the
 deparse-substitute idiom:
 
 dat - data.frame(Col1=1:10, Col2=rnorm(10))
 myPlotFun - function(x, y) {
plot(y ~ x, xlab=deparse(substitute(x)),
 ylab=deparse(substitute(y)))
 }
 myPlotFun(dat$Col1, dat$Col2)
 
 
 Pat
 
 On 17/01/2013 18:53, mtb...@gmail.com wrote:
 
 Hello R-helpers,
 
 I have run the following line of code:
 
 x-dat$col
 
 and now I would like to assign names(x) to be dat$col (e.g., a
 character
 string equal to the column name that I assigned to x).
 
 What I am trying to do is to assign columns in my dataframe to new
 objects
 called x and y. Then I will use x and y within a new function to make
 plots
 with informative axis labels (e.g., dat$col instead of x. So, for
 example, I would like to plot (y~x,xlab=names(x)) and have dat$col
 printed in the x-axis label. I can do this all manually, by typing
 
 names(x)- dat$col)
 
 but I'd like to do it with non-specific code within my function so I
 don't
 have to type the variable names manually each time.
 
 Many thanks,
 
 Mark Na
 
[[alternative HTML version deleted]]
 
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 --
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 (home of 'Some hints for the R beginner'
 and 'The R Inferno')
 
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 PLEASE do read the posting guide
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 and provide commented, minimal, self-contained, reproducible code.
 
 
 
 --
 
 Bert Gunter
 Genentech Nonclinical Biostatistics
 
 Internal Contact Info:
 Phone: 467-7374
 Website:
 
 http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
 
 
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Re: [R] Issue while installing Hmisc package

2013-01-17 Thread Peter Langfelder

On Thu, Jan 17, 2013 at 4:57 AM, vaseem shaikh vsma...@gmail.com wrote:


 But still, i am also trying to install the package by locally giving
 absolute path with repos= Null and type = Source, will CRAN have any role
 to play here???

To compile a package on Windows, you need to install R tools (Rtools)
that contain the compilers and other software necessary for the
compilation. The error you saw with the installation from local
.tar.gz file indeed has nothing to do with CRAN but with the fact that
you are missing the tools. See
http://cran.r-project.org/doc/manuals/r-release/R-admin.html#The-Windows-toolset
for details.

If you don't want to install the Windows toolset, you can go to the
Hmisc CRAN web page at
http://cran.r-project.org/web/packages/Hmisc/index.html , download the
Windows binary and install that using repos = NULL (but not type =
source).

HTH

Peter

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Re: [R] How to convert a string to the column it represents in a dataframe, with a reproducible example


On Jan 17, 2013, at 1:36 PM, mtb...@gmail.com wrote:

 Hello R-helpers,
 
 I have run the following lines of code:
 
 x-cars$dist
 y-noquote(x)
 
 
 Now y is a string containing the characters cars$dist
 
 My questionis there an R function (or combination of functions) that I
 can apply to y that will cause y to contain the numbers in cars$dist? Even
 better, can I do it without using noquote()?

What is the goal of this effort?

-- 

David Winsemius
Alameda, CA, USA

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Re: [R] How to delete repeated values in MCMC sampling and get index of unique values?

2013-01-17 Thread C W

Exactly what I am looking for.

Thanks a lot!

Mike

On Thu, Jan 17, 2013 at 4:59 PM, Duncan Murdoch murdoch.dun...@gmail.comwrote:

 On 13-01-17 4:50 PM, C W wrote:

 Dear list,
 How do you delete repeated samples?  In MCMC, when your candidate value
 has
 been reject, so you remain on the same point, so you keep that value.

 Say I have this toy example,

  c(1,6,6,6,3,5,4,4,2,3,5)


 The 6 and 4 are repeated, I only want the index of the non-repeated
 values.

 I thought of using which() and unique(), but that does not give you the
 index of the unique values.


 You could use x[!duplicated(x)] or rle(x)$values, depending on your
 definition of repeated.  I hope you're aware that you can't use either
 for things like quantiles and moments of the limiting distribution.

  x - c(1,6,6,6,3,5,4,4,2,3,5)
  x[!duplicated(x)]
 [1] 1 6 3 5 4 2
  rle(x)$values
 [1] 1 6 3 5 4 2 3 5

 Duncan Murdoch



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Re: [R] How to delete repeated values in MCMC sampling and get index of unique values?


On Jan 17, 2013, at 1:50 PM, C W wrote:

 Dear list,
 How do you delete repeated samples?  In MCMC, when your candidate value has
 been reject, so you remain on the same point, so you keep that value.
 
 Say I have this toy example,
 
 c(1,6,6,6,3,5,4,4,2,3,5)

 c(1,6,6,6,3,5,4,4,2,3,5)[!duplicated(c(1,6,6,6,3,5,4,4,2,3,5))]
[1] 1 6 3 5 4 2
 x - c(1,6,6,6,3,5,4,4,2,3,5)
 x[!duplicated(x)]
[1] 1 6 3 5 4 2
 !duplicated(x)
 [1]  TRUE  TRUE FALSE FALSE  TRUE  TRUE  TRUE FALSE  TRUE FALSE FALSE
 which(!duplicated(x))
[1] 1 2 5 6 7 9


 

 The 6 and 4 are repeated, I only want the index of the non-repeated values.
 
 I thought of using which() and unique(), but that does not give you the
 index of the unique values.
 
 Thanks in advance.
 
 Mike
 
   [[alternative HTML version deleted]]
 
 __
 R-help@r-project.org mailing list
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 and provide commented, minimal, self-contained, reproducible code.

David Winsemius
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Re: [R] How to delete repeated values in MCMC sampling and get index of unique values?

What answer is wanted for

c(1,1,1,2,3,1)   ?

Note that Duncan's  two suggestions below give different answers for this.

-- Bert

On Thu, Jan 17, 2013 at 1:59 PM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
 On 13-01-17 4:50 PM, C W wrote:

 Dear list,
 How do you delete repeated samples?  In MCMC, when your candidate value
 has
 been reject, so you remain on the same point, so you keep that value.

 Say I have this toy example,

 c(1,6,6,6,3,5,4,4,2,3,5)


 The 6 and 4 are repeated, I only want the index of the non-repeated
 values.

 I thought of using which() and unique(), but that does not give you the
 index of the unique values.


 You could use x[!duplicated(x)] or rle(x)$values, depending on your
 definition of repeated.  I hope you're aware that you can't use either for
 things like quantiles and moments of the limiting distribution.

 x - c(1,6,6,6,3,5,4,4,2,3,5)
 x[!duplicated(x)]
 [1] 1 6 3 5 4 2
 rle(x)$values
 [1] 1 6 3 5 4 2 3 5

 Duncan Murdoch

 __
 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.



-- 

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

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

Re: [R] How to convert a string to the column it represents in a dataframe, with a reproducible example

Hi David,

I would like to have two objects, one containing the values in a column and
the other containing the column's name.

Of course, that's easy to do manually, but I don't want to have to type out
the name of the column more than once (thus, below, I have typed it once in
quotes, and I am trying to find a programatic way to create the other
object, without typing the column name again).

Thank you for your help.

Mark Na


On Thu, Jan 17, 2013 at 4:06 PM, David Winsemius dwinsem...@comcast.netwrote:


 On Jan 17, 2013, at 1:36 PM, mtb...@gmail.com wrote:

  Hello R-helpers,
 
  I have run the following lines of code:
 
  x-cars$dist
  y-noquote(x)
 
 
  Now y is a string containing the characters cars$dist
 
  My questionis there an R function (or combination of functions) that
 I
  can apply to y that will cause y to contain the numbers in cars$dist?
 Even
  better, can I do it without using noquote()?

 What is the goal of this effort?

 --

 David Winsemius
 Alameda, CA, USA



[[alternative HTML version deleted]]

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

Re: [R] How to delete repeated values in MCMC sampling and get index of unique values?

2013-01-17 Thread C W

I was looking for the first answer.

In MCMC, at time t, when the candidate sample is rejected,

 candidate_sample[t] - current_sample

say, at time t+1, the sample is rejected AGAIN, we have

 candidate_sample[t+1] - current_sample

so, at time t, and t+1, we have the same value.  When I calculate the monte
carlo mean, I don't want repeated value.

Mike

On Thu, Jan 17, 2013 at 5:20 PM, Bert Gunter gunter.ber...@gene.com wrote:

 What answer is wanted for

 c(1,1,1,2,3,1)   ?

 Note that Duncan's  two suggestions below give different answers for this.

 -- Bert

 On Thu, Jan 17, 2013 at 1:59 PM, Duncan Murdoch
 murdoch.dun...@gmail.com wrote:
  On 13-01-17 4:50 PM, C W wrote:
 
  Dear list,
  How do you delete repeated samples?  In MCMC, when your candidate value
  has
  been reject, so you remain on the same point, so you keep that value.
 
  Say I have this toy example,
 
  c(1,6,6,6,3,5,4,4,2,3,5)
 
 
  The 6 and 4 are repeated, I only want the index of the non-repeated
  values.
 
  I thought of using which() and unique(), but that does not give you the
  index of the unique values.
 
 
  You could use x[!duplicated(x)] or rle(x)$values, depending on your
  definition of repeated.  I hope you're aware that you can't use either
 for
  things like quantiles and moments of the limiting distribution.
 
  x - c(1,6,6,6,3,5,4,4,2,3,5)
  x[!duplicated(x)]
  [1] 1 6 3 5 4 2
  rle(x)$values
  [1] 1 6 3 5 4 2 3 5
 
  Duncan Murdoch
 
  __
  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.



 --

 Bert Gunter
 Genentech Nonclinical Biostatistics

 Internal Contact Info:
 Phone: 467-7374
 Website:

 http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm


[[alternative HTML version deleted]]

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Re: [R] How to convert a string to the column it represents in a dataframe, with a reproducible example


On Jan 17, 2013, at 2:26 PM, mtb...@gmail.com wrote:

 Hi David,
 
 I would like to have two objects, one containing the values in a column and 
 the other containing the column's name. 

You have not addressed the question ... why? Where are you going with this?

 Of course, that's easy to do manually, but I don't want to have to type out 
 the name of the column more than once (thus, below, I have typed it once in 
 quotes, and I am trying to find a programatic way to create the other object, 
 without typing the column name again).

I would think that this is be best way to proceed:

x - cars[ , dist,  drop=FALSE]

Now x is a data.frame (and inherits from the list-class)  and names(x) will 
return dist and the usual access methods would work. 

-- 
David.
 
 Thank you for your help.
 
 Mark Na
 
 
 On Thu, Jan 17, 2013 at 4:06 PM, David Winsemius dwinsem...@comcast.net 
 wrote:
 
 On Jan 17, 2013, at 1:36 PM, mtb...@gmail.com wrote:
 
  Hello R-helpers,
 
  I have run the following lines of code:
 
  x-cars$dist
  y-noquote(x)
 
 
  Now y is a string containing the characters cars$dist
 
  My questionis there an R function (or combination of functions) that I
  can apply to y that will cause y to contain the numbers in cars$dist? Even
  better, can I do it without using noquote()?
 
 What is the goal of this effort?
 
 --
 
 David Winsemius
 Alameda, CA, USA
 
 

David Winsemius
Alameda, CA, USA

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Re: [R] How to convert a string to the column it represents in a dataframe, with a reproducible example

Inline below.

-- Bert

On Thu, Jan 17, 2013 at 3:02 PM, David Winsemius dwinsem...@comcast.net wrote:

On Jan 17, 2013, at 2:26 PM, mtb...@gmail.com wrote:

Hi David,

I would like to have two objects, one containing the values in a column and
the other containing the column's name.

You have not addressed the question ... why? Where are you going with this?

Of course, that's easy to do manually, but I don't want to have to type out
the name of the column more than once (thus, below, I have typed it once in
quotes, and I am trying to find a programatic way to create the other
object, without typing the column name again).

I would think that this is be best way to proceed:

x - cars[ , dist, drop=FALSE]

Now x is a data.frame (and inherits from the list-class) and names(x) will
return dist and the usual access methods would work.

... which, of course, begs the question: why bother. as you would do
the same with the original data frame.

-- Bert

--
David.

Thank you for your help.

Mark Na

On Thu, Jan 17, 2013 at 4:06 PM, David Winsemius dwinsem...@comcast.net
wrote:

On Jan 17, 2013, at 1:36 PM, mtb...@gmail.com wrote:

Hello R-helpers,

I have run the following lines of code:

x-cars$dist
y-noquote(x)

Now y is a string containing the characters cars$dist

My questionis there an R function (or combination of functions) that I
can apply to y that will cause y to contain the numbers in cars$dist? Even
better, can I do it without using noquote()?

What is the goal of this effort?

David Winsemius
Alameda, CA, USA

__
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.

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

Re: [R] How to delete repeated values in MCMC sampling and get index of unique values?


On 13-01-17 5:33 PM, C W wrote:

I was looking for the first answer.

In MCMC, at time t, when the candidate sample is rejected,

  candidate_sample[t] - current_sample

say, at time t+1, the sample is rejected AGAIN, we have

  candidate_sample[t+1] - current_sample

so, at time t, and t+1, we have the same value.  When I calculate the
monte carlo mean, I don't want repeated value.


Then you will get the wrong answer, unless you want something very strange.

Duncan Murdoch



Mike

On Thu, Jan 17, 2013 at 5:20 PM, Bert Gunter gunter.ber...@gene.com
mailto:gunter.ber...@gene.com wrote:

What answer is wanted for

c(1,1,1,2,3,1)   ?

Note that Duncan's  two suggestions below give different answers for
this.

-- Bert

On Thu, Jan 17, 2013 at 1:59 PM, Duncan Murdoch
murdoch.dun...@gmail.com mailto:murdoch.dun...@gmail.com wrote:
  On 13-01-17 4:50 PM, C W wrote:
 
  Dear list,
  How do you delete repeated samples?  In MCMC, when your
candidate value
  has
  been reject, so you remain on the same point, so you keep that
value.
 
  Say I have this toy example,
 
  c(1,6,6,6,3,5,4,4,2,3,5)
 
 
  The 6 and 4 are repeated, I only want the index of the non-repeated
  values.
 
  I thought of using which() and unique(), but that does not give
you the
  index of the unique values.
 
 
  You could use x[!duplicated(x)] or rle(x)$values, depending on your
  definition of repeated.  I hope you're aware that you can't use
either for
  things like quantiles and moments of the limiting distribution.
 
  x - c(1,6,6,6,3,5,4,4,2,3,5)
  x[!duplicated(x)]
  [1] 1 6 3 5 4 2
  rle(x)$values
  [1] 1 6 3 5 4 2 3 5
 
  Duncan Murdoch
 
  __
  R-help@r-project.org mailto: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.



--

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:

http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm




__
R-help@r-project.org mailing list
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Re: [R] How to convert a string to the column it represents in a dataframe, with a reproducible example

2013-01-17 Thread Nordlund, Dan (DSHS/RDA)

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of mtb...@gmail.com
 Sent: Thursday, January 17, 2013 2:27 PM
 To: David Winsemius; r-help@r-project.org
 Subject: Re: [R] How to convert a string to the column it represents in
 a dataframe, with a reproducible example

 Hi David,

 I would like to have two objects, one containing the values in a column
 and
 the other containing the column's name.

 Of course, that's easy to do manually, but I don't want to have to type
 out
 the name of the column more than once (thus, below, I have typed it
 once in
 quotes, and I am trying to find a programatic way to create the other
 object, without typing the column name again).

 Thank you for your help.

 Mark Na

Something like this

eval(parse(text=y))

could be what you want. But even if it is, I am not sure it is what you should 
want.  Without more context, it is hard to say.

Hope this is at least somewhat helpful, 

Dan

Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204

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Re: [R] How can i keep only latest entries?

2013-01-17 Thread p_connolly


Quoting Mat matthias.we...@fnt.de:


Hello togehter,

i have a data.frame like this one:

No. Date last change
1  1 2012-10-04 change settings
2  1 2012-10-20 bug fix
3  1 2012-11-05 final
4  2 2013-01-15new task
5  2 2013-01-16Bug fix
6  2 2013-01-17final



Calling your data frame, x and assuming the `last change` column is character,

x[x$`last change` == final,]

will give what you want.  If `last change` is a factor column (likely),


x[as.character(x$`last change`) == final,]


If you don't have spaces in your column names, you'd avoid using the 
backticks.


HTH




now i want a new data.frame, who show me only the newest entries according
to the No.
The solution look like this one:

   No. Datelast change
1  1 2012-11-05   final
2  2 2013-01-17   final

can anyone help me?

thanks.

Mat



--
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Re: [R] Sweave, Texshop, and sync with included Rnw file

2013-01-17 Thread michele caseposta

Hi,
I just updated R and patchDVI (from CRAN).
Now I can reverse search from the pdf to the included.Rnw.
However, I cannot forward search from the included to the pdf. Is this how it 
is expected to work?
Forward and inverse search work between main Rnw and pdf.

I am pasting below the code in the two files.

% Sweavetest.Rnw 

\documentclass{article}

\begin{document}

\SweaveOpts{concordance=TRUE}

\title{an Sweave inclusion test}
\author{Michele}

\maketitle

This is an example for the inclusion of RNW files and forward-inverse search.

following is a chunk of sweave code:

=
n - 5
for(i in 1:n){
print(i)
}
@

\section{text from an included file}
\SweaveInput{included.Rnw}

\end{document}

% Included.Rnw 

% !Rnw root = Sweavetest.Rnw

this is some text included in an Rnw file

=
for(i in 1:3){
print('included')
}
@



On Jan 11, 2013, at 5:38 AM, Duncan Murdoch wrote:

 On 13-01-10 4:54 PM, michele caseposta wrote:
 Hi everybody,
 thanks for the replies.
 I might have not explained the problem completely.
 Duncan Mackay:
 Yes, I am already having a master file and separate Rnw files.
 Duncan Murdock:
 I am using patchDVI in the TexShop Sweave engine.
 Sync works flawlessly between the master file and the pdf produced by 
 pdflatex.
 
 My problem is that I don't seem to be able to obtain sync between the 
 *included* Rnws and the pdf, either way.
 
 The sweave engine is as follows:
 
 #!/bin/bash
 
 R CMD Sweave $1
 latexmk -pdf -silent -pdflatex=‘pdflatex –shell-escape –synctex=1′${1%.*}
 Rscript -e library(‘patchDVI’);patchSynctex(‘${1%.*}.synctex.gz’)
 
 
 Funny thing is that the sync works in texworks, using the following Rscript 
 line
 
 patchDVI::SweavePDF('$fullname',stylepath=FALSE)
 
 I tried to mix and match configurations between texshop and texworks but I 
 had no luck
 
 I just tried a simple example in TeXShop and it worked for me.  My Sweave 
 engine is
 
 #!/bin/tcsh
 
 # set path= ($path /usr/local/bin)
 Rscript -e patchDVI::SweavePDF(  '$1' )
 
 
 So it seems to be work listing versions:  On the Mac, I'm using R 2.15.0 
 patched, rev 59478, with patchDVI version 1.8.1584 (I just uploaded 1.9 to 
 CRAN, by the way), TeXShop version 2.43.
 
 It also works on Windows, where I have current releases of R and patchDVI 
 installed.
 
 If you've got current versions of everything installed and it's still not 
 working, could you try putting together a small reproducible example?
 
 Duncan Murdoch
 
 
 
 
 
 
 
 On Jan 10, 2013, at 11:23 AM, Duncan Murdoch wrote:
 
 On 13-01-09 9:09 PM, Duncan Murdoch wrote:
 On 13-01-09 3:25 PM, michele caseposta wrote:
 Hello everyone.
 I am in the process of writing a book in Latex with Texshop, on Mac.
 This book contains a lot of R code, hence the need to use Sweave.
 I was able to compile Rnw files, and to sync back and forth from the pdf 
 to the source Rnw.
 My problem now is that the book is divided in Chapters, and every chapter 
 is in its own Rnw file.
 I can compile them from the main one (book.Rnw) using the directive
 
 \SweaveInput{chapter1.Rnw}
 
 The problem stands in the fact that like this I am missing 
 synchronization between the pdf and the source Rnw. If part of text is in 
 book.Rnw I can synchronize, but if the text is in one of the included 
 files, it just doesn't work.
 I am using the sweave engine found in the following webpage:
 
 http://cameron.bracken.bz/synctex-with-sweavepgfsweave-in-texshoptexworks
 
 Has anybody succeeded in synchronizing with included Rnw files?
 
 This is a problem addressed by my patchDVI package, available on
 R-forge.  You have a main file (which can be .tex or .Rnw), and put code
 at the start of each .Rnw file to indicate where to find it.  Then you
 just run Sweave on one of the chapters, and it automatically produces
 the full document.
 
 The sample document here:
 
 http://www.umanitoba.ca/statistics/seminars/2011/3/4/duncan-murdoch-using-sweave-R/
 
 includes an appendix describing how to set this up with TeXShop.
 
 I just committed an update to the vignette in patchDVI giving a quick 
 version of the instructions for basic use. Version 1.8.1585 has the new 
 vignette.
 
 I should get around to pushing it to CRAN one of these days...
 
 Duncan Murdoch
 
 
 
 

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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] Line breaks in documentation \useage

2013-01-17 Thread Rolf Turner


On 01/18/2013 12:33 AM, Markku Karhunen wrote:



Hi again, R community.

I wonder how you do line breaks in \useage{} section in .Rd files. I 
am sure there's some tutorial for this somewhere, but I just haven't 
found it.


I have tried \\, \cr, \br and \newline, admittedly arbitrarily, but 
all of these produce warnings or errors.


The \usage{} section appears to be treated verbatim or as if it were 
preformated.
So just put the line break literally in your *.Rd and it will 
automagically appear in the

resulting help.

cheers,

Rolf Turner

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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] how to use ...

2013-01-17 Thread Rolf Turner




The help facility is applicable to functions and data sets.  It is not 
designed

or intended to give help with respect to R syntax (with the exception of
the basic syntax of the operators --- unary and binary --- and the 
associated

rules of precedence).

cheers,

Rolf Turner

On 01/18/2013 09:17 AM, Steve Taylor wrote:

The ellipsis object is not listed in the base help pages!

help(`+`) # this works - help on arithmetic operators
help(+) # also works
help(`...`) # fails with Error: '...' used in an incorrect context
help(...) # fails also with No documentation for '...' in specified packages 
and libraries: you could try '??...'


-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of Ivan Calandra
Sent: Friday, 18 January 2013 4:43a
To: R. Michael Weylandt
Cc: R list
Subject: Re: [R] how to use ...

Do you know where I can find some documentation about it?


__
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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.

[R] powerTransform Warning Message

2013-01-17 Thread Beatriz González

Hello,

I would like to perform a Box-Cox (âbcPowerâ) transformation on my data. 
For this, I am determining lambda using the âpowerTransformâ function. 
However, with one of my variables I get the following Warning Message:
In estimateTransform(x, y, NULL, ...) :
  Convergence failure: return code = 52
My variable is: 
 x
 [1] 0.0001031130 0.0001029480 0.0001040010 0.0001037940 0.0001046280 
0.927650 0.942960
 [8] 0.950250 0.949780 0.988210 0.975860 0.962850 
0.0001015800 0.0001009190
[15] 0.0001035350 0.0001033210 0.0001028420 0.0001023150 0.0001033000 
0.0001036910 0.0001026890
[22] 0.0001032480 0.0001012160 0.0001031270 0.0001034390 0.0001028180 
0.0001034820 0.0001022700
[29] 0.0001042640 0.0001046720 0.0001045860 0.0001045160 0.0001045850 
0.0001038040 0.0001037020
[36] 0.0001034050 0.0001048150 0.0001017180 0.0001017920 0.0001032730 
0.0001029420 0.0001025220
[43] 0.0001031490 0.0001036995 0.0001039735 0.0001014877 0.0001027960 And my 
code is:dlambda - powerTransform(x)   #It is here where I get the warning 
message

x_BCx - bcPower(x, lambda= dlambda$lambda, jacobian.adjusted= FALSE)
  Does anyone have an idea of why I could be getting this Warning Message?  If 
you require more details please get in touch.

Thanks a lot!

Bea
[[alternative HTML version deleted]]

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

Re: [R] Sweave, Texshop, and sync with included Rnw file

2013-01-17 Thread huang min

Anybody know if it is possible to use texshop and knitr with the sync
working? I add a knitr engine but cannot sync.

PS, I am comfortable with texshop but not RStudio.

Huang

On Fri, Jan 18, 2013 at 7:33 AM, michele caseposta mic.c...@gmail.comwrote:

 Hi,
 I just updated R and patchDVI (from CRAN).
 Now I can reverse search from the pdf to the included.Rnw.
 However, I cannot forward search from the included to the pdf. Is this how
 it is expected to work?
 Forward and inverse search work between main Rnw and pdf.

 I am pasting below the code in the two files.

 % Sweavetest.Rnw 

 \documentclass{article}

 \begin{document}

 \SweaveOpts{concordance=TRUE}

 \title{an Sweave inclusion test}
 \author{Michele}

 \maketitle

 This is an example for the inclusion of RNW files and forward-inverse
 search.

 following is a chunk of sweave code:

 =
 n - 5
 for(i in 1:n){
 print(i)
 }
 @

 \section{text from an included file}
 \SweaveInput{included.Rnw}

 \end{document}

 % Included.Rnw 

 % !Rnw root = Sweavetest.Rnw

 this is some text included in an Rnw file

 =
 for(i in 1:3){
 print('included')
 }
 @



 On Jan 11, 2013, at 5:38 AM, Duncan Murdoch wrote:

  On 13-01-10 4:54 PM, michele caseposta wrote:
  Hi everybody,
  thanks for the replies.
  I might have not explained the problem completely.
  Duncan Mackay:
  Yes, I am already having a master file and separate Rnw files.
  Duncan Murdock:
  I am using patchDVI in the TexShop Sweave engine.
  Sync works flawlessly between the master file and the pdf produced by
 pdflatex.
 
  My problem is that I don't seem to be able to obtain sync between the
 *included* Rnws and the pdf, either way.
 
  The sweave engine is as follows:
 
  #!/bin/bash
 
  R CMD Sweave $1
  latexmk -pdf -silent -pdflatex=âpdflatex âshell-escape
 âsynctex=1â²${1%.*}
  Rscript -e library(âpatchDVIâ);patchSynctex(â${1%.*}.synctex.gzâ)
 
 
  Funny thing is that the sync works in texworks, using the following
 Rscript line
 
  patchDVI::SweavePDF('$fullname',stylepath=FALSE)
 
  I tried to mix and match configurations between texshop and texworks
 but I had no luck
 
  I just tried a simple example in TeXShop and it worked for me.  My
 Sweave engine is
 
  #!/bin/tcsh
 
  # set path= ($path /usr/local/bin)
  Rscript -e patchDVI::SweavePDF(  '$1' )
 
 
  So it seems to be work listing versions:  On the Mac, I'm using R 2.15.0
 patched, rev 59478, with patchDVI version 1.8.1584 (I just uploaded 1.9 to
 CRAN, by the way), TeXShop version 2.43.
 
  It also works on Windows, where I have current releases of R and
 patchDVI installed.
 
  If you've got current versions of everything installed and it's still
 not working, could you try putting together a small reproducible example?
 
  Duncan Murdoch
 
 
 
 
 
 
 
  On Jan 10, 2013, at 11:23 AM, Duncan Murdoch wrote:
 
  On 13-01-09 9:09 PM, Duncan Murdoch wrote:
  On 13-01-09 3:25 PM, michele caseposta wrote:
  Hello everyone.
  I am in the process of writing a book in Latex with Texshop, on Mac.
  This book contains a lot of R code, hence the need to use Sweave.
  I was able to compile Rnw files, and to sync back and forth from the
 pdf to the source Rnw.
  My problem now is that the book is divided in Chapters, and every
 chapter is in its own Rnw file.
  I can compile them from the main one (book.Rnw) using the directive
 
  \SweaveInput{chapter1.Rnw}
 
  The problem stands in the fact that like this I am missing
 synchronization between the pdf and the source Rnw. If part of text is in
 book.Rnw I can synchronize, but if the text is in one of the included
 files, it just doesn't work.
  I am using the sweave engine found in the following webpage:
 
 
 http://cameron.bracken.bz/synctex-with-sweavepgfsweave-in-texshoptexworks
 
  Has anybody succeeded in synchronizing with included Rnw files?
 
  This is a problem addressed by my patchDVI package, available on
  R-forge.  You have a main file (which can be .tex or .Rnw), and put
 code
  at the start of each .Rnw file to indicate where to find it.  Then you
  just run Sweave on one of the chapters, and it automatically produces
  the full document.
 
  The sample document here:
 
 
 http://www.umanitoba.ca/statistics/seminars/2011/3/4/duncan-murdoch-using-sweave-R/
 
  includes an appendix describing how to set this up with TeXShop.
 
  I just committed an update to the vignette in patchDVI giving a quick
 version of the instructions for basic use. Version 1.8.1585 has the new
 vignette.
 
  I should get around to pushing it to CRAN one of these days...
 
  Duncan Murdoch
 
 
 
 

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


[[alternative HTML version deleted]]

Re: [R] Sweave, Texshop, and sync with included Rnw file

2013-01-17 Thread Yihui Xie

Sweave produces foo-concordance.tex from foo.Rnw, and writes
\input{foo-concordance.tex} in the LaTeX output. You can turn on the
concordance option in knitr as well. Since you do not use RStudio, you
have a couple of more steps to go:

1. borrow \Sconcordance from Sweave.sty;
2. manually \input{foo-concordance.tex};
3. set the option concordance=TRUE in opts_knit;

The rest of steps are the same as Sweave -- you compile the output by
`pdflatex --synctex=1 foo.tex`, and run patchDVI on the results. I do
not use TeXShop, so I'm not sure if it really works.

\documentclass{article}
% from Sweave.sty in R
\newcommand{\Sconcordance}[1]{%
  \ifx\pdfoutput\undefined%
  \csname newcount\endcsname\pdfoutput\fi%
  \ifcase\pdfoutput\special{#1}%
  \else%
   \begingroup%
 \pdfcompresslevel=0%
 \immediate\pdfobj stream{#1}%
 \pdfcatalog{/SweaveConcordance \the\pdflastobj\space 0 R}%
   \endgroup%
  \fi}
\input{foo-concordance.tex}

setup, include=FALSE=
opts_knit$set(concordance=TRUE)
@

\begin{document}

=
for (i in 1:5) {
  print(i)
}
@

\end{document}

Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA


On Thu, Jan 17, 2013 at 7:45 PM, huang min minhua...@gmail.com wrote:
 Anybody know if it is possible to use texshop and knitr with the sync
 working? I add a knitr engine but cannot sync.

 PS, I am comfortable with texshop but not RStudio.

 Huang

 On Fri, Jan 18, 2013 at 7:33 AM, michele caseposta mic.c...@gmail.comwrote:

 Hi,
 I just updated R and patchDVI (from CRAN).
 Now I can reverse search from the pdf to the included.Rnw.
 However, I cannot forward search from the included to the pdf. Is this how
 it is expected to work?
 Forward and inverse search work between main Rnw and pdf.

 I am pasting below the code in the two files.

 % Sweavetest.Rnw 

 \documentclass{article}

 \begin{document}

 \SweaveOpts{concordance=TRUE}

 \title{an Sweave inclusion test}
 \author{Michele}

 \maketitle

 This is an example for the inclusion of RNW files and forward-inverse
 search.

 following is a chunk of sweave code:

 =
 n - 5
 for(i in 1:n){
 print(i)
 }
 @

 \section{text from an included file}
 \SweaveInput{included.Rnw}

 \end{document}

 % Included.Rnw 

 % !Rnw root = Sweavetest.Rnw

 this is some text included in an Rnw file

 =
 for(i in 1:3){
 print('included')
 }
 @



 On Jan 11, 2013, at 5:38 AM, Duncan Murdoch wrote:

  On 13-01-10 4:54 PM, michele caseposta wrote:
  Hi everybody,
  thanks for the replies.
  I might have not explained the problem completely.
  Duncan Mackay:
  Yes, I am already having a master file and separate Rnw files.
  Duncan Murdock:
  I am using patchDVI in the TexShop Sweave engine.
  Sync works flawlessly between the master file and the pdf produced by
 pdflatex.
 
  My problem is that I don't seem to be able to obtain sync between the
 *included* Rnws and the pdf, either way.
 
  The sweave engine is as follows:
 
  #!/bin/bash
 
  R CMD Sweave $1
  latexmk -pdf -silent -pdflatex=‘pdflatex –shell-escape
 –synctex=1′${1%.*}
  Rscript -e library(‘patchDVI’);patchSynctex(‘${1%.*}.synctex.gz’)
 
 
  Funny thing is that the sync works in texworks, using the following
 Rscript line
 
  patchDVI::SweavePDF('$fullname',stylepath=FALSE)
 
  I tried to mix and match configurations between texshop and texworks
 but I had no luck
 
  I just tried a simple example in TeXShop and it worked for me.  My
 Sweave engine is
 
  #!/bin/tcsh
 
  # set path= ($path /usr/local/bin)
  Rscript -e patchDVI::SweavePDF(  '$1' )
 
 
  So it seems to be work listing versions:  On the Mac, I'm using R 2.15.0
 patched, rev 59478, with patchDVI version 1.8.1584 (I just uploaded 1.9 to
 CRAN, by the way), TeXShop version 2.43.
 
  It also works on Windows, where I have current releases of R and
 patchDVI installed.
 
  If you've got current versions of everything installed and it's still
 not working, could you try putting together a small reproducible example?
 
  Duncan Murdoch
 
 
 
 
 
 
 
  On Jan 10, 2013, at 11:23 AM, Duncan Murdoch wrote:
 
  On 13-01-09 9:09 PM, Duncan Murdoch wrote:
  On 13-01-09 3:25 PM, michele caseposta wrote:
  Hello everyone.
  I am in the process of writing a book in Latex with Texshop, on Mac.
  This book contains a lot of R code, hence the need to use Sweave.
  I was able to compile Rnw files, and to sync back and forth from the
 pdf to the source Rnw.
  My problem now is that the book is divided in Chapters, and every
 chapter is in its own Rnw file.
  I can compile them from the main one (book.Rnw) using the directive
 
  \SweaveInput{chapter1.Rnw}
 
  The problem stands in the fact that like this I am missing
 synchronization between the pdf and the source Rnw. If part of text is in
 book.Rnw I can synchronize, but if the text is in one of the included
 files, it just doesn't work.
  I am using the sweave

Re: [R] Colors in interaction plots

But I don't want to plot random colors.
...
 That's why I have this vector with length 24 - each one matches one line in 
 the npk dataset.
... which is not what interaction.plot, or matplot, needs; it needs one per 
line on the plot.

How can I inform to the interaction.plot function the color
corresponding to the block it will plot?

You need a colour vector the same length as the number of levels you're 
plotting. One easy way to do that would be to do something like
col=1:nlevels(fac), for example as in 
 interaction.plot(npk$N, npk$block, fit, xlab=N,  
ylab=yield,col=1:nlevels(fac))

If you have several levels that correspond to the same level of a third factor, 
you need to provide a cross-reference of sorts. In your toy example, fac 
corresponds to three levels of block, so one could specify manually. One way of 
doing it in code, though, could be to use table to identify levels of fac 
corresponding to levels of block. That sounds a bit complicated, but let's see:

npk$fac - fac #just so it's in the same data frame
fac.by.block - with(npk, table(fac, block)) #cross-reference fac levels by 
block
fac.index.by.block.level - apply(fac.by.block, 2, function(x) which(x0)[1])
   #Assumes that you want the first nonzero table entry and 
that numerical indices are OK
  # you could also use which directly: which(fac.by.block 1, 
arr.ind=TRUE)[,1] gives the same result IF there's a 1:1 fac:block matching

Then
 interaction.plot(npk$N, npk$block, fit, xlab=N,  
ylab=yield,col=(1:nlevels(fac))[fac.index.by.block.level])
... which I think is something like what you;re after?

S

***
This email and any attachments are confidential. Any use...{{dropped:8}}

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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] Using table to get frequencies of several factors at once

2013-01-17 Thread Greg Snow

The tables package may be of use to you for this.


On Fri, Jan 11, 2013 at 4:17 AM, Pancho Mulongeni 
p.mulong...@namibia.pharmaccess.org wrote:

 Hi, I have a dataframe with n columns, but I am only looking at five of
 them. And lots of rows, over 700.
 So I would like to find frequencies for each of the numeric columns
 (variables) using the table function. However, is there a fast way to
 produce a frequency table where the 5 rows represent the 5 numeric
 variables and the columns refer to the values (levels) of the respective
 numeric variables, which in this case are 0 and 1.
 The only way I have figured it out is via a for loop:
 m-seq(218,222,1) #these are columns of the variables in the larger
 dataframe
 tm-m[1:5] #I need this for the for loop
 l.tm-length(tm)
 B-matrix(nrow=l.tm,ncol=2)  #the matrix to hold the freqs
 for (p in 1:l.tm) {
 var.num-m[p]
 B[p,]-table(DATA[,var.num])
 }

  B
  [,1] [,2]
 [1,]  6979
 [2,]  512  194
 [3,]  604  102
 [4,]  7006
 [5,]  706  706
 So the rows represent my five variables (columns) that occupy columns 218
 through 222 in the DATA dataframe.
 So the second column represents my frequencies of the value 1, which is
 what I am interested in. The last row has a double entry, because there was
 only one value, 0, with a freq of 706 and so R duplicated in the two
 columns, but that's ok, I can just ignore it.

 So is there are better way to do this? Is there a way to use the so called
 tapply function? I struggle to understand the help doc for this.function.


 Pancho Mulongeni
 Research Assistant
 PharmAccess Foundation
 1 Fouché Street
 Windhoek West
 Windhoek
 Namibia

 Tel:   +264 61 419 000
 Fax:  +264 61 419 001/2
 Mob: +264 81 4456 286

 __
 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.




-- 
Gregory (Greg) L. Snow Ph.D.
538...@gmail.com

[[alternative HTML version deleted]]

__
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] how to use ...

2013-01-17 Thread John Sorkin

Rolf
Perhaps the philosophy of the help system needs to change . . .
John

Sent from my iPhone

On Jan 17, 2013, at 7:11 PM, Rolf Turner rolf.tur...@xtra.co.nz 
rolf.tur...@xtra.co.nz wrote:

 
 
 The help facility is applicable to functions and data sets.  It is not 
 designed
 or intended to give help with respect to R syntax (with the exception of
 the basic syntax of the operators --- unary and binary --- and the 
 associated
 rules of precedence).
 
 cheers,
 
 Rolf Turner
 
 On 01/18/2013 09:17 AM, Steve Taylor wrote:
 The ellipsis object is not listed in the base help pages!
 
 help(`+`) # this works - help on arithmetic operators
 help(+) # also works
 help(`...`) # fails with Error: '...' used in an incorrect context
 help(...) # fails also with No documentation for '...' in specified 
 packages and libraries: you could try '??...'
 
 
 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf Of Ivan Calandra
 Sent: Friday, 18 January 2013 4:43a
 To: R. Michael Weylandt
 Cc: R list
 Subject: =?utf-8?B?UmU6IFtSXSBob3cgdG8gdXNlICIuLi4i==? 
 Do you know where I can find some documentation about 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.
 

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

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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] how to use ...