date:20090423



On Apr 24, 2009, at 12:16 AM, David Winsemius wrote:

I cannot figure out what you are trying to do, but I'm reasonably  
sure that the construction:


xxx[[index,index2]] ... is going to fail. If you want to create a  
two dimensional structure, most people would use either a dataframe  
or a matrix rather than a list, although lists have the advantage  
that they do not need to be "pre-dimensioned". If you know what your  
dimension are in advance they are probably the way to go.


If you want to access multiple "dimensions" of a list, do not use  
[[ , ]] but rather [[ idx ]] to get the desired first "dimension"  
and then [ idx2 ] to the right of the expression for the first  
"dimension" to get or change the second "dimension", although  
"level" or "depth" might be the better terminology.


m <- list(sin, 1:3, letters[1:3], expression(a+b))
m[[2]][3] # ]

# that "]" shouldn't be there


m[[2]][3] <- 8
m
m[[2]][3]

You have a better chance of getting on target advice if you explain  
what you want to do rather than showing us code patterned on another  
language that obviously doesn't work in R.


On Apr 23, 2009, at 10:29 PM, Toby wrote:

I must be going about my idea really backwards.  I have functions  
like so:


Nasa_PART_Bounds <- rbind(Nasa_PART_Bounds, c("Nasa_PART_B1"))
Nasa_PART_B1 <- function(l,u) {
  l[["Nasa_PART_B1", "CYCLOMATIC_COMPLEXITY"]] <- 8
  u[["Nasa_PART_B1", "CYCLOMATIC_COMPLEXITY"]] <- 60
  l[["Nasa_PART_B1", "LOC_TOTAL"]] <- 73
}

Where I add the function names of each "bounds" function to a list,
which later on get called with two variables.

dispatch2 <- function(f, x, y) {
  eval(call(f, x, y))
}

for (i in bounds) {
  bb <- eval(substitute(foo[,1], list(foo=as.name(i
  rr <- lapply(bb, dispatch2, lb, ub)
}

However, I don't know how to do this in a better "R" way.  I think  
my C and

pearl experience is showing through... :(

-Toby.


On Thu, Apr 23, 2009 at 8:20 PM, Gabor Grothendieck 
wrote:



Matrix made from a list:

m <- list(sin, 1:3, letters[1:3], expression(a+b))
dim(m) <- c(2, 2)
dimnames(m) <- list(letters[1:2], LETTERS[1:2])
class(m) # matrix

or

M <- structure(list(sin, 1:3, letters[1:3], expression(a+b)), .Dim  
= c(2,

2),
.Dimnames = list(c("a", "b"), c("A", "B")))
class(M) # matrix

On Thu, Apr 23, 2009 at 10:03 PM, Toby  


wrote:

I'm trying to figure out how I can get a generalized 2D
list/array/matrix/whatever
working.  Seems I can't figure out how to make the variables the  
right

type.  I
always seem to get some sort of error... out of bounds, wrong  
type, wrong

dim, etc.
Very confused... :)

x[["some label", "some other index"]] <- 3
x[["some other label", "something else"]] <- 4

I don't know the indexes/label ahead of time... they get  
generated...

Any

thoughts?




David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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

Re: [R] Generalized 2D list/array/whatever?

I cannot figure out what you are trying to do, but I'm reasonably sure  
that the construction:


xxx[[index,index2]] ... is going to fail. If you want to create a two  
dimensional structure, most people would use either a dataframe or a  
matrix rather than a list, although lists have the advantage that they  
do not need to be "pre-dimensioned". If you know what your dimension  
are in advance they are probably the way to go.


If you want to access multiple "dimensions" of a list, do not use  
[[ , ]] but rather [[ idx ]] to get the desired first "dimension" and  
then [ idx2 ] to the right of the expression for the first "dimension"  
to get or change the second "dimension", although "level" or "depth"  
might be the better terminology.


m <- list(sin, 1:3, letters[1:3], expression(a+b))
m[[2]][3]  ]
m[[2]][3] <- 8
m
m[[2]][3]

You have a better chance of getting on target advice if you explain  
what you want to do rather than showing us code patterned on another  
language that obviously doesn't work in R.


On Apr 23, 2009, at 10:29 PM, Toby wrote:

I must be going about my idea really backwards.  I have functions  
like so:


Nasa_PART_Bounds <- rbind(Nasa_PART_Bounds, c("Nasa_PART_B1"))
Nasa_PART_B1 <- function(l,u) {
   l[["Nasa_PART_B1", "CYCLOMATIC_COMPLEXITY"]] <- 8
   u[["Nasa_PART_B1", "CYCLOMATIC_COMPLEXITY"]] <- 60
   l[["Nasa_PART_B1", "LOC_TOTAL"]] <- 73
}

Where I add the function names of each "bounds" function to a list,
which later on get called with two variables.

dispatch2 <- function(f, x, y) {
   eval(call(f, x, y))
}

for (i in bounds) {
   bb <- eval(substitute(foo[,1], list(foo=as.name(i
   rr <- lapply(bb, dispatch2, lb, ub)
}

However, I don't know how to do this in a better "R" way.  I think  
my C and

pearl experience is showing through... :(

-Toby.


On Thu, Apr 23, 2009 at 8:20 PM, Gabor Grothendieck 
wrote:



Matrix made from a list:

m <- list(sin, 1:3, letters[1:3], expression(a+b))
dim(m) <- c(2, 2)
dimnames(m) <- list(letters[1:2], LETTERS[1:2])
class(m) # matrix

or

M <- structure(list(sin, 1:3, letters[1:3], expression(a+b)), .Dim  
= c(2,

2),
.Dimnames = list(c("a", "b"), c("A", "B")))
class(M) # matrix

On Thu, Apr 23, 2009 at 10:03 PM, Toby 
wrote:

I'm trying to figure out how I can get a generalized 2D
list/array/matrix/whatever
working.  Seems I can't figure out how to make the variables the  
right

type.  I
always seem to get some sort of error... out of bounds, wrong  
type, wrong

dim, etc.
Very confused... :)

x[["some label", "some other index"]] <- 3
x[["some other label", "something else"]] <- 4

I don't know the indexes/label ahead of time... they get  
generated...

Any

thoughts?


David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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

Re: [R] deleting rows provisionally

2009-04-23 Thread Ben Bolker




onyourmark wrote:
> 
> I have an object. I think it is a matrix, called 'answer2'
>  str(answer2)
>  int [1:1537, 1:2] 1 399 653 2 3 600 4 5 271 870 ...
>  - attr(*, "dimnames")=List of 2
>   ..$ : chr [1:1537] "a4.1" "hirschsprung.399" "peritoneal.653"
> "abdomen.2" ...
>   ..$ : chr [1:2] "row" "col"
> 
> 
> I want to delete rows that have the same entries.
> 
> 

Your "for" loop didn't make any assignments.
How about 

answer3 <- answer2[-(answer2[,1]==answer2[,2]),]  

or

answer3 <- answer2[answer2[,1]!=answer2[,2],]  

(which will be much faster than a loop anyway)

?
-- 
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[R] deleting rows provisionally

2009-04-23 Thread onyourmark


I have an object. I think it is a matrix, called 'answer2'
 str(answer2)
 int [1:1537, 1:2] 1 399 653 2 3 600 4 5 271 870 ...
 - attr(*, "dimnames")=List of 2
  ..$ : chr [1:1537] "a4.1" "hirschsprung.399" "peritoneal.653" "abdomen.2"
...
  ..$ : chr [1:2] "row" "col"


I want to delete rows that have the same entries.
For example:
> answer2[4,1]
[1] 2
> answer2[4,2]
[1] 2
> 
And so I would like to delete row 4.

I tried the following but it does not work.
I defined answer3 to hold the result by first:
>answer3<-answer2
and then:
> for(i in 1:1537){if(answer2[i,1]==answer2[i,2]){answer3[-i,]}}

Why doesn't this work?
Thanks.
-- 
View this message in context: 
http://www.nabble.com/deleting-rows-provisionally-tp23209365p23209365.html
Sent from the R help mailing list archive at Nabble.com.

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

Re: [R] Error building package: LaTeX error when creating PDF version

2009-04-23 Thread Peter Dunn

Duncan suggested R CMD check was not finding  pdflatex, because there is no 
.log file in the same directory as the .tex file.
 
However, the first line of output from R CMD check says:  
> * checking for working pdflatex ... OK
Indeed:
pdu...@pdunnubuntu:~/DSdata/GLMsData.Rchecks$ R CMD pdflatex
This is pdfTeXk, Version 3.141592-1.40.3 (Web2C 7.5.6)
 %&-line parsing enabled.
**
 
And it waits for input.
 
So that doesn't seem to be the problem.
 
Further suggestions?
 
P.
 
 
 
Peter Dunn: Biostatistician
School of Health and Sport Science
Faculty of Science, Health and Education  ML-34
University of the Sunshine Coast, Locked Bag 4
Maroochydore DC Qld 4558
 
Tel: +61 7 5456 5085
Fax: +61 7 5430 2896
Email: pdu...@usc.edu.au 
www.usc.edu.au ( http://www.usc.edu.au/ )

>>> Duncan Murdoch  24/04/2009 10:40 am >>>
On 23/04/2009 8:24 PM, Peter Dunn wrote:
> Hi all
>  
> I am trying to build an R package, which I have successfully done many times 
> before, but have an error I cannot trace.  I hope someone can help me.
>  
> Here's is some edited output (full output below if it is useful):
>  
> pdu...@pdunnubuntu:~/DSdata$ R CMD build GLMsData
> * checking for file 'GLMsData/DESCRIPTION' ... OK
> * preparing 'GLMsData':
> * checking DESCRIPTION meta-information ... OK
> * removing junk files
> * checking for LF line-endings in source and make files
> * checking for empty or unneeded directories
> * building 'GLMsData_0.7.tar.gz'
> 
> pdu...@pdunnubuntu:~/DSdata$ R CMD check GLMsData
> * checking for working pdflatex ... OK
> * using log directory '/home/pdunn2/DSdata/GLMsData.Rcheck'
> * using R version 2.9.0 (2009-04-17)
> 
> 
>  
> * checking data for non-ASCII characters ... OK
> * checking examples ... OK
> * checking PDF version of manual ... WARNING
> LaTeX errors when creating PDF version.
> This typically indicates Rd problems.
> * checking PDF version of manual without index ... ERROR
>  
>  
> So there is an error making the PDf version of the package manual.  That's 
> fine; I am happy to fix it.  But first I need to find it.
>  
> So I enter the directory GLMsData.Rcheck and find the corresponding tex file 
> for the manual, and run  pdflatex  on this tex file:
>  
> pdu...@pdunnubuntu:~/DSdata$ cd GLMsData.Rcheck/
> pdu...@pdunnubuntu:~/DSdata/GLMsData.Rcheck$ pdflatex GLMsData-manual.tex
> 
> 
> 
> Output written on GLMsData-manual.pdf (79 pages, 405091 bytes).
> Transcript written on GLMsData-manual.log.
> pdu...@pdunnubuntu:~/DSdata/GLMsData.Rcheck$ 
>  
>  
> So from what I gather:
> * R CMD check  finds an error when using  pdflatex  to make the PDF version 
> of the manual
> * But when I manually run  pdflatex  on the tex file, it works OK.
>  
> Obviously, I am missing something.  Can someone tell me what it is?  I can't 
> find enlightenment on the archives or the manuals.

You should also have a .log file in the same directory as the .tex file. 
  You probably have a new one now that pdflatex works; please move that 
one aside and run R CMD check again.  If you get a .log file, it should 
list whatever the error was; you could compare it to the .log that 
worked if you can't spot it.  If you don't get a new log, that means R 
CMD check couldn't run pdflatex.

Duncan Murdoch

> 
> Thanks everyone.
> 
> P.
> 
> 
> 
>> sessionInfo()
> R version 2.9.0 (2009-04-17)
> i486-pc-linux-gnu
> 
> locale:
> LC_CTYPE=en_AU.UTF-8;LC_NUMERIC=C;LC_TIME=en_AU.UTF-8;LC_COLLATE=en_AU.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_AU.UTF-8;LC_PAPER=en_AU.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_AU.UTF-8;LC_IDENTIFICATION=C
> 
> attached base packages:
> [1] stats graphics  grDevices utils datasets  methodsbase
> 
> 
> 
> 
> 
> FULL OUTPUT:
> 
> pdu...@pdunnubuntu:~/DSdata$ R CMD build GLMsData
> * checking for file 'GLMsData/DESCRIPTION' ... OK
> * preparing 'GLMsData':
> * checking DESCRIPTION meta-information ... OK
> * removing junk files
> * checking for LF line-endings in source and make files
> * checking for empty or unneeded directories
> * building 'GLMsData_0.7.tar.gz'
> 
> pdu...@pdunnubuntu:~/DSdata$ R CMD check GLMsData
> * checking for working pdflatex ... OK
> * using log directory '/home/pdunn2/DSdata/GLMsData.Rcheck'
> * using R version 2.9.0 (2009-04-17)
> * using session charset: UTF-8
> * checking for file 'GLMsData/DESCRIPTION' ... OK
> * checking extension type ... Package
> * this is package 'GLMsData' version '0.7'
> * checking package dependencies ... OK
> * checking if this is a source package ... OK
> * checking for executable files ... OK
> * checking whether package 'GLMsData' can be installed ... OK
> * checking package directory ... OK
> * checking for portable file names ... OK
> * checking for sufficient/correct file permissions ... OK
> * checking DESCRIPTION meta-information ... OK
> * checking top-level files ... OK
> * checking index information ... OK
> * checking package subdirectories ... OK
> * checking whether the package can be loade

Re: [R] Generalized 2D list/array/whatever?

2009-04-23 Thread Toby

I must be going about my idea really backwards.  I have functions like so:

Nasa_PART_Bounds <- rbind(Nasa_PART_Bounds, c("Nasa_PART_B1"))
Nasa_PART_B1 <- function(l,u) {
l[["Nasa_PART_B1", "CYCLOMATIC_COMPLEXITY"]] <- 8
u[["Nasa_PART_B1", "CYCLOMATIC_COMPLEXITY"]] <- 60
l[["Nasa_PART_B1", "LOC_TOTAL"]] <- 73
}

Where I add the function names of each "bounds" function to a list,
which later on get called with two variables.

dispatch2 <- function(f, x, y) {
eval(call(f, x, y))
}

for (i in bounds) {
bb <- eval(substitute(foo[,1], list(foo=as.name(i
rr <- lapply(bb, dispatch2, lb, ub)
}

However, I don't know how to do this in a better "R" way.  I think my C and
pearl experience is showing through... :(

-Toby.


On Thu, Apr 23, 2009 at 8:20 PM, Gabor Grothendieck  wrote:

> Matrix made from a list:
>
> m <- list(sin, 1:3, letters[1:3], expression(a+b))
> dim(m) <- c(2, 2)
> dimnames(m) <- list(letters[1:2], LETTERS[1:2])
> class(m) # matrix
>
> or
>
> M <- structure(list(sin, 1:3, letters[1:3], expression(a+b)), .Dim = c(2,
> 2),
>  .Dimnames = list(c("a", "b"), c("A", "B")))
> class(M) # matrix
>
> On Thu, Apr 23, 2009 at 10:03 PM, Toby 
> wrote:
> > I'm trying to figure out how I can get a generalized 2D
> > list/array/matrix/whatever
> > working.  Seems I can't figure out how to make the variables the right
> > type.  I
> > always seem to get some sort of error... out of bounds, wrong type, wrong
> > dim, etc.
> > Very confused... :)
> >
> > x[["some label", "some other index"]] <- 3
> > x[["some other label", "something else"]] <- 4
> >
> > I don't know the indexes/label ahead of time... they get generated...
>  Any
> > thoughts?
> >
> > -Toby.
> >
> >[[alternative HTML version deleted]]
> >
> > __
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>

[[alternative HTML version deleted]]

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Re: [R] joining multiple lists

I'm not sure that merge will do what you originally requested. At  
least look at rbind before deciding to go with merge.



On Apr 23, 2009, at 9:30 PM, Daniel Bradley wrote:


Merge()!

Thanks!

On Fri, Apr 24, 2009 at 11:26 AM, Daniel Bradley >wrote:



Hi!

I'm reading and cleaning data from multiple excel spreadsheets.   
All the
data has the same column names and the natural next step is to join  
the
lists of data together to make one list for reporting purposes.  I  
had
thought that c(file1data, file2data) would do the trick but that  
seems to
append the new columns to the list so I end up with duplicate  
column names
"col1" "col2" "col3" "col1" col2" "col3" rather than consolidated  
column
data.  It's obviously a straight forward operation, but damned if I  
can find
a straight forward explanation on the web.  Any assistance would be  
most

appreciated!



David Winsemius, MD
Heritage Laboratories
West Hartford, CT

__
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Re: [R] Help with for/if loop



On Apr 23, 2009, at 8:07 PM, Giggles_Fairy wrote:



I have a set of data that includes various data columns. One if the  
survival
time and another if a continuous variable of ages. I want to put the  
ages
into intervals so that I can then perform the Kalpan Meier test. I  
am trying
to use the following code to build a column with the age group  
numbers in


agecatagory<-c( )
for (i in 1:137)
{
{
if(age[i]<=46)  {agecat[i]<-1}
if(age[i]>46 & age[i]<= 58) {agecat[i]<-2}
if(age[i]>58) {agecat[i]<-3}
}
agecatagory<-c(agecatagory, agecat[i])
}


My help with your loop will be to eliminate it. Try:
# age= sample(20:120, 100, replace=T)
# use your own data

 agecatagory<- cut(age, breaks=c(0, 46, 58, max(age) ) )

 table(agecatagory)
#agecatagory
#  (0,46]  (46,58] (58,120]   #agecatagory is now a factor
#  319   60

 age.n <- as.numeric(agecatagory)
 table(age.n)
#age.n
# 1  2  3
#31  9 60

so a one liner would be:
agecatagory <- as.numeric( cut(age, breaks=c(0, 46, 58, max(age) ) ) )



I have been getting various errors for various things and have  
finally got

it so that only one error comes up
Error in if (age[i] <= 46) { : missing value where TRUE/FALSE needed

Could anyone pleaseee shed some light on this for me and tell em  
where I
am going wrong. I am sure it is just a minor thing but I cant for  
the life

of me figure it out!



David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Help with for/if loop

2009-04-23 Thread andrew

or perhaps

agec <- 0*age
age[age<=46] <- 1
age[age>46 & age<=58 <- 2
age[age>58] <- 3

or perhaps a one liner

cut(x = age, breaks= c(0,46, 58,Inf), labels = c(1,2,3))



On Apr 24, 11:24 am, Jorge Ivan Velez 
wrote:
> Dear Hollie,
> ifelse() is one alternative in this particular case:
>
> # Some data
> age<- c(46,47,43,46,47,59,50,54,59,60)
>
> ifelse(age<=46, 1,
>         ifelse( age>46 & age<=58 , 2, 3) )
> [1] 1 2 1 1 2 3 2 2 3 3
>
> See ?ifelse for more details.
>
> HTH,
>
> Jorge
>
> On Thu, Apr 23, 2009 at 8:07 PM, Giggles_Fairy
> wrote:
>
>
>
>
>
> > I have a set of data that includes various data columns. One if the
> > survival
> > time and another if a continuous variable of ages. I want to put the ages
> > into intervals so that I can then perform the Kalpan Meier test. I am
> > trying
> > to use the following code to build a column with the age group numbers in
>
> > agecatagory<-c( )
> > for (i in 1:137)
> > {
> > {
> > if(age[i]<=46)  {agecat[i]<-1}
> > if(age[i]>46 & age[i]<= 58) {agecat[i]<-2}
> > if(age[i]>58) {agecat[i]<-3}
> > }
> > agecatagory<-c(agecatagory, agecat[i])
> > }
>
> > I have been getting various errors for various things and have finally got
> > it so that only one error comes up
> > Error in if (age[i] <= 46) { : missing value where TRUE/FALSE needed
>
> > Could anyone pleaseee shed some light on this for me and tell em where
> > I
> > am going wrong. I am sure it is just a minor thing but I cant for the life
> > of me figure it out!
>
> > Your replies will be very much appreciated!
>
> > Hollie
> > --
> > View this message in context:
> >http://www.nabble.com/Help-with-for-if-loop-tp23207428p23207428.html
> > Sent from the R help mailing list archive at Nabble.com.
>
> > __
> > r-h...@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]]
>
> __
> r-h...@r-project.org mailing listhttps://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Generalized 2D list/array/whatever?

Matrix made from a list:

m <- list(sin, 1:3, letters[1:3], expression(a+b))
dim(m) <- c(2, 2)
dimnames(m) <- list(letters[1:2], LETTERS[1:2])
class(m) # matrix

or

M <- structure(list(sin, 1:3, letters[1:3], expression(a+b)), .Dim = c(2, 2),
  .Dimnames = list(c("a", "b"), c("A", "B")))
class(M) # matrix

On Thu, Apr 23, 2009 at 10:03 PM, Toby  wrote:
> I'm trying to figure out how I can get a generalized 2D
> list/array/matrix/whatever
> working.  Seems I can't figure out how to make the variables the right
> type.  I
> always seem to get some sort of error... out of bounds, wrong type, wrong
> dim, etc.
> Very confused... :)
>
> x[["some label", "some other index"]] <- 3
> x[["some other label", "something else"]] <- 4
>
> I don't know the indexes/label ahead of time... they get generated...  Any
> thoughts?
>
> -Toby.
>
>        [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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

[R] Generalized 2D list/array/whatever?

2009-04-23 Thread Toby

I'm trying to figure out how I can get a generalized 2D
list/array/matrix/whatever
working.  Seems I can't figure out how to make the variables the right
type.  I
always seem to get some sort of error... out of bounds, wrong type, wrong
dim, etc.
Very confused... :)

x[["some label", "some other index"]] <- 3
x[["some other label", "something else"]] <- 4

I don't know the indexes/label ahead of time... they get generated...  Any
thoughts?

-Toby.

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Re: [R] joining multiple lists

2009-04-23 Thread Daniel Bradley

Merge()!



Thanks!

On Fri, Apr 24, 2009 at 11:26 AM, Daniel Bradley wrote:

> Hi!
>
> I'm reading and cleaning data from multiple excel spreadsheets.  All the
> data has the same column names and the natural next step is to join the
> lists of data together to make one list for reporting purposes.  I had
> thought that c(file1data, file2data) would do the trick but that seems to
> append the new columns to the list so I end up with duplicate column names
> "col1" "col2" "col3" "col1" col2" "col3" rather than consolidated column
> data.  It's obviously a straight forward operation, but damned if I can find
> a straight forward explanation on the web.  Any assistance would be most
> appreciated!
>
> Thanks!
> Dan
>

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[R] joining multiple lists

2009-04-23 Thread Daniel Bradley

Hi!

I'm reading and cleaning data from multiple excel spreadsheets.  All the
data has the same column names and the natural next step is to join the
lists of data together to make one list for reporting purposes.  I had
thought that c(file1data, file2data) would do the trick but that seems to
append the new columns to the list so I end up with duplicate column names
"col1" "col2" "col3" "col1" col2" "col3" rather than consolidated column
data.  It's obviously a straight forward operation, but damned if I can find
a straight forward explanation on the web.  Any assistance would be most
appreciated!

Thanks!
Dan

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Re: [R] Help with for/if loop

2009-04-23 Thread Jorge Ivan Velez

Dear Hollie,
ifelse() is one alternative in this particular case:


# Some data
age<- c(46,47,43,46,47,59,50,54,59,60)

ifelse(age<=46, 1,
ifelse( age>46 & age<=58 , 2, 3) )
[1] 1 2 1 1 2 3 2 2 3 3


See ?ifelse for more details.

HTH,

Jorge


On Thu, Apr 23, 2009 at 8:07 PM, Giggles_Fairy
wrote:

>
> I have a set of data that includes various data columns. One if the
> survival
> time and another if a continuous variable of ages. I want to put the ages
> into intervals so that I can then perform the Kalpan Meier test. I am
> trying
> to use the following code to build a column with the age group numbers in
>
> agecatagory<-c( )
> for (i in 1:137)
> {
> {
> if(age[i]<=46)  {agecat[i]<-1}
> if(age[i]>46 & age[i]<= 58) {agecat[i]<-2}
> if(age[i]>58) {agecat[i]<-3}
> }
> agecatagory<-c(agecatagory, agecat[i])
> }
>
> I have been getting various errors for various things and have finally got
> it so that only one error comes up
> Error in if (age[i] <= 46) { : missing value where TRUE/FALSE needed
>
> Could anyone pleaseee shed some light on this for me and tell em where
> I
> am going wrong. I am sure it is just a minor thing but I cant for the life
> of me figure it out!
>
> Your replies will be very much appreciated!
>
> Hollie
> --
> View this message in context:
> http://www.nabble.com/Help-with-for-if-loop-tp23207428p23207428.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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

[R] Can't install package "glmnet"

2009-04-23 Thread Liang Zhang

Hi, I was trying to install package glmnet in R, but failed and it show such 
messages:

* Installing *source* package glmnet ...
 This package has only been tested with gfortran.
 So some checks are needed.
 R_HOME is /home/username/R/R-2.9.0
Attempting to determine R_ARCH...
R_ARCH is
Attempting to detect how R was configured for Fortran 90
Unsupported Fortran 90 compiler or Fortran 90
compilers unavailable! Stop!
ERROR: configuration failed for package glmnet
* Removing /home/username/R/R-2.9.0/library/glmnet

The downloaded packages are in
/tmp/RtmpwsLWSc/downloaded_packages
Updating HTML index of packages in '.Library'
Warning message:
In install.packages("glmnet") :
  installation of package 'glmnet' had non-zero exit status


It seems that it needs fortran 90 complier. What can I do to solve this 
problem? (I am not the administrator, only a user in the linux system).

Thank you very much!
Liang



  
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[R] Help with for/if loop

2009-04-23 Thread Giggles_Fairy


I have a set of data that includes various data columns. One if the survival
time and another if a continuous variable of ages. I want to put the ages
into intervals so that I can then perform the Kalpan Meier test. I am trying
to use the following code to build a column with the age group numbers in

agecatagory<-c( )
for (i in 1:137)
{
{
if(age[i]<=46)  {agecat[i]<-1}
if(age[i]>46 & age[i]<= 58) {agecat[i]<-2}
if(age[i]>58) {agecat[i]<-3} 
}
agecatagory<-c(agecatagory, agecat[i])
}

I have been getting various errors for various things and have finally got
it so that only one error comes up
Error in if (age[i] <= 46) { : missing value where TRUE/FALSE needed

Could anyone pleaseee shed some light on this for me and tell em where I
am going wrong. I am sure it is just a minor thing but I cant for the life
of me figure it out!

Your replies will be very much appreciated!

Hollie
-- 
View this message in context: 
http://www.nabble.com/Help-with-for-if-loop-tp23207428p23207428.html
Sent from the R help mailing list archive at Nabble.com.

__
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Re: [R] Error building package: LaTeX error when creating PDF version

2009-04-23 Thread Duncan Murdoch


On 23/04/2009 8:24 PM, Peter Dunn wrote:

Hi all
 
I am trying to build an R package, which I have successfully done many times before, but have an error I cannot trace.  I hope someone can help me.
 
Here's is some edited output (full output below if it is useful):
 
pdu...@pdunnubuntu:~/DSdata$ R CMD build GLMsData

* checking for file 'GLMsData/DESCRIPTION' ... OK
* preparing 'GLMsData':
* checking DESCRIPTION meta-information ... OK
* removing junk files
* checking for LF line-endings in source and make files
* checking for empty or unneeded directories
* building 'GLMsData_0.7.tar.gz'

pdu...@pdunnubuntu:~/DSdata$ R CMD check GLMsData
* checking for working pdflatex ... OK
* using log directory '/home/pdunn2/DSdata/GLMsData.Rcheck'
* using R version 2.9.0 (2009-04-17)


 
* checking data for non-ASCII characters ... OK

* checking examples ... OK
* checking PDF version of manual ... WARNING
LaTeX errors when creating PDF version.
This typically indicates Rd problems.
* checking PDF version of manual without index ... ERROR
 
 
So there is an error making the PDf version of the package manual.  That's fine; I am happy to fix it.  But first I need to find it.
 
So I enter the directory GLMsData.Rcheck and find the corresponding tex file for the manual, and run  pdflatex  on this tex file:
 
pdu...@pdunnubuntu:~/DSdata$ cd GLMsData.Rcheck/

pdu...@pdunnubuntu:~/DSdata/GLMsData.Rcheck$ pdflatex GLMsData-manual.tex



Output written on GLMsData-manual.pdf (79 pages, 405091 bytes).
Transcript written on GLMsData-manual.log.
pdu...@pdunnubuntu:~/DSdata/GLMsData.Rcheck$ 
 
 
So from what I gather:

* R CMD check  finds an error when using  pdflatex  to make the PDF version of 
the manual
* But when I manually run  pdflatex  on the tex file, it works OK.
 
Obviously, I am missing something.  Can someone tell me what it is?  I can't find enlightenment on the archives or the manuals.


You should also have a .log file in the same directory as the .tex file. 
 You probably have a new one now that pdflatex works; please move that 
one aside and run R CMD check again.  If you get a .log file, it should 
list whatever the error was; you could compare it to the .log that 
worked if you can't spot it.  If you don't get a new log, that means R 
CMD check couldn't run pdflatex.


Duncan Murdoch



Thanks everyone.

P.




sessionInfo()

R version 2.9.0 (2009-04-17)
i486-pc-linux-gnu

locale:
LC_CTYPE=en_AU.UTF-8;LC_NUMERIC=C;LC_TIME=en_AU.UTF-8;LC_COLLATE=en_AU.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_AU.UTF-8;LC_PAPER=en_AU.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_AU.UTF-8;LC_IDENTIFICATION=C

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





FULL OUTPUT:

pdu...@pdunnubuntu:~/DSdata$ R CMD build GLMsData
* checking for file 'GLMsData/DESCRIPTION' ... OK
* preparing 'GLMsData':
* checking DESCRIPTION meta-information ... OK
* removing junk files
* checking for LF line-endings in source and make files
* checking for empty or unneeded directories
* building 'GLMsData_0.7.tar.gz'

pdu...@pdunnubuntu:~/DSdata$ R CMD check GLMsData
* checking for working pdflatex ... OK
* using log directory '/home/pdunn2/DSdata/GLMsData.Rcheck'
* using R version 2.9.0 (2009-04-17)
* using session charset: UTF-8
* checking for file 'GLMsData/DESCRIPTION' ... OK
* checking extension type ... Package
* this is package 'GLMsData' version '0.7'
* checking package dependencies ... OK
* checking if this is a source package ... OK
* checking for executable files ... OK
* checking whether package 'GLMsData' can be installed ... OK
* checking package directory ... OK
* checking for portable file names ... OK
* checking for sufficient/correct file permissions ... OK
* checking DESCRIPTION meta-information ... OK
* checking top-level files ... OK
* checking index information ... OK
* checking package subdirectories ... OK
* checking whether the package can be loaded ... OK
* checking whether the package can be loaded with stated dependencies ... OK
* checking Rd files ... OK
* checking Rd files against version 2 parser ... OK
* checking Rd cross-references ... OK
* checking for missing documentation entries ... OK
* checking for code/documentation mismatches ... OK
* checking Rd \usage sections ... OK
* checking data for non-ASCII characters ... OK
* checking examples ... OK
* checking PDF version of manual ... WARNING
LaTeX errors when creating PDF version.
This typically indicates Rd problems.
* checking PDF version of manual without index ... ERROR
pdu...@pdunnubuntu:~/DSdata$ cd GLMsData.Rcheck/
pdu...@pdunnubuntu:~/DSdata/GLMsData.Rcheck$ pdflatex GLMsData-manual.tex
This is pdfTeXk, Version 3.141592-1.40.3 (Web2C 7.5.6)
 %&-line parsing enabled.
entering extended mode
(./GLMsData-manual.tex
LaTeX2e <2005/12/01>
Babel  and hyphenation patterns for english, usenglishmax, dumylang, noh
yphenation, ukenglish, loaded.

(/usr/share/texmf-texlive/tex/latex/base/book.cls
Docume

[R] Error building package: LaTeX error when creating PDF version

2009-04-23 Thread Peter Dunn

Hi all
 
I am trying to build an R package, which I have successfully done many times 
before, but have an error I cannot trace.  I hope someone can help me.
 
Here's is some edited output (full output below if it is useful):
 
pdu...@pdunnubuntu:~/DSdata$ R CMD build GLMsData
* checking for file 'GLMsData/DESCRIPTION' ... OK
* preparing 'GLMsData':
* checking DESCRIPTION meta-information ... OK
* removing junk files
* checking for LF line-endings in source and make files
* checking for empty or unneeded directories
* building 'GLMsData_0.7.tar.gz'

pdu...@pdunnubuntu:~/DSdata$ R CMD check GLMsData
* checking for working pdflatex ... OK
* using log directory '/home/pdunn2/DSdata/GLMsData.Rcheck'
* using R version 2.9.0 (2009-04-17)


 
* checking data for non-ASCII characters ... OK
* checking examples ... OK
* checking PDF version of manual ... WARNING
LaTeX errors when creating PDF version.
This typically indicates Rd problems.
* checking PDF version of manual without index ... ERROR
 
 
So there is an error making the PDf version of the package manual.  That's 
fine; I am happy to fix it.  But first I need to find it.
 
So I enter the directory GLMsData.Rcheck and find the corresponding tex file 
for the manual, and run  pdflatex  on this tex file:
 
pdu...@pdunnubuntu:~/DSdata$ cd GLMsData.Rcheck/
pdu...@pdunnubuntu:~/DSdata/GLMsData.Rcheck$ pdflatex GLMsData-manual.tex



Output written on GLMsData-manual.pdf (79 pages, 405091 bytes).
Transcript written on GLMsData-manual.log.
pdu...@pdunnubuntu:~/DSdata/GLMsData.Rcheck$ 
 
 
So from what I gather:
* R CMD check  finds an error when using  pdflatex  to make the PDF version of 
the manual
* But when I manually run  pdflatex  on the tex file, it works OK.
 
Obviously, I am missing something.  Can someone tell me what it is?  I can't 
find enlightenment on the archives or the manuals.

Thanks everyone.

P.



> sessionInfo()
R version 2.9.0 (2009-04-17)
i486-pc-linux-gnu

locale:
LC_CTYPE=en_AU.UTF-8;LC_NUMERIC=C;LC_TIME=en_AU.UTF-8;LC_COLLATE=en_AU.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_AU.UTF-8;LC_PAPER=en_AU.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_AU.UTF-8;LC_IDENTIFICATION=C

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





FULL OUTPUT:

pdu...@pdunnubuntu:~/DSdata$ R CMD build GLMsData
* checking for file 'GLMsData/DESCRIPTION' ... OK
* preparing 'GLMsData':
* checking DESCRIPTION meta-information ... OK
* removing junk files
* checking for LF line-endings in source and make files
* checking for empty or unneeded directories
* building 'GLMsData_0.7.tar.gz'

pdu...@pdunnubuntu:~/DSdata$ R CMD check GLMsData
* checking for working pdflatex ... OK
* using log directory '/home/pdunn2/DSdata/GLMsData.Rcheck'
* using R version 2.9.0 (2009-04-17)
* using session charset: UTF-8
* checking for file 'GLMsData/DESCRIPTION' ... OK
* checking extension type ... Package
* this is package 'GLMsData' version '0.7'
* checking package dependencies ... OK
* checking if this is a source package ... OK
* checking for executable files ... OK
* checking whether package 'GLMsData' can be installed ... OK
* checking package directory ... OK
* checking for portable file names ... OK
* checking for sufficient/correct file permissions ... OK
* checking DESCRIPTION meta-information ... OK
* checking top-level files ... OK
* checking index information ... OK
* checking package subdirectories ... OK
* checking whether the package can be loaded ... OK
* checking whether the package can be loaded with stated dependencies ... OK
* checking Rd files ... OK
* checking Rd files against version 2 parser ... OK
* checking Rd cross-references ... OK
* checking for missing documentation entries ... OK
* checking for code/documentation mismatches ... OK
* checking Rd \usage sections ... OK
* checking data for non-ASCII characters ... OK
* checking examples ... OK
* checking PDF version of manual ... WARNING
LaTeX errors when creating PDF version.
This typically indicates Rd problems.
* checking PDF version of manual without index ... ERROR
pdu...@pdunnubuntu:~/DSdata$ cd GLMsData.Rcheck/
pdu...@pdunnubuntu:~/DSdata/GLMsData.Rcheck$ pdflatex GLMsData-manual.tex
This is pdfTeXk, Version 3.141592-1.40.3 (Web2C 7.5.6)
 %&-line parsing enabled.
entering extended mode
(./GLMsData-manual.tex
LaTeX2e <2005/12/01>
Babel  and hyphenation patterns for english, usenglishmax, dumylang, noh
yphenation, ukenglish, loaded.

(/usr/share/texmf-texlive/tex/latex/base/book.cls
Document Class: book 2005/09/16 v1.4f Standard LaTeX document class
(/usr/share/texmf-texlive/tex/latex/base/bk10.clo))
(/usr/share/texmf/tex/latex/R/Rd.sty
(/usr/share/texmf-texlive/tex/latex/base/ifthen.sty)
(/usr/share/texmf-texlive/tex/latex/tools/longtable.sty)
(/usr/share/texmf-texlive/tex/latex/tools/bm.sty)
(/usr/share/texmf-texlive/tex/latex/base/alltt.sty)
(/usr/share/texmf-texlive/tex/latex/tools/verbatim.sty)
(/usr/share/texmf-texlive/tex/latex/l

Re: [R] conditional grouping of variables: ave or tapply or by or???

Try this:

> df$v4 <- ave(1:nrow(df), df$v1, FUN = function(i) with(df[i,], v2[!v3]))
> df
  v1 v2 v3 v4
1 10  7  0  7
2 10  5  1  7
3 10  7  2  7
4 11  9  0  9
5 11  7  2  9

> # If as is the case here that 0 is always the first in each v1 group
> # then it can be simplified further to:

> df$v4 <- with(df, ave(v2, v1, FUN = function(x) x[1]))


On Thu, Apr 23, 2009 at 6:11 PM, ozan bakis  wrote:
> Dear R Users,
> I have the following data frame:
>
> v1 <- c(rep(10,3),rep(11,2))
> v2 <- sample(5:10, 5, replace = T)
> v3 <- c(0,1,2,0,2)
> df <- data.frame(v1,v2,v3)
>> df
>  v1 v2 v3
> 1 10  9  0
> 2 10  5  1
> 3 10  6  2
> 4 11  7  0
> 5 11  5  2
>
> I want to add a new column v4 such that its values are equal to the value
> of v2 conditional on v3=0 for each subgroup of v1. In the above example,
> the final result should be like
>
> df$v4 <- c(9,9,9,7,7)
>> df
>  v1 v2 v3 v4
> 1 10  9  0  9
> 2 10  5  1  9
> 3 10  6  2  9
> 4 11  7  0  7
> 5 11  5  2  7
>
>
> I tried the following commands without success.
>
> df$v4 <- ave(df$v2, df$v1, FUN=function(x) x[df$v3==0])
> tapply(df$v2, df$v1, FUN=function(x) x[df$v3==0])
> by(df$v2, df$v1, FUN=function(x) x[df$v3==0])
>
> Any help? Thanks in advance!
> Ozan
>
>        [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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Re: [R] conditional grouping of variables: ave or tapply or by or???

2009-04-23 Thread hadley wickham

On Thu, Apr 23, 2009 at 5:11 PM, ozan bakis  wrote:
> Dear R Users,
> I have the following data frame:
>
> v1 <- c(rep(10,3),rep(11,2))
> v2 <- sample(5:10, 5, replace = T)
> v3 <- c(0,1,2,0,2)
> df <- data.frame(v1,v2,v3)
>> df
>  v1 v2 v3
> 1 10  9  0
> 2 10  5  1
> 3 10  6  2
> 4 11  7  0
> 5 11  5  2
>
> I want to add a new column v4 such that its values are equal to the value
> of v2 conditional on v3=0 for each subgroup of v1. In the above example,
> the final result should be like
>
> df$v4 <- c(9,9,9,7,7)
>> df
>  v1 v2 v3 v4
> 1 10  9  0  9
> 2 10  5  1  9
> 3 10  6  2  9
> 4 11  7  0  7
> 5 11  5  2  7
>
>
> I tried the following commands without success.
>
> df$v4 <- ave(df$v2, df$v1, FUN=function(x) x[df$v3==0])
> tapply(df$v2, df$v1, FUN=function(x) x[df$v3==0])
> by(df$v2, df$v1, FUN=function(x) x[df$v3==0])
>
> Any help? Thanks in advance!

Here's one approach with the plyr package, http://had.co.nz/plyr

library(plyr)
ddply(df, .(v1), transform, v4 = v2[v3 == 0])

Hadley


-- 
http://had.co.nz/

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Re: [R] suppress output from step function

2009-04-23 Thread Jorge Ivan Velez

Dear Pierre,

See argument trace in ?step. Setting up trace=FALSE in your code should hide
the steps:

step(linear_model, trace=FALSE)

HTH,

Jorge


On Thu, Apr 23, 2009 at 6:34 PM, Pierre Moffard  wrote:

> Dear all,
>
> Is there a way of using the step function on a linear model such that all
> the steps are not printed out?
>
> Say I have a matrix X of 50 explanatory variables and a binary response Y.
>
> linear_model<-glm(Y~X, data=my_data, family="binomial")
> SS<-step(linear_model)
>
> This last step produces all the output that would have been produced even
> if I hadn't named the result SS.
>
> SS<-step(linear_model,print=FALSE) - this doesnt work either.
>
> Does anyone have any ideas?
>
> Thank you all in advance.
>
> Best,
> Pierre
>
>
>
>
>[[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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[R] suppress output from step function

2009-04-23 Thread Pierre Moffard

Dear all,

Is there a way of using the step function on a linear model such that all the 
steps are not printed out?

Say I have a matrix X of 50 explanatory variables and a binary response Y.

linear_model<-glm(Y~X, data=my_data, family="binomial")
SS<-step(linear_model)

This last step produces all the output that would have been produced even if I 
hadn't named the result SS.

SS<-step(linear_model,print=FALSE) - this doesnt work either.

Does anyone have any ideas?

Thank you all in advance.

Best,
Pierre



  
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Re: [R] Parenthesis around date/time using chron?

Try this:

> x <- as.chron(Sys.time())
> format(x, enclosed = c("", ""))
[1] "04/23/09 22:26:23"


On Thu, Apr 23, 2009 at 6:03 PM, Kenneth Takagi  wrote:
> Hi,
>
>
>
> I've been using the chron package to convert excel time into
> month/day/year and h:m:s formats, specifically for use as axis labels.
> I've come across something I don't quite understand.
>
>
>
> ### Here are some excel times:
>
> dat = c(39083, 39083.00694, 39083.01389, 39083.02083, 39083.02778,
>
> 39083.03472, 39083.04167, 39083.04861, 39083.05556, 39083.0625
>
> )
>
>
>
> ### I create date/time vectors from excel times:
>
> orig =chron("01/01/1900");
>
> # month/day/year h:m:s format
>
> date.time = orig + dat-2;
>
> # h:m:s format
>
> time= date.time - floor(date.time);
>
> # month/day/year format
>
> date = orig + floor(dat)-2;
>
>
>
> #I can easily recreate the date.time vector without the parenthesis (as
> characters though):
>
> paste(date, time)
>
>
>
> But, I don't see why the values in the ORIGINAL date.time vector have
> parenthesis around them?  Is there as specific reason?
>
>
>
> Thanks!
>
>
>
> ken
>
>
>
> kat...@psu.edu
>
>
>
>
>        [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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Re: [R] Stuck using constrOptim

My bad.  It skipped my mind that constrOptim requires that you specify the
gradient function for using BFGS.  I have writtten a constrOPtim function
that computes the gradient numerically and also can incorporate non-linear
inequality constraints.  I will send it to you, if you are interested.

Ravi.  

---

Ravi Varadhan, Ph.D.

Assistant Professor, The Center on Aging and Health

Division of Geriatric Medicine and Gerontology 

Johns Hopkins University

Ph: (410) 502-2619

Fax: (410) 614-9625

Email: rvarad...@jhmi.edu

Webpage:  http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of dre968
Sent: Thursday, April 23, 2009 5:23 PM
To: r-help@r-project.org
Subject: Re: [R] Stuck using constrOptim

if i run the same program with
constrOptim(p0,minsquare,keys=keys,benKeys=benKeys,NULL,ui=A,ci=B,method="BF
GS")

i get the following error:
Error in dR(theta, theta.old, ...) : could not find function "grad"

i'm not very mathematically inclined, i dont even know what a gradient is. 
any suggestions?

dre968 wrote:
> 
> Trying to use constrOptim to minimize the sum of squared deviations.  
> I put the objective function in as: sum((x %*% Y - Z)^2) so i'm trying 
> to get values for x to minimize the sum of the squared deviations 
> between the product of x and Y and Z.
> 
> Anyways i have no problem using this when x is a 3x1 test variable.  
> it works great with the constraints and everything.  when i actually 
> use it on my data x is 3x900 or so and it wont optimize it just gives 
> me back my initial guessthis is the output i'm getting:
> 
> $value
> [1] 22.7438
> 
> $counts
> function gradient 
>  906   NA 
> 
> $convergence
> [1] 1
> 
> $message
> NULL
> 
> $outer.iterations
> [1] 1
> 
> $barrier.value
> [1] 1.381551e-06
> 
> this is definitely not the right answer and it is just spitting back 
> my initial guess back.
> 
> Any ideas?  are there limits as to how big the variables can be for 
> this function?
> 

--
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[R] conditional grouping of variables: ave or tapply or by or???

2009-04-23 Thread ozan bakis

Dear R Users,
I have the following data frame:

v1 <- c(rep(10,3),rep(11,2))
v2 <- sample(5:10, 5, replace = T)
v3 <- c(0,1,2,0,2)
df <- data.frame(v1,v2,v3)
> df
  v1 v2 v3
1 10  9  0
2 10  5  1
3 10  6  2
4 11  7  0
5 11  5  2

I want to add a new column v4 such that its values are equal to the value
of v2 conditional on v3=0 for each subgroup of v1. In the above example,
the final result should be like

df$v4 <- c(9,9,9,7,7)
> df
  v1 v2 v3 v4
1 10  9  0  9
2 10  5  1  9
3 10  6  2  9
4 11  7  0  7
5 11  5  2  7


I tried the following commands without success.

df$v4 <- ave(df$v2, df$v1, FUN=function(x) x[df$v3==0])
tapply(df$v2, df$v1, FUN=function(x) x[df$v3==0])
by(df$v2, df$v1, FUN=function(x) x[df$v3==0])

Any help? Thanks in advance!
Ozan

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Re: [R] My surprising experience in trying out REvolution's R

2009-04-23 Thread David M Smith

We've taken a look at this in a bit more detail; it's a very
interesting example.  Although the code uses several functions that
exploit the parallel processing in REvolution R (notably %*% and
chol), this was one of those situations where the overheads of
threading pretty much balanced any performance gains: the individual
matrices for the operations were too small.

For some examples where the performance gains do show, see:
http://www.revolution-computing.com/products/r-performance.php

A more promising avenue for speeding up this code lies in
parallelizing the outer for(i in 1:200) loop...

# David Smith

On Tue, Apr 21, 2009 at 9:26 AM, Jason Liao  wrote:
> I care a lot about R's speed. So I decided to give REvolution's R
> (http://revolution-computing.com/) a try, which bills itself as an
> optimized R. Note that I used the free version.
>
> My machine is a Intel core 2 duo under Windows XP professional. The code
> I run is in the end of this post.
>
> First, the regular R 1.9. It takes 2 minutes and 6 seconds, CPU usage
> 50%
>
> Next, REvolution's R. It takes 2 minutes and 10 seconds, CPU usage 100%.
>
>
> In other words, REvolution's R consumes double the CPU with slightly
> less speed.
>
> The above has been replicated a few times (as a statistician of course).
>
>
> Anyone has any insight on this? Anyway, my high hope was dashed.

-- 
David M Smith 
Director of Community, REvolution Computing www.revolution-computing.com
Tel: +1 (206) 577-4778 x3203 (San Francisco, USA)

Check out our upcoming events schedule at www.revolution-computing.com/events

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[R] Parenthesis around date/time using chron?

2009-04-23 Thread Kenneth Takagi

Hi,

 

I've been using the chron package to convert excel time into
month/day/year and h:m:s formats, specifically for use as axis labels.
I've come across something I don't quite understand.

 

### Here are some excel times:

dat = c(39083, 39083.00694, 39083.01389, 39083.02083, 39083.02778, 

39083.03472, 39083.04167, 39083.04861, 39083.05556, 39083.0625

)

 

### I create date/time vectors from excel times:

orig =chron("01/01/1900");

# month/day/year h:m:s format

date.time = orig + dat-2; 

# h:m:s format

time= date.time - floor(date.time); 

# month/day/year format

date = orig + floor(dat)-2; 

 

#I can easily recreate the date.time vector without the parenthesis (as
characters though):

paste(date, time)

 

But, I don't see why the values in the ORIGINAL date.time vector have
parenthesis around them?  Is there as specific reason?

 

Thanks!

 

ken

 

kat...@psu.edu

 


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[R] Interpreting the results of Friedman test

2009-04-23 Thread Doerte

Hello,

I have problems interpreting the results of a Friedman test. It seems
to me that the p-value resulting from a Friedman test and with it the
"significance" has to be interpreted in another way than the p-value
resulting from e.g. ANOVA?

Let me describe the problem with some detail: I'm testing a lot of
different hypotheses in my observer study and only for some the
premises for performing an ANOVA are fulfilled (tested with Shapiro
Wilk and Bartlett). For the others I perform a Friedman test.

To my surprise, the p-value of the Friedman test is < 0.05 for all my
tested hypotheses. Thus, I tried to "compare" the results with the
results of an ANOVA by performing both test methods (Friedman, ANOVA)
to a given set of data.
While ANOVA results in p = 0.34445 (--> no significant difference
between the groups), the Friedman test results in p = 1.913e-06 (-->
significant difference between the groups?).

How can this be?

Or am I doing something wrong? I have three measured values for each
condition. For ANOVA I use them all, for the Friedman test I
calculated the geometric mean of the three values, since this test
does not work with replicated values. Is this a crude mistake?

Thanks in advance for any help.

Doerte

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Re: [R] Stuck using constrOptim


so i set gr=Null:
> constrOptim(p0,minsquare,Y=Y,Z=Z,NULL,gr=NULL,ui=A,ci=B,method="BFGS")
and i recieved the following error:
Error in optim(theta.old, fun, gradient, control = control, method = method, 
: 
  objective function in optim evaluates to length 0 not 1
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Re: [R] R 2.8.1 change user input color of R scrpt too

2009-04-23 Thread Greg Snow

The short answer is to upgrade to R version 2.9.0 where you have more control 
over the colors in the different types of windows.


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


> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of matildet
> Sent: Thursday, April 23, 2009 11:25 AM
> To: r-help@r-project.org
> Subject: [R] R 2.8.1 change user input color of R scrpt too
> 
> 
> Hello R community
> 
> I'm using R 2.8.1 version and would like to change the color of user
> input
> for script pages. Once I have changed the color of console background
> and
> user input (by means of Interface Preferences menu), the color of
> script
> background has been automatically changed but the color of user input
> has
> not.
> 
> Has someone any useful advice for my (little) problem?
> 
> Thanks in advance
> 
> Matilde
> --
> View this message in context: http://www.nabble.com/R-2.8.1-change-
> user-input-color-of-R-scrpt-too-tp23197473p23197473.html
> Sent from the R help mailing list archive at Nabble.com.
> 
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-
> guide.html
> and provide commented, minimal, self-contained, reproducible code.

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Re: [R] Stuck using constrOptim


if i run the same program with 
constrOptim(p0,minsquare,keys=keys,benKeys=benKeys,NULL,ui=A,ci=B,method="BFGS")

i get the following error:
Error in dR(theta, theta.old, ...) : could not find function "grad"

i'm not very mathematically inclined, i dont even know what a gradient is. 
any suggestions?


dre968 wrote:
> 
> Trying to use constrOptim to minimize the sum of squared deviations.  I
> put the objective function in as: sum((x %*% Y - Z)^2) so i'm trying to
> get values for x to minimize the sum of the squared deviations between the
> product of x and Y and Z.  
> 
> Anyways i have no problem using this when x is a 3x1 test variable.  it
> works great with the constraints and everything.  when i actually use it
> on my data x is 3x900 or so and it wont optimize it just gives me back my
> initial guessthis is the output i'm getting:
> 
> $value
> [1] 22.7438
> 
> $counts
> function gradient 
>  906   NA 
> 
> $convergence
> [1] 1
> 
> $message
> NULL
> 
> $outer.iterations
> [1] 1
> 
> $barrier.value
> [1] 1.381551e-06
> 
> this is definitely not the right answer and it is just spitting back my
> initial guess back.  
> 
> Any ideas?  are there limits as to how big the variables can be for this
> function?
> 

-- 
View this message in context: 
http://www.nabble.com/Stuck-using-constrOptim-tp23197912p23203595.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] simple for loop question - how do you exit?

2009-04-23 Thread Greg Snow

To "break" out of a loop early, use the "break" command (or rethink your logic 
and use a different type of loop, often if you will be commonly leaving the 
loop, a while loop may make more sense than a for loop).

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


> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of dre968
> Sent: Thursday, April 23, 2009 12:29 PM
> To: r-help@r-project.org
> Subject: [R] simple for loop question - how do you exit?
> 
> 
> I have a loop and an if statement in the loop.  once the if statement
> is true
> for 1 value in the loop i'd like to exit the loop.  is there a command
> to do
> this?  i know its going to be something like exit and i feel stupid
> asking
> this question
> --
> View this message in context: http://www.nabble.com/simple-for-loop-
> question---how-do-you-exit--tp23197504p23197504.html
> Sent from the R help mailing list archive at Nabble.com.
> 
> __
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide http://www.R-project.org/posting-
> guide.html
> and provide commented, minimal, self-contained, reproducible code.

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Re: [R] Stuck using constrOptim

Hi,

The constrOptim function used "Nelder-Mead" for the inner iterations, when
gradient is not specified.  Nelder-Mead is not good when you have a
non-small number of parameters.  I would sugest that you specify method =
"BFGS" in the call to constrOptim.  This might help.  

Ravi.



---

Ravi Varadhan, Ph.D.

Assistant Professor, The Center on Aging and Health

Division of Geriatric Medicine and Gerontology 

Johns Hopkins University

Ph: (410) 502-2619

Fax: (410) 614-9625

Email: rvarad...@jhmi.edu

Webpage:  http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html

 





-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of dre968
Sent: Thursday, April 23, 2009 4:22 PM
To: r-help@r-project.org
Subject: Re: [R] Stuck using constrOptim


Sorry for the not enough info.  the code seems to work for 10 or so
variables but stops optimizing if i give it anymore than that.

#Load data
Data <- read.table..csv',head=TRUE,sep=",",stringsAsFactors = FALSE)
#Load goal variables.  this is what i'm trying to minimize to
Z<-read.tablecsv',head=TRUE,sep=",",stringsAsFactors = FALSE)

#set array
Y<-array(0,dim=c(nrow(Data),15))

#load applicable values into array
for(j in 1:15){
for(k in 1:nrow(Data)){
Y[k,j] = Data[k,3+j]
}
}
#set constraint

A <- array(0,dim=c(2+nrow(Y),nrow(Y)))
B <- array(0,dim=c(2+nrow(Y),1))

# Constraint for total = 100%
for(w in 1:nrow(Data)){
A[1,w] = 1
A[2,w] = -1
}
#each value in x must be greater than 0
for(i in 1:nrow(Data)){
A[i+2,i] = 1
B[i+2] = -.1
}

B[1,1] = .999
B[2,1] = -1.001
#initial guess
p0<- array(0,dim=c(nrow(Data)))

p0[1] = .5
p0[2] = .501

#objective function
minsquare<-function(x,Y,Z){sum((x %*% Y - Z)^2)}


#Optimization with constraints
constrOptim(p0,minsquare,Y=Y,Z=Z,NULL,ui=A,ci=B)


here is the code.  



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Re: [R] function output with for loop and if statement

2009-04-23 Thread Richard M. Heiberger

tmp.out.sort <- tmp.out[, order(names(tmp.out))]

tmp.out.sort <- tmp.out[, order(names(tmp.out)), drop=FALSE]

>From your description of misbehavior with a single column,
I think the drop=FALSE argument will provide the protection you need.
Then you will not need the if clause.

See

?`[.data.frame`
for the story and examples.


Rich

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Re: [R] Stuck using constrOptim


Sorry for the not enough info.  the code seems to work for 10 or so variables
but stops optimizing if i give it anymore than that.

#Load data
Data <- read.table..csv',head=TRUE,sep=",",stringsAsFactors = FALSE)
#Load goal variables.  this is what i'm trying to minimize to
Z<-read.tablecsv',head=TRUE,sep=",",stringsAsFactors = FALSE)

#set array
Y<-array(0,dim=c(nrow(Data),15))

#load applicable values into array
for(j in 1:15){
for(k in 1:nrow(Data)){
Y[k,j] = Data[k,3+j]
}
}
#set constraint

A <- array(0,dim=c(2+nrow(Y),nrow(Y)))
B <- array(0,dim=c(2+nrow(Y),1))

# Constraint for total = 100% 
for(w in 1:nrow(Data)){
A[1,w] = 1
A[2,w] = -1
}
#each value in x must be greater than 0
for(i in 1:nrow(Data)){
A[i+2,i] = 1
B[i+2] = -.1
}

B[1,1] = .999
B[2,1] = -1.001
#initial guess
p0<- array(0,dim=c(nrow(Data)))

p0[1] = .5
p0[2] = .501

#objective function
minsquare<-function(x,Y,Z){sum((x %*% Y - Z)^2)}


#Optimization with constraints
constrOptim(p0,minsquare,Y=Y,Z=Z,NULL,ui=A,ci=B)


here is the code.  



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Re: [R] function output with for loop and if statement

2009-04-23 Thread aaron wells


Gavin, thank you for the suggestions.  Unfortunately the function is still not 
working correctly.  Below are the dummy datasets that you requested. In the 
function dummy.vegdata = vegetation; and dummy.spplist = specieslist.  A little 
clarification on why the if statement is in the function.  I am using the apply 
function to sum columns of data that correspond with different lifeforms in 
order to derive a total cover value for each  lifeform in each plot (plt).  
When only one species occurs in a lifeform the apply function doesn't work 
since there is only one column of data. So, the if statement is an attempt to 
include the column of data from the dummy.vegdata into the output when there is 
only one species in a given lifeform.  Examples of this condition in the 
dummy.vegdata include water (Bare_Ground) and popbal (Deciduous_Tree).
 
 
 Aaron
 
 
> dummy.vegdata

plt water salarb salpul popbal leddec picgla picmar arcuva zygele epiang calpur 
poaarc pelaph flacuc tomnit hylspl carvag caraqu calcan carsax
T1  0  0   10.00.0 200.0 35  00.00.0  0 
   0.05.00.0 20 550.0  15.0  0
T2  0  00.00.0  30.0 62  00.00.0  0 
   0.08.02.0  5 650.0  03.0  0
T3  0  00.00.0  00.0  1 802.00.0  4 
   0.00.00.0  0  00.0  00.0  0
T4  0  10.0   30.0  00.1  0  00.00.0  0 
   0.00.00.0  0  00.0  00.0  0
T5  0  00.00.0  00.0  0  00.00.0  0 
   0.00.00.0  0  00.0  00.0  0
T6  0  00.00.0  0   20.0  0  00.01.0  0 
   0.02.00.1  0 200.0  00.0  0
T7  0  00.00.0  50.0 40  00.00.0  0 
   0.05.00.0  0 850.0  00.0  0
T8  0  03.00.0 400.0  5  00.00.0  0 
   0.00.00.0  1  00.0  20.0  0
T9  0  01.00.0 200.0 10  00.00.0  0 
   0.00.10.0  0  30.0  10.0  0
T10 0  00.0   65.0  02.0  0  00.00.0  0 
   0.00.00.0  0  00.0  05.0  0
T11 0  11.00.0 300.0 45  00.00.0  0 
   0.11.00.0  0 101.0  00.0  0
T12 0  57.00.1  03.0  5  00.00.0  0 
   0.00.00.0  0  51.0  02.0  5
T13 0  0   20.00.0  52.0  5  00.00.0  0 
   0.00.00.0 25  02.0  23.0 15
T14 0  0   35.00.0  00.0  0  00.00.0  0 
   0.00.00.0  0  00.1  03.0  2
T15 0  00.01.0  0   20.0  0  50.10.0  0 
   0.00.00.0  0 100.0  00.0  0
T16 0  00.00.0  1   80.0  0  00.00.1  0 
   0.01.00.0  0 500.0  00.1  0
T17   100  00.00.0  00.0  0  00.00.0  0 
   0.00.00.0  0  00.0  00.0  0
T18 0  00.00.0  0   15.0  0  50.02.0  0 
   0.01.00.0  0 200.0  01.0  0
T19 0  00.0   10.0  0   50.0  0  00.00.0  0 
   0.00.10.0  0 550.0  0   20.0  0
T20 0  00.00.0 450.0 15  00.00.0  0 
   0.01.00.0  0 500.0  00.0  0
T21 0  00.00.0  00.0  0  00.00.0  0 
   0.00.00.0  0  00.0  0   35.0  0
T22   100  00.00.0  00.0  0  00.00.0  0 
   0.00.00.0  0  00.0  00.0  0
T23 0  00.00.0  0   15.0  0  10.00.0  0 
   0.00.00.0  0  20.0  00.0  0
T24 0  00.10.0  00.0  0  00.00.0  0 
   0.00.00.0  0  00.0  0   25.0  0
T25 0  00.00.0  0   25.0  0  00.01.0  0 
   0.00.00.0  0 150.0  02.0  0
 
> dummy.spplist

code  SciName LifeFormClass
water   Water  Bare_GroundOther
salarbSalix_arbusculoides Deciduous_Shrubs Vascular
salpul   Salix_planifolia_pulchra Deciduous_Shrubs Vascular
popbalPopulus_balsamifera   Deciduous_Tree Vascular
leddecLed

[R] iteration limit error in gamm and notExp2

2009-04-23 Thread isidora k

hi,
I am trying to run a mixed effect gam using gamm (mgcv)
and I get the following error:
"Error in lme.formula(y ~ X - 1, random = rand, data = strip.offset(mf),  : 
nlminb problem, convergence error code = 1
message = iteration limit reached without convergence (9)"

I've read all about  'notExp2' but since I am not an expert I didn't understand 
that much. I was wondering if anyone has an example of the use of notExp2, or 
another suggestion on solving this problem.

many thanks
Isidora


  
[[alternative HTML version deleted]]

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[R] latex(Hmisc): cgroup + rownames shifts column names

2009-04-23 Thread Michael Erickson

I have submitted this as a bug
(http://biostat.mc.vanderbilt.edu/trac/Hmisc/ticket/29) but I am
wondering if anyone else has seen it or perhaps developed a
workaround.  I could certainly fix the LaTeX by hand, but I am using
this inside Sweave, so it is a bit cumbersome.  The exact same code
used to work fine, but something changed underneath it.  Even so,
perhaps I am using the latex() command incorrectly.

When I do a latex table with rownames but without cgroup, everything
is fine, but when I use cgroup (as in the second case), the rownames
column gets the first column's name and subsequent column names are
shifted. This worked fine for me a year ago, so something has changed
in my setup or in Hmisc. I have another machine with Hmisc_3.4-3
installed, and its cgroup()-ed table was correct as well.  I included
the output from sessionInfo() at the bottom.

test.df <- data.frame(row.names=letters[1:4], col1=1:4, col2=4:1, col3=4:7)

latex(test.df, file="", table.env=FALSE, center="none")

% latex.default(test.df, file = "", table.env = FALSE, center = "none") %
\begin{tabular}{lrrr}
\hline\hline
%% Comment added by ME -- this next line is the one that is fine without cgroup
\multicolumn{1}{l}{test.df}& \multicolumn{1}{c}{col1}&
\multicolumn{1}{c}{col2}& \multicolumn{1}{c}{col3} \tabularnewline
\hline a&$1$&$4$&$4$\tabularnewline
b&$2$&$3$&$5$\tabularnewline
c&$3$&$2$&$6$\tabularnewline
d&$4$&$1$&$7$\tabularnewline
\hline
\end{tabular}

latex(test.df, file="", n.cgroup=c(1,2), cgroup=c("","95\\% Conf. Limits"),
table.env=FALSE, center="none")

% latex.default(test.df, file = "", n.cgroup = c(1, 2), cgroup = c("",
"95\\% Conf. Limits"), table.env = FALSE, center = "none")
%
\begin{tabular}{lrcrr}
\hline\hline
\multicolumn{1}{l}{\bfseries test.df}& \multicolumn{1}{c}{\bfseries }&
\multicolumn{1}{c}{\bfseries }&
\multicolumn{2}{c}{\bfseries 95\% Conf. Limits} \tabularnewline
 \cline{4-5}
%% Comment added by ME -- this next line is the one that is broken with cgroup
\multicolumn{1}{l}{col1}& \multicolumn{1}{c}{}&
\multicolumn{1}{c}{col2}& \multicolumn{1}{c}{col3}&
\multicolumn{1}{c}{col1} \tabularnewline
\hline
a&$1$&&$4$&$4$\tabularnewline
b&$2$&&$3$&$5$\tabularnewline c&$3$&&$2$&$6$\tabularnewline
d&$4$&&$1$&$7$\tabularnewline
\hline \end{tabular}

I am attaching a .pdf of the two tables.  (I hope it goes through.)

sessionInfo()

R version 2.8.1 (2008-12-22) i486-pc-linux-gnu

locale: 
LC_CTYPE=en_US;LC_NUMERIC=C;LC_TIME=en_US;LC_COLLATE=en_US;LC_MONETARY=C;LC_MESSAGES=en_US;LC_PAPER=en_US;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US;LC_IDENTIFICATION=C

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

other attached packages: [1] Hmisc_3.5-2

loaded via a namespace (and not attached): [1] cluster_1.11.13
grid_2.9.0 lattice_0.17-22 tcltk_2.9.0 [5] tools_2.9.0

I hope this is enough information to make the problem clear and that I
haven't missed something obvious.  Finally, I will subscribe to the
help list (I usually just search the archives), but I would appreciate
replies sent directly to me as well!

Thanks!

Michael


x.pdf
Description: Adobe PDF document
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Re: [R] Stuck using constrOptim

The information you have provided is not enough.  Please read the posting
guide on how ask for help.  Provide a reproducible example so we can help
you.

Ravi.

---

Ravi Varadhan, Ph.D.

Assistant Professor, The Center on Aging and Health

Division of Geriatric Medicine and Gerontology 

Johns Hopkins University

Ph: (410) 502-2619

Fax: (410) 614-9625

Email: rvarad...@jhmi.edu

Webpage:  http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html

 





-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of dre968
Sent: Thursday, April 23, 2009 3:20 PM
To: r-help@r-project.org
Subject: [R] Stuck using constrOptim


Trying to use constrOptim to minimize the sum of squared deviations.  I put
the objective function in as: sum((x %*% Y - Z)^2) so i'm trying to get
values for x to minimize the sum of the squared deviations between the
product of x and Y and Z.  

Anyways i have no problem using this when x is a 3x1 test variable.  it
works great with the constraints and everything.  when i actually use it on
my data x is 3x900 or so and it wont optimize it just gives me back my
initial guessthis is the output i'm getting:

$value
[1] 22.7438

$counts
function gradient 
 906   NA 

$convergence
[1] 1

$message
NULL

$outer.iterations
[1] 1

$barrier.value
[1] 1.381551e-06

this is definitely not the right answer and it is just spitting back my
initial guess back.  

Any ideas?  are there limits as to how big the variables can be for this
function?
--
View this message in context:
http://www.nabble.com/Stuck-using-constrOptim-tp23197912p23197912.html
Sent from the R help mailing list archive at Nabble.com.

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[R] Loess over split data

2009-04-23 Thread Luc Villandre


Dear R users,

I am having trouble devising an efficient way to run a loess() function 
on all columns of a data.frame (with the x factor remaining the same for 
all columns) and I was hoping that someone here could help me fix my 
code so that I won't have to resort to using a for loop.


(You'll find the upcoming lines of code in a single block near the end 
of the message.)


Here's a small sample of my data :
  
my.data=data.frame(birth.vec=c(274, 259, 288, 284, 276, 274, 284, 288, 
273, 278), final.weight= c(2609.328, 2955.701, 4159.837, 4591.258, 
5144.559, 2877.159, 4384.498, 3348.454, 3323.391, 2918.055)) ;


"birth.vec" is in days and "final.weight" is in grams. Now, what I'm 
trying to do is to output smoothed curves for the 10th, 50th and 90th 
percentiles over completed weeks. In other words, I transform 
"birth.vec" into a factor vector of completed weeks with


completed.weeks = as.factor(floor(my.data$birth.vec/7)) ;

I use these factors to get weight quantiles by completed weeks with

quan.list = tapply(my.data$final.weight,INDEX = 
completed.weeks,FUN=quantile,probs=c(0.1,0.5,0.9)) ;


I then collapse the list into a data.frame with

quan.frame = data.frame(do.call(rbind,quan.list)) ;

Now comes the time to apply the loess() function to each percentile 
curve (i.e. to each column in quan.frame). Note that the x values for 
all percentile columns (i.e. X10., X50., X90.)


x.values = as.numeric(rownames(quan.frame)) ;

Once again, using tapply() (and transposing beforehand):

t.quan.frame = t(quan.frame) ;

## The following command doesn't work

smooth.result = 
tapply(t.quan.frame,INDEX=as.factor(rep(1:nrow(t.quan.frame),each=ncol(t.quan.frame))),FUN=loess) 
;


The "formula" argument of loess() is of course missing, since I have no 
idea how I could specify it in a way that the function could understand it.


Now, for your convenience, here's the code patched together:


my.data=data.frame(birth.vec=c(274, 259, 288, 284, 276, 274, 284, 288, 
273, 278), final.weight= c(2609.328, 2955.701, 4159.837, 4591.258, 
5144.559, 2877.159, 4384.498, 3348.454, 3323.391, 2918.055)) ;

completed.weeks = as.factor(floor(my.data$birth.vec/7)) ;
quan.list = tapply(my.data$final.weight,INDEX = 
completed.weeks,FUN=quantile,probs=c(0.1,0.5,0.9)) ;

quan.frame = data.frame(do.call(rbind,quan.list)) ;
x.values = as.numeric(rownames(quan.frame)) ;
t.quan.frame = t(quan.frame) ;

## The following command doesn't work

smooth.result = 
tapply(t.quan.frame,INDEX=as.factor(rep(1:nrow(t.quan.frame),each=ncol(t.quan.frame))),FUN=loess) 
;   
###


So, is there a way to modify this code so that I'll get smoothed 
percentile curves?  Any suggestion as to how it could be improved is 
also welcome.


Although the forum archive is fairly comprehensive, phrasing search 
queries in a way that will yield the desired result is sometimes hard, 
hence this request for help that to some of you will most likely seem 
redundant.


I thank you sincerely for your time,

Luc

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Re: [R] SAGE: was large factorials

2009-04-23 Thread Wacek Kusnierczyk

Albyn Jones wrote:
> On Wed, Apr 22, 2009 at 08:26:51PM -0700, Ben Bolker wrote:
>
>   
>>   ???  octave is a Matlab clone, not a Mathematica clone (I would be
>> interested in an open source Mathematica clone ...) ???
>>
>> 
>
> You might take a look at Sage.  It is not a mathematica clone, but 
> open source mathematical software which has a development community 
> similar to that of R.   
>
>   

... except for that ws has quite a different attitude than bdr.

vQ

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Re: [R] R 2.9 binaries for redhat entreprise 4 / X86_64

2009-04-23 Thread José Matos

On Thursday 23 April 2009 15:08:26 Marc Schwartz wrote:
> I can't speak to the timeline for the CRAN RPMs, but you might also  
> want to keep an eye on the EPEL repos, which provide add-on RPMs for  
> RHEL.
>
> More info here:
>
>http://fedoraproject.org/wiki/EPEL
>
> and the specific link for R for RHEL 4/x84_64 is:
>
>http://download.fedora.redhat.com/pub/epel/4/x86_64/repoview/R.html
>
> Not there yet either.

The package is already built and it will be out with the next batch of 
packages. EPEL has a regular schedule for these pushes.

> HTH,
>
> Marc Schwartz

-- 
José Abílio

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Re: [R] Stuck using constrOptim


Sorry i Mean 900x1 not 3x900 below

dre968 wrote:
> 
> Trying to use constrOptim to minimize the sum of squared deviations.  I
> put the objective function in as: sum((x %*% Y - Z)^2) so i'm trying to
> get values for x to minimize the sum of the squared deviations between the
> product of x and Y and Z.  
> 
> Anyways i have no problem using this when x is a 3x1 test variable.  it
> works great with the constraints and everything.  when i actually use it
> on my data x is 3x900 or so and it wont optimize it just gives me back my
> initial guessthis is the output i'm getting:
> 
> $value
> [1] 22.7438
> 
> $counts
> function gradient 
>  906   NA 
> 
> $convergence
> [1] 1
> 
> $message
> NULL
> 
> $outer.iterations
> [1] 1
> 
> $barrier.value
> [1] 1.381551e-06
> 
> this is definitely not the right answer and it is just spitting back my
> initial guess back.  
> 
> Any ideas?  are there limits as to how big the variables can be for this
> function?
> 

-- 
View this message in context: 
http://www.nabble.com/Stuck-using-constrOptim-tp23197912p23198151.html
Sent from the R help mailing list archive at Nabble.com.

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[R] Stuck using constrOptim


Trying to use constrOptim to minimize the sum of squared deviations.  I put
the objective function in as: sum((x %*% Y - Z)^2) so i'm trying to get
values for x to minimize the sum of the squared deviations between the
product of x and Y and Z.  

Anyways i have no problem using this when x is a 3x1 test variable.  it
works great with the constraints and everything.  when i actually use it on
my data x is 3x900 or so and it wont optimize it just gives me back my
initial guessthis is the output i'm getting:

$value
[1] 22.7438

$counts
function gradient 
 906   NA 

$convergence
[1] 1

$message
NULL

$outer.iterations
[1] 1

$barrier.value
[1] 1.381551e-06

this is definitely not the right answer and it is just spitting back my
initial guess back.  

Any ideas?  are there limits as to how big the variables can be for this
function?
-- 
View this message in context: 
http://www.nabble.com/Stuck-using-constrOptim-tp23197912p23197912.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] qqnorm.lme & pairs.lme

2009-04-23 Thread Kingsford Jones

The model has multiple levels of random effects, so you need to tell
qqnorm.lme which one you're interested in.

e.g.

qqnorm(m1, ~ranef(., level=1))



hth,
Kingsford Jones



On Thu, Apr 23, 2009 at 11:03 AM, Patrick Zimmerman  wrote:
> Hello,
>
> I am trying to do some plotting to check random effect assumptions for a
> model I fit using lme.
>
> I want to use qqnorm and pairs (similarly to examples given in Pinheiro &
> Bates p. 188), but it's not working.  Here's some relevant code and the
> error message:
>
> library(nlme)
> data(Machines)
> m1 <- lme(fixed=score~Machine,random=~1|Worker/Machine, data=Machines)
> qqnorm(m1, ~ranef(.))
>
> Error in rep(names(fData), rep(nr, nc)) : invalid 'times' argument
> In addition: Warning message:
> In data.frame(Worker = list("(Intercept)" = c(-7.51429457896346,  :
>  row names were found from a short variable and have been discarded
>
> I also tried changing the last line to qqnorm(m1, form=~ranef(.))
>
> Any advice?
>
> Thanks,
> Pat Zimmerman
>
>        [[alternative HTML version deleted]]
>
> __
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> and provide commented, minimal, self-contained, reproducible code.
>

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Re: [R] surface interpolating 3d

2009-04-23 Thread Kingsford Jones

There are a variety of interplation tools to choose from (kriging,
splines, inverse-distance-weighting, ...), and a variety of ways to do
each of these in R.

For example, have a look at the help page and examples for krige in
package gstat, krige.conv in geoR, or Krig in package fields.

hth,
Kingsford Jones

On Thu, Apr 23, 2009 at 7:15 AM, calpeda  wrote:
>
> Hi all,
>
> When you want to draw a surface with a mathematics program you need of two
> vectors x and y and a matrix z. Then you plot the surface.
> For example:
> x=[1 2 3];
> y=[5 6 7];
> z=[1 2 3
>  4 5 6
>  7 8 9];
>
> For my applications I have a 3 vectors x, y, z, that are the coordinates x,
> y, and z of points in the space. I would find the surface that interpolate
> this point. How can I do?
> For example:
> x=[1 2 3];
> y=[5 6 7];
> z=[6 9 8];
>
> Thank you very much
> --
> View this message in context: 
> http://www.nabble.com/surface-interpolating-3d-tp23196713p23196713.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
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> and provide commented, minimal, self-contained, reproducible code.
>

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Re: [R] simple for loop question - how do you exit?

Nevermind i got it

dre968 wrote:
> 
> I have a loop and an if statement in the loop.  once the if statement is
> true for 1 value in the loop i'd like to exit the loop.  is there a
> command to do this?  i know its going to be something like exit and i feel
> stupid asking this question
> 

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Re: [R] simple for loop question - how do you exit?

2009-04-23 Thread Chuck Cleland

On 4/23/2009 2:29 PM, dre968 wrote:
> I have a loop and an if statement in the loop.  once the if statement is true
> for 1 value in the loop i'd like to exit the loop.  is there a command to do
> this?  i know its going to be something like exit and i feel stupid asking
> this question

  Type ?"if" at the R prompt, which should load the help page for
"Control Flow".

-- 
Chuck Cleland, Ph.D.
NDRI, Inc. (www.ndri.org)
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
tel: (732) 512-0171 (M, W, F)
fax: (917) 438-0894

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Re: [R] simple for loop question - how do you exit?

2009-04-23 Thread Duncan Murdoch


On 4/23/2009 2:29 PM, dre968 wrote:

I have a loop and an if statement in the loop.  once the if statement is true
for 1 value in the loop i'd like to exit the loop.  is there a command to do
this?  i know its going to be something like exit and i feel stupid asking
this question


You want "break".  To see the help on this, you need the quotes around 
it:  ?"break", but you don't use them in code, of course.


Duncan Murdoch

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[R] simple for loop question - how do you exit?


I have a loop and an if statement in the loop.  once the if statement is true
for 1 value in the loop i'd like to exit the loop.  is there a command to do
this?  i know its going to be something like exit and i feel stupid asking
this question
-- 
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http://www.nabble.com/simple-for-loop-question---how-do-you-exit--tp23197504p23197504.html
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[R] suffix convention?

2009-04-23 Thread Sebastian P. Luque

Hi,

Is there some convention for choosing 'RData' or 'rda' for binary files
written by save() or save.image()?  The docs treat these
interchangeably.  Thanks.


Cheers,

-- 
Seb

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[R] How to construct confidence bands from a gls fit?

2009-04-23 Thread Maik Renner

Dear R-list,

I would like to show the implications of estimating a linear trend to
time series,
which contain significant serial correlation.

I want to demonstrate this, comparing lm() and an gls() fits, using
the LakeHuron
data set, available in R.
Now in my particular case I would like to draw confidence bands on the plot and
show that there are differences. Unfortunately, I do not know how to
construct those bands
for the gls fit object.
Does anyone know how to do that?

Here is the code:

library(nlme)
library(car)
lm.hu  <- lm(LakeHuron~time(LakeHuron))
## use durbin-watson test to estimate the AR(p) model
durbin.watson(lm.hu,max.lag=7)
# I decided to use a AR(2) model
gls.hu <- gls(LakeHuron ~ time(LakeHuron), correlation=corARMA(p=2),
method="ML")

plot(LakeHuron, main="Lake level time series (Lake Huron)")
lines(as.numeric(time(LakeHuron)),predict(lm.hu,interval=c("conf"))[,1],lty=1,col=2)
lines(as.numeric(time(LakeHuron)),predict(lm.hu,interval=c("conf"))[,2],lty=2,col=2)
lines(as.numeric(time(LakeHuron)),predict(lm.hu,interval=c("conf"))[,3],lty=2,col=2)
## predict of the gls class does not contain confidence intervals
lines(as.numeric(time(LakeHuron)),predict(gls.hu,interval=c("conf")),lty=1,col=4)

Thank you,

Maik

--
Maik Renner
Institut für Hydrologie und Meteorologie
Technische Universität Dresden

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[R] R 2.8.1 change user input color of R scrpt too

2009-04-23 Thread matildet


Hello R community

I'm using R 2.8.1 version and would like to change the color of user input
for script pages. Once I have changed the color of console background and
user input (by means of Interface Preferences menu), the color of script
background has been automatically changed but the color of user input has
not.

Has someone any useful advice for my (little) problem?   

Thanks in advance

Matilde
-- 
View this message in context: 
http://www.nabble.com/R-2.8.1-change-user-input-color-of-R-scrpt-too-tp23197473p23197473.html
Sent from the R help mailing list archive at Nabble.com.

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[R] qqnorm.lme & pairs.lme

2009-04-23 Thread Patrick Zimmerman

Hello,

I am trying to do some plotting to check random effect assumptions for a
model I fit using lme.

I want to use qqnorm and pairs (similarly to examples given in Pinheiro &
Bates p. 188), but it's not working.  Here's some relevant code and the
error message:

library(nlme)
data(Machines)
m1 <- lme(fixed=score~Machine,random=~1|Worker/Machine, data=Machines)
qqnorm(m1, ~ranef(.))

Error in rep(names(fData), rep(nr, nc)) : invalid 'times' argument
In addition: Warning message:
In data.frame(Worker = list("(Intercept)" = c(-7.51429457896346,  :
  row names were found from a short variable and have been discarded

I also tried changing the last line to qqnorm(m1, form=~ranef(.))

Any advice?

Thanks,
Pat Zimmerman

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Re: [R] Load a data from oracle database to R

2009-04-23 Thread Marc Schwartz



On Apr 23, 2009, at 12:18 PM, Gagan Pabla wrote:


Hello,

I am have trying to load data in R by connecting R to the database the
following way:

library(RODBC)
channel<-odbcConnect("gagan")


now after I connect to the server by putting pwd. I want to load  
table from

the database named "temp" in to R so that I can do some descriptive
statistics with it.

now when I try to do "data(temp)" it gives error saying "In  
data(temp) :

data set 'temp' not found".

So, how do I load the table to R.

Your help will be greatly appreciated.

Thanks,

Gagan


Once you have established the connection to the Oracle server, you  
then need to perform a query against the database to get the data into  
R. So something like the following:


  temp <- sqlQuery(channel, "select * from temp")
  close(channel)

Now you will have a data frame in R called temp.

See ?sqlQuery after loading RODBC.

BTW, there is a r-sig-db list. More info here:

  https://stat.ethz.ch/mailman/listinfo/r-sig-db

HTH,

Marc Schwartz

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[R] SAGE: was large factorials

2009-04-23 Thread Albyn Jones

On Wed, Apr 22, 2009 at 08:26:51PM -0700, Ben Bolker wrote:

> 
>   ???  octave is a Matlab clone, not a Mathematica clone (I would be
> interested in an open source Mathematica clone ...) ???
> 

You might take a look at Sage.  It is not a mathematica clone, but 
open source mathematical software which has a development community 
similar to that of R.   

http://www.sagemath.org/

albyn

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[R] Load a data from oracle database to R

2009-04-23 Thread Gagan Pabla

Hello,

I am have trying to load data in R by connecting R to the database the
following way:
> library(RODBC)
> channel<-odbcConnect("gagan")

now after I connect to the server by putting pwd. I want to load table from
the database named "temp" in to R so that I can do some descriptive
statistics with it.

now when I try to do "data(temp)" it gives error saying "In data(temp) :
data set 'temp' not found".

So, how do I load the table to R.

Your help will be greatly appreciated.

Thanks,

Gagan

[[alternative HTML version deleted]]

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Re: [R] fitting assimptotic decaing with - and + on X

2009-04-23 Thread milton ruser

Hi Christian,

Thank you very much for the help.
Now I can run. But my main interest is on the region of x[-200 to +200]
and as you can see this logistic regression not fit very well.

You think that using other package / function I can get better fit of these
data?

Best wishes,

miltinho
brazil-toronto

sp.coredep.attr<-read.table(stdin(), head=T, sep=",")
species,dst,preference,probdye
coredep,-7500,1,0.001
coredep,-1000,0.95,0.002
coredep,-500,0.9,0.003
coredep,-250,0.8,0.005
coredep,-120,0.7,0.01
coredep,-100,0.5,0.02
coredep,-90,0.3,0.03
coredep,-60,0.2,0.05
coredep,-30,0.15,0.065
coredep,0,0.1,0.08
coredep,30,0.09,0.1
coredep,60,0.08,0.15
coredep,90,0.07,0.2
coredep,120,0.06,0.3
coredep,240,0.05,0.4
coredep,500,0.04,0.5
coredep,1000,0.02,0.6
coredep,7500,0.01,0.7


plot(preference~as.factor(dst), data=sp.coredep.attr, type="n", cex=0.001,
col="transparent", xlab="Distance (m)", main="CORE Dependent")
lines(sp.coredep.attr$preference, type="b", lwd=2, col="blue")
abline(v=10, col="blue", lty=3)
text(10, 1, "EDGE", cex=1.5)
text(7, 1, "CORE", cex=1.5, col="green")
text(13, 1, "MATRIX", cex=1.5, col="red")
text(4,0.65, "Preference", col="blue")
text(16,0.65, "Prob.Die", col="red")
lines(sp.coredep.attr$probdye, type="b", lwd=2, col="red", lty=2)
require(drc)
sp.coredep.attr.m1 <- drm(preference~dst, data=sp.coredep.attr, fct=L.4())
x11()
plot(sp.coredep.attr.m1, log = "")
x11()
plot(sp.coredep.attr.m1, log = "", xlim = c(-500, 500), ylim = c(0, 1))
summary(sp.coredep.attr.m1)






On Thu, Apr 23, 2009 at 1:41 AM, Christian Ritz  wrote:

> Hi Milton,
>
> you're right that most of the functions in the package "drc" are suited for
> sigmoidal/s-shaped curves defined on the positive axis.
>
> However, there is one exception: the logistic model:
>
> library(drc)
>
> sp.coredep.attr.m1 <- drm(preference~dst, data=sp.coredep.attr, fct=L.4())
>
> plot(sp.coredep.attr.m1, log = "")
> plot(sp.coredep.attr.m1, log = "", xlim = c(-500, 500), ylim = c(0, 1))
>
> summary(sp.coredep.attr.m1)
>
>
> The fit isn't very good, possibly because the curve you've observed seems
> to have a kind
> of bent at or close to 0.
>
> Christian
>
>

[[alternative HTML version deleted]]

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Re: [R] Floating simulation error

2009-04-23 Thread Nordlund, Dan (DSHS/RDA)

> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of Brendan Morse
> Sent: Thursday, April 23, 2009 9:12 AM
> To: r-help@r-project.org
> Subject: [R] Floating simulation error
> 
> 
> Hi all, I am running a simulation and a curious error keeps coming up
> that stops the whole process. The error is a subscript out of bounds
> error, and it seems to happen at different points (floating around)
> throughout the looping simulation. Say, for example, it crashes on
> sample 1 - iteration 200. I can force it to start again on iteration
> 202 with all of the same settings, and it is fine. I can also force it
> to start again on iteration 201 if I modify the starting seed value
> and it goes on its merry way. What I think is happening is that
> certain seed values are disrupting something and causing the subscript
> error.
> 
> My seeds are set as follows: The simulation has x number of
> conditions. The starting seeds are set to equal the condition number
> (1 to x). The program runs 500 iterations within each condition, and
> the seed values for each iteration are set as x+iteration number. So,
> for condition 1, iteration 1, the starting seed value would be 2, then
> 3 etc. etc. At some point, I will get the subscript error but it seems
> unpredictable.
> 
> Has anyone had a similar problem or an idea as to what might be
> happening?
> 
> - Brendan
> 
> 

Brendan,

This is one of those situations where a self-contained, reproducible example 
would be really helpful.  In the absence of that, seeing your actual code is 
absolutely necessary.  I can't imagine anyone being able to provide any useful 
help here otherwise.

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] Floating simulation error

2009-04-23 Thread David M Smith

Cosmic rays, perhaps? :)
http://blog.revolution-computing.com/2009/04/blame-it-on-cosmic-rays.html

Seriously though, my guess is that the simulation data from your 201st
iteration is tickling some subtle bug in your code. This has happened
to me before where I generate a random matrix that happens to be
non-invertible, for example.

I'd suggest grabbing the data from your 201st iteration (a simple way
would be to run your RNG 200 times at the start of your code) and
debug from there.

# David Smith

On Thu, Apr 23, 2009 at 11:12 AM, Brendan Morse  wrote:
>
> Hi all, I am running a simulation and a curious error keeps coming up
> that stops the whole process. The error is a subscript out of bounds
> error, and it seems to happen at different points (floating around)
> throughout the looping simulation. Say, for example, it crashes on
> sample 1 - iteration 200. I can force it to start again on iteration
> 202 with all of the same settings, and it is fine. I can also force it
> to start again on iteration 201 if I modify the starting seed value
> and it goes on its merry way. What I think is happening is that
> certain seed values are disrupting something and causing the subscript
> error.
>
> My seeds are set as follows: The simulation has x number of
> conditions. The starting seeds are set to equal the condition number
> (1 to x). The program runs 500 iterations within each condition, and
> the seed values for each iteration are set as x+iteration number. So,
> for condition 1, iteration 1, the starting seed value would be 2, then
> 3 etc. etc. At some point, I will get the subscript error but it seems
> unpredictable.
>
> Has anyone had a similar problem or an idea as to what might be
> happening?
>
> - Brendan
>
>
> Brendan Morse, M.S.
> Industrial/Organizational Psychology
> Ohio University
> Office: 335 Porter Hall
> Website: http://oak.cats.ohiou.edu/~bm123504
>
>
>
>
>        [[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.
>



-- 
David M Smith 
Director of Community, REvolution Computing www.revolution-computing.com
Tel: +1 (206) 577-4778 x3203 (San Francisco, USA)

Check out our upcoming events schedule at www.revolution-computing.com/events

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Re: [R] Floating simulation error

2009-04-23 Thread Bert Gunter

?traceback
options("error")  ( ?options)
?debug
?try
?tryCatch

R has facilities to help you with such problems. Please use them -- and then
repost with more specific info if you still cannot solve it.

-- Bert Gunter

 

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Brendan Morse
Sent: Thursday, April 23, 2009 9:12 AM
To: r-help@r-project.org
Subject: [R] Floating simulation error


Hi all, I am running a simulation and a curious error keeps coming up  
that stops the whole process. The error is a subscript out of bounds  
error, and it seems to happen at different points (floating around)  
throughout the looping simulation. Say, for example, it crashes on  
sample 1 - iteration 200. I can force it to start again on iteration  
202 with all of the same settings, and it is fine. I can also force it  
to start again on iteration 201 if I modify the starting seed value  
and it goes on its merry way. What I think is happening is that  
certain seed values are disrupting something and causing the subscript  
error.

My seeds are set as follows: The simulation has x number of  
conditions. The starting seeds are set to equal the condition number  
(1 to x). The program runs 500 iterations within each condition, and  
the seed values for each iteration are set as x+iteration number. So,  
for condition 1, iteration 1, the starting seed value would be 2, then  
3 etc. etc. At some point, I will get the subscript error but it seems  
unpredictable.

Has anyone had a similar problem or an idea as to what might be  
happening?

- Brendan


Brendan Morse, M.S.
Industrial/Organizational Psychology
Ohio University
Office: 335 Porter Hall
Website: http://oak.cats.ohiou.edu/~bm123504




[[alternative HTML version deleted]]

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Re: [R] power.t.test formula

2009-04-23 Thread Peter Dalgaard

Usuario R wrote:
> Hi,
> 
> Does anyone of you knows a reference for the formula used in power.t.test
> function? And also why it uses the Student's distribution instead of Normal.
> (I know both of them can be used but don't see whether choose one or the
> other)

It is  a straightforward first-principles calculation. The t
distribution calculation is exact for normally distributed data with the
same unknown variance in both groups. Formulas based on the normal
distribution assumes the variance to be known, which overestimates small n.

-- 
   O__   Peter Dalgaard Øster Farimagsgade 5, Entr.B
  c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
 (*) \(*) -- University of Copenhagen   Denmark  Ph:  (+45) 35327918
~~ - (p.dalga...@biostat.ku.dk)  FAX: (+45) 35327907

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[R] update.packages(checkBuilt=TRUE) returns Error: invalid version specification NA in 2.9.0

2009-04-23 Thread Peter_Keller



I installed R-2.9.0 yesterday, and followed the instructions in the R for
Windows FAQ, 2.8 What's the best way to upgrade?  It reads "run
update.packages(checkBuilt=TRUE, ask=FALSE) in the new R and then delete
anything left of the old installation."  I did this but it returned an
error.  It's the checkBuilt=TRUE which causes it, though I don't know why:
leaving it out or changing it to checkBuilt=FALSE do not return errors.

> update.packages(checkBuilt=TRUE, ask=FALSE)
--- Please select a CRAN mirror for use in this session ---
Error: invalid version specification NA

> update.packages(checkBuilt=TRUE)
Error: invalid version specification NA

Peter Keller
Wildlife Biologist
Tetlin National Wildlife Refuge
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[R] Floating simulation error

2009-04-23 Thread Brendan Morse


Hi all, I am running a simulation and a curious error keeps coming up  
that stops the whole process. The error is a subscript out of bounds  
error, and it seems to happen at different points (floating around)  
throughout the looping simulation. Say, for example, it crashes on  
sample 1 - iteration 200. I can force it to start again on iteration  
202 with all of the same settings, and it is fine. I can also force it  
to start again on iteration 201 if I modify the starting seed value  
and it goes on its merry way. What I think is happening is that  
certain seed values are disrupting something and causing the subscript  
error.

My seeds are set as follows: The simulation has x number of  
conditions. The starting seeds are set to equal the condition number  
(1 to x). The program runs 500 iterations within each condition, and  
the seed values for each iteration are set as x+iteration number. So,  
for condition 1, iteration 1, the starting seed value would be 2, then  
3 etc. etc. At some point, I will get the subscript error but it seems  
unpredictable.

Has anyone had a similar problem or an idea as to what might be  
happening?

- Brendan


Brendan Morse, M.S.
Industrial/Organizational Psychology
Ohio University
Office: 335 Porter Hall
Website: http://oak.cats.ohiou.edu/~bm123504




[[alternative HTML version deleted]]

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

[R] power.t.test formula

2009-04-23 Thread Usuario R

Hi,

Does anyone of you knows a reference for the formula used in power.t.test
function? And also why it uses the Student's distribution instead of Normal.
(I know both of them can be used but don't see whether choose one or the
other)

Thank you.
Regards

[[alternative HTML version deleted]]

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Re: [R] boxplot of two variables

2009-04-23 Thread Mike Lawrence

Check out ggplot2:

http://had.co.nz/ggplot2

especially:
http://had.co.nz/ggplot2/geom_boxplot.html

But you are strongly advised to read the book:
http://had.co.nz/ggplot2/book/

On Thu, Apr 23, 2009 at 12:13 PM, Gabriel R. Rodriguez
 wrote:
> Hello !
>
>
>
> I have a dataframe with 6 variables (A1,A2,B1,B2,C1,C2) and 1 factor (F).
>
>
>
> I would like to produce a graph consisting of 3 boxplots sets, one for every
> two  variables (i.e A1 &A2) by the factor (F).
>
> I was looking around and I cannot figure it out, any suggestions?
>
>
>
> Best Regards,
>
> Gabriel
>
>
>
>
>        [[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.
>



-- 
Mike Lawrence
Graduate Student
Department of Psychology
Dalhousie University

Looking to arrange a meeting? Check my public calendar:
http://tr.im/mikes_public_calendar

~ Certainty is folly... I think. ~

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[R] Course announcement: R for Financial Data Analysis in New York (July 16-18th)

2009-04-23 Thread Francisco Gochez

Dear userRs,
 
Mango Solutions are pleased to announce that we will deliver a 3-day
introductory R course focused on financial analysis in New York on the
16-18th of July.  The course topics are as follows:
 
*   Introduction
*   The R Environment 
*   Data Objects 
*   Functions 
*   Important R Functions 
*   Traditional Graphics 
*   Basic R Statistical and Mathematical Functions used in the
Finance Industry 
*   Time Series Analysis, Manipulation and Visualization 
*   Performance and Risk Measurement 

As with all our courses, attendees are provided with comprehensive
training manuals complete with detailed examples and laminated tip
sheets for future reference.
The cost of this course is $1800 for commercial attendees and $1000 for
academic attendees.   Should you wish to book a place on this course or
have any questions please contact us at train...@mango-solutions.com.

Further information is available on our web-site at
http://www.mango-solutions.com/services/rtraining/r_finance.html

Kind regards,

Francisco

mango solutions

S & R Consulting and Training

+44 (0)1249 767 700

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Re: [R] R-User groups in North America (SF, LA, NYC, Ottawa)

2009-04-23 Thread Michael E. Driscoll

Folks, Here is the correct URL for the NYC R Users group:

New York City
  http://www.meetup.com/nyhackr  (61 members)
  Organized by Joshua Reich 

(Thanks to Johnathan Boysielal for pointing out this error, and Curt
for the OSU correction as well).

On Wed, Apr 22, 2009 at 12:01 PM, Michael E. Driscoll
 wrote:
> Los Angeles
>  http://www.meetup.com/LAarea-R-usergroup/ (79 members)
>  May 7th Event: "Four case studies of using R" at 7pm, Boelter Hall, UCLA
>  Presenters:  J. Leeuw, R. Gould, & B. Brett-Esborn (UCLA), S. Pafka (Epoch)
>  Organized by Szilard Pafka 
>
> San Francisco / Bay Area
>  http://www.meetup.com/R-Users (228 members)
>  May 13th Event:  Parallel Computing with R using ParallelR at Microsoft SF HQ
>  Presenter:  David M Smith (Revolutions Computing)
>  Organized by Jim Porzak  and yours truly
>

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[R] boxplot of two variables

2009-04-23 Thread Gabriel R. Rodriguez

Hello !

 

I have a dataframe with 6 variables (A1,A2,B1,B2,C1,C2) and 1 factor (F).

 

I would like to produce a graph consisting of 3 boxplots sets, one for every
two  variables (i.e A1 &A2) by the factor (F).

I was looking around and I cannot figure it out, any suggestions?

 

Best Regards, 

Gabriel

 


[[alternative HTML version deleted]]

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Re: [R] R 2.9 binaries for redhat entreprise 4 / X86_64

2009-04-23 Thread Mathieu Van der Haegen


Hi !

Thank you for the info.

Is this a little bit stable ?

Mathieu

On 23 Apr 2009, at 16:01, R P Herrold wrote:


On Thu, 23 Apr 2009, Mathieu Van der Haegen wrote:

I'd like to ask if someone could tell me when the binaries for R  
2.9 /

redhat entreprise 4 / X86_64 will be available ?
I don't need a precise date but some indication like 1 month, 6  
month,

never ?


How about: Now, in Red Hat's RawHide archive?

[herr...@centos-5 ~]$ uname -a ; rpm -q R
Linux centos-5.first.lan 2.6.18-128.1.6.el5xen #1 SMP Wed Apr
1 09:53:14 EDT 2009 x86_64 x86_64 x86_64 GNU/Linux
R-2.9.0-1orc
[herr...@centos-5 ~]$

I built it earlier this week from SRPM on CentOS (a rebuild of RHEL)

-- Russ herrold


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Re: [R] Adobe FLEX interacting with R for rich visualization over the Web

2009-04-23 Thread Jeffrey Horner

Harsh wrote on 04/23/2009 09:49 AM:

Hi R users,
I am looking to create a rich internet application using R as the
analytical back-end with a GUI written entirely in Adobe FLEX.
I have come across this paper
http://www.bioconductor.org/packages/2.3/bioc/vignettes/RWebServices/inst/doc/RelatedWork.pdf
which provides relevant information.

From what I understand, FLEX can read xml files. The StatDataML

library in R creates xml files which can be read by FLEX. Since I
would need to implement the GUI on a web server and the R server would
be on a different system, I felt using StatDataML to write xml files
would be the best approach.

I have also heard of the RSOAP library which possibly can be another
alternative to the above library.

I am trying to create a way for users to be able to visualize the
results of regressions and other statistical functions in an
interactive manner over the web.

If anyone has worked with the above tools or related scenarios, I
would greatly appreciate your feedback.

You will be interested in the following links then:

http://blog.revolution-computing.com/2009/04/find-a-safer-place-in-the-bay-area.html
http://blog.revolution-computing.com/2009/04/comparing-baseball-pitcher-styles-with-lattice-graphics.html

They provide examples of building web dashboards with R, which probably
falls in the category of rich internet applications.

Best,

Jeff

--
http://biostat.mc.vanderbilt.edu/JeffreyHorner

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Re: [R] large factorials

One more improvement.  Perhaps it would be best just to return a numeric
so sum1 inputs and outputs numerics:

library(rSymPy)

# define factorial to return a Sym object
factorial.Sym <- function(n) Sym("factorial(", n, ")")

sum1 <- function(l,u,t,i,n,w) {
 v <- 0
 for (m in 0 :w) {
v1 <- ((u^(1/2))*(l^(1/2))*t)^(i-n+2*m)
v2 <- (factorial.Sym(i-n+m))*(factorial.Sym(m))
v3 <- v1/v2
v <- v+v3
 }

 as.numeric(sympy(v)) # send it to SymPy, make result a Sym obj

}

s <- sum1(1,2,10,80,3,80)
s # numeric


On Thu, Apr 23, 2009 at 10:44 AM, Gabor Grothendieck
 wrote:
> The code in my prior post works (except one comment was wrong)
> but try this instead.  The only change is the last line of the sum1
> function.   This way it produces a Sym object rather than a character
> string.
>
> library(rSymPy)
>
> # define factorial to return a Sym object
> factorial.Sym <- function(n) Sym("factorial(", n, ")")
>
> sum1 <- function(l,u,t,i,n,w) {
>  v <- 0
>  for (m in 0 :w) {
>         v1 <- ((u^(1/2))*(l^(1/2))*t)^(i-n+2*m)
>         v2 <- (factorial.Sym(i-n+m))*(factorial.Sym(m))
>         v3 <- v1/v2
>         v <- v+v3
>  }
>  Sym(sympy(v)) # send it to SymPy, make result a Sym obj
> }
>
> s <- sum1(1,2,10,80,3,80)
> s # Sym
> as.numeric(s) # numeric
>
>
> On Thu, Apr 23, 2009 at 10:37 AM, Gabor Grothendieck
>  wrote:
>> sympy() returns a character string, not an R numeric -- it shouldn't
>> automatically return an R numeric since R can't represent all
>> the numbers that sympy can.
>>
>> The development version of rSymPy has a second class which
>> produces objects of class c("Sym", "character") and those
>> can be manipulated with +, -, *, / producing other Sym
>> objects so try this:
>>
>>
>> library(rSymPy)
>>
>> # next line pulls in code to handle Sym objects
>> source("http://rsympy.googlecode.com/svn/trunk/R/Sym.R";)
>>
>> # define factorial to return a Sym object
>> factorial.Sym <- function(n) Sym("factorial(", n, ")")
>>
>> sum1 <- function(l,u,t,i,n,w) {
>>  v <- 0
>>  for (m in 0 :w) {
>>         v1 <- ((u^(1/2))*(l^(1/2))*t)^(i-n+2*m)
>>         v2 <- (factorial.Sym(i-n+m))*(factorial.Sym(m))
>>         v3 <- v1/v2
>>         v <- v+v3
>>  }
>>  sympy(v)
>> }
>>
>> s <- sum1(1,2,10,80,3,80)
>> s # Sym object
>> as.numeric(s) # numeric
>>
>>
>> On Thu, Apr 23, 2009 at 10:00 AM, molinar  wrote:
>>>
>>> Here is what I did:
>>> library(rSymPy)
>>> factorial.sympy <- function(n) sympy(paste("factorial(", n, ")"))
>>> factorial.sympy(171)
>>> [1]
>>> "124101807021766782342484052410310399261660557750169318538895180361199607522169175299275197812048758557646495950167038705280988985869071076733124203221848436431047357788996854827829075454156196485215346831804429323959817369689965723590394761615227855818006117636510842880"

>>> Which work perfectly.
>>>
>>> Here is one of my summation functions:
>>>
>>> sum1 <- function(l,u,t,i,n,w) {
>>> + v <- 0
>>> + for (m in 0 :w) {
>>> + v1 <- ((u^(1/2))*(l^(1/2))*t)^(i-n+2*m)
>>> + v2 <- (factorial.sympy(i-n+m))*(factorial.sympy(m))
>>> + v3 <- v1/v2
>>> + v <- v+v3
>>> + }
>>> + return(v)
>>> + }
>>>
>>> sum1(1,2,10,80,3,80)
>>> Error in (factorial.sympy(i - n + m)) * (factorial.sympy(m)) :
>>>  non-numeric argument to binary operator
>>>
>>> I'm not sure why it works when I do the factorial normally but when I call
>>> my function it doesn't work?
>>>
>>>
>>>
>>>
>>>
>>>
>>>
>>> molinar wrote:

 Thank you everyone all of your posts were very helpful.  I tried each one
 and I think I have about 10 new packages installed.  The formula I need to
 calculate did not involve any logarithms but infinite summations of
 factorials, I'm sorry for not specifying.  I read some things about using
 logarithms but I thought in my case I would have to do an e to the log and
 by doing that R still gave me the same problems with numbers over 170.

 But I was able to get it to work by using the last post about the rsympy
 packages.

 I tried downloading bc but I didn't know how to connect it to R, so R said
 "could not find function bc".

 Thanks again for all of your help.
 Samantha





 Gabor Grothendieck wrote:
>
> Also the R sympy package can handle this:
>
>> library(rSymPy)
> Loading required package: rJava
>
>> factorial.sympy <- function(n) sympy(paste("factorial(", n, ")"))
>
>> # note that first time sympy is called it loads java, jython and sympy
>> # but on subsequent calls its faster.  So make a dummy call first.
>> factorial.sympy(10)
> [1] "3628800"
>
>> # code from earlier post defining factorial.bc to be inserted here
>
>>    benchmark(replications=10, columns=c('test', 'elapsed'),
> +       bc=factorial.bc(500),
> +       sympy = factorial.sympy(500))
>    test elapsed
> 1    bc    2.17
> 2 sympy    0.09
>
> See the rSymPy

Re: [R] Plots - several pages per pdf - quality/size issue

2009-04-23 Thread Richard . Cotton

> I hope that question will not be too redundant (sorry if it is) but 
> i don't seem able to find the answer i need in the archives...
> 
> I try to create a file which would have 1.several pages and 2.
> several plots by page. I know how to make a pdf file this way, but 
> my problem is that the pdf size gets way too big and cannot be open.
> 
> So my question is:
> - is there a way to diminish the pdf size/quality?
> - does any other file format exists that would be of smaller quality
> but still be able to print plots on several pages for the same file?

You could try postscript instead. e.g.

p1 <- xyplot(runif(1)~rnorm(1)|sample(letters[1:3], 1, 
replace=TRUE), layout=c(1,1))

pdf("test.pdf")
print(p1)
dev.off()
# File size 623kb

postscript("test.ps")
print(p1)
dev.off()
# File size 257kb

The other alternative is to print to a raster format, e.g. png.  this 
creates different files for different pages, but you could always find a 
way to view multiple plot files at once.  Some possibilities:
1. Use, e.g.,  Picasa to browse your plot files.
2. Create a simple webpage containing each image

# R code
png("test%d.png")
print(p1)
dev.off()

# Web page code
http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd";>
http://www.w3.org/1999/xhtml";>



 




# File sizes 3*13kb + 1kb

Open and save this file in a word processor of your choice, and you can 
probably drop the file size even further.

Regards,
Richie.

Mathematical Sciences Unit
HSL




ATTENTION:

This message contains privileged and confidential inform...{{dropped:20}}

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Re: [R] transposing a matrix - row by row?

2009-04-23 Thread Dimitri Liakhovitski

Thank you very much, Thierry!
Really amazing, the magic melt from reshape. And, in fact, a bit
dangerous - it just do what I need without any efforts on my part!
Dimitri

On Thu, Apr 23, 2009 at 10:40 AM, ONKELINX, Thierry
 wrote:
> Dear Dimitri,
>
> Have a look at melt() from the reshape package.
>
> X<-matrix(c(10,20,30,40,50,60),2,3)
> dimnames(X)<-list(c("1","2"),c("1","2","3"))
> library(reshape)
> melt(X)
>
> HTH,
>
> Thierry
>
>
> 
> 
> ir. Thierry Onkelinx
> Instituut voor natuur- en bosonderzoek / Research Institute for Nature
> and Forest
> Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
> methodology and quality assurance
> Gaverstraat 4
> 9500 Geraardsbergen
> Belgium
> tel. + 32 54/436 185
> thierry.onkel...@inbo.be
> www.inbo.be
>
> To call in the statistician after the experiment is done may be no more
> than asking him to perform a post-mortem examination: he may be able to
> say what the experiment died of.
> ~ Sir Ronald Aylmer Fisher
>
> The plural of anecdote is not data.
> ~ Roger Brinner
>
> The combination of some data and an aching desire for an answer does not
> ensure that a reasonable answer can be extracted from a given body of
> data.
> ~ John Tukey
>
> -Oorspronkelijk bericht-
> Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> Namens Dimitri Liakhovitski
> Verzonden: donderdag 23 april 2009 16:31
> Aan: R-Help List
> Onderwerp: [R] transposing a matrix - row by row?
>
> Hello,
> I have a matrix that is a product of tapply on a larger data set.
> Let's assume it looks like this:
>
> X<-matrix(c(10,20,30,40,50,60),2,3)
> dimnames(X)<-list(c("1","2"),c("1","2","3"))
> (X)
>
>   1   2   3
> 1 10 30 50
> 2 20 40 60
>
> Is there an efficient way of transforming this matrix into the following
> matrix:
>
> rows columns entries
> 1          1          10
> 1          2          30
> 1          3          50
> 2          1          20
> 2          2          40
> 2          3          60
>
>
> Thank you very much!
> --
> Dimitri Liakhovitski
> MarketTools, Inc.
> dimitri.liakhovit...@markettools.com
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
> Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer
> en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd 
> is
> door een geldig ondertekend document. The views expressed in  this message
> and any annex are purely those of the writer and may not be regarded as 
> stating
> an official position of INBO, as long as the message is not confirmed by a 
> duly
> signed document.
>



-- 
Dimitri Liakhovitski
MarketTools, Inc.
dimitri.liakhovit...@markettools.com

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Adobe FLEX interacting with R for rich visualization over the Web

2009-04-23 Thread Harsh

Hi R users,
I am looking to create a rich internet application using R as the
analytical back-end with a GUI written entirely in Adobe FLEX.
I have come across this paper
http://www.bioconductor.org/packages/2.3/bioc/vignettes/RWebServices/inst/doc/RelatedWork.pdf
which provides relevant information.

>From what I understand, FLEX can read xml files. The StatDataML
library in R creates xml files which can be read by FLEX. Since I
would need to implement the GUI on a web server and the R server would
be on a different system, I felt using StatDataML to write xml files
would be the best approach.

I have also heard of the RSOAP library which possibly can be another
alternative to the above library.

I am trying to create a way for users to be able to visualize the
results of regressions and other statistical functions in an
interactive manner over the web.

If anyone has worked with the above tools or related scenarios, I
would greatly appreciate your feedback.

Thanks
Harsh Singhal

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] transposing a matrix - row by row?

2009-04-23 Thread Dimitris Rizopoulos


one way is the following:

X <- matrix(c(10,20,30,40,50,60), 2, 3,
dimnames = list(c("1","2"), c("1","2","3")))

tX <- t(X)
cbind(
rows = c(col(tX)),
columns = c(row(tX)),
entries = c(tX)
)


I hope it helps.

Best,
Dimitris


Dimitri Liakhovitski wrote:

Hello,
I have a matrix that is a product of tapply on a larger data set.
Let's assume it looks like this:

X<-matrix(c(10,20,30,40,50,60),2,3)
dimnames(X)<-list(c("1","2"),c("1","2","3"))
(X)

   1   2   3
1 10 30 50
2 20 40 60

Is there an efficient way of transforming this matrix into the following matrix:

rows columns entries
1  1  10
1  2  30
1  3  50
2  1  20
2  2  40
2  3  60


Thank you very much!


--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center

Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014

__
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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] large factorials

The code in my prior post works (except one comment was wrong)
but try this instead.  The only change is the last line of the sum1
function.   This way it produces a Sym object rather than a character
string.

library(rSymPy)

# define factorial to return a Sym object
factorial.Sym <- function(n) Sym("factorial(", n, ")")

sum1 <- function(l,u,t,i,n,w) {
 v <- 0
 for (m in 0 :w) {
 v1 <- ((u^(1/2))*(l^(1/2))*t)^(i-n+2*m)
 v2 <- (factorial.Sym(i-n+m))*(factorial.Sym(m))
 v3 <- v1/v2
 v <- v+v3
 }
 Sym(sympy(v)) # send it to SymPy, make result a Sym obj
}

s <- sum1(1,2,10,80,3,80)
s # Sym
as.numeric(s) # numeric


On Thu, Apr 23, 2009 at 10:37 AM, Gabor Grothendieck
 wrote:
> sympy() returns a character string, not an R numeric -- it shouldn't
> automatically return an R numeric since R can't represent all
> the numbers that sympy can.
>
> The development version of rSymPy has a second class which
> produces objects of class c("Sym", "character") and those
> can be manipulated with +, -, *, / producing other Sym
> objects so try this:
>
>
> library(rSymPy)
>
> # next line pulls in code to handle Sym objects
> source("http://rsympy.googlecode.com/svn/trunk/R/Sym.R";)
>
> # define factorial to return a Sym object
> factorial.Sym <- function(n) Sym("factorial(", n, ")")
>
> sum1 <- function(l,u,t,i,n,w) {
>  v <- 0
>  for (m in 0 :w) {
>         v1 <- ((u^(1/2))*(l^(1/2))*t)^(i-n+2*m)
>         v2 <- (factorial.Sym(i-n+m))*(factorial.Sym(m))
>         v3 <- v1/v2
>         v <- v+v3
>  }
>  sympy(v)
> }
>
> s <- sum1(1,2,10,80,3,80)
> s # Sym object
> as.numeric(s) # numeric
>
>
> On Thu, Apr 23, 2009 at 10:00 AM, molinar  wrote:
>>
>> Here is what I did:
>> library(rSymPy)
>> factorial.sympy <- function(n) sympy(paste("factorial(", n, ")"))
>> factorial.sympy(171)
>> [1]
>> "124101807021766782342484052410310399261660557750169318538895180361199607522169175299275197812048758557646495950167038705280988985869071076733124203221848436431047357788996854827829075454156196485215346831804429323959817369689965723590394761615227855818006117636510842880"
>>>
>> Which work perfectly.
>>
>> Here is one of my summation functions:
>>
>> sum1 <- function(l,u,t,i,n,w) {
>> + v <- 0
>> + for (m in 0 :w) {
>> + v1 <- ((u^(1/2))*(l^(1/2))*t)^(i-n+2*m)
>> + v2 <- (factorial.sympy(i-n+m))*(factorial.sympy(m))
>> + v3 <- v1/v2
>> + v <- v+v3
>> + }
>> + return(v)
>> + }
>>
>> sum1(1,2,10,80,3,80)
>> Error in (factorial.sympy(i - n + m)) * (factorial.sympy(m)) :
>>  non-numeric argument to binary operator
>>
>> I'm not sure why it works when I do the factorial normally but when I call
>> my function it doesn't work?
>>
>>
>>
>>
>>
>>
>>
>> molinar wrote:
>>>
>>> Thank you everyone all of your posts were very helpful.  I tried each one
>>> and I think I have about 10 new packages installed.  The formula I need to
>>> calculate did not involve any logarithms but infinite summations of
>>> factorials, I'm sorry for not specifying.  I read some things about using
>>> logarithms but I thought in my case I would have to do an e to the log and
>>> by doing that R still gave me the same problems with numbers over 170.
>>>
>>> But I was able to get it to work by using the last post about the rsympy
>>> packages.
>>>
>>> I tried downloading bc but I didn't know how to connect it to R, so R said
>>> "could not find function bc".
>>>
>>> Thanks again for all of your help.
>>> Samantha
>>>
>>>
>>>
>>>
>>>
>>> Gabor Grothendieck wrote:

 Also the R sympy package can handle this:

> library(rSymPy)
 Loading required package: rJava

> factorial.sympy <- function(n) sympy(paste("factorial(", n, ")"))

> # note that first time sympy is called it loads java, jython and sympy
> # but on subsequent calls its faster.  So make a dummy call first.
> factorial.sympy(10)
 [1] "3628800"

> # code from earlier post defining factorial.bc to be inserted here

>    benchmark(replications=10, columns=c('test', 'elapsed'),
 +       bc=factorial.bc(500),
 +       sympy = factorial.sympy(500))
    test elapsed
 1    bc    2.17
 2 sympy    0.09

 See the rSymPy, r-bc and rbenchmark home pages:
 http://rsympy.googlecode.com
 http://r-bc.googlecode.com
 http://rbenchmark.googlecode.com

 On Wed, Apr 22, 2009 at 3:21 PM, molinar  wrote:
>
> I am working on a project that requires me to do very large factorial
> evaluations.  On R the built in factorial function and the one I created
> both are not able to do factorials over 170.  The first gives an error
> and
> mine return Inf.
>
> Is there a way to have R do these larger calculations (the calculator in
> accessories can do 1 factorial and Maple can do even larger)
> --
> View this message in context:
> http://www.nabble.com/large-factorials-tp23175816p23175816.html
> Sent from the R he

Re: [R] transposing a matrix - row by row?

2009-04-23 Thread ONKELINX, Thierry

Dear Dimitri,

Have a look at melt() from the reshape package.

X<-matrix(c(10,20,30,40,50,60),2,3)
dimnames(X)<-list(c("1","2"),c("1","2","3"))
library(reshape)
melt(X)

HTH,

Thierry 




ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
methodology and quality assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium 
tel. + 32 54/436 185
thierry.onkel...@inbo.be 
www.inbo.be 

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

The plural of anecdote is not data.
~ Roger Brinner

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

-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
Namens Dimitri Liakhovitski
Verzonden: donderdag 23 april 2009 16:31
Aan: R-Help List
Onderwerp: [R] transposing a matrix - row by row?

Hello,
I have a matrix that is a product of tapply on a larger data set.
Let's assume it looks like this:

X<-matrix(c(10,20,30,40,50,60),2,3)
dimnames(X)<-list(c("1","2"),c("1","2","3"))
(X)

   1   2   3
1 10 30 50
2 20 40 60

Is there an efficient way of transforming this matrix into the following
matrix:

rows columns entries
1  1  10
1  2  30
1  3  50
2  1  20
2  2  40
2  3  60


Thank you very much!
-- 
Dimitri Liakhovitski
MarketTools, Inc.
dimitri.liakhovit...@markettools.com

__
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Re: [R] large factorials

sympy() returns a character string, not an R numeric -- it shouldn't
automatically return an R numeric since R can't represent all
the numbers that sympy can.

The development version of rSymPy has a second class which
produces objects of class c("Sym", "character") and those
can be manipulated with +, -, *, / producing other Sym
objects so try this:


library(rSymPy)

# next line pulls in code to handle Sym objects
source("http://rsympy.googlecode.com/svn/trunk/R/Sym.R";)

# define factorial to return a Sym object
factorial.Sym <- function(n) Sym("factorial(", n, ")")

sum1 <- function(l,u,t,i,n,w) {
 v <- 0
 for (m in 0 :w) {
 v1 <- ((u^(1/2))*(l^(1/2))*t)^(i-n+2*m)
 v2 <- (factorial.Sym(i-n+m))*(factorial.Sym(m))
 v3 <- v1/v2
 v <- v+v3
 }
 sympy(v)
}

s <- sum1(1,2,10,80,3,80)
s # Sym object
as.numeric(s) # numeric


On Thu, Apr 23, 2009 at 10:00 AM, molinar  wrote:
>
> Here is what I did:
> library(rSymPy)
> factorial.sympy <- function(n) sympy(paste("factorial(", n, ")"))
> factorial.sympy(171)
> [1]
> "124101807021766782342484052410310399261660557750169318538895180361199607522169175299275197812048758557646495950167038705280988985869071076733124203221848436431047357788996854827829075454156196485215346831804429323959817369689965723590394761615227855818006117636510842880"
>>
> Which work perfectly.
>
> Here is one of my summation functions:
>
> sum1 <- function(l,u,t,i,n,w) {
> + v <- 0
> + for (m in 0 :w) {
> + v1 <- ((u^(1/2))*(l^(1/2))*t)^(i-n+2*m)
> + v2 <- (factorial.sympy(i-n+m))*(factorial.sympy(m))
> + v3 <- v1/v2
> + v <- v+v3
> + }
> + return(v)
> + }
>
> sum1(1,2,10,80,3,80)
> Error in (factorial.sympy(i - n + m)) * (factorial.sympy(m)) :
>  non-numeric argument to binary operator
>
> I'm not sure why it works when I do the factorial normally but when I call
> my function it doesn't work?
>
>
>
>
>
>
>
> molinar wrote:
>>
>> Thank you everyone all of your posts were very helpful.  I tried each one
>> and I think I have about 10 new packages installed.  The formula I need to
>> calculate did not involve any logarithms but infinite summations of
>> factorials, I'm sorry for not specifying.  I read some things about using
>> logarithms but I thought in my case I would have to do an e to the log and
>> by doing that R still gave me the same problems with numbers over 170.
>>
>> But I was able to get it to work by using the last post about the rsympy
>> packages.
>>
>> I tried downloading bc but I didn't know how to connect it to R, so R said
>> "could not find function bc".
>>
>> Thanks again for all of your help.
>> Samantha
>>
>>
>>
>>
>>
>> Gabor Grothendieck wrote:
>>>
>>> Also the R sympy package can handle this:
>>>
 library(rSymPy)
>>> Loading required package: rJava
>>>
 factorial.sympy <- function(n) sympy(paste("factorial(", n, ")"))
>>>
 # note that first time sympy is called it loads java, jython and sympy
 # but on subsequent calls its faster.  So make a dummy call first.
 factorial.sympy(10)
>>> [1] "3628800"
>>>
 # code from earlier post defining factorial.bc to be inserted here
>>>
    benchmark(replications=10, columns=c('test', 'elapsed'),
>>> +       bc=factorial.bc(500),
>>> +       sympy = factorial.sympy(500))
>>>    test elapsed
>>> 1    bc    2.17
>>> 2 sympy    0.09
>>>
>>> See the rSymPy, r-bc and rbenchmark home pages:
>>> http://rsympy.googlecode.com
>>> http://r-bc.googlecode.com
>>> http://rbenchmark.googlecode.com
>>>
>>> On Wed, Apr 22, 2009 at 3:21 PM, molinar  wrote:

 I am working on a project that requires me to do very large factorial
 evaluations.  On R the built in factorial function and the one I created
 both are not able to do factorials over 170.  The first gives an error
 and
 mine return Inf.

 Is there a way to have R do these larger calculations (the calculator in
 accessories can do 1 factorial and Maple can do even larger)
 --
 View this message in context:
 http://www.nabble.com/large-factorials-tp23175816p23175816.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.

>>>
>>> __
>>> 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.
>>>
>>>
>>
>>
>
> --
> View this message in context: 
> http://www.nabble.com/large-factorials-tp23175816p23197344.html
> Sent from the R help mailing list archive at Nabble.com.
>
> _

[R] Plots - several pages per pdf - quality/size issue

2009-04-23 Thread Sarah Bonnin


Dear R-experts,

I hope that question will not be too redundant (sorry if it is) but i 
don't seem able to find the answer i need in the archives...


I try to create a file which would have 1.several pages and 2.several 
plots by page. I know how to make a pdf file this way, but my problem is 
that the pdf size gets way too big and cannot be open.


So my question is:
- is there a way to diminish the pdf size/quality?
- does any other file format exists that would be of smaller quality but 
still be able to print plots on several pages for the same file?


Thank you for your help!

Sarah

__
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] transposing a matrix - row by row?

2009-04-23 Thread Dimitri Liakhovitski

Hello,
I have a matrix that is a product of tapply on a larger data set.
Let's assume it looks like this:

X<-matrix(c(10,20,30,40,50,60),2,3)
dimnames(X)<-list(c("1","2"),c("1","2","3"))
(X)

   1   2   3
1 10 30 50
2 20 40 60

Is there an efficient way of transforming this matrix into the following matrix:

rows columns entries
1  1  10
1  2  30
1  3  50
2  1  20
2  2  40
2  3  60


Thank you very much!
-- 
Dimitri Liakhovitski
MarketTools, Inc.
dimitri.liakhovit...@markettools.com

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

[R] Problem to get a simple Analysis of Variance table with lmer()

2009-04-23 Thread Julien Beguin

Dear R users,

Is someone know how to get a simple analysis of variance table using other
random distribution than normal (ex: Poisson or Binomial)? When I try, I
recieved this message: (See my R code below)

- R response 
Erreur dans anova(fit.poisson) :
  single argument anova for GLMMs not yet implemented

OR

Erreur dans anova(fit.binomial) :
  single argument anova for GLMMs not yet implemented
-

Here is the code that I used for a quick simulation of my data and for a
gaussian distribution I have no problem to get a simple table but not when I
use poisson or binomial distribution... Is it possible that it is effectively
not yet implemented in the lmer package? or did I make something wrong?

- My R code -
z1<-as.factor(rep(seq(1:2), each=48)) # first random factor with 2 levels
z2<-as.factor(rep(seq(1:16), each=6)) # second random factor (= repetition)
x1<-as.factor(rep(seq(1:3), 32)) # first fixed factor with three levels
x2<-as.character(rep(rep(seq(c("O","N")), each=3), 16)) # second fixed factor
  with 2 levels
y.integer<-as.integer(runif(96, 0, 25)) # response variable (integer)
y.binary<-as.integer(rbinom(96, 1,0.5)) # response variable (binary)
data.test<-as.data.frame(cbind(z1,z2,x1,x2,y))
data.test$y<-as.integer(data.test$y)

The models that I have tested:

fit.normal<-lmer(y.integer~x1*x2 + (1|z1/z2/x1), data=data.test,
family=gaussian)
anova(fit.normal)
summary(fit.normal)

fit.poisson<-lmer(y.integer~x1*x2 + (1|z1/z2/x1), data=data.test,
family=poisson(link = "log"))
anova(fit.poisson)
summary(fit.poisson)

fit.binomial<-lmer(y.binary~x1*x2 + (1|z1/z2/x1), data=data.test,
family=binomial(link = "logit"))
anova(fit.binomial)
summary(fit.binomial)
---

Thank's in advance for you amswer.

Julien

--
Julien Beguin
Etudiant au doctorat
Faculté de Foresterie et de Géomatique
Université Laval, local 2113
2405 rue de la Terrasse, G1V 0A6 Québec (Qc)
Tel: (418) 656-2131 poste 2620
http://www.cen.ulaval.ca/anticosti/jbeguin.html

__
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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] Plots - several pages per pdf - quality/size issue

2009-04-23 Thread Sarah Bonnin

Dear R-experts,

I hope that question will not be too redundant (sorry if it is) but i don't 
seem able to find the answer i need in the archives...

I try to create a file which would have 1.several pages and 2.several plots by 
page. I know how to make a pdf file this way, but my problem is that the pdf 
size gets way too big and cannot be open.

So my question is:
- is there a way to diminish the pdf size/quality?
- does any other file format exists that would be of smaller quality but still 
be able to print plots on several pages for the same file?

Thank you for your help!

Sarah

[[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] R 2.9 binaries for redhat entreprise 4 / X86_64

2009-04-23 Thread Marc Schwartz


On Apr 23, 2009, at 8:18 AM, Mathieu Van der Haegen wrote:


Hi everybody.

I'd like to ask if someone could tell me when the binaries for R 2.9 /
redhat entreprise 4 / X86_64 will be available ?
I don't need a precise date but some indication like 1 month, 6 month,
never ?

Thank you for any help.

Mathieu


I can't speak to the timeline for the CRAN RPMs, but you might also  
want to keep an eye on the EPEL repos, which provide add-on RPMs for  
RHEL.


More info here:

  http://fedoraproject.org/wiki/EPEL

and the specific link for R for RHEL 4/x84_64 is:

  http://download.fedora.redhat.com/pub/epel/4/x86_64/repoview/R.html

Not there yet either.

HTH,

Marc Schwartz

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Setting lattice par parameters

2009-04-23 Thread Steve_Friedman


Sundar,

Thank you very much.  I see my mistake.  Much appreciated.

Steve Friedman Ph. D.
Spatial Statistical Analyst
Everglades and Dry Tortugas National Park
950 N Krome Ave (3rd Floor)
Homestead, Florida 33034

steve_fried...@nps.gov
Office (305) 224 - 4282
Fax (305) 224 - 4147


   
 Sundar Dorai-Raj  
To 
   steve_fried...@nps.gov  
 04/23/2009 06:56   cc 
 AM MSTr-help Help   
   Subject 
   Re: [R] Setting lattice par 
   parameters  
   
   
   
   
   
   




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

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

However, I usually do things like this:

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

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

HTH,

--sundar

On Thu, Apr 23, 2009 at 6:25 AM,   wrote:
>
> Hello
>
> I'm plotting a large suite of barcharts and need to modify the size of
the
> text for both the yaxis and xaxis labels.
>
> I've tried using the following:
>
>> trellis.par.set(list(par.ylab.text = list(cex = 0.65)),
> trellis.par.set(list = par.xlab.text = list(cex = 0.65
>
> On inspection,  however after I invoke this line,
>
>>  trellis.par.get("par.ylab.text")
>>  trellis.par.get("par.xlab.text")
>
> It looks like the parameters have not been modified.
>
> Do I need to do something different than this approach ?
>
> Thank you very much in advance.
>
> Steve
>
> Steve Friedman Ph. D.
> Spatial Statistical Analyst
> Everglades and Dry Tortugas National Park
> 950 N Krome Ave (3rd Floor)
> Homestead, Florida 33034
>
> steve_fried...@nps.gov
> Office (305) 224 - 4282
> Fax     (305) 224 - 4147
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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

Re: [R] large factorials