date:20120917

Re: [R] eval(parse(...)) only once in a function

2012-09-17 Thread David Winsemius

On Sep 16, 2012, at 11:27 PM, Christof Kluß wrote:

> Hi
> 
> I would like to have something like
> 
> str <- "df$JT == 12"
> 
> fun <- function(df) {
> 
>  b <- eval(parse(str))
> 
>  return(b)
> }
> 

What are you trying to do?

str <- quote(df$JT == 12)

> but for performance "eval(parse(a))" should not be evaluated at each
> function call, but should work as
> 
> fun <- function(df) {
> 
>  b <- df$JT == 12
> 
>  return(b)
> }
> 
> Do you have an idea how I can implement this?

Implement what?  And what issues regarding "performance" are relevant? What 
does it mean to "not evaluate at each function call"?

str <- quote(df$JT == 12)
fun <- function(df) {
b<-eval(str)
return(b)
   }

df3 <- data.frame(JT=12)
fun(df3)
#[1] TRUE

-- 

David Winsemius, MD
Alameda, CA, USA

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Re: [R] Using paste to create and evaluate a variable expression

2012-09-17 Thread David Winsemius

On Sep 16, 2012, at 9:04 PM, Bryan Keller wrote:

> Is it possible to use "paste" to write out an expression and evaluate it?

Of course.

?as.expression
?parse
?eval

> Suppose I want to add two vectors X1 and X2, defined as follows:
> 
> X1 <- 1:6
> X2 <- 6:1
> 
> If I write the following it looks like what I want but is a character:
> noquote(paste(paste("X", 1, sep = ""), paste("X", 2, sep = ""), sep = "+"))

Instead of noquote, you might want to look at bquote.

> 
> Is there a way to tell R that I want to evaluate the text, not just print
> it out as a character?

Evaluate in what context? It looks as though you are building a formula.

parse(text= paste(paste("X", 1, sep = ""), paste("X", 2, sep = ""), sep = "+") )
expression(X1+X2)

-- 
David Winsemius, MD
Alameda, CA, USA

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Re: [R] Boxplot lattice vs standard graphics

2012-09-17 Thread Massimo Bressan


thank you for the help, bert

unfortunately, for reasons I can not understand (yet) I can not put to 
wortk it all

(I'm always in trouble with the panel functions);

max

Il 14/09/2012 18:38, Bert Gunter ha scritto:

Thanks for the example. Makes it easy to see what you mean.

Yes, if I understand you correctly, you are right:
boxplot() (base) transforms the axes, so ?boxplot.stats, which is the
function that essentially computes the boxplot, does so on the
original data.
bwplot(lattice) transforms the data first, as the documentation for
the "log" component of the scales list makes clear, and **then** calls
boxplot.stats.

Although I think the latter makes more sense then the former, I think
the way to do it is to modify the "stats" function in an explicit call
to panel.bwplot to something like (UNTESTED!)
mystats <- function(x){
out <- boxplot.stats(10^x)
out$stats <- log10(out$stats)
out$conf <- log10(out$conf) ## Omit if you don't want notches
out$out <- log10(out$out)
out ## With the boxplot statistics converted to the log10 scale
}

I leave it to you to test and modify as necessary.

Cheers,
Bert

On Fri, Sep 14, 2012 at 2:37 AM, maxbre  wrote:

Given my reproducible example

test<-structure(list(site = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L,
3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L), .Label = c("A",
"B", "C", "D", "E"), class = "factor"), conc = c(2.32, 0.902,
0.468, 5.51, 1.49, 0.532, 0.72, 0.956, 0.887, 20, 30, 2.12, 0.442,
10, 50, 110, 3.36, 2.41, 20, 70, 3610, 100, 4.79, 20, 0.0315,
30, 60, 1, 3.37, 80, 1.21, 0.302, 0.728, 1.29, 30, 40, 90, 30,
0.697, 6.25, 0.576, 0.335, 20, 10, 620, 40, 9.98, 4.76, 2.61,
3.39, 20, 4.59)), .Names = c("site", "conc"), row.names = c(NA,
52L), class = "data.frame")



And the following code

#standard graphics
with(test,boxplot(conc~site, log="y"))

#lattice
bwplot(conc~site, data=test,
scales=list(y=list(log=10))
)

There is an evident difference for site A, B, D in the way some outliers are
plotted by comparing the plot produced by lattice vs. the standard graphics

I think to understand this might be due to the different treatment of data:
i.e. log transformation (before or after the plotting?)

Is it possible to achieve the same plotting result with both graphic
facilities?
I would like to show the outliers also in lattice…

Thank you

http://r.789695.n4.nabble.com/file/n4643121/standard.png

http://r.789695.n4.nabble.com/file/n4643121/lattice.png





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Re: [R] eval(parse(...)) only once in a function

2012-09-17 Thread Heramb Gadgil

If you have a data frame "df" with a column "JT"

Try this one:
str <- "df$JT == 12"

fun<-function(str){b<-eval(parse(text=str))
return(b)}

fun(str)

On Mon, Sep 17, 2012 at 11:57 AM, Christof Kluß wrote:

> Hi
>
> I would like to have something like
>
> str <- "df$JT == 12"
>
> fun <- function(df) {
>
>   b <- eval(parse(str))
>
>   return(b)
> }
>
> but for performance "eval(parse(a))" should not be evaluated at each
> function call, but should work as
>
> fun <- function(df) {
>
>   b <- df$JT == 12
>
>   return(b)
> }
>
> Do you have an idea how I can implement this?
>
> Thx
> Christof
>
> __
> 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] Zip a file

2012-09-17 Thread Rantony

Hi Arun,

 

I tried this way, but getting an error like

 

Using Eclips:

âError in .rj.originals$help(..., help_type = "html") : 'topic' should be a 
name, length-one character vector or reserved wordâ

 

Using R:

Error: could not find function "zip"

 

So, is it required to install any package for zip functioning ?

 

 

 

From: arun kirshna [via R] [mailto:ml-node+s789695n4643049...@n4.nabble.com] 
Sent: Thursday, September 13, 2012 11:55 PM
To: Akkara, Antony (GE Energy, Non-GE)
Subject: Re: Zip a file

 

HI, 

If you look ?zip(), the usage is: 
 zip(zipfile, files, flags = "-r9X", extras = "", 
 zip = Sys.getenv("R_ZIPCMD", "zip")) 
 
A.K. 




- Original Message - 
From: Rantony <[hidden email]> 
To: [hidden email] 
Cc: 
Sent: Thursday, September 13, 2012 8:10 AM 
Subject: [R] Zip a file 

Hi, 

How to zip a csv file ? i tried with "zipfile.ZipFile" - but it showing 
function could not find. 
Is there have any other way to zip files without installing any package ? 

- Thanks in advance 
Antony 



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[R] LoadLibrary failure: : %1 is not a valid Win32 application although arch x64 has been specified

2012-09-17 Thread Franckx Laurent

Dear all,

I have problems loading a library in the 64 version of R.

The following happens:

· I have a file, equildistC.cpp, which contains a few lines of C++ code.

· I have compiled the file using the following commands:
PATH =
c:\Rtools\bin;C:\Rtools\gcc-4.6.3\bin;C:\Rtools\gcc-4.6.3\bin64;C:\Rtools\gcc-4.6.3\i686-w64-mingw32;C:\R-2.15.1\bin\i386;C:\R-2.15.1\bin\x64;c:\windows;c:\windows\system32;C:\Windows\SysWOW64
R --arch x64 CMD SHLIB equildistC.cpp

· I have received no messages indicating problem during compilation,
and both the equildistC.o and the equildistC.dll have been created.

· When I load the dll in the 32bit version of R with
dyn.load("C:\\Ccodefortransortmodel\\equildistC.dll"), things appear to go well.

· However, in the 64 bit version, I get the message:
Error in inDL(x, as.logical(local), as.logical(now), ...) :
unable to load shared object 'C:/Ccodefortransortmodel/equildistC.dll':
LoadLibrary failure: %1 is not a valid Win32 application.

I have googled the error message. Initially I had omitted --arch x64 when
compiling the file, but even after specifying the architecture, the error
persists.
However, I definitely need to be able to run my code on the 64 version.

Kind regards
Laurent

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[R] Adding legends on plots using Lattice package

2012-09-17 Thread jpm miao

To dear Dr Sarkar and anyone that knows about Lattice package,

   I make 4 graphs by Lattice package. Each of the graphs has two time
series. All the series are plotted in plain lines by default, and I would
like one series to be in plain line and the other to be in dotted line in
each graph. How can I modify the command of xyplot in the following line to
achieve this? It seems that "key" or "auto.key"  parameters are needed, but
I don't know how to do it. Thank you very much!


require(graphics)
library(lattice)
data1<-read.csv(file="G_Results_3FX1.csv", header=TRUE)
xts<-ts(data1[,2:9],frequency = 4, start = c(1983, 1))
xyplot(xts, screen=list("a","a","b","b","c","c","d","d"), layout=c(2,2),
scales=list(x="same",y="same"))



Miao

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Re: [R] LoadLibrary failure: : %1 is not a valid Win32 application although arch x64 has been specified

2012-09-17 Thread Duncan Murdoch

On 12-09-17 3:56 AM, Franckx Laurent wrote:

Dear all,

I have problems loading a library in the 64 version of R.

The following happens:

· I have a file, equildistC.cpp, which contains a few lines of C++ code.

That runs the 32 bit version of R (since you list it first in your
path). I don't see any docs that state that it will pay attention to
the --arch directive. (The docs say that syntax is supported under
Unix-alikes, not under Windows.)

· I have received no messages indicating problem during compilation,
and both the equildistC.o and the equildistC.dll have been created.

· When I load the dll in the 32bit version of R with
dyn.load("C:\\Ccodefortransortmodel\\equildistC.dll"), things appear to go well.

I have googled the error message. Initially I had omitted “--arch x64” when
compiling the file, but even after specifying the architecture, the error
persists.
However, I definitely need to be able to run my code on the 64 version.

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