Re: [R] windows vs. linux code

[Wacek Kusnierczyk]

(Ted Harding) wrote:
 There is one MAJOR issue you will have to watch out for, far more
 likely to turn up than calls like system().

 This is that, if you want to have two or more plotting windows
 in use at the same time, while the first one is autoatically
 opened by the plot() command, you will have to open additional
 ones explcitily.

 In Linux, the command is X11() [possibly with paramaters, though
 usually you don't need to bother].

 In Windows, it is windows() [ditto].
   

as far as i understand, you can just dev.new() instead, if not in all
then in most circumstances, which will do the system-dependent stuff for
you, making your code less system-dependent.

vQ

 I run R on Linux, so use the X11() command. However, If I write
 a script which would also be run on a Windows system, I write
 using windows() in the first instance, but with a conditional
 alias to X11():

  if(length(grep(linux,R.Version()$os))){
   windows - function( ... ) X11( ... )
 }

 and put this at the beginning of the code file. Then, if the code
 is run on a Windows machine, the function call windows() does the
 Windows thing; but if the code is run on Linux then the above test
 detects that, and defines a function windows() which does the same
 as X11().

 Ted.

 On 26-Feb-09 01:25:36, Sherri Heck wrote:
   
 i am asking if, in general, r code can be written on a linux-based 
 system and then run on a windows-based system. 

 Rolf Turner wrote:
 
 On 26/02/2009, at 2:08 PM, Sherri Heck wrote:

   
 Dear All-

 I have been given some Rcode that was written using a Linux OS, but I
 use Windows-based R.  The person that is giving it to me said that it
 needs to run on a Linux system.  Does anyone have any insight and/or
 can
 verify this.  I haven't yet obtained the code, so I haven't been able
 to
 try it yet.
 
 Despite the knowledge, wisdom, insight, skill, good looks, and other
 admirable characteristics of the members of the R-help list, few of
 us are skilled in telepathy or clairvoyance.

 cheers,

 Rolf Turner

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 E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
 Fax-to-email: +44 (0)870 094 0861
 Date: 26-Feb-09   Time: 03:58:35
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Email: w...@idi.ntnu.no
Phone: +47 73591875, +47 72574609

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Faculty of Information Technology, Mathematics and Electrical Engineering (IME)
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[R] AsciiGridPredict returns error in unionDataJoin

[Frauke Barthold]

Dear list, 

I am using AsciiGridPredict from the yaImpute package to make a spatial 
prediction of a randomForest model. When I call AsciiGirdPredict I am getting 
an error message, that I do not understand:

AsciiGridPredict(st.rf.fit, xlist, ylist, xtypes=NULL, rows=NULL, 
myPredFunc=NULL)

Rows per dot:  8  Rows to do: 884 
ToDo: 
..
Done: Error in unionDataJoin(m1, m2) : 
  row names are requried within all input matrices
In addition: Warning message:
In AsciiGridImpute(object, xfiles, outfiles, xtypes = xtypes, lon = lon,  :
  NA's generated due to illegal level(s).

Does someone know what the error means and how I can solve this problem?

Thank you very much in advance,
Frauke



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[R] Incorporating cumsum in for loop

[joe1985]

Hello

Hello i have this DF:

 c
 CHR_NR 

diffdato1 x
1022   10395 1994-05-3011.0   39   0 1994 1 1 
0  024  11  24
29100  11377 2003-05-2217.0   24   0 2003 1 1 
0  015  11  29
29409  11377 2003-06-0618.8   15   0 2003 1 0 
0  014  01  29
28964  11377 2003-07-0420.1   14   0 2003 1 1 
0  017  11  59
29422  11377 2003-07-2132.9   17   0 2003 1 0 
0  014  01  59
28605  11377 2003-08-0419.9   14   0 2003 1 0 
0  014  01  59
28996  11377 2003-08-1810.5   14   0 2003 1 0 
0  014  01  59
29932  11377 2003-12-0820.5   27   0 2003 1 1 
0  029  11  78
30393  11377 2004-01-0628.6   29   0 2004 1 0 
0  0 8  01  78
36100  11377 2004-01-1428.68   0 2004 1 0 
0  0 8  01  78
30847  11377 2004-01-2219.08   0 2004 1 0 
0  0 7  01  78
34549  11377 2004-01-2919.07   0 2004 1 0 
0  0 7  01  78
30035  11377 2004-02-0514.47   0 2004 1 0 
0  0 7  01  78
34550  11377 2004-02-1214.47   0 2004 1 0 
0  012  01  78
42629  11493 2007-05-3111.8   31   0 2007 1 1 
0  025  11  25
20900  12558 2000-06-3049.8   38   0 2000 1 1 
0  027  11 118
21618  12558 2000-07-2751.0   27   0 2000 1 0 
0  014  01 118
22014  12558 2000-08-1036.6   14   0 2000 1 0 
0  014  01 118
21405  12558 2000-08-2418.6   14   0 2000 1 0 
0  014  01 118
21790  12558 2000-09-0721.9   14   0 2000 1 0 
0  014  01 118
22185  12558 2000-09-2114.4   14   0 2000 1 0 
0  035  01 118
16695  13018 1999-02-0919.0   33   0 1999 1 1 
0  014  11  14
22315  13018 2000-08-1821.8   36   0 2000 1 1 
0  028  11  81
22029  13018 2000-09-1526.2   28   0 2000 1 0 
0  014  01  81
21280  13018 2000-09-2920.2   14   0 2000 1 0 
0  014  01  81
21738  13018 2000-10-1310.1   14   0 2000 1 0 
0  025  01  81
24610  13018 2001-09-1731.3   27   0 2001 1 1 
0  017  11  32
23958  13018 2001-10-0415.2   17   0 2001 1 0 
0  015  01  32
43479  13168 2007-10-0315.6   33   0 2007 1 1 
0  035  11  35
44755  13168 2008-04-0418.3   23   0 2008 1 1 
0  025  11  25
45355  13168 2008-07-0415.2   36   0 2008 1 1 
0  032  11  66
45540  13168 2008-08-0510.3   32   0 2008 1 0 
0  034  01  66

And I want like this for every x-value, illustrated with x=118:

 a
CHR_NR   diffdato diffCHR  aar ind10 diffind10 dato63 diffdato63
 
diffdato1  x  akkusum
12558 2000-06-3049.8   38   0 2000 1 1  0   
  
027  11 118  27 12558 2000-07-2751.0   27  
0 2000 1 0  0  014  01 118  41
12558 2000-08-1036.6   14   0 2000 1 0  0   
  
014  01 118  55
12558 2000-08-2418.6   14   0 2000 1 0  0   
  
014  01 118  69
12558 2000-09-0721.9   14   0 2000 1 0  0   
  
014  01 118  83
12558 2000-09-2114.4   14   0 2000 1 0  0   
  
035  01 118 118

So I want the vector akkusum just for the whole dataset, i thought of
using af for loop with cumsum (which i have used in the example with x=118

Re: [R] Problems with ARIMA models?

[Wacek Kusnierczyk]

that's interesting, marie.

on his webpage [1], david s. stoffer provides concrete arguments for why
the r arima function in the stats package should be considered to
produce misleading, if not just wrong results. 

david (cc:ed for reference) reported the issue in october 2005, bug
report #8231 [2], and then again in april 2006, same bug report id [3]. 
a search through the maling list's archives reveals that there are only
two messages matching the pattern '8231', namely the two mentioned
above.  that is, there has been *no public response* posted.

after a few lessons, i have learned that you can't demand anything from
r developers (which isn't quite accurate:  you can demand, but they're
likely to ignore you), so it's hard to demand that this issue be
appropriately addressed.  however, one can, and should, expect that such
bug reports, correct or not, are addressed in public, leaving no doubt
as to the accuracy of the design and implementation of any functionality
in r.

responding to bug reports is an important part of the open source
culture (in this particular case, it's not really about sources) which r
folks do not hesitate to refer to.  it is surprising that david's
concrete and polite (please prove me wrong) post hasn't received the
treatment it deserved.  it hasn't any! 

a brief look at some of the respective sources:

svn log http://svn.r-project.org/R/trunk/src/library/stats/R/arima.R
svn log http://svn.r-project.org/R/trunk/src/library/stats/src/arima.c

hints that they've been edited mostly by one single developer.  if the
responsibility, as it seems, is not diffused among many, why wouldn't
the author provide a clear and convincing answer to david's comments? 
as far as i can see, david has sufficient expertise in both the
particular field and r in general, not to be treated as a naive user
asking dumb questions (see fortune('dumb'), for example).

is ignoring concrete critique part of the official strategy?  can't see
this documented on r's website.

vQ


[1] http://www.stat.pitt.edu/stoffer/tsa2/Rissues.htm
[2] http://tolstoy.newcastle.edu.au/R/devel/05/10/2885.html
[3] http://tolstoy.newcastle.edu.au/R/devel/06/04/4756.html



Marie Sivertsen wrote:
 Dear R,

 I have find a website where they report problem with ARIMA models in R.  I
 run the examples there and they give result as shown on the website.  Does
 this mean that nothing has corrected in R?  Maybe you not have seen the
 page, but the author said he contacted you.

 Here is the URL: http://www.stat.pitt.edu/stoffer/tsa2/Rissues.htm

 I like to know your opinion.

 Mvh.
 Marie

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Re: [R] Hi, Coding problem

[Uwe Ligges]

Ssophia wrote:


  Hi, there 

 

  Below is my code to one Homework question. I couldn't come up with the reasonable answer. 



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

could you please help me to figure out what is the problem with my code? 
thank you 



Question is Coding P{X=j} =(1/2)^(j+1) + (1/2) *2^(j-1)/3^j
my code is 
sim - function(n.gen){

urandom - runif(n.gen)
sim.vector - rep(0,n.gen)
for(j in 1:n.gen){
 i - 1
 p - 5/12
 F - p
 while(urandom[j] = F){
  p - p*((1/2)^(i+1)+1/3*(2/3)^i)/((1/2)^i+(1/2)*(2/3)^i)
  F - F+p
  i-i+1
  }
 sim.vector[j] - i
}
# output
sim.vector
}

 

result is 

   12345678   11 
0.37 0.22 0.16 0.13 0.05 0.02 0.03 0.01 0.01 

 

always, there are some numbers missing, it should be continuous. 

why 9 and 10 are missing 

thank you 



sophia

  


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Re: [R] Strange behavior of savePlot

[Uwe Ligges]

Christophe Genolini wrote:

Hi all,

I am using savePlot in a loop for saving several graph but I get some 
graph in 553x552, some other in 1920x1119. How comes ?
My data are almost all the same (same label, same xlim / ylim, almost 
same data. Only the color changes).  I save them in bmp.

Thanks for your help.



Rather than using a Windows Device and copying the contents, just use 
the bmp() device directly if you really want bitmaps.


Uwe Ligges



Christophe

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Re: [R] Place independent labels between values on x-axis

[Uwe Ligges]

T.D.Rudolph wrote:

I conducted a frequency averaging procedure which left me with the data frame
below (Bin is an artifact of a cut() procedure and can be either
as.character or as.factor):

   Bin Freq
1  (-180,-160] 7.904032
2  (-160,-140] 5.547901
3  (-140,-120] 4.522542
4  (-120,-100] 4.784184
5   (-100,-80] 4.490083
6(-80,-60] 4.754268
7(-60,-40] 5.597407
8(-40,-20] 5.964031
9  (-20,0] 7.266519
10  (0,20] 6.947202
11 (20,40] 6.168730
12 (40,60] 4.918232
13 (60,80] 4.638589
14(80,100] 4.288087
15   (100,120] 4.091052
16   (120,140] 4.451199
17   (140,160] 5.869740
18   (160,180] 7.796204

I am able to present the data as a histogram using barplot(), but on the
x-axis I would like to see the values that separate (i.e. that are located
between) my various bins (i.e. seq(-180,180,20)).  Thus 0 would appear
directly under the centre line that separates (-20,0] and (0,20], etc.

barplot(angledist$Freq, xlab=Turning Angle, ylab=Average frequency (%),
ylim=c(0,10))

?names.arg, ?xlim don't seem to do it but I could be wrong
Tyler




Two comments:

1. In fact you probably want a histpgram on your un-cut()-ted data and 
specify breaks. A barplot seems to be misleading for originally 
continuous data.


2. You can suppress x-axis labels by specifying the argument xaxt=n 
and add your own x-axis later on by a call to axis() which allows to 
specify the tick mark positions.


Uwe Ligges

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Re: [R] survival::survfit,plot.survfit

[Bernhard Reinhardt]

Jeff Xu wrote:

I am confused when trying the function survfit.
my question is:  what does the survival curve given by plot.survfit mean?
is it the survival curve with different covariates at different points?
or just the baseline survival curve?

for example, I run the following code and get the survival curve


library(survival)
fit-coxph(Surv(futime,fustat)~resid.ds+rx+ecog.ps,data=ovarian)
plot(survfit(fit,type=breslow))
summary(survfit(fit,type=breslow))


for the first two failure points, we have s(59|x1)=0.971, s(115|x2)=0.942
how can we guarantee that s(59|x1) is always greater than s(115|x2)?
since s(59|x1)=s_0(59)^exp(\beta'x1) and s(115|x2)=s_0(115)^exp(\beta'x2),
we can manipulate covariates to make s(59|x1)  s(115|x2), right?
do I miss anything?


In advance: I´m a beginner in survival analysis, too. But I think I can 
help you with this.


plot(survfit(fit)) should plot the survival-function for x=0 or 
equivalently beta'=0. This curve is independent of any covariates.


If you want to see the impact of residual-status=2 you could add 
something like:


attach(ovarian)
ovarian_new - data.frame(resid.ds=2, 
rx=(mean(rx)),ecog.ps=mean(ecog.ps))

detach()

plot(survfit(fit), newdata=ovarian_new)

This should give you the survival-function for an average patient with 
residual-status 2.


Regards

Bernhard

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Re: [R] Strange behavior of savePlot

[Christophe Genolini]

Uwe Ligges a écrit :



Christophe Genolini wrote:

Hi all,

I am using savePlot in a loop for saving several graph but I get some 
graph in 553x552, some other in 1920x1119. How comes ?
My data are almost all the same (same label, same xlim / ylim, almost 
same data. Only the color changes).  I save them in bmp.

Thanks for your help.



Rather than using a Windows Device and copying the contents, just use 
the bmp() device directly if you really want bitmaps.


Uwe Ligges
I do not want a bitmaps. I use savePlot in a library, so I let the user 
decide whith export he wants.

Christophe

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Re: [R] Importing zoo object (index contains NAs)

[Achim Zeileis]

On Wed, 25 Feb 2009, Rob Denniker wrote:


Dear list,


I have an irregular time series saved and exported as a zoo object. What 
is the trick to force zoo to ignore the missing dates when reading it 
back in? Thanks.


Almost certainly there are no NAs in your index (although we can't say for 
sure as Gabor pointed out).


 write.zoo(g, file = gdata.txt, index.name = date, append = F, 
quote = T, sep = ,)



 h - read.zoo(gdata.txt, sep = ,, format = Y-%m-%d)

   ^^^
First problem: This should be %Y.
Second problem: You need header = TRUE.

Thus, I would guess that
  h - read.zoo(gdata.txt, sep = ,, format = %Y-%m-%d,
header = TRUE)
should do what you want.
Z

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Re: [R] Using package ROCR

[wiener30]

Thank you very much for the response!

The plot(1,1) helped to resolve the first problem.
But I am still getting a second error message when running demo(ROCR)

Error in as.double(y) : 
  cannot coerce type 'S4' to vector of type 'double'

It seems it has something to do with compatibility of S4 objects.

My versions of R and ROCR package are the same as you listed.
But it seems something other is missing in my installation.


William Doane wrote:
 
 
 Responding to question 1... it seems the demo assumes you already have a
 plot window open.
 
   library(ROCR)
   plot(1,1)
   demo(ROCR)
 
 seems to work.
 
 For question 2, my environment produces the expected results... plot
 doesn't generate an error:
   * R 2.8.1 GUI 1.27 Tiger build 32-bit (5301)
   * OS X 10.5.6
   * ROCR 1.0-2
 
 -Wil
 
 
 
 wiener30 wrote:
 
 I am trying to use package ROCR to analyze classification accuracy,
 unfortunately there are some problems right at the beginning.
 
 Question 1) 
 When I try to run demo I am getting the following error message
 library(ROCR)
 demo(ROCR)
 if(dev.cur() = 1)  [TRUNCATED] 
 Error in get(getOption(device)) : wrong first argument
 When I issue the command
 dev.cur() 
 it returns
 null device 
   1
 It seems something is wrong with my R-environment ?
 Could somebody provide a hint, what is wrong.
 
 Question 2)
 When I run an example commands from the manual
 library(ROCR)
 data(ROCR.simple)
 pred - prediction( ROCR.simple$predictions, ROCR.simple$labels )
 perf - performance( pred, tpr, fpr )
 plot( perf )
 
 the plot command issues the following error message
 Error in as.double(y) : 
   cannot coerce type 'S4' to vector of type 'double'
 
 How this could be fixed ?
 
 Thanks for the support
 
 
 
 

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[R] (no subject)

[Tanja Srebotnjak]

Hello,

I'm trying to fit a generalized linear mixed model to estimate diabetes 
prevalence at US county level. To do this I'm using the glmer() function in 
package lme4. I can fit relatively simple models (i.e. few covariates) but when 
expanding the number of covariates I usually encounter the following error 
message.

gm8 - 
glmer(DIAB05F~AGE+as.factor(SEX)+poolt+poolx+poverty+fastfood+(1|as.factor(diab$fips)),
 family = binomial(link=logit), data = diab, doFit=TRUE)
Error in validObject(.Object) :
  invalid class mer object: Slot Zt must by dims['q']  by dims['n']*dims['s']

In the above, the response is person-level diabetes status as a function of 
AGE=age, SEX=sex, poolt=average county diabetes prevalence for previous years, 
poolx=pooled county diabetes prevalence for counties with similar age, sex, 
race, and income structure, poverty=county poverty rate, fastfood=number of 
fastfood places per 100,000 people in the county, and a county random effect.

If I leave out fastfood, the model gets at least fitted - although it doesn't 
converge (yet):

Warning message:
In mer_finalize(ans) : false convergence (8)

I would be grateful for any advice on what the problem could be and how to 
resolve it.

Thanks,
Tanja


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Re: [R] Using package ROCR

[William Doane]

Responding to question 1... it seems the demo assumes you already have a
plot window open.

  library(ROCR)
  plot(1,1)
  demo(ROCR)

seems to work.

For question 2, my environment produces the expected results... plot doesn't
generate an error:
  * R 2.8.1 GUI 1.27 Tiger build 32-bit (5301)
  * OS X 10.5.6
  * ROCR 1.0-2

-Wil



wiener30 wrote:
 
 I am trying to use package ROCR to analyze classification accuracy,
 unfortunately there are some problems right at the beginning.
 
 Question 1) 
 When I try to run demo I am getting the following error message
 library(ROCR)
 demo(ROCR)
 if(dev.cur() = 1)  [TRUNCATED] 
 Error in get(getOption(device)) : wrong first argument
 When I issue the command
 dev.cur() 
 it returns
 null device 
   1
 It seems something is wrong with my R-environment ?
 Could somebody provide a hint, what is wrong.
 
 Question 2)
 When I run an example commands from the manual
 library(ROCR)
 data(ROCR.simple)
 pred - prediction( ROCR.simple$predictions, ROCR.simple$labels )
 perf - performance( pred, tpr, fpr )
 plot( perf )
 
 the plot command issues the following error message
 Error in as.double(y) : 
   cannot coerce type 'S4' to vector of type 'double'
 
 How this could be fixed ?
 
 Thanks for the support
 
 

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Re: [R] Strange behavior of savePlot

[Uwe Ligges]

Christophe Genolini wrote:

Uwe Ligges a écrit :



Christophe Genolini wrote:

Hi all,

I am using savePlot in a loop for saving several graph but I get some 
graph in 553x552, some other in 1920x1119. How comes ?
My data are almost all the same (same label, same xlim / ylim, almost 
same data. Only the color changes).  I save them in bmp.

Thanks for your help.



Rather than using a Windows Device and copying the contents, just use 
the bmp() device directly if you really want bitmaps.


Uwe Ligges
I do not want a bitmaps. I use savePlot in a library, so I let the user 
decide whith export he wants.



Then, why not let the user choose a proper device?
Anyway, the resulting bitmap is copied from the screen device, that 
means the size is also copied.
That means if you have some 1600x1200 screen and your windows device is 
resized to fullscreen, than you will almost get 1600x1200 pixels in your 
bitmap...


Uwe Ligges



Christophe

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[R] R communication

[ARDIA David]

Dear all,
Imagine that you have a small LAN with two computers. I would like to run R on 
both, and possible to run computations from one computer to the other one. TCP 
IP protocole would be preferable. Which package would you use for that? I would 
be very glad if you could also provide me with some lines of code, e.g., create 
a matrix X in computer 1, transfer its value to the second computer, make some 
calculation, and get the value back.
Thanks for your help,
David

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Re: [R] Strange behavior of savePlot

[Christophe Genolini]

Uwe Ligges a écrit :



Christophe Genolini wrote:

Uwe Ligges a écrit :



Christophe Genolini wrote:

Hi all,

I am using savePlot in a loop for saving several graph but I get 
some graph in 553x552, some other in 1920x1119. How comes ?
My data are almost all the same (same label, same xlim / ylim, 
almost same data. Only the color changes).  I save them in bmp.

Thanks for your help.



Rather than using a Windows Device and copying the contents, just 
use the bmp() device directly if you really want bitmaps.


Uwe Ligges
I do not want a bitmaps. I use savePlot in a library, so I let the 
user decide whith export he wants.



Then, why not let the user choose a proper device?
Because I program a graphical interface that let the user to chose the 
graph he want to export (that can be one but also 50). Then he sets 
dynamicaly the parameters for all the graphics he wants to export. Then 
all the graphs are exported at once.
Anyway, the resulting bitmap is copied from the screen device, that 
means the size is also copied.
That means if you have some 1600x1200 screen and your windows device 
is resized to fullscreen, than you will almost get 1600x1200 pixels in 
your bitmap...
Ok, the strange behevior was occuring because I change the size of the 
windows during the export...

Thanks a lot

Christophe


Uwe Ligges



Christophe

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Re: [R] Modifying Names from (x,y] into x

[baptiste auguie]

Hi,


I got this problem once, and Prof. Ripley kindly added an example in  
the help page of ?cut,


 aaa - c(1,2,3,4,5,2,3,4,5,6,7)

 ## one way to extract the breakpoints
 labs - levels(cut(aaa, 3))
 cbind(lower = as.numeric( sub(\\((.+),.*, \\1, labs) ),
   upper = as.numeric( sub([^,]*,([^]]*)\\], \\1, labs) ))
  lower upper
 [1,] 0.994  3.00
 [2,] 3.000  5.00
 [3,] 5.000  7.01

Hope this helps,

baptiste

On 26 Feb 2009, at 02:55, Gundala Viswanath wrote:

 Hi,

 I have the following data that looks like this:

 names(dat)
 [1] (-2329,-2319] (-1399,-1389] (-669.4,-659.4]

 How can I modify those names into just this?

 [1] -2329   -1399 -669.4




 - Gundala Viswanath
 Jakarta - Indonesia

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_

Baptiste Auguié

School of Physics
University of Exeter
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Exeter, Devon,
EX4 4QL, UK

Phone: +44 1392 264187

http://newton.ex.ac.uk/research/emag
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Re: [R] T-test by groups

[Uwe Ligges]

Ingrid Tohver wrote:
I would like to run a t-test within a by group function. My  
dataset, error,  is organized as the following (I have 133 Sites):


Site week Dataset Region  lat_map  long_map mean_tsim diff20 diff40  
diff80
ALFI   15   USACE UC 48.15625 -117.0938  8.87   1.34   1.90
2.98
ALFI   16   USACE UC 48.15625 -117.0938 10.28   0.57   1.08
2.27
ALFI   17   USACE UC 48.15625 -117.0938 11.08   0.74   1.30
2.52
ALFI   18   USACE UC 48.15625 -117.0938 12.23   0.42   1.11
2.42
ALFI   19   USACE UC 48.15625 -117.0938 13.19   1.00   1.73
3.14
ALFI   20   USACE UC 48.15625 -117.0938 14.31   1.77   2.62
3.78


I am interested in running the t-test by the Site index. My code  
looks like this:


t_test-by(error, error['Site'], function(dat) t.test(subset(error 
$diff20),subset(error$diff80), data=dat))


This code runs the t-test, but over the whole dataset without  
discriminating by Site, so each Site's result is the same. Could  
someone help determine a better approach or why mine is not working.


I guess you want

by(error, error['Site'], function(dat) t.test(dat$diff20, dat$diff40))

Uwe Ligges



Thank you,
Ingrid


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Re: [R] (no subject)

[ONKELINX, Thierry]

Dear Tanja,

R-Sig-Mixed-models is a better list for questions about lme4 and nlme.
There you are much more likely to get an answer from the mixed models
specialists.

First of all I would recommend you to write the random effect as
(1|fips) instead of (1|as.factor(diab$fips)). You will run into troubles
when you change the dataset as only the random effect explicitly refers
to the dataset.

I can think of two things that may cause the errors: a lack of data
points or an overspecified model. If you have a lot of data points then
you should have a look at the correlations between the covariates.
Highly correlated covariates can lead to unstable models with false
convergences as a result.

HTH,

Thierry

PS An informative subject line is recommended by the posting guide.



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 Tanja Srebotnjak
Verzonden: donderdag 26 februari 2009 9:17
Aan: r-help@r-project.org
Onderwerp: [R] (no subject)

Hello,

I'm trying to fit a generalized linear mixed model to estimate diabetes
prevalence at US county level. To do this I'm using the glmer() function
in package lme4. I can fit relatively simple models (i.e. few
covariates) but when expanding the number of covariates I usually
encounter the following error message.

gm8 -
glmer(DIAB05F~AGE+as.factor(SEX)+poolt+poolx+poverty+fastfood+(1|as.fact
or(diab$fips)), family = binomial(link=logit), data = diab,
doFit=TRUE)
Error in validObject(.Object) :
  invalid class mer object: Slot Zt must by dims['q']  by
dims['n']*dims['s']

In the above, the response is person-level diabetes status as a function
of AGE=age, SEX=sex, poolt=average county diabetes prevalence for
previous years, poolx=pooled county diabetes prevalence for counties
with similar age, sex, race, and income structure, poverty=county
poverty rate, fastfood=number of fastfood places per 100,000 people in
the county, and a county random effect.

If I leave out fastfood, the model gets at least fitted - although it
doesn't converge (yet):

Warning message:
In mer_finalize(ans) : false convergence (8)

I would be grateful for any advice on what the problem could be and how
to resolve it.

Thanks,
Tanja


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en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is
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Re: [R] Incorporating cumsum in for loop

[Uwe Ligges]

joe1985 wrote:

Hello

Hello i have this DF:


c
 CHR_NR 
diffdato1 x
1022   10395 1994-05-3011.0   39   0 1994 1 1 
0  024  11  24
29100  11377 2003-05-2217.0   24   0 2003 1 1 
0  015  11  29
29409  11377 2003-06-0618.8   15   0 2003 1 0 
0  014  01  29
28964  11377 2003-07-0420.1   14   0 2003 1 1 
0  017  11  59
29422  11377 2003-07-2132.9   17   0 2003 1 0 
0  014  01  59
28605  11377 2003-08-0419.9   14   0 2003 1 0 
0  014  01  59
28996  11377 2003-08-1810.5   14   0 2003 1 0 
0  014  01  59
29932  11377 2003-12-0820.5   27   0 2003 1 1 
0  029  11  78
30393  11377 2004-01-0628.6   29   0 2004 1 0 
0  0 8  01  78
36100  11377 2004-01-1428.68   0 2004 1 0 
0  0 8  01  78
30847  11377 2004-01-2219.08   0 2004 1 0 
0  0 7  01  78
34549  11377 2004-01-2919.07   0 2004 1 0 
0  0 7  01  78
30035  11377 2004-02-0514.47   0 2004 1 0 
0  0 7  01  78
34550  11377 2004-02-1214.47   0 2004 1 0 
0  012  01  78
42629  11493 2007-05-3111.8   31   0 2007 1 1 
0  025  11  25
20900  12558 2000-06-3049.8   38   0 2000 1 1 
0  027  11 118
21618  12558 2000-07-2751.0   27   0 2000 1 0 
0  014  01 118
22014  12558 2000-08-1036.6   14   0 2000 1 0 
0  014  01 118
21405  12558 2000-08-2418.6   14   0 2000 1 0 
0  014  01 118
21790  12558 2000-09-0721.9   14   0 2000 1 0 
0  014  01 118
22185  12558 2000-09-2114.4   14   0 2000 1 0 
0  035  01 118
16695  13018 1999-02-0919.0   33   0 1999 1 1 
0  014  11  14
22315  13018 2000-08-1821.8   36   0 2000 1 1 
0  028  11  81
22029  13018 2000-09-1526.2   28   0 2000 1 0 
0  014  01  81
21280  13018 2000-09-2920.2   14   0 2000 1 0 
0  014  01  81
21738  13018 2000-10-1310.1   14   0 2000 1 0 
0  025  01  81
24610  13018 2001-09-1731.3   27   0 2001 1 1 
0  017  11  32
23958  13018 2001-10-0415.2   17   0 2001 1 0 
0  015  01  32
43479  13168 2007-10-0315.6   33   0 2007 1 1 
0  035  11  35
44755  13168 2008-04-0418.3   23   0 2008 1 1 
0  025  11  25
45355  13168 2008-07-0415.2   36   0 2008 1 1 
0  032  11  66
45540  13168 2008-08-0510.3   32   0 2008 1 0 
0  034  01  66


And I want like this for every x-value, illustrated with x=118:


a
CHR_NR   diffdato diffCHR  aar ind10 diffind10 dato63 diffdato63 
diffdato1  x  akkusum
12558 2000-06-3049.8   38   0 2000 1 1  0 
027  11 118  27 12558 2000-07-2751.0   27  
0 2000 1 0  0  014  01 118  41
12558 2000-08-1036.6   14   0 2000 1 0  0 
014  01 118  55
12558 2000-08-2418.6   14   0 2000 1 0  0 
014  01 118  69
12558 2000-09-0721.9   14   0 2000 1 0  0 
014  01 118  83
12558 2000-09-2114.4   14   0 2000 1 0  0 
035  01 118 118


So I want the vector akkusum just for the whole dataset, i thought of
using af for loop with cumsum (which i have used in the example 

Re: [R] T-test by groups

[Peter Dalgaard]

Uwe Ligges wrote:
 
 
 Ingrid Tohver wrote:
 I would like to run a t-test within a by group function. My 
 dataset, error,  is organized as the following (I have 133 Sites):

 Site week Dataset Region  lat_map  long_map mean_tsim diff20 diff40 
 diff80
 ALFI   15   USACE UC 48.15625 -117.0938  8.87   1.34   1.90   
 2.98
 ALFI   16   USACE UC 48.15625 -117.0938 10.28   0.57   1.08   
 2.27
 ALFI   17   USACE UC 48.15625 -117.0938 11.08   0.74   1.30   
 2.52
 ALFI   18   USACE UC 48.15625 -117.0938 12.23   0.42   1.11   
 2.42
 ALFI   19   USACE UC 48.15625 -117.0938 13.19   1.00   1.73   
 3.14
 ALFI   20   USACE UC 48.15625 -117.0938 14.31   1.77   2.62   
 3.78

 I am interested in running the t-test by the Site index. My code 
 looks like this:

 t_test-by(error, error['Site'], function(dat) t.test(subset(error
 $diff20),subset(error$diff80), data=dat))

 This code runs the t-test, but over the whole dataset without 
 discriminating by Site, so each Site's result is the same. Could 
 someone help determine a better approach or why mine is not working.
 
 I guess you want
 
 by(error, error['Site'], function(dat) t.test(dat$diff20, dat$diff40))


...and most likely also, paired=TRUE. Notice that the data= argument
only works for the formula interface, which is not (yet) defined for
paired data, but only for the y~group type two-sample case.


 Uwe Ligges
 
 
 Thank you,
 Ingrid


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-- 
   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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Re: [R] biplot.princomp - changing score labels

[Axel Strauß]

Thanks for the hints! They seem to help me.

Axel

Héctor Villalobos schrieb:

Perhaps this may help you.

Regards




data(iris3)

ir - rbind(iris3[ , , 1], iris3[ , , 2], iris3[ , , 3])

  ir.pca - princomp(ir)

  biplot(ir.pca)

# Rehacer Biplot
# Calcular Factor para re-escalar scores y eigenvectores
  lambda - ir.pca$sdev[1:2] * sqrt(ir.pca$n.obs)

  scores - t( t(ir.pca$scores[ , c(1, 2)]) / lambda 
)
  variables - t( t(ir.pca$loadings[ , c(1, 2)]) * lambda )
 
 
  x11()

  plot(scores, type=p, xlim=c(-0.22, 0.24), ylim=c(-0.2, 0.2),
  pch=c(rep(16, 50), rep(21, 50), rep(14, 50)),
  col=c(rep(red, 50), rep(blue, 50), rep(green, 50)))
abline(v=0, h=0, lty=3)
  par(new=TRUE)
  plot(variables, type=n, xaxt=n, yaxt=n, xlim=c(-22, 24), 
ylim=c(-20, 20))

arrows(0, 0, variables[ , 1], variables[ , 2], len=0.1, col=red)
text(2*variables, rownames(variables), col=red, xpd=TRUE)
axis(3); axis(4)



On 25 Feb 2009 at 9:52, Axel Strauß wrote:

Date sent:  Wed, 25 Feb 2009 09:52:54 +0100
From: Axel Strauß a.stra...@tu-bs.de
To:R-help@r-project.org
Subject: Re: [R] biplot.princomp - changing score 
labels


 Prof Brian Ripley schrieb:
  On Tue, 24 Feb 2009, Axel Strauß wrote:
 
  OK, the one thing I figured out:
  Is should be like:
  biplot(test.pca, cex=c(2,1), col=c(red,green)...
  to change size, colours etc separately. But I still don't know how
  change lables of observations to symbols properly.
 
  That's not part of the design of the function, so just make a copy
  and edit to meet your fancies.
 
  The designer of biplot.princomp.
 
 The idea behind my question was actually not a styling one but to
 provide additional information in the graph. My observations have
 different species richness and I wanted to adopt symbol size to the
 number of species to show the change of species richness along the
 PCs. Anyway, thanks for the comment - and for designing
 biplot.princomp.

 Axel



--
Héctor Villalobos hvill...@ipn.mx
 CICIMAR - IPN
 A.P. 592. Col. Centro
 La Paz, Baja California Sur, MÉXICO. 23000
 Tels. (+52 612) 122 53 44; 123 46 58; 123 47 34  ext. 82425
 Fax.  (+52 612) 122 53 22



--
Gravity is a habit that is hard to shake off.
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[R] survival::predict.coxph

[Bernhard Reinhardt]

Hi,

I just came across another question concerning predict.coxph

Terry Therneau states in A Package for Survival Analysis in S that

term - predict(fit, type=terms)

yields predicted values for the individual components of the linear 
predictor X*beta


My coxmodel looks like:

   coef exp(coef) se(coef)  z  p
S0   -3.106  4.48e-02 2.88 -1.080 0.2800
S16.365  5.81e+02 5.20  1.224 0.2200
S2  -14.009  8.24e-07 5.32 -2.636 0.0084
[..]


The first line of my input-data looks like: 
S0   S1  S2  S3
1   -1.030 -0.9500 -1.0950 -1.0700

So I thought the first line of term should be calculated by

-1.030*-3.106,  -0.9500*6.365,  -1.0950*-14.009 [..] which is
3.20,   -6.04,  15.34

Actually the first line of term contains:
  S0  S1   S2
1   3.36737346 -6.36032595  15.73846097

which is quite similar but not the same.

Can anyone shed some light on this? I guess there must be tons of 
literature on this topic but I find it quite hard to find the good one.


I´d also appreciate literature on how to choose the appropriate number 
of covariates for a coxmodel and overfitting.


Regards

Bernhard

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Re: [R] Moving Average

[Gabor Grothendieck]

See rollapply in zoo or filter or embed in the core of R.

On Thu, Feb 26, 2009 at 7:07 AM,  mau...@alice.it wrote:
 I am looking for some help at removing low-frequency components from a 
 signal, through Moving Average on a sliding window.
 I understand thiis is a smoothing procedure that I never done in my life 
 before .. sigh.

 I searched R archives and found rollmean, MovingAverages {TTR}, 
 SymmetricMA.
 None of the above mantioned functions seems to accept the smoothing 
 polynomial order and the sliding window with as input parameters. Maybe I am 
 missing something.

 I wonder whether there is some building blocks in R if not even a function 
 which does it all (I do not expect that much,though).
 Even some literature references and/or tutorials are very welcome.

 Thank you so much,
 Maura



 tutti i telefonini TIM!


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[R] ggplot2: labels points with different colors or idnumbers

[Tom Cohen]

Dear list,
 
Using ggplot2 I could produce both boxplot and points in the same plot but 
instead of points I would like to label the different
subjects with different colors or their idnumbers. Is there away to do it? Also 
how can I put three plots on the same graph with ggplot2? 
mfrow=c(3,1) did not do the job. 
 
 dat
   group time   id  freq
1  1   00 0018  5.21
2  1   00 3026  3.13
3  1   00 5030  5.04
4  1   00 5108  3.23
5  1   00 5152  3.97
6  1   00 6080  0.16
7  1   01 0018  4.89
8  1   01 3026  6.58
9  1   01 5030  7.42
10 1   01 5108 10.10
11 1   01 5152  3.74
12 1   01 6080  0.81
 
library(ggplot2)
qplot(factor(dat$time),dat$freq,dat,geom=c(boxplot,jitter),
    ylab=names(dat[,4]),xlab=time) 



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Re: [R] ggplot2: labels points with different colors or idnumbers

[ONKELINX, Thierry]

Dear Tom,

I think you better switch to ggplot() instead of qplot().

library(ggplot2)
dat - expand.grid(id = 1:10, time = 1:3, group = 1:3)
dat$Freq - rnorm(nrow(dat), mean = dat$id * dat$time)
ggplot(dat, aes(x = factor(time), y = Freq)) + geom_boxplot() + 
geom_jitter(aes(colour = factor(id)))

But you can do it with qplot()

qplot(factor(dat$time), dat$Freq, dat, geom = boxplot) + 
geom_jitter(aes(colour = factor(dat$id))) 

IMO the ggplot version gives more readable code. Especially when you are doing 
complex plots.

Splitting plots according to a factor is easy with facet_grid() or facet_wrap().

dat - expand.grid(id = 1:10, time = 1:3, Group = 1:3)
dat$Freq - rnorm(nrow(dat), mean = dat$id * dat$time)
ggplot(dat, aes(x = factor(time), y = Freq)) + geom_boxplot() + 
geom_jitter(aes(colour = factor(id))) + facet_grid(. ~ Group)

If you want three different plots then you will need to do some reading on 
viewports, grobs and stuff. The ggplot2 book on Hadley's website give a good 
intro.

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 
Tom Cohen
Verzonden: donderdag 26 februari 2009 13:05
Aan: r-help@r-project.org
Onderwerp: [R] ggplot2: labels points with different colors or idnumbers

Dear list,
 
Using ggplot2 I could produce both boxplot and points in the same plot but 
instead of points I would like to label the different
subjects with different colors or their idnumbers. Is there away to do it? Also 
how can I put three plots on the same graph with ggplot2? 
mfrow=c(3,1) did not do the job. 
 
 dat
   group time   id  freq
1  1   00 0018  5.21
2  1   00 3026  3.13
3  1   00 5030  5.04
4  1   00 5108  3.23
5  1   00 5152  3.97
6  1   00 6080  0.16
7  1   01 0018  4.89
8  1   01 3026  6.58
9  1   01 5030  7.42
10 1   01 5108 10.10
11 1   01 5152  3.74
12 1   01 6080  0.81
 
library(ggplot2)
qplot(factor(dat$time),dat$freq,dat,geom=c(boxplot,jitter),
    ylab=names(dat[,4]),xlab=time) 



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[R] kmeans: invalid length argument

[Lowell Gould]

Dear R Experts,

I am running a cluter analysis using kmeans and have come across an error to 
which I am unable to find a solution.  First, let me describe the problem:

THE R CODE IS:
--
# NRM is a 100 x 100 numerical matrix
infile = 't:\\NRM\\NRM'
groups = 7

outfile = 't:\\NRM\\cluster.groups'

print( paste(infile, groups, outfile, sep='  '))

pairs - read.table(file=infile, header=TRUE, sep='')
names(pairs) - row.names(pairs)
dist.pairs - dist(pairs)
clust - kmeans(dist.pairs, groups)
write.table(clust$cluster, file=outfile, quote=FALSE)

THE ERROR IS:

Error in vector(integer, length) : invalid 'length' argument
Calls: kmeans - do_one - switch - integer - vector Execution halted

I ran this code on a WindowsXP R6.2.6 host and it ran fine with acceptable 
results.  However, the error ocurrs when I run it on CentOS (redhat 4.1.1-52) 
R2.7.2.  The following search queries yielded no pertinent results from the 
web(Google), Google groups, R-help archives (Nabble and Namazu), and R-FAQ:
1. kmeans
2. invalid length argument
3. vector

?vector says that the length argument must be non-negative integer.  But I 
don't know how to access that call to vector to see what is actually being used 
as an argument.

Any help on this matter will be greatly appreciated.  Thank you for your time.

Lowell

---
Lowell Gould, Ph.D.
Smithfield Premium Genetics
316 W. Charity Rd.
Rose Hill, NC  28458

v. 910.282.4292
f. 910.289.6466

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Re: [R] statistical significance of accuracy increase in classification

[Max Kuhn]

 Do you know about any good reference that discusses kappa for classification 
 and maybe CI for kappa???

I don't, but googling on kappa and confusion matrix etc should get you
there. Kappa works very well when the true classes are skewed. For
example, if 10% of you samples are class A and 90% class B, you can
optimize accuracy by calling everything B.

Max

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Re: [R] Moving Average

[David Winsemius]

I saw Gabor's reply but have a clarification to request. You say you  
want to remove low frequency components but then you request smoothing  
functions. The term smoothing implies removal of high-frequency  
components of a series.


If smoothing really is your goal then additional R resource would be  
smooth.spline, loess (or lowess), ksmooth, or using smoothing terms in  
regressions. Venables and Ripley have quite a few worked examples of  
such in MASS.


--
David Winsemius


On Feb 26, 2009, at 7:07 AM, mau...@alice.it wrote:

I am looking for some help at removing low-frequency components from  
a signal, through Moving Average on a sliding window.
I understand thiis is a smoothing procedure that I never done in my  
life before .. sigh.


I searched R archives and found rollmean, MovingAverages {TTR},  
SymmetricMA.
None of the above mantioned functions seems to accept the smoothing  
polynomial order and the sliding window with as input parameters.  
Maybe I am missing something.


I wonder whether there is some building blocks in R if not even a  
function which does it all (I do not expect that much,though).

Even some literature references and/or tutorials are very welcome.

Thank you so much,
Maura



tutti i telefonini TIM!


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Re: [R] C-index for models fitted using start, stop in Surv?

[Terry Therneau]

One of my colleagues has written a technical report on how to do this, but I 
have not yet implemented it in the survival package.  

Terry Therneau

http://mayoresearch.mayo.edu/mayo/research/biostat/techreports.cfm
  #80
  Concordance for Survival Time Data: Fixed and Time-Dependent Covariates and 
Possible Ties in Predictor and Time
Concordance, or synonymously the C-statistic, is a valuable measure of model 
discrimination in analyses involving survival time data. This report provides a 
definition of concordance in the case of survival data, allowing for 
time-dependent covariates with the counting process data representation and 
accounting for ties in the covariates and times.
Walter K Kremers and The William J. von Liebig Transplant Center [April 2007]

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Re: [R] survival::predict.coxph

[Terry Therneau]

You are mostly correct.
Because of the censoring issue, there is no good estimate of the mean survival 
time.  The survival curve either does not go to zero, or gets very noisy near 
the right hand tail (large standard error); a smooth parametric estimate is 
what 
is really needed to deal with this.
  For this reason the mean survival, though computed (but see the 
survfit.print.mean option, help(print.survfit)) is not highly regarded.  It is 
not an option in predict.coxph.
  
Terry T.

 begin included message --
 
Hi,

if I got it right then the survival-time we expect for a subject is the 
integral over the specific survival-function of the subject from 0 to t_max.

If I have a trained cox-model and want to make a prediction of the 
survival-time for a new subject I could use
survfit(coxmodel, newdata=newSubject) to estimate a new 
survival-function which I have to integrate thereafter.

Actually I thought predict(coxmodel, newSubject) would do this for me, 
but I?m confused which type I have to declare. If I understand the 
little pieces of documentation right then none of the available types is 
exactly the predicted survival-time.
I think I have to use the mean survival-time of the baseline-function 
times exp(the result of type linear predictor).

Am I right?

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Re: [R] plot.survfit

[Terry Therneau]

 For a fitted Cox model, one can either produce the predicted survival curve 
for 
a particular hypothetical subject (survfit), or the predicted curve for a 
particular cohort of subjects (survexp).  See chapter 10 of Therneau and 
Grambsch for a long discussion of the differences between these, and the 
various 
pitfalls.
 
 By default, survfit produces the curve for a hypothetical average subject 
whose covariate values are the respective means of the data set.  I'm not very 
keen on this estimate --- what is sex=.453, a hermaphrodite?  But it is the 
historical default.
 
Terry Therenau

  begin included message -
 
 I am confused when trying the function survfit.
my question is:  what does the survival curve given by plot.survfit mean?
is it the survival curve with different covariates at different points?
or just the baseline survival curve?

for example, I run the following code and get the survival curve


library(survival)
fit-coxph(Surv(futime,fustat)~resid.ds+rx+ecog.ps,data=ovarian)
plot(survfit(fit,type=breslow))
summary(survfit(fit,type=breslow))


for the first two failure points, we have s(59|x1)=0.971, s(115|x2)=0.942
how can we guarantee that s(59|x1) is always greater than s(115|x2)?
since s(59|x1)=s_0(59)^exp(\beta'x1) and s(115|x2)=s_0(115)^exp(\beta'x2),
we can manipulate covariates to make s(59|x1)  s(115|x2), right?
do I miss anything?

thanks in advance

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Re: [R] statistical significance of accuracy increase in classification

[Stefan Evert]

On 26 Feb 2009, at 14:14, Max Kuhn wrote:

Do you know about any good reference that discusses kappa for  
classification and maybe CI for kappa???


You might also want to take a look at this survey article on kappa and  
its alternatives:


	Artstein, Ron and Poesio, Massimo (2008). Survey article: Inter-coder  
agreement for computational linguistics. Computational Linguistics,  
34(4), 555–596.


which you can download from

http://www.aclweb.org/anthology-new/J/J08/

Alternatives to the standard Fleiss-Cohen asymptotic confidence  
intervals in the elementary 2x2 case are discussed in


	Lee, J.J., Tu, Z. N.:A Better Confidence for Kappa on Measuring  
Agreement Between Two Raters with Binary Outcomes Journal of  
Computational and Graphical Statistics, 3:301-321, 1994.


which is available from JSTOR:

http://www.jstor.org/stable/1390914

An S implementation of their approximations can be downloaded here:

http://lib.stat.cmu.edu/S/kappa

I've started to evaluate the accuracy of these approximations with  
simulation experiments some time ago, but haven't found the time to  
follow up on it.


Hope this helps,
Stefan

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Re: [R] survival::survfit,plot.survfit

[Terry Therneau]

 plot(survfit(fit)) should plot the survival-function for x=0 or 
 equivalently beta'=0. This curve is independent of any covariates.

  This is not correct.  It plots the curve for a hypothetical subject with x= 
mean of each covariate.  
  This is NOT the average survival of the data set.  Imagine a cohort made up 
of 60 year old men and their 10 year old grandsons: the expected survival of 
this cohort does not look that for a 35 year old male.
  
Terry T

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[R] R: Moving Average

[mauede]

I apologize for my messy post which stems from my own confusion ... and 
depression as well. In fact I though I was done with a big chunk of a project 
and to my dismay I found out there is more to do.
I am trying to adapt an algorithm, based on advanced wavelet analysis, to my 
respiration signals. The original algorithm was implemented in Fortran by the 
mathematician author of the underlying theory.
I implemented it in R language with some minor changes due to the nature o fthe 
phenomenon I am studying.
Since my results and the mathematician's results mostly disagree on the same 
sample signal, I was suggested to remove the low frequencies components in 
advance of starting wavelet analysis. Upon rereading his suggestion I got more 
and more confused. 
As far as I know Moving Average is one of the simplest DSP low-pass filters. 
Therefore I cannot understand how MA can be used to remove low-frequency 
components. Nor I can understand hi suggestion that I paste in the following. 

You have a rather short signal - 120 samples only. I avoid regarding wavelet 
coeffcients on senior detail levels because their main support interval is of 
the same order as the whole time interval and the circular effect of discrete 
finite wavelet transform is too strong for them. Thus, if the length of time 
series equals N=2^k then I work with detail levels from 1 to (k-3). It means 
that for this time series k=7 and the working detail levels are 1,2,3. Besides 
that you use periodic extension of the signal whereas I use zero padding till 
the length nearest N=2^k and I do not include into analysis zero wavelet 
coefficients which arises due to zero padding. Moreover, the SpAn removes 
automatically before wavelet analysis low-frequency components from the signal 
(which are the main source of circular effect) by moving average within time 
window of the radius 2^(k-3). I advise you to remove low-frequency components 
as well, for example by local polynomials of the 2-nd order wit!
 hin moving time window of the radius 8 samples (the length of moving window 
equals 17, i.e. slightly more than 16 - maximum scale for the 3-rd detail 
level).

Thank you so much,
Maura








-Messaggio originale-
Da: David Winsemius [mailto:dwinsem...@comcast.net]
Inviato: gio 26/02/2009 14.54
A: mau...@alice.it
Cc: r-help@r-project.org
Oggetto: Re: [R] Moving Average
 
I saw Gabor's reply but have a clarification to request. You say you  
want to remove low frequency components but then you request smoothing  
functions. The term smoothing implies removal of high-frequency  
components of a series.

If smoothing really is your goal then additional R resource would be  
smooth.spline, loess (or lowess), ksmooth, or using smoothing terms in  
regressions. Venables and Ripley have quite a few worked examples of  
such in MASS.

-- 
David Winsemius


On Feb 26, 2009, at 7:07 AM, mau...@alice.it wrote:

 I am looking for some help at removing low-frequency components from  
 a signal, through Moving Average on a sliding window.
 I understand thiis is a smoothing procedure that I never done in my  
 life before .. sigh.

 I searched R archives and found rollmean, MovingAverages {TTR},  
 SymmetricMA.
 None of the above mantioned functions seems to accept the smoothing  
 polynomial order and the sliding window with as input parameters.  
 Maybe I am missing something.

 I wonder whether there is some building blocks in R if not even a  
 function which does it all (I do not expect that much,though).
 Even some literature references and/or tutorials are very welcome.

 Thank you so much,
 Maura



 tutti i telefonini TIM!


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[R] Singularity in a regression?

[Bob Gotwals]

R friends,

In a matrix of 1s and 0s, I'm getting a singularity error.  Any helpful ideas?

lm(formula = activity ~ metaF + metaCl + metaBr + metaI + metaMe +
paraF + paraCl + paraBr + paraI + paraMe)

Residuals:
   Min 1Q Median 3QMax
-4.573e-01 -7.884e-02  3.469e-17  6.616e-02  2.427e-01

Coefficients: (1 not defined because of singularities)
Estimate Std. Error t value Pr(|t|)
(Intercept)   7.9173 0.1129  70.135   2e-16 ***
metaF-0.3973 0.2339  -1.698 0.115172
metaClNA NA  NA   NA
metaBr0.3454 0.1149   3.007 0.010929 *
metaI 0.4827 0.2339   2.063 0.061404 .
metaMe0.3654 0.1149   3.181 0.007909 **
paraF 0.7675 0.1449   5.298 0.000189 ***
paraCl0.3400 0.1449   2.347 0.036925 *
paraBr1.0200 0.1449   7.040 1.36e-05 ***
paraI 1.3327 0.2339   5.697 9.96e-05 ***
paraMe1.2191 0.1573   7.751 5.19e-06 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.2049 on 12 degrees of freedom
Multiple R-squared: 0.9257, Adjusted R-squared: 0.8699
F-statistic: 16.61 on 9 and 12 DF,  p-value: 1.811e-05

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Re: [R] Moving Average

[Ted Harding]

On 26-Feb-09 13:54:51, David Winsemius wrote:
 I saw Gabor's reply but have a clarification to request. You say you  
 want to remove low frequency components but then you request smoothing 
 functions. The term smoothing implies removal of high-frequency  
 components of a series.

If you produce a smoothed series, your result of course contains
the low-frequency comsponents, with the high-frequency components
removed.

But if you then subtract that from the original series, your result
contains the high-frequency components, with the low-frequency
compinents removed.

Moving-average is one way of smoothing (but can introduce periodic
components which were not there to start with).

Filtering a time-series is a very open-ended activity! In many
cases a useful start is exploration of the spectral properties
of the series, for which R has several functions. 'spectrum()'
in the stats package (loaded bvy default) is one basic function.
help.search(time series) will throw up a lot of functions.

You might want to look at package 'ltsa' (linear time series
analysis).

Alternatively, if yuou already have good information about the
frequency-structure of the series, or (for instance) know that
it has a will-defined seasonal component, then you could embark
on designing a transfer function specifically tuned to the job.
Have a look at RSiteSearch({transfer function})

Hoping this helps,
Ted.



 If smoothing really is your goal then additional R resource would be  
 smooth.spline, loess (or lowess), ksmooth, or using smoothing terms in 
 regressions. Venables and Ripley have quite a few worked examples of  
 such in MASS.
 
 -- 
 David Winsemius
 
 
 On Feb 26, 2009, at 7:07 AM, mau...@alice.it wrote:
 
 I am looking for some help at removing low-frequency components from  
 a signal, through Moving Average on a sliding window.
 I understand thiis is a smoothing procedure that I never done in my  
 life before .. sigh.

 I searched R archives and found rollmean, MovingAverages {TTR},  
 SymmetricMA.
 None of the above mantioned functions seems to accept the smoothing  
 polynomial order and the sliding window with as input parameters.  
 Maybe I am missing something.

 I wonder whether there is some building blocks in R if not even a  
 function which does it all (I do not expect that much,though).
 Even some literature references and/or tutorials are very welcome.

 Thank you so much,
 Maura



 tutti i telefonini TIM!


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E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 26-Feb-09   Time: 14:54:43
-- XFMail --

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Re: [R] Moving Average

[stephen sefick]

I wrote a little code using Fourier filtering if you would like to
take a look at this:

library(StreamMetabolism)
library(mFilter)
x - read.production(file.choose())
#contiguous.zoo(data.frame(x[,RM202DO.Conc], coredata(x[,RM202DO.Conc])))
#contiguous.zoo(data.frame(x[,RM61DO.Conc], coredata(x[,RM61DO.Conc])))
short - x[42685:48535,RM202DO.Conc]
#short - x[53909:59957,RM61DO.Conc]
short.ts - ts(coredata(short), frequency=96)
#fourier filtering
short.fft - fft(short.ts)
plot(Re(short.fft), xlim=c(0,10), ylim=c(-1000, 1000))
short.fft[789:5563] = 0+0i
short.ifft = fft(short.fft, inverse = TRUE)/length(short.fft)
#zoo series
filt - zoo(coredata(Re(short.ifft)) , index(short))



par(mfrow=c(2,1))
plot(short)
plot(filt)

window.plot - function(x, y, a, b, s, d){
par(mfrow=c(2,1))
plot(window.chron(x, a, b, s, d))
plot(window.chron(y, a, b, s, d))
}
window.plot(short, filt, 04/17/2007, 00:01:00, 04/17/2007, 23:46:00)


plot.e - function(b, w, x, y, z){
a - window.chron(b, w, x, y, z)
plot(a, ylim=range(a)+0.06*c(-1, 1))
lines(a*0.98, col=blue)
lines(a*1.02, col=red)
}

it may not be exactly what you want, but you will have a handle on
what spectral properties that you have removed.

On Thu, Feb 26, 2009 at 9:54 AM, Ted Harding
ted.hard...@manchester.ac.uk wrote:
 On 26-Feb-09 13:54:51, David Winsemius wrote:
 I saw Gabor's reply but have a clarification to request. You say you
 want to remove low frequency components but then you request smoothing
 functions. The term smoothing implies removal of high-frequency
 components of a series.

 If you produce a smoothed series, your result of course contains
 the low-frequency comsponents, with the high-frequency components
 removed.

 But if you then subtract that from the original series, your result
 contains the high-frequency components, with the low-frequency
 compinents removed.

 Moving-average is one way of smoothing (but can introduce periodic
 components which were not there to start with).

 Filtering a time-series is a very open-ended activity! In many
 cases a useful start is exploration of the spectral properties
 of the series, for which R has several functions. 'spectrum()'
 in the stats package (loaded bvy default) is one basic function.
 help.search(time series) will throw up a lot of functions.

 You might want to look at package 'ltsa' (linear time series
 analysis).

 Alternatively, if yuou already have good information about the
 frequency-structure of the series, or (for instance) know that
 it has a will-defined seasonal component, then you could embark
 on designing a transfer function specifically tuned to the job.
 Have a look at RSiteSearch({transfer function})

 Hoping this helps,
 Ted.



 If smoothing really is your goal then additional R resource would be
 smooth.spline, loess (or lowess), ksmooth, or using smoothing terms in
 regressions. Venables and Ripley have quite a few worked examples of
 such in MASS.

 --
 David Winsemius


 On Feb 26, 2009, at 7:07 AM, mau...@alice.it wrote:

 I am looking for some help at removing low-frequency components from
 a signal, through Moving Average on a sliding window.
 I understand thiis is a smoothing procedure that I never done in my
 life before .. sigh.

 I searched R archives and found rollmean, MovingAverages {TTR},
 SymmetricMA.
 None of the above mantioned functions seems to accept the smoothing
 polynomial order and the sliding window with as input parameters.
 Maybe I am missing something.

 I wonder whether there is some building blocks in R if not even a
 function which does it all (I do not expect that much,though).
 Even some literature references and/or tutorials are very welcome.

 Thank you so much,
 Maura



 tutti i telefonini TIM!


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 E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
 Fax-to-email: +44 (0)870 094 0861
 Date: 26-Feb-09                                       Time: 14:54:43
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-- 
Stephen Sefick

Let's not spend our time and resources 

Re: [R] Singularity in a regression?

[Ted Harding]

On 26-Feb-09 12:58:49, Bob Gotwals wrote:
 R friends,
 
 In a matrix of 1s and 0s, I'm getting a singularity error.  Any helpful
 ideas?

From the degress of freedom in your output, it seems you are fitting
10 binary variables to a total of 23 observations. In such circumstances,
it is not unlikely that the matrix of 0s and 1s representing the
binary variables would have at least 1 column which can be represented
as a linear combination of the others (which is what the 1 not defined
because of singularities means). Get more data, or use fewer variables!
Or, also worth considering, check whether there are relationahips
in the real world between your 10 variables which would tend to
generate such linear dependence.

Ted.

 lm(formula = activity ~ metaF + metaCl + metaBr + metaI + metaMe +
  paraF + paraCl + paraBr + paraI + paraMe)
 
 Residuals:
 Min 1Q Median 3QMax
 -4.573e-01 -7.884e-02  3.469e-17  6.616e-02  2.427e-01
 
 Coefficients: (1 not defined because of singularities)
  Estimate Std. Error t value Pr(|t|)
 (Intercept)   7.9173 0.1129  70.135   2e-16 ***
 metaF-0.3973 0.2339  -1.698 0.115172
 metaClNA NA  NA   NA
 metaBr0.3454 0.1149   3.007 0.010929 *
 metaI 0.4827 0.2339   2.063 0.061404 .
 metaMe0.3654 0.1149   3.181 0.007909 **
 paraF 0.7675 0.1449   5.298 0.000189 ***
 paraCl0.3400 0.1449   2.347 0.036925 *
 paraBr1.0200 0.1449   7.040 1.36e-05 ***
 paraI 1.3327 0.2339   5.697 9.96e-05 ***
 paraMe1.2191 0.1573   7.751 5.19e-06 ***
 ---
 Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
 
 Residual standard error: 0.2049 on 12 degrees of freedom
 Multiple R-squared: 0.9257,   Adjusted R-squared: 0.8699
 F-statistic: 16.61 on 9 and 12 DF,  p-value: 1.811e-05
 
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 https://stat.ethz.ch/mailman/listinfo/r-help
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 http://www.R-project.org/posting-guide.html
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E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 26-Feb-09   Time: 15:07:40
-- XFMail --

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Re: [R] Moving Average

[David Winsemius]

On Feb 26, 2009, at 9:54 AM, (Ted Harding) wrote:


On 26-Feb-09 13:54:51, David Winsemius wrote:

I saw Gabor's reply but have a clarification to request. You say you
want to remove low frequency components but then you request  
smoothing

functions. The term smoothing implies removal of high-frequency
components of a series.


If you produce a smoothed series, your result of course contains
the low-frequency comsponents, with the high-frequency components
removed.

But if you then subtract that from the original series, your result
contains the high-frequency components, with the low-frequency
compinents removed.


Yes. The time series term would be detrending or de-trending.



Moving-average is one way of smoothing (but can introduce periodic
components which were not there to start with).

Filtering a time-series is a very open-ended activity! In many
cases a useful start is exploration of the spectral properties
of the series, for which R has several functions. 'spectrum()'
in the stats package (loaded bvy default) is one basic function.
help.search(time series) will throw up a lot of functions.

You might want to look at package 'ltsa' (linear time series
analysis).

Alternatively, if yuou already have good information about the
frequency-structure of the series, or (for instance) know that
it has a will-defined seasonal component, then you could embark
on designing a transfer function specifically tuned to the job.
Have a look at RSiteSearch({transfer function})


As the OP's reply indicates, she is already using wavelet analysis.
My question at this point is whether she should just be advised to  
ignore

the low frequency components and concentrate on the middle and high
frequency components. If you already have some sort of spectral
decomposition, there should be no necessity of a subtraction or
de-trending step.

--
David Winsemius



Hoping this helps,
Ted.




If smoothing really is your goal then additional R resource would be
smooth.spline, loess (or lowess), ksmooth, or using smoothing terms  
in

regressions. Venables and Ripley have quite a few worked examples of
such in MASS.

--
David Winsemius


On Feb 26, 2009, at 7:07 AM, mau...@alice.it wrote:


I am looking for some help at removing low-frequency components from
a signal, through Moving Average on a sliding window.
I understand thiis is a smoothing procedure that I never done in my
life before .. sigh.

I searched R archives and found rollmean, MovingAverages {TTR},
SymmetricMA.
None of the above mantioned functions seems to accept the smoothing
polynomial order and the sliding window with as input parameters.
Maybe I am missing something.

I wonder whether there is some building blocks in R if not even a
function which does it all (I do not expect that much,though).
Even some literature references and/or tutorials are very welcome.

Thank you so much,
Maura



tutti i telefonini TIM!


[[alternative HTML version deleted]]

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E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 26-Feb-09   Time: 14:54:43
-- XFMail --


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Re: [R] Singularity in a regression?

[ONKELINX, Thierry]

It looks like your data has not enough information to estimate the
parameter for metaCl. Maybe because metaCL is identical to one of the
other variables or a constant.
 
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 Bob Gotwals
Verzonden: donderdag 26 februari 2009 13:59
Aan: r-help@r-project.org
Onderwerp: [R] Singularity in a regression?

R friends,

In a matrix of 1s and 0s, I'm getting a singularity error.  Any helpful
ideas?

lm(formula = activity ~ metaF + metaCl + metaBr + metaI + metaMe +
 paraF + paraCl + paraBr + paraI + paraMe)

Residuals:
Min 1Q Median 3QMax
-4.573e-01 -7.884e-02  3.469e-17  6.616e-02  2.427e-01

Coefficients: (1 not defined because of singularities)
 Estimate Std. Error t value Pr(|t|)
(Intercept)   7.9173 0.1129  70.135   2e-16 ***
metaF-0.3973 0.2339  -1.698 0.115172
metaClNA NA  NA   NA
metaBr0.3454 0.1149   3.007 0.010929 *
metaI 0.4827 0.2339   2.063 0.061404 .
metaMe0.3654 0.1149   3.181 0.007909 **
paraF 0.7675 0.1449   5.298 0.000189 ***
paraCl0.3400 0.1449   2.347 0.036925 *
paraBr1.0200 0.1449   7.040 1.36e-05 ***
paraI 1.3327 0.2339   5.697 9.96e-05 ***
paraMe1.2191 0.1573   7.751 5.19e-06 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.2049 on 12 degrees of freedom
Multiple R-squared: 0.9257, Adjusted R-squared: 0.8699
F-statistic: 16.61 on 9 and 12 DF,  p-value: 1.811e-05

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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.

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[R] layout of igraph

[kevinchang]

Dear R users, 

I am trying to draw a network using igraph package. I intend to place the
hub nodes (the ones with the relatively more connection with other nodes) in
the center of the graph.  Also, the graph need to be in the fashion that the
higher the correlation between two nodes is , the closer the two nodes will
be.  Is there any layout that can help or any other way to do this?  

Thanks in advance. 
Shukai 
-- 
View this message in context: 
http://www.nabble.com/layout-of-igraph-tp6348p6348.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Singularity in a regression?

[David Winsemius]

I saw Ted's reply and it is certainly sensible. I would wonder whether  
to model ought to be recast so that the scientific question is more  
clear? You are obviously studying the effect of different  
substitutions (F, Cl, Br, I, Me) and different positions around an  
aromatic ring (meta, para). Why not consider the order of  
electrophilicity (or possibly size) and the position as two different  
variables, one ordered and the other binomial?


After recoding, your formula might then look like  activity ~ electro  
+ position ... or possibly activity ~ electro + size + position, and  
you would be less likely to run into difficulties with collinearity.  
You would also have some science in your model rather than casting  
aimlessly about in the data. If your ordering is sensible, you end up  
testing with 2 or 3 degrees of freedom.


--
David Winsemius

On Feb 26, 2009, at 7:58 AM, Bob Gotwals wrote:


R friends,

In a matrix of 1s and 0s, I'm getting a singularity error.  Any  
helpful ideas?


lm(formula = activity ~ metaF + metaCl + metaBr + metaI + metaMe +
   paraF + paraCl + paraBr + paraI + paraMe)

Residuals:
  Min 1Q Median 3QMax
-4.573e-01 -7.884e-02  3.469e-17  6.616e-02  2.427e-01

Coefficients: (1 not defined because of singularities)
   Estimate Std. Error t value Pr(|t|)
(Intercept)   7.9173 0.1129  70.135   2e-16 ***
metaF-0.3973 0.2339  -1.698 0.115172
metaClNA NA  NA   NA
metaBr0.3454 0.1149   3.007 0.010929 *
metaI 0.4827 0.2339   2.063 0.061404 .
metaMe0.3654 0.1149   3.181 0.007909 **
paraF 0.7675 0.1449   5.298 0.000189 ***
paraCl0.3400 0.1449   2.347 0.036925 *
paraBr1.0200 0.1449   7.040 1.36e-05 ***
paraI 1.3327 0.2339   5.697 9.96e-05 ***
paraMe1.2191 0.1573   7.751 5.19e-06 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.2049 on 12 degrees of freedom
Multiple R-squared: 0.9257, Adjusted R-squared: 0.8699
F-statistic: 16.61 on 9 and 12 DF,  p-value: 1.811e-05

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Re: [R] layout of igraph

[Gábor Csárdi]

Shukai,

the force based layout algorithms (layout.drl,
layout.fruchterman.reingold, layout.graphopt, layout.kamada.kawai) are
likely to do this; although they are not explicitly required to place
hubs in the center, usually they do.

I am not sure what is the correlation between two nodes. You mean
that the graph is weighted?

If you have a small graph, then you can refine the layout
interactively by using 'tkplot'.

Gabor

On Thu, Feb 26, 2009 at 4:20 PM, kevinchang shu...@seas.upenn.edu wrote:

 Dear R users,

 I am trying to draw a network using igraph package. I intend to place the
 hub nodes (the ones with the relatively more connection with other nodes) in
 the center of the graph.  Also, the graph need to be in the fashion that the
 higher the correlation between two nodes is , the closer the two nodes will
 be.  Is there any layout that can help or any other way to do this?

 Thanks in advance.
 Shukai
 --
 View this message in context: 
 http://www.nabble.com/layout-of-igraph-tp6348p6348.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.




-- 
Gabor Csardi gabor.csa...@unil.ch UNIL DGM

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Re: [R] Using very large matrix

[Jay Emerson]

Corrado,

Package bigmemory has undergone a major re-engineering and will be available
soon (available now in Beta version upon request).  The version
currently on CRAN
is probably of limited use unless you're in Linux.

bigmemory may be useful to you for data management, at the very least, where

x - filebacked.big.matrix(8, 8, init=n, type=double)

would accomplish what you want using filebacking (disk space) to hold
the object.
But even this requires 64-bit R (Linux or Mac, or perhaps a Beta
version of Windows 64-bit
R that REvolution Computing is working on).

Subsequent operations (e.g. extraction of a small portion for analysis) are then
easy enough:

y - x[1,]

would give you the first row of x as an object y in R.  Note that x is
not itself an R matrix,
and most existing R analytics can't work on x directly (and would max
out the RAM if they
tried, anyway).

Feel free to email me for more information (and this invitation
applies to anyone who is
interested in this).

Cheers,

Jay

#Dear friends,
#
#I have to use a very large matrix. Something of the sort of
#matrix(8,8,n)  where n is something numeric of the sort 0.xx
#
#I have not found a way of doing it. I keep getting the error
#
#Error in matrix(nrow = 8, ncol = 8, 0.2) : too many elements specified
#
#Any suggestions? I have searched the mailing list, but to no avail.
#
#Best,
#--
#Corrado Topi
#
#Global Climate Change  Biodiversity Indicators
#Area 18,Department of Biology
#University of York, York, YO10 5YW, UK
#Phone: + 44 (0) 1904 328645, E-mail: ct...@york.ac.uk



-- 
John W. Emerson (Jay)
Assistant Professor of Statistics
Department of Statistics
Yale University
http://www.stat.yale.edu/~jay

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Re: [R] layout of igraph

[kevinchang]

Thanks Gabor's fast reply. 

In my research, every node has it's own vector of scores. So I can compute
correlation between every pair of nodes. I used the width and color of the
edge for this purpose. But visualizing the correlations by distance may be
clearer. 

Best,
Shukai


Gábor Csárdi-2 wrote:
 
 Shukai,
 
 the force based layout algorithms (layout.drl,
 layout.fruchterman.reingold, layout.graphopt, layout.kamada.kawai) are
 likely to do this; although they are not explicitly required to place
 hubs in the center, usually they do.
 
 I am not sure what is the correlation between two nodes. You mean
 that the graph is weighted?
 
 If you have a small graph, then you can refine the layout
 interactively by using 'tkplot'.
 
 Gabor
 
 On Thu, Feb 26, 2009 at 4:20 PM, kevinchang shu...@seas.upenn.edu wrote:

 Dear R users,

 I am trying to draw a network using igraph package. I intend to place the
 hub nodes (the ones with the relatively more connection with other nodes)
 in
 the center of the graph.  Also, the graph need to be in the fashion that
 the
 higher the correlation between two nodes is , the closer the two nodes
 will
 be.  Is there any layout that can help or any other way to do this?

 Thanks in advance.
 Shukai
 --
 View this message in context:
 http://www.nabble.com/layout-of-igraph-tp6348p6348.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.

 
 
 
 -- 
 Gabor Csardi gabor.csa...@unil.ch UNIL DGM
 
 __
 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/layout-of-igraph-tp6348p6854.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] layout of igraph

[Gábor Csárdi]

Shukai,

layout.drl supports edge weights, so you could try that. An
alternative is doing MDS, see ?cmdscale and maybe help.search(MDS).

But both these methods are approximate, obviously, most often you
cannot embed an n-dimensional (n2) graph into the 2-dimensional
plane and keep all the distances between the vertices.

Best,
Gabor

On Thu, Feb 26, 2009 at 4:44 PM, kevinchang shu...@seas.upenn.edu wrote:

 Thanks Gabor's fast reply.

 In my research, every node has it's own vector of scores. So I can compute
 correlation between every pair of nodes. I used the width and color of the
 edge for this purpose. But visualizing the correlations by distance may be
 clearer.

 Best,
 Shukai


 Gábor Csárdi-2 wrote:

 Shukai,

 the force based layout algorithms (layout.drl,
 layout.fruchterman.reingold, layout.graphopt, layout.kamada.kawai) are
 likely to do this; although they are not explicitly required to place
 hubs in the center, usually they do.

 I am not sure what is the correlation between two nodes. You mean
 that the graph is weighted?

 If you have a small graph, then you can refine the layout
 interactively by using 'tkplot'.

 Gabor

 On Thu, Feb 26, 2009 at 4:20 PM, kevinchang shu...@seas.upenn.edu wrote:

 Dear R users,

 I am trying to draw a network using igraph package. I intend to place the
 hub nodes (the ones with the relatively more connection with other nodes)
 in
 the center of the graph.  Also, the graph need to be in the fashion that
 the
 higher the correlation between two nodes is , the closer the two nodes
 will
 be.  Is there any layout that can help or any other way to do this?

 Thanks in advance.
 Shukai
 --
 View this message in context:
 http://www.nabble.com/layout-of-igraph-tp6348p6348.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.




 --
 Gabor Csardi gabor.csa...@unil.ch     UNIL DGM

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 --
 View this message in context: 
 http://www.nabble.com/layout-of-igraph-tp6348p6854.html
 Sent from the R help mailing list archive at Nabble.com.

 __
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-- 
Gabor Csardi gabor.csa...@unil.ch UNIL DGM

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[R] which zip file is the emulator package?

[eric lee]

Hi.  I'm trying to run the elliptic package on my computer (windows
platform, version 2.7.2).  I downloaded the elliptic package zip file
from

http://lib.stat.cmu.edu/R/CRAN/

and installed it, but it says that it needs the emulator package.
Can you tell me where to download this?  The only similar package by
name is 'emu_4.0.zip' which is not 'emulator.'  I also can't have R
automatically install this by clicking on 'packages' and 'install
package' because it gives me an error message.

Thank you.

eric




version 2.7.2 on a windows platform

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Re: [R] survival::survfit,plot.survfit

[Jeff Xu]

Thanks very much, Bernhard and Terry.
It clarify my confusion and really helps a lot.

Regards
Jeff Xu

Terry Therneau wrote:
 
 plot(survfit(fit)) should plot the survival-function for x=0 or 
 equivalently beta'=0. This curve is independent of any covariates.
 
   This is not correct.  It plots the curve for a hypothetical subject with
 x= 
 mean of each covariate.  
   This is NOT the average survival of the data set.  Imagine a cohort
 made up 
 of 60 year old men and their 10 year old grandsons: the expected survival
 of 
 this cohort does not look that for a 35 year old male.
   
   Terry T
 
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Re: [R] which zip file is the emulator package?

[Duncan Murdoch]

On 2/26/2009 11:08 AM, eric lee wrote:

Hi.  I'm trying to run the elliptic package on my computer (windows
platform, version 2.7.2).  I downloaded the elliptic package zip file
from

http://lib.stat.cmu.edu/R/CRAN/

and installed it, but it says that it needs the emulator package.
Can you tell me where to download this?  The only similar package by
name is 'emu_4.0.zip' which is not 'emulator.'  I also can't have R
automatically install this by clicking on 'packages' and 'install
package' because it gives me an error message.



The current release gives no error if you ask it to install elliptic, 
and it automatically finds that emulator is in the BACCO bundle.  You 
didn't say what the error message was that you saw, so I can't tell you 
whether it was due to a problem on CRAN or with what you asked for.


Generally speaking it's easier and less error-prone to just ask R to 
install a package, rather than manually downloading and installing one.


Duncan Murdoch

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

Re: [R] which zip file is the emulator package?

[eric lee]

Hi, Duncan.  I choose 'package' then ''install package' and tried 6
different U.S. mirrors.  The message I always get is:

Warning: unable to access index for repository
http://lib.stat.cmu.edu/R/CRAN/bin/windows/contrib/2.7
Warning: unable to access index for repository
http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.7
Error in install.packages(NULL, .libPaths()[1], dependencies = NA,
type = type) :
  no packages were specified
In addition: Warning message:
In open.connection(con, r) :
  unable to connect to 'cran.r-project.org' on port 80.

I assume it's something on my end because I've been having
connectivity issues here.

Thanks again for your help.

eric


On Thu, Feb 26, 2009 at 11:22 AM, Duncan Murdoch murd...@stats.uwo.ca wrote:
 On 2/26/2009 11:08 AM, eric lee wrote:

 Hi.  I'm trying to run the elliptic package on my computer (windows
 platform, version 2.7.2).  I downloaded the elliptic package zip file
 from

 http://lib.stat.cmu.edu/R/CRAN/

 and installed it, but it says that it needs the emulator package.
 Can you tell me where to download this?  The only similar package by
 name is 'emu_4.0.zip' which is not 'emulator.'  I also can't have R
 automatically install this by clicking on 'packages' and 'install
 package' because it gives me an error message.


 The current release gives no error if you ask it to install elliptic, and it
 automatically finds that emulator is in the BACCO bundle.  You didn't say
 what the error message was that you saw, so I can't tell you whether it was
 due to a problem on CRAN or with what you asked for.

 Generally speaking it's easier and less error-prone to just ask R to install
 a package, rather than manually downloading and installing one.

 Duncan Murdoch


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

Re: [R] which zip file is the emulator package?

[David Winsemius]

Reading this article, announcing the elliptic package (which was a  
distinct pleasure) and also looking at the R-FAQ,  my guess is that  
you need the BACCO bundle.


http://www.jstatsoft.org/v15/i07/paper

--
David Winsemius

On Feb 26, 2009, at 11:08 AM, eric lee wrote:


Hi.  I'm trying to run the elliptic package on my computer (windows
platform, version 2.7.2).  I downloaded the elliptic package zip file
from

http://lib.stat.cmu.edu/R/CRAN/

and installed it, but it says that it needs the emulator package.
Can you tell me where to download this?  The only similar package by
name is 'emu_4.0.zip' which is not 'emulator.'  I also can't have R
automatically install this by clicking on 'packages' and 'install
package' because it gives me an error message.

Thank you.

eric




version 2.7.2 on a windows platform

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


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

Re: [R] which zip file is the emulator package?

[eric lee]

Thanks, Duncan.  The BACCO package worked.

On Thu, Feb 26, 2009 at 11:22 AM, Duncan Murdoch murd...@stats.uwo.ca wrote:
 On 2/26/2009 11:08 AM, eric lee wrote:

 Hi.  I'm trying to run the elliptic package on my computer (windows
 platform, version 2.7.2).  I downloaded the elliptic package zip file
 from

 http://lib.stat.cmu.edu/R/CRAN/

 and installed it, but it says that it needs the emulator package.
 Can you tell me where to download this?  The only similar package by
 name is 'emu_4.0.zip' which is not 'emulator.'  I also can't have R
 automatically install this by clicking on 'packages' and 'install
 package' because it gives me an error message.


 The current release gives no error if you ask it to install elliptic, and it
 automatically finds that emulator is in the BACCO bundle.  You didn't say
 what the error message was that you saw, so I can't tell you whether it was
 due to a problem on CRAN or with what you asked for.

 Generally speaking it's easier and less error-prone to just ask R to install
 a package, rather than manually downloading and installing one.

 Duncan Murdoch


__
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] windows vs. linux code

[Paul Gilbert]

Gabor Grothendieck wrote:


Try

if (.Platform$OS.type == windows) ... else ...


Gabor has suggested what I think is the best way to do this check, but 
in my experience, if you are doing this check then you are almost 
certainly missing some feature of R that will let you avoid doing it.


For graphs, dev.new() has been mentioned.  I would be curious to know 
when people find it necessary to do this check, other than to issue a 
system() command for some very specialized local system reason (i.e. 
something that would never be used in a package and rarely in code run 
on different computers - not just different OSs).


Paul


La version française suit le texte anglais.



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Re: [R] R communication

[Greg Snow]

This could depend somewhat on which OS you have on the computers as to which 
packages will work or work best for you.  A couple of packages to look at 
include, Rmpi, nws, and snow (no relation), and the other packages in the 
suggests field for snow.

-- 
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 ARDIA David
 Sent: Thursday, February 26, 2009 3:05 AM
 To: r-help@r-project.org
 Subject: [R] R communication
 
 Dear all,
 Imagine that you have a small LAN with two computers. I would like to
 run R on both, and possible to run computations from one computer to
 the other one. TCP IP protocole would be preferable. Which package
 would you use for that? I would be very glad if you could also provide
 me with some lines of code, e.g., create a matrix X in computer 1,
 transfer its value to the second computer, make some calculation, and
 get the value back.
 Thanks for your help,
 David
 
   [[alternative HTML version deleted]]
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-
 guide.html
 and provide commented, minimal, self-contained, reproducible code.

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

[R] ftp fetch using RCurl?

[CHD850]

Hi everyone,

I have to fetch about 300 to 500 zipped archives from a remote ftp server.
Each of the archive is about 1Mb. I know I can get it done by using
download.file() in R, but I am curious that is there a faster way to do this
using RCurl. For example, are there some parameters that I can set so that
the connection does not need to be rebuiltetc. 

A even simpler question is, how can I fetch an archive from the server and
place it somewhere locally? I have spent a lot of time reading RCurl
documents and curl web pages but in vain. Can someone show me an example of
the syntax? Pardon me if this is trivial to you.

Thanks
Stanley

-- 
View this message in context: 
http://www.nabble.com/ftp-fetch-using-RCurl--tp8067p8067.html
Sent from the R help mailing list archive at Nabble.com.

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

[R] error message and convergence issues in fitting glmer in package lme4

[Tanja Srebotnjak]

I'm resending this message because I did not include a subject line in my first 
posting.

Apologies for the inconvenience!



Tanja



 Hello,



 I'm trying to fit a generalized linear mixed model to estimate diabetes 
 prevalence at US county level. To do this I'm using the glmer() function in 
 package lme4. I can fit relatively simple models (i.e. few covariates) but 
 when expanding the number of covariates I usually encounter the following 
 error message.



 gm8 -

 glmer(DIAB05F~AGE+as.factor(SEX)+poolt+poolx+poverty+fastfood+(1|as.factor(diab$fips)),
  family = binomial(link=logit), data = diab, doFit=TRUE) Error in 
 validObject(.Object) :

   invalid class mer object: Slot Zt must by dims['q']  by

 dims['n']*dims['s']



 In the above, the response is person-level diabetes status as a function of 
 AGE=age, SEX=sex, poolt=average county diabetes prevalence for previous 
 years, poolx=pooled county diabetes prevalence for counties with similar age, 
 sex, race, and income structure, poverty=county poverty rate, fastfood=number 
 of fastfood places per 100,000 people in the county, and a county random 
 effect.



 If I leave out fastfood, the model gets at least fitted - although it doesn't 
 converge (yet):



 Warning message:

 In mer_finalize(ans) : false convergence (8)



 I would be grateful for any advice on what the problem could be and how to 
 resolve it.



 Thanks,

 Tanja


Tanja Srebotnjak, PhD, MSc, Dipl. Stat.
Postgraduate Fellow
Institute for Health Metrics and Evaluation
University of Washington
2301 5th Ave, Suite 600
Seattle, WA 98121
Email: tan...@u.washington.edumailto:tan...@u.washington.edu
Tel: +1-206-897-2866
www.healthmetricsandevaluation.orghttp://www.healthmetricsandevaluation.org

From: Tanja Srebotnjak
Sent: Thursday, February 26, 2009 12:17 AM
To: 'r-help@r-project.org'
Subject:

Hello,

I'm trying to fit a generalized linear mixed model to estimate diabetes 
prevalence at US county level. To do this I'm using the glmer() function in 
package lme4. I can fit relatively simple models (i.e. few covariates) but when 
expanding the number of covariates I usually encounter the following error 
message.

gm8 - 
glmer(DIAB05F~AGE+as.factor(SEX)+poolt+poolx+poverty+fastfood+(1|as.factor(diab$fips)),
 family = binomial(link=logit), data = diab, doFit=TRUE)
Error in validObject(.Object) :
  invalid class mer object: Slot Zt must by dims['q']  by dims['n']*dims['s']

In the above, the response is person-level diabetes status as a function of 
AGE=age, SEX=sex, poolt=average county diabetes prevalence for previous years, 
poolx=pooled county diabetes prevalence for counties with similar age, sex, 
race, and income structure, poverty=county poverty rate, fastfood=number of 
fastfood places per 100,000 people in the county, and a county random effect.

If I leave out fastfood, the model gets at least fitted - although it doesn't 
converge (yet):

Warning message:
In mer_finalize(ans) : false convergence (8)

I would be grateful for any advice on what the problem could be and how to 
resolve it.

Thanks,
Tanja


[[alternative HTML version deleted]]

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

Re: [R] Random Forest confusion matrix

[Gabor Grothendieck]

randomForest output is based on predict(iris.rf) whereas the
code shown below uses predict(iris.rf, iris).  See ?predict.randomForest
for an explanation.

On Thu, Feb 26, 2009 at 11:10 AM, Li GUO guol...@yahoo.com wrote:
 Dear R users,

 I have a question on the confusion matrix generated by function randomForest.
 I used the entire data
 set to generate the forest, for example:
 print(iris.rf)

 Call:
  randomForest(formula = Species ~ ., data = iris, importance = TRUE,
 keep.forest = TRUE)

 confusion
           setosa versicolor virginica class.error
 setosa         50          0         0        0.00
 versicolor      0         47         3        0.06
 virginica       0          3        47        0.06

 then I classified the same data set with this forest:

 iris.pred - predict(iris.rf, iris)
 table(observed = iris[,Species], predicted = iris.pred)
            predicted
 observed     setosa versicolor virginica
  setosa         50          0         0
  versicolor      0         50         0
  virginica       0          0        50
 Why the two matrices are different?
 Thinks,

 Li




        [[alternative HTML version deleted]]

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


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

Re: [R] Hi, Coding problem

[Greg Snow]

It looks like what you did was something like:

 tmp - sim(100)
 table(tmp)/100

To get your results (it is easier for us to help if you tell us what you 
actually did, also setting a seed and telling us that seed helps us reproduce 
what you did exactly).

The reason that you did not see 9 and 10 is just due to chance, the 
probabilities for those values is small and it is not surprising that you did 
not see a couple of unlikely values.  You can either run the sim function more 
times or tell the computer exactly what values you want counted so it knows to 
include them, for example:

 tmp - factor(tmp, levels= 1:max(tmp) )
 table(tmp)/100

Hope this helps,

-- 
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 Ssophia
 Sent: Wednesday, February 25, 2009 7:28 PM
 To: r-help@r-project.org
 Subject: [R] Hi, Coding problem
 
 
 
   Hi, there
 
 
 
   Below is my code to one Homework question. I couldn't come up with
 the reasonable answer.
 could you please help me to figure out what is the problem with my
 code?
 thank you
 
 
 Question is Coding P{X=j} =(1/2)^(j+1) + (1/2) *2^(j-1)/3^j
 my code is
 sim - function(n.gen){
 urandom - runif(n.gen)
 sim.vector - rep(0,n.gen)
 for(j in 1:n.gen){
  i - 1
  p - 5/12
  F - p
  while(urandom[j] = F){
   p - p*((1/2)^(i+1)+1/3*(2/3)^i)/((1/2)^i+(1/2)*(2/3)^i)
   F - F+p
   i-i+1
   }
  sim.vector[j] - i
 }
 # output
 sim.vector
 }
 
 
 
 result is
 
12345678   11
 0.37 0.22 0.16 0.13 0.05 0.02 0.03 0.01 0.01
 
 
 
 always, there are some numbers missing, it should be continuous.
 
 why 9 and 10 are missing
 
 thank you
 
 
 
 sophia
 
 
 
 _
 [[elided Hotmail spam]]
 
   [[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] use R Group SFBA March meeting reminder

[Jim Porzak]

All San Francisco Bay Area useRs,

On March 11th, Spencer Graves  Sundar Dorai-Raj will talk about
Creating R Packages, and related issues.

This is the first of our (now) regular monthly meetings, details here
http://www.meetup.com/R-Users/calendar/9718957/

Last week Wednesday, we had over 60 people attend our 2009 kick-off
meeting in cooperation with Predictive Analytics World (see Group link
below).

Based on that response, we are now scheduling out the rest of the year.

PRESENTERS needed! Please contact Mike or myself if you are interested
in presenting.

Best,
Jim Porzak
TGN.com
San Francisco, CA
http://www.linkedin.com/in/jimporzak
use R! Group SF: http://www.meetup.com/R-Users/

__
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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] which zip file is the emulator package?

[Greg Snow]

Are you behind a firewall?

See the help for download.file for details on setting proxy information if this 
is the case.

-- 
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 eric lee
 Sent: Thursday, February 26, 2009 9:33 AM
 To: Duncan Murdoch
 Cc: r-h...@stat.math.ethz.ch
 Subject: Re: [R] which zip file is the emulator package?
 
 Hi, Duncan.  I choose 'package' then ''install package' and tried 6
 different U.S. mirrors.  The message I always get is:
 
 Warning: unable to access index for repository
 http://lib.stat.cmu.edu/R/CRAN/bin/windows/contrib/2.7
 Warning: unable to access index for repository
 http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.7
 Error in install.packages(NULL, .libPaths()[1], dependencies = NA,
 type = type) :
   no packages were specified
 In addition: Warning message:
 In open.connection(con, r) :
   unable to connect to 'cran.r-project.org' on port 80.
 
 I assume it's something on my end because I've been having
 connectivity issues here.
 
 Thanks again for your help.
 
 eric
 
 
 On Thu, Feb 26, 2009 at 11:22 AM, Duncan Murdoch murd...@stats.uwo.ca
 wrote:
  On 2/26/2009 11:08 AM, eric lee wrote:
 
  Hi.  I'm trying to run the elliptic package on my computer (windows
  platform, version 2.7.2).  I downloaded the elliptic package zip
 file
  from
 
  http://lib.stat.cmu.edu/R/CRAN/
 
  and installed it, but it says that it needs the emulator package.
  Can you tell me where to download this?  The only similar package by
  name is 'emu_4.0.zip' which is not 'emulator.'  I also can't have R
  automatically install this by clicking on 'packages' and 'install
  package' because it gives me an error message.
 
 
  The current release gives no error if you ask it to install elliptic,
 and it
  automatically finds that emulator is in the BACCO bundle.  You didn't
 say
  what the error message was that you saw, so I can't tell you whether
 it was
  due to a problem on CRAN or with what you asked for.
 
  Generally speaking it's easier and less error-prone to just ask R to
 install
  a package, rather than manually downloading and installing one.
 
  Duncan Murdoch
 
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-
 guide.html
 and provide commented, minimal, self-contained, reproducible code.
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and provide commented, minimal, self-contained, reproducible code.

[R] Merge question

[Vadlamani, Subrahmanyam {FLNA}]

Hi:
I am a new R user. I have the following question and would appreciate your input

Data1 (data frame 1)
p1,d1,d2 (p1 is text and d1 and d2 are numeric)
xyz,10,25

Data2 (data frame 2)
p1,d1,d2
xyz,11,15

Now I want to create a new data frame that looks like so below. The fields d1 
and s2 are summed by the product key.
Data3
p1,d1,d2
xyz,21 (sum of 10 from Data1 and 11 from Data2),40 (sum of 25 from Data1 and 15 
from Data2)

Any other examples of merge you may have will be appreciated. Thanks.
Satish

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[R] generalized linear mixed models with a beta distribution

[Jeff Evans]

Has there been any follow up to this question? I have found myself wondering
the same thing: How then does SAS fit a beta distributed GLMM? It also fits
the negative binomial distribution. 

Both of these would be useful in glmer/lmer if they aren't 'illegal' as
Brian suggested. Especially as SAS indicates a favorable delta BIC of over
1000 when I fit the beta to my data (could be the beginning of a great
song..) versus my original binomial fit.

Jeff Evans
Michigan State University

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[R] glm with large datasets

[Julio Gonzalez Diaz]

Hi all,

I have to run a logit regresion over a large dataset and I am not sure 
about the best option to do it. The dataset is about 20x2000 and R 
runs out of memory when creating it.


After going over help archives and the mailing lists, I think there are 
two main options, though I am not sure about which one will be better. 
Of course, any alternative will be welcome as well.


Actually, I am not quite sure about whether any of these options will 
work but, before getting into it, I would like to get some advice.


-A first option is to use the package ff, that allows to work with the 
dataset without loading it into the RAM. This, combined with the bigglm 
function should do the job.


-The dataset contains a lot of sparse variables, so I was wondering 
whether creating the model matrix as a sparse matrix might deliver good 
results. In this case, I am not sure about the capabilities of glm or 
some extension of it to deal with sparse matrices (I could not find any 
documentation about this). If possible, this second option seems more 
efficient since R might be capable of using the fact that matrices are 
sparse to speed up the computations.



Thanks in advance.
All the best!
Julio.

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Re: [R] windows vs. linux code

[Prof Brian Ripley]

On Thu, 26 Feb 2009, Paul Gilbert wrote:


Gabor Grothendieck wrote:


Try

if (.Platform$OS.type == windows) ... else ...


Gabor has suggested what I think is the best way to do this check, but in my 
experience, if you are doing this check then you are almost certainly missing 
some feature of R that will let you avoid doing it.


For graphs, dev.new() has been mentioned.  I would be curious to know when 
people find it necessary to do this check, other than to issue a system() 
command for some very specialized local system reason (i.e. something that 
would never be used in a package and rarely in code run on different 
computers - not just different OSs).


Grep-ing the R and package sources is informative.

One place that such a test is needed is for system/shell differences, 
and the fact that some Windows commands need \ as the file separator 
(and some need / or shell-specific quoting).  Quite a few packages are 
calling e.g. gs (where the name is platform-dependent).


Some packages seem to treat tk differently by platform, but that seems 
to me to be largely a legacy of Tk 8.4: 8.5 is more portable.


But many uses in packages are no longer necessary, e.g. flush.console 
exists everywhere, and R 2.9.0 will have an unzip() function.


--
Brian D. Ripley,  rip...@stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

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

Re: [R] two colors and putting lines on bar plots

[Greg Snow]

For the first question you can use %in% rather than ==, for example:

ifelse( dataframe$vector_o_number %in% c('00','01), 'red', 'black') 

for the reference line, the abline function will draw a line the full 
width/height of a graph for a general reference, if you want separate lines for 
each group of bars then the segments function will draw the shorter lines.  
Note that the return value of the barplot function gives information on the x 
coordinates of the bars that could be useful in drawing the lines (see the 
examples for barplot).

Hope this helps,


-- 
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 John Malone
 Sent: Wednesday, February 25, 2009 7:05 PM
 To: r-help@r-project.org
 Subject: [R] two colors and putting lines on bar plots
 
 Hi all,
 
 I want to represent two categories with one color and have other
 categories
 a different color on a bar plot.
 
 I can do this using for one category/number using the ifelse call in
 col but
 how to extend to two categories/numbers?
 
 barplot(dataframe$vector_o_numbers,
 col=ifelse(dataframe$vector_o_numbers ==
 00, red, black), names.arg=dataframe$labels)
 
 For example if I wanted to label 00 and 01 red and have the rest of the
 categories colored blackhow to do this?
 
 FinallyI'd Iike to put a line on the bar plot representing an
 expected number to compare to the observed number for a particular
 category.
 In the past I've done this by drawing a line manually on a graph using
 photoshop or something similar.  Is there way to do this in R?
 
 Many thanks all in advance for your help!!
 
 John
 
   [[alternative HTML version deleted]]
 
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 guide.html
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Re: [R] Merge question

[milton ruser]

Hi there,

something like this?


Data1-read.table(stdin(), head=T, sep=,)
p1,d1,d2
xyz,10,25
kmz,100,250

Data2-read.table(stdin(), head=T, sep=,)
p1,d1,d2
xyz,11,15
kmz,110,150

Data1
Data2
Data3-data.frame(rbind(Data1,Data2))
Data3
Data3.sum-aggregate(Data3[,c(d1,d2)], list(Data3$p1), sum)
Data3.sum

Best wishes

miltinho astronauta
brazil
On Thu, Feb 26, 2009 at 2:52 PM, Vadlamani, Subrahmanyam {FLNA} 
subrahmanyam.vadlam...@fritolay.com wrote:

 Hi:
 I am a new R user. I have the following question and would appreciate your
 input

 Data1 (data frame 1)
 p1,d1,d2 (p1 is text and d1 and d2 are numeric)
 xyz,10,25

 Data2 (data frame 2)
 p1,d1,d2
 xyz,11,15

 Now I want to create a new data frame that looks like so below. The fields
 d1 and s2 are summed by the product key.
 Data3
 p1,d1,d2
 xyz,21 (sum of 10 from Data1 and 11 from Data2),40 (sum of 25 from Data1
 and 15 from Data2)

 Any other examples of merge you may have will be appreciated. Thanks.
 Satish

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 http://www.R-project.org/posting-guide.htmlhttp://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] Bug in predict function for naiveBayes?

[joris meys]

Dear all,

I've sent a mail last week already. I've been trying to get this predict
function work, but somehow I keep on getting the same error. Read the help
files, searched the internet, but I don't seem to get what I'm doing wrong.

Anybody who has experience with this function? It's contained in the package
e1071.
Thank you in advance

On Thu, Feb 19, 2009 at 2:16 PM, joris meys jorism...@gmail.com wrote:

 Dear all,

 I tried a simple naive Bayes classification on an artificial dataset, but I
 have troubles getting the predict function to work with the type=class
 specification. With type= raw, it works perfectly, but with type=class I
 get following error :

 Error in as.vector(x, mode) : invalid 'mode' argument

 Data : mixture.train is a training set with 100 points originating from 2
 multivariate gaussian distributions (class 0 and class 1), with X1 and X2 as
 coordinates in a 2-dimensional space. Mixture.test is a grid going from -15
 to +15 in both dimensions. Stupid data, but it's just to test.

 Code :
 Sigma - matrix(c(10,3,3,2),2,2)
 mixture.train - cbind(mvrnorm(n=50, c(0, 2), Sigma),rep(0,50))
 mixture.train - as.data.frame(rbind(mixture.train,cbind(mvrnorm(n=50, c(2,
 0), Sigma),rep(1,50
 names(mixture.train) -c(X1,X2,Class)
 X1 - rep(seq(-15,15,by=1),31)
 X2 - rep(seq(-15,15,by=1),each = 31)
 mixture.test - data.frame(X1,X2)

 Bayes.res - naiveBayes(Class ~ X1 + X2, data=mixture.train)
 pred.bayes -predict(Bayes.res, cbind(mixture.test$X1,
 mixture.test$X2),type=class)

 Tried it also with pred.bayes -predict(Bayes.res,
 mixture.test,type=class), but that gives the same effect. Is this a bug or
 am I missing something?

 Kind regards
 Joris Meys
 University Ghent




[[alternative HTML version deleted]]

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

Re: [R] Merge question

[Marc Schwartz]

on 02/26/2009 11:52 AM Vadlamani, Subrahmanyam {FLNA} wrote:
 Hi:
 I am a new R user. I have the following question and would appreciate your 
 input
 
 Data1 (data frame 1)
 p1,d1,d2 (p1 is text and d1 and d2 are numeric)
 xyz,10,25
 
 Data2 (data frame 2)
 p1,d1,d2
 xyz,11,15
 
 Now I want to create a new data frame that looks like so below. The fields d1 
 and s2 are summed by the product key.
 Data3
 p1,d1,d2
 xyz,21 (sum of 10 from Data1 and 11 from Data2),40 (sum of 25 from Data1 and 
 15 from Data2)
 
 Any other examples of merge you may have will be appreciated. Thanks.
 Satish

Given the nature of your data, having the same column structure with
repeated keys, I would not use merge(), but rbind() the two data frames
together and then use aggregate():

DF - rbind(Data1, Data2)

 DF
   p1 d1 d2
1 xyz 10 25
2 xyz 11 15

 aggregate(DF[-1], list(p1 = DF$p1), sum)
   p1 d1 d2
1 xyz 21 40


See ?rbind and ?aggregate

If you search the list archives:

  RSiteSearch(merge)

you will yield hundreds of posts showing the use of that particular
function.

HTH,

Marc Schwartz

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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] bottom legends in ggplot2 ?

[Avram Aelony]

Has anyone had success with producing legends to a qplot graph such that the 
legend is placed on the bottom, under the abcissa rather than to the right hand 
side ?

The following doesn't move the legend:
   library(ggplot2)
   qplot(mpg, wt, data=mtcars, colour=cyl, gpar(legend.position=bottom) )


I am using ggplot2_0.8.2.

Thanks in advance,

Avram

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

Re: [R] Bug in predict function for naiveBayes?

[Liaw, Andy]

You need to set Class as factor before you call naiveBayes(); i.e.,

mixture.train$Class - factor(mixture.train$Class)

Then you can just do:

pred.bayes -predict(Bayes.res, mixture.test, type=class)

Andy 

 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of joris meys
 Sent: Thursday, February 26, 2009 1:36 PM
 To: r-help@r-project.org
 Subject: Re: [R] Bug in predict function for naiveBayes?
 
 Dear all,
 
 I've sent a mail last week already. I've been trying to get 
 this predict
 function work, but somehow I keep on getting the same error. 
 Read the help
 files, searched the internet, but I don't seem to get what 
 I'm doing wrong.
 
 Anybody who has experience with this function? It's contained 
 in the package
 e1071.
 Thank you in advance
 
 On Thu, Feb 19, 2009 at 2:16 PM, joris meys 
 jorism...@gmail.com wrote:
 
  Dear all,
 
  I tried a simple naive Bayes classification on an 
 artificial dataset, but I
  have troubles getting the predict function to work with the 
 type=class
  specification. With type= raw, it works perfectly, but 
 with type=class I
  get following error :
 
  Error in as.vector(x, mode) : invalid 'mode' argument
 
  Data : mixture.train is a training set with 100 points 
 originating from 2
  multivariate gaussian distributions (class 0 and class 1), 
 with X1 and X2 as
  coordinates in a 2-dimensional space. Mixture.test is a 
 grid going from -15
  to +15 in both dimensions. Stupid data, but it's just to test.
 
  Code :
  Sigma - matrix(c(10,3,3,2),2,2)
  mixture.train - cbind(mvrnorm(n=50, c(0, 2), Sigma),rep(0,50))
  mixture.train - 
 as.data.frame(rbind(mixture.train,cbind(mvrnorm(n=50, c(2,
  0), Sigma),rep(1,50
  names(mixture.train) -c(X1,X2,Class)
  X1 - rep(seq(-15,15,by=1),31)
  X2 - rep(seq(-15,15,by=1),each = 31)
  mixture.test - data.frame(X1,X2)
 
  Bayes.res - naiveBayes(Class ~ X1 + X2, data=mixture.train)
  pred.bayes -predict(Bayes.res, cbind(mixture.test$X1,
  mixture.test$X2),type=class)
 
  Tried it also with pred.bayes -predict(Bayes.res,
  mixture.test,type=class), but that gives the same effect. 
 Is this a bug or
  am I missing something?
 
  Kind regards
  Joris Meys
  University Ghent
 
 
 
 
   [[alternative HTML version deleted]]
 
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 http://www.R-project.org/posting-guide.html
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Notice:  This e-mail message, together with any attachme...{{dropped:12}}

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Re: [R] windows vs. linux code

[Jeffrey J. Hallman]

Rolf Turner r.tur...@auckland.ac.nz writes:
 Despite the knowledge, wisdom, insight, skill, good looks, and other
 admirable characteristics of the members of the R-help list, few of
 us are skilled in telepathy or clairvoyance.

Oh, yeah?  Then how did I know you were going to say that, huh?

-- 
Jeff

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

[R] plot with pairwise joined points

[William Simpson]

I would like to do as follows
plot(a,b)
points(c,d,pch=19)

Now join with a line segment point a[1], b[1] to c[1], d[1]; a[2],
b[2] to c[2], d[2] ... a[n], b[n] to c[n], d[n]

All corresponding points from the two data sets are joined by line segments.

Thanks very much for any tips on how to do this.

Bill

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

Re: [R] error message and convergence issues in fitting glmer in package lme4

[Douglas Bates]

On Thu, Feb 26, 2009 at 10:58 AM, Tanja Srebotnjak
tan...@u.washington.edu wrote:
 I'm resending this message because I did not include a subject line in my 
 first posting.

Also, it is generally more effective to send questions about
lmer/glmer to the R-SIG-Mixed-Models list, which I am cc:ing on this
reply.

 Hello,

 I'm trying to fit a generalized linear mixed model to estimate diabetes 
 prevalence at US county level. To do this I'm using the glmer() function in 
 package lme4. I can fit relatively simple models (i.e. few covariates) but 
 when expanding the number of covariates I usually encounter the following 
 error message.

 gm8 -
 glmer(DIAB05F~AGE+as.factor(SEX)+poolt+poolx+poverty+fastfood+(1|as.factor(diab$fips)),
  family = binomial(link=logit), data = diab, doFit=TRUE)

Error in validObject(.Object) :   invalid class mer object: Slot Zt
must by dims['q']  by dims['n']*dims['s']

Getting that error message from this model is peculiar.  I couldn't
actually say what might be happening without trying the fit myself.  I
would suggest setting doFit = FALSE but I think that this error would
be encountered even with doFit = FALSE.  Again, it would be hard to
say exactly what is happening here.

 In the above, the response is person-level diabetes status as a function of 
 AGE=age, SEX=sex, poolt=average county diabetes prevalence for previous 
 years, poolx=pooled county diabetes prevalence for counties with similar 
 age, sex, race, and income structure, poverty=county poverty rate, 
 fastfood=number of fastfood places per 100,000 people in the county, and a 
 county random effect.

 If I leave out fastfood, the model gets at least fitted - although it 
 doesn't converge (yet):

The version of lmer currently under development tries to address that
problem.  The optimization of the parameter estimates is performed in
a slightly different way that will, I hope, provide smoother
convergence.  If your data are not restricted and you would be willing
to send me a copy of the diab data frame I could check what happens on
that version (or you could install the development version yourself
but that is a non-trivial undertaking).  If you can send the data the
best way to send it is to create an R data file as

save(diab, file = diab.rda)

and send the file diab.rda

 Warning message:
 In mer_finalize(ans) : false convergence (8)

Frequently that is a sign of an overspecified model.



 I would be grateful for any advice on what the problem could be and how to 
 resolve it.



 Thanks,

 Tanja


 Tanja Srebotnjak, PhD, MSc, Dipl. Stat.
 Postgraduate Fellow
 Institute for Health Metrics and Evaluation
 University of Washington
 2301 5th Ave, Suite 600
 Seattle, WA 98121
 Email: tan...@u.washington.edumailto:tan...@u.washington.edu
 Tel: +1-206-897-2866
 www.healthmetricsandevaluation.orghttp://www.healthmetricsandevaluation.org

 From: Tanja Srebotnjak
 Sent: Thursday, February 26, 2009 12:17 AM
 To: 'r-help@r-project.org'
 Subject:

 Hello,

 I'm trying to fit a generalized linear mixed model to estimate diabetes 
 prevalence at US county level. To do this I'm using the glmer() function in 
 package lme4. I can fit relatively simple models (i.e. few covariates) but 
 when expanding the number of covariates I usually encounter the following 
 error message.

 gm8 - 
 glmer(DIAB05F~AGE+as.factor(SEX)+poolt+poolx+poverty+fastfood+(1|as.factor(diab$fips)),
  family = binomial(link=logit), data = diab, doFit=TRUE)
 Error in validObject(.Object) :
  invalid class mer object: Slot Zt must by dims['q']  by dims['n']*dims['s']

 In the above, the response is person-level diabetes status as a function of 
 AGE=age, SEX=sex, poolt=average county diabetes prevalence for previous 
 years, poolx=pooled county diabetes prevalence for counties with similar age, 
 sex, race, and income structure, poverty=county poverty rate, fastfood=number 
 of fastfood places per 100,000 people in the county, and a county random 
 effect.

 If I leave out fastfood, the model gets at least fitted - although it doesn't 
 converge (yet):

 Warning message:
 In mer_finalize(ans) : false convergence (8)

 I would be grateful for any advice on what the problem could be and how to 
 resolve it.

 Thanks,
 Tanja


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

[R] Download daily weather data

[Thomas Levine]

I'm writing a program that will tell me whether I should wear a coat,
so I'd like to be able to download daily weather forecasts and daily
reports of recent past weather conditions.

The NOAA has very promising tabular forecasts
(http://forecast.weather.gov/MapClick.php?CityName=Ithacastate=NYsite=BGMtextField1=42.4422textField2=-76.5002e=0FcstType=digital),
but I can't figure out how to import them.

Someone must have needed to do this before. Suggestions?

Thomas Levine!

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Re: [R] generalized linear mixed models with a beta distribution

[Douglas Bates]

On Thu, Feb 26, 2009 at 12:04 PM, Jeff Evans evans...@msu.edu wrote:
 Has there been any follow up to this question? I have found myself wondering
 the same thing: How then does SAS fit a beta distributed GLMM? It also fits
 the negative binomial distribution.

When SAS decides to open-source their code we'll be able to find out.

 Both of these would be useful in glmer/lmer if they aren't 'illegal' as
 Brian suggested. Especially as SAS indicates a favorable delta BIC of over
 1000 when I fit the beta to my data (could be the beginning of a great
 song..) versus my original binomial fit.

Definitions of generalized linear mixed models are not entirely
straightforward, at least for me.  I'm making some progress but, as
always, it is slower than one would like it to be.

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

Re: [R] plot with pairwise joined points

[Greg Snow]

Try:
 segments(a,b,c,d)

-- 
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 William Simpson
 Sent: Thursday, February 26, 2009 1:46 PM
 To: r-help@r-project.org
 Subject: [R] plot with pairwise joined points
 
 I would like to do as follows
 plot(a,b)
 points(c,d,pch=19)
 
 Now join with a line segment point a[1], b[1] to c[1], d[1]; a[2],
 b[2] to c[2], d[2] ... a[n], b[n] to c[n], d[n]
 
 All corresponding points from the two data sets are joined by line
 segments.
 
 Thanks very much for any tips on how to do this.
 
 Bill
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-
 guide.html
 and provide commented, minimal, self-contained, reproducible code.

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

Re: [R] plot with pairwise joined points

[Dimitris Rizopoulos]

have a look at segments(), e.g.,

x - rnorm(5)
y - rnorm(5)
z - rnorm(5)
w - rnorm(5)

r1 - range(x, z)
r2 - range(y, w)
plot(r1, r2, type = n)
points(x, y)
points(z, w, pch = 19)
segments(x, y, z, w)


I hope it helps.

Best,
Dimitris


William Simpson wrote:

I would like to do as follows
plot(a,b)
points(c,d,pch=19)

Now join with a line segment point a[1], b[1] to c[1], d[1]; a[2],
b[2] to c[2], d[2] ... a[n], b[n] to c[n], d[n]

All corresponding points from the two data sets are joined by line segments.

Thanks very much for any tips on how to do this.

Bill

__
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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.



--
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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Re: [R] plot with pairwise joined points

[Rolf Turner]

On 27/02/2009, at 9:46 AM, William Simpson wrote:


I would like to do as follows
plot(a,b)
points(c,d,pch=19)

Now join with a line segment point a[1], b[1] to c[1], d[1]; a[2],
b[2] to c[2], d[2] ... a[n], b[n] to c[n], d[n]

All corresponding points from the two data sets are joined by line  
segments.


?segments

cheers,

Rolf Turner

##
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Re: [R] Download daily weather data

[James Muller]

Thomas,

Have a look at the source code for the webpage (ctrl-u in firefox,
don't know in internet explorer, etc.). That is what you'd have to
parse in order to get the forecast from this page. Typically when I
parse webpages such as this I use regular expressions to do so (and I
would never downplay the usefulness of regular expressions, but they
take a little getting used to). There are two parts to the task: find
patterns that allow you to pull out the datum/data you're after; and
then write a program to pull it/them out. Also, of course, download
the webpage (but that's no issue).

I bet you'd be able to find a comma separated value (CSV) file
containing the weather report somewhere, which would probably involve
a little less labor in order to produce your automatic wardrobe
advice.

James



On Thu, Feb 26, 2009 at 3:47 PM, Thomas Levine thomas.lev...@gmail.com wrote:
 I'm writing a program that will tell me whether I should wear a coat,
 so I'd like to be able to download daily weather forecasts and daily
 reports of recent past weather conditions.

 The NOAA has very promising tabular forecasts
 (http://forecast.weather.gov/MapClick.php?CityName=Ithacastate=NYsite=BGMtextField1=42.4422textField2=-76.5002e=0FcstType=digital),
 but I can't figure out how to import them.

 Someone must have needed to do this before. Suggestions?

 Thomas Levine!

 __
 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] plot with pairwise joined points

[William Simpson]

Thanks very much Rolf, Dimitris,  Greg!

Bill

On Thu, Feb 26, 2009 at 8:56 PM, Rolf Turner r.tur...@auckland.ac.nz wrote:

 On 27/02/2009, at 9:46 AM, William Simpson wrote:

 I would like to do as follows
 plot(a,b)
 points(c,d,pch=19)

 Now join with a line segment point a[1], b[1] to c[1], d[1]; a[2],
 b[2] to c[2], d[2] ... a[n], b[n] to c[n], d[n]

 All corresponding points from the two data sets are joined by line
 segments.

 ?segments

cheers,

Rolf Turner

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Re: [R] generalized linear mixed models with a beta distribution

[Jeff Evans]

Thanks for responding Doug. I'm sure SAS just hasn't gotten around to
releasing their code yet.

lme4 does have a leg up on GLIMMIX in other areas, though.
The latest SAS release (9.2) is now able to compute the Laplace
approximation of the likelihood, but it will only fit an overdispersion
parameter using pseudo-likelihoods which can't be used for model selection.
I'm not sure what lme4 is doing differently through the quasi-distributions
that allows this, but it's enormously useful.

Jeff

-Original Message-
From: dmba...@gmail.com [mailto:dmba...@gmail.com] On Behalf Of Douglas
Bates
Sent: Thursday, February 26, 2009 3:50 PM
To: Jeff Evans
Cc: r-help@r-project.org
Subject: Re: [R] generalized linear mixed models with a beta distribution

On Thu, Feb 26, 2009 at 12:04 PM, Jeff Evans evans...@msu.edu wrote:
 Has there been any follow up to this question? I have found myself
wondering
 the same thing: How then does SAS fit a beta distributed GLMM? It also
fits
 the negative binomial distribution.

When SAS decides to open-source their code we'll be able to find out.

 Both of these would be useful in glmer/lmer if they aren't 'illegal' as
 Brian suggested. Especially as SAS indicates a favorable delta BIC of over
 1000 when I fit the beta to my data (could be the beginning of a great
 song..) versus my original binomial fit.

Definitions of generalized linear mixed models are not entirely
straightforward, at least for me.  I'm making some progress but, as
always, it is slower than one would like it to be.

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

[R] Substituting in a variable file name in a Windows system command

[Elaine Jones]

I am running R version 2.8.1 on  Windows XP OS.

I generate and write a .csv file from my R script. Then the following
command works to upload it to a remote server using a windows batch file
that carries out the ftp (among other things).

   system(C:/upload_data/uploadq8.bat
C:/upload_data/out_2009-02-26.csv, wait=FALSE)

I want to set this up to run daily and create a unique filename each day it
runs. I write  the .csv file with a unique filename by

   fname - paste(out_,Sys.Date(),.csv,sep=)

   write.table(config_all5,file=fname,row.names=FALSE,
quote=FALSE,sep=,);

 I can build the string (including quotes) that is the first argument in
the system command:

   com - paste('C:/upload_data/uploadq8table.bat
C:/upload_data/',fname,'', sep=)
   com
  [1] \C:/upload_data/uploadq8table.bat
C:/upload_data/out_2009-02-26.csv\,

But when I substitute it into the system command I get an error:
   system(com, wait=FALSE)
  Warning in system(com, wait = FALSE) :
  C:/upload_data/uploadq8table.bat  C:/upload_data/out_2009-02-26.csv
not found
  

Any suggestions for how to resolve are appreciated!

 Elaine McGovern Jones 

 ISC Tape and DASD Storage Products
 Characterization and Failure Analysis Engineering
   Phone:  408  284 4853  Internal: 3-4853
   jon...@us.ibm.com




[[alternative HTML version deleted]]

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[R] User manual or tutorial for GAM package

[Neotropical bat risk assessments]

Hi all,

I would like to run the gam package, but can not quite seem to get 
the syntax correct for my data format as simple as it is.


The gamlss package has a user manual but other than the help with gam 
package I can not find a user manual per se.


I did install the gamair that has loads of sample data so perhaps 
trail and error with my data I can get something to work.


Anyone who can point me to a link for a download tutorial?

Climbing the steep learning curve and spending a hour or so per day 
is beginning to pay off with most areas of R


Tnx

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Re: [R] User manual or tutorial for GAM package

[David Winsemius]

On Feb 26, 2009, at 4:44 PM, Neotropical bat risk assessments wrote:


Hi all,

I would like to run the gam package, but can not quite seem to get  
the syntax correct for my data format as simple as it is.


The gamlss package has a user manual but other than the help with  
gam package I can not find a user manual per se.


Maybe you need to buy the book:

http://www.amazon.com/Generalized-Additive-Models-Introduction-Statistical/dp/1584884746

--
David Winsemius





I did install the gamair that has loads of sample data so perhaps  
trail and error with my data I can get something to work.


Anyone who can point me to a link for a download tutorial?

Climbing the steep learning curve and spending a hour or so per day  
is beginning to pay off with most areas of R


Tnx

__
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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-help@r-project.org mailing list
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Download daily weather data

[Scillieri, John]

Looks like you can sign up to get XML feed data from Weather.com

http://www.weather.com/services/xmloap.html

Hope it works out!

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of James Muller
Sent: Thursday, February 26, 2009 3:57 PM
To: r-help@r-project.org
Subject: Re: [R] Download daily weather data

Thomas,

Have a look at the source code for the webpage (ctrl-u in firefox,
don't know in internet explorer, etc.). That is what you'd have to
parse in order to get the forecast from this page. Typically when I
parse webpages such as this I use regular expressions to do so (and I
would never downplay the usefulness of regular expressions, but they
take a little getting used to). There are two parts to the task: find
patterns that allow you to pull out the datum/data you're after; and
then write a program to pull it/them out. Also, of course, download
the webpage (but that's no issue).

I bet you'd be able to find a comma separated value (CSV) file
containing the weather report somewhere, which would probably involve
a little less labor in order to produce your automatic wardrobe
advice.

James



On Thu, Feb 26, 2009 at 3:47 PM, Thomas Levine thomas.lev...@gmail.com wrote:
 I'm writing a program that will tell me whether I should wear a coat,
 so I'd like to be able to download daily weather forecasts and daily
 reports of recent past weather conditions.

 The NOAA has very promising tabular forecasts
 (http://forecast.weather.gov/MapClick.php?CityName=Ithacastate=NYsite=BGMtextField1=42.4422textField2=-76.5002e=0FcstType=digital),
 but I can't figure out how to import them.

 Someone must have needed to do this before. Suggestions?

 Thomas Levine!

 __
 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.
 This e-mail and any attachments are confidential, may contain legal, 
 professional or other privileged information, and are intended solely for 
 the addressee.  If you are not the intended recipient, do not use the 
 information in this e-mail in any way, delete this e-mail and notify the 
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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] When download.file fails...

[CHD850]

Hi everyone, 

In the situation that the remote file does not exist, download.file fails,
yet it still creates a file as provided for destfile argument. I tried to
delete this bad file but got the message that it is still being used by
other programs, which I assume is R. 

Does anyone know how to avoid this problem? Potential solutions I can think
of now is either checking the existence on remote server before using
download.file, or change download.file's behavior, or use other functions.
Any suggestions are appreciated. 

Stanley 
-- 
View this message in context: 
http://www.nabble.com/When-download.file-fails...-tp22234662p22234662.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] User manual or tutorial for GAM package

[David Winsemius]

On Feb 26, 2009, at 4:54 PM, David Winsemius wrote:



On Feb 26, 2009, at 4:44 PM, Neotropical bat risk assessments wrote:


Hi all,

I would like to run the gam package, but can not quite seem to get  
the syntax correct for my data format as simple as it is.


The gamlss package has a user manual but other than the help with  
gam package I can not find a user manual per se.


Maybe you need to buy the book:

http://www.amazon.com/Generalized-Additive-Models-Introduction-Statistical/dp/1584884746


OR

Statistical Models in S chapter 7 by Hastie
http://www.amazon.com/Statistical-Models-Chapman-Computer-Science/dp/0412052911


--
David Winsemius





I did install the gamair that has loads of sample data so perhaps  
trail and error with my data I can get something to work.


Anyone who can point me to a link for a download tutorial?

Climbing the steep learning curve and spending a hour or so per day  
is beginning to pay off with most areas of R


Tnx

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


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

Re: [R] Download daily weather data

[Wacek Kusnierczyk]

Scillieri, John wrote:
 Looks like you can sign up to get XML feed data from Weather.com

 http://www.weather.com/services/xmloap.html
   

... and use the excellent R package XML by Duncan Temple Lang to parse
the document and easily access the data with, e.g.., XPath rather than
regular expressions.

vQ


 Hope it works out!

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf Of James Muller
 Sent: Thursday, February 26, 2009 3:57 PM
 To: r-help@r-project.org
 Subject: Re: [R] Download daily weather data

 Thomas,

 Have a look at the source code for the webpage (ctrl-u in firefox,
 don't know in internet explorer, etc.). That is what you'd have to
 parse in order to get the forecast from this page. Typically when I
 parse webpages such as this I use regular expressions to do so (and I
 would never downplay the usefulness of regular expressions, but they
 take a little getting used to). There are two parts to the task: find
 patterns that allow you to pull out the datum/data you're after; and
 then write a program to pull it/them out. Also, of course, download
 the webpage (but that's no issue).

 I bet you'd be able to find a comma separated value (CSV) file
 containing the weather report somewhere, which would probably involve
 a little less labor in order to produce your automatic wardrobe
 advice.

 James



 On Thu, Feb 26, 2009 at 3:47 PM, Thomas Levine thomas.lev...@gmail.com 
 wrote:
   
 I'm writing a program that will tell me whether I should wear a coat,
 so I'd like to be able to download daily weather forecasts and daily
 reports of recent past weather conditions.

 The NOAA has very promising tabular forecasts
 (http://forecast.weather.gov/MapClick.php?CityName=Ithacastate=NYsite=BGMtextField1=42.4422textField2=-76.5002e=0FcstType=digital),
 but I can't figure out how to import them.

 Someone must have needed to do this before. Suggestions?

 Thomas Levine!

 __
 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.
   
 This e-mail and any attachments are confidential, may contain legal, 
 professional or other privileged information, and are intended solely for 
 the addressee.  If you are not the intended recipient, do not use the 
 information in this e-mail in any way, delete this e-mail and notify the 
 sender. CEG-IP1
 

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

Re: [R] Download daily weather data

[James Muller]

Yes, as a general thing go to regular expressions if you don't have an
existing library available to do the same thing (or you're lazy like
me:).

Jame

On Thu, Feb 26, 2009 at 5:16 PM, Wacek Kusnierczyk
waclaw.marcin.kusnierc...@idi.ntnu.no wrote:
 Scillieri, John wrote:
 Looks like you can sign up to get XML feed data from Weather.com

 http://www.weather.com/services/xmloap.html


 ... and use the excellent R package XML by Duncan Temple Lang to parse
 the document and easily access the data with, e.g.., XPath rather than
 regular expressions.

 vQ

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

Re: [R] Download daily weather data

[Wacek Kusnierczyk]

James Muller wrote:
 Yes, as a general thing go to regular expressions if you don't have an
 existing library available to do the same thing (or you're lazy like
 me:).

   

many things are simply *much* easier with xpath than with regexes, and
with the XML package you got it for free.

vQ


 Jame

 On Thu, Feb 26, 2009 at 5:16 PM, Wacek Kusnierczyk
 waclaw.marcin.kusnierc...@idi.ntnu.no wrote:
   
 Scillieri, John wrote:
 
 Looks like you can sign up to get XML feed data from Weather.com

 http://www.weather.com/services/xmloap.html

   
 ... and use the excellent R package XML by Duncan Temple Lang to parse
 the document and easily access the data with, e.g.., XPath rather than
 regular expressions.

 vQ
 

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

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

Re: [R] User manual or tutorial for GAM package

[Bert Gunter]

 
-- or perhaps even the section on Additive Models in the latest edition of
VR's MASS (still a useul book to have in one's library, IMHO, although,
like me, it's getting grayer)


Bert Gunter
Genentech Nonclinical Biostatistics
650-467-7374

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of David Winsemius
Sent: Thursday, February 26, 2009 2:05 PM
To: David Winsemius
Cc: r-help@r-project.org; Neotropical bat risk assessments
Subject: Re: [R] User manual or tutorial for GAM package


On Feb 26, 2009, at 4:54 PM, David Winsemius wrote:


 On Feb 26, 2009, at 4:44 PM, Neotropical bat risk assessments wrote:

 Hi all,

 I would like to run the gam package, but can not quite seem to get  
 the syntax correct for my data format as simple as it is.

 The gamlss package has a user manual but other than the help with  
 gam package I can not find a user manual per se.

 Maybe you need to buy the book:


http://www.amazon.com/Generalized-Additive-Models-Introduction-Statistical/d
p/1584884746

OR

Statistical Models in S chapter 7 by Hastie
http://www.amazon.com/Statistical-Models-Chapman-Computer-Science/dp/0412052
911

 -- 
 David Winsemius




 I did install the gamair that has loads of sample data so perhaps  
 trail and error with my data I can get something to work.

 Anyone who can point me to a link for a download tutorial?

 Climbing the steep learning curve and spending a hour or so per day  
 is beginning to pay off with most areas of R

 Tnx

 __
 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.

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Download daily weather data

[Barry Rowlingson]

2009/2/26 Thomas Levine thomas.lev...@gmail.com:
 I'm writing a program that will tell me whether I should wear a coat,
 so I'd like to be able to download daily weather forecasts and daily
 reports of recent past weather conditions.

 The NOAA has very promising tabular forecasts
 (http://forecast.weather.gov/MapClick.php?CityName=Ithacastate=NYsite=BGMtextField1=42.4422textField2=-76.5002e=0FcstType=digital),
 but I can't figure out how to import them.

 Someone must have needed to do this before. Suggestions?

 You could use my geonames package that uses the GeoNames query
service. There's a sample queries here:

http://geonames.r-forge.r-project.org/

 Easiest is probably to use GNfindNearByWeather:

  as.data.frame(GNfindNearByWeather(57,-2))
 clouds weatherCondition
1 broken clouds  n/a
 observation windDirection ICAO
1 EGPD 262120Z 25003KT 9000 -RA BKN018 06/05 Q1012 NOSIG   250 EGPD
  elevation countryCode   lng temperature dewPoint windSpeed humidity
165  GB -2.216667   6503   93
  stationNamedatetime  lat hectoPascAltimeter
1 Aberdeen / Dyce 2009-02-26 21:20:00 57.2   1012

 The package is on CRAN.

 There is of course an easier way to decide if you need to wear a
coat, and that is to look out the window :)

Barry

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[R] Inefficiency of SAS Programming

[Frank E Harrell Jr]

If anyone wants to see a prime example of how inefficient it is to 
program in SAS, take a look at the SAS programs provided by the US 
Agency for Healthcare Research and Quality for risk adjusting and 
reporting for hospital outcomes at 
http://www.qualityindicators.ahrq.gov/software.htm .  The PSSASP3.SAS 
program is a prime example.  Look at how you do a vector product in the 
SAS macro language to evaluate predictions from a logistic regression 
model.  I estimate that using R would easily cut the programming time of 
this set of programs by a factor of 4.


Frank
--
Frank E Harrell Jr   Professor and Chair   School of Medicine
 Department of Biostatistics   Vanderbilt University

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Re: [R] left truncated data survival analysis package

[Göran Broström]

On Thu, Feb 26, 2009 at 7:02 AM,  phguard...@aol.com wrote:

 Hello,
 I d like to run a survival analysis with left truncated data. Could you
 recommend me a package to do this please ?

The 'eha' package if you want parametric or discrete time models.

Göran

 Thanks
 Philippe Guardiola

 
 Reçevez AOL Mail sur votre téléphone. Vos e-mails accessibles à tout moment!
 Créez un e-mail gratuit aujourd’hui.

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-- 
Göran Broström

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Re: [R] Download daily weather data

[hadley wickham]

See also http://umbrellatoday.com/
Hadley

On Thu, Feb 26, 2009 at 2:47 PM, Thomas Levine thomas.lev...@gmail.com wrote:
 I'm writing a program that will tell me whether I should wear a coat,
 so I'd like to be able to download daily weather forecasts and daily
 reports of recent past weather conditions.

 The NOAA has very promising tabular forecasts
 (http://forecast.weather.gov/MapClick.php?CityName=Ithacastate=NYsite=BGMtextField1=42.4422textField2=-76.5002e=0FcstType=digital),
 but I can't figure out how to import them.

 Someone must have needed to do this before. Suggestions?

 Thomas Levine!

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

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[R] User manual or tutorial for GAM package

[Neotropical bat risk assessments]

Hi all,

Tnx for the number of suggestion to buy one of the books.
Unfortunately being a non profit conservation project now w/o funding 
since all Central American projects were shut down 2 years ago due to 
lack of donor funding I have nothing now at all for books or even my 
support, to cover expenses.


I am trying to wrap up several risk assessments w/o funding.


Tnx again everyone is very helpful on the list as are those on the ggplot list.

Cheers from the jungles of Belize,

Bruce

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[R] Something similar to data.frame, but allows variables with unique lengths?

[Jason Rupert]

I have a tightly coupled collection of variables with different lengths and 
types (some characters and others numerics).  I looked at the documentation for 
data.frame, but indicates that it expects all variables to have the same 
length, i.e. number of elements.  
 
I was hoping the See Also documentation for data.frame would have helped, but 
it did not have any good leads.  

Here is a list of data I would like to store together:
(FileName (character string), 
 HighValue (numeric),
 LowValue (numeric), 
 Units (character), 
 OthersValues (array of numerics), 
 Descriptor (character string))
 
By any chance is there a generic object type feature within R to store such 
data together in one variable?  
 
Thank you for any details and leads you can provide. 
 
 


  
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[R] complex data summary

[James J. Roper]

Dear All,

I am interested in using R to summarize data so that I can THEN do some
additional analyses using behavioral data.

I study birds, mostly, and often behavioral data that includes timing.  So,
for example, each line may be the occurrence of some behavior, and it would
include identifiers for species, individual, date, time of observation and
the various behavioral observations that may be important.  Thus, if
observing one individual, the difference in the time of one line to the
subsequent line could be the time it took to carry out a behavior.  Or,
perhaps several lines include all the behaviors of a bout of behaviors.
For example, I have a project on hummingbirds, in which an individual can be
observed for a time interval, during which it may do several things.  So,
each thing becomes a line to identify what it was.  Then, I would like to
count behaviors from the start to the end of his bout.  Basically, I want
to calculate a few things - the difference between the times of the
behaviors of one line and the subsequent or some number of subsequent lines.
I want to count the number of other things in a time interval, and so on.

In JMP, to calculate the interval from one line to the next, I would
generate a new variable, and that basically subtracts the time in line x
from the time in line x+1.  I can't figure out how to do that in R.

Next, in JMP (or SAS), I would do a summary, in which I based it on date,
species, individual, etc., and which took a Maximum and Minimum for time,
the sum of say variable X, the Maximum of variable Y and so on.

Example:

Date Begin   EndSpecies  Individual   Feeding
Interaction  Interactsp
2008-09-11   11:15:2111:15:58   CHAU 1
3
2008-09-11   11:16:0111:16:12   CHAU 1
ChaseCHAU
2008-09-11   11:16:1711:16:28   CHAU 15
2008-09-11   11:16:3211:16:49   CHAU 1
Flee THGL
2008-09-11   11:16:5311:17:15   THGL 18

Clearly, to make another variably by just subtracting one variable from
another, like time begin time end, is no problem.  Also, since the species
changed, the individual number can remain. What I would like to do, for
example, is count the number of feedings, interactions, and so on in an
interval, as well as the number of species, and calculate the total time the
individual was at the food source.

Well, all that is very simple in JMP and I just can't figure out how to do
it in R.  In fact, just figuring out how to make time and data TIME and DATE
was a bit complex, but I have managed to do that. I hope I have explained my
problem well enough, because all the other problems I have hinge upon
summarizing my data in a way that the summary can then be used in analysis.
If I could figure out this one part, I would probably switch to R
completely!

If you have any helpful suggestions, I am sure that there are other
behavioral biologists out there that would be able to benefit from this kind
of information.  I know my future students will!

Sincerely,

Jim


-- 
James J. Roper
Smithsonian Tropical Research Institute
Bocas del Toro Marine Research Station
MRC 0580-03
Unit 9100, Box 0948
DPO AA 34002-9998

Skype-in (USA):+1 706 5501064
Skype-in (Brazil): 41 39415715

E-mail - personal: jjro...@gmail.com
E-mail - consulting: arsart...@gmail.com

9 21.122' N, and 82 15.390' W
In Google Earth, copy and paste - 9 21.122' N, 82 15.390' W

Ecologia e Conservação na UFPR
http://www.bio.ufpr.br/ecologia/

Personal Pages
http://jjroper.googlespages.com
http://arsartium.googlepages.com

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Re: [R] Substituting in a variable file name in a Windows system command

[Duncan Murdoch]

Elaine Jones wrote:

I am running R version 2.8.1 on  Windows XP OS.

I generate and write a .csv file from my R script. Then the following
command works to upload it to a remote server using a windows batch file
that carries out the ftp (among other things).

   system(C:/upload_data/uploadq8.bat
C:/upload_data/out_2009-02-26.csv, wait=FALSE)

I want to set this up to run daily and create a unique filename each day it
runs. I write  the .csv file with a unique filename by

   fname - paste(out_,Sys.Date(),.csv,sep=)

   write.table(config_all5,file=fname,row.names=FALSE,
quote=FALSE,sep=,);

 I can build the string (including quotes) that is the first argument in
the system command:
  


Don't include the quotes.

   com - paste('C:/upload_data/uploadq8table.bat
C:/upload_data/',fname,'', sep=)
  


Should be

com - paste('C:/upload_data/uploadq8table.bat
C:/upload_data/',fname, sep=)

Duncan Murdoch


   com
  [1] \C:/upload_data/uploadq8table.bat
C:/upload_data/out_2009-02-26.csv\,

But when I substitute it into the system command I get an error:
   system(com, wait=FALSE)
  Warning in system(com, wait = FALSE) :
  C:/upload_data/uploadq8table.bat  C:/upload_data/out_2009-02-26.csv
not found
  

Any suggestions for how to resolve are appreciated!

 Elaine McGovern Jones 

 ISC Tape and DASD Storage Products
 Characterization and Failure Analysis Engineering
   Phone:  408  284 4853  Internal: 3-4853
   jon...@us.ibm.com




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