Re: [R] help get character output R in Delphi

[Dieter Menne]

mahendra mahfood mahendra025 at yahoo.com writes:

 i cant get the character output of R in delphi.
   i have got the graphic of R in delphi. i wanna get the character output in 
R for my delphi application. anybody knows???
   the example source please?

I have written a little wrapper for Delphi and R (see http:www.menne-
biomed.de/download/download.html). I never tried to capture the full output, 
because it's better to pick the items you want selectively. If you really want 
the text, I suggest redirecting selected output to a file (?sink()) and reading 
it in from Delphi.

Dieter

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Re: [R] Systematic treatment of missing values

[David Soloveichik]

Thank you very much for your prompt reply and for adding the comments  
to the help pages for match and ==.  I think the source of my  
confusion was that by looking at the current documentation (v 2.3.0)  
I did not realize that matching is different from equality testing.   
(Obviously in the case of using regular expressions, etc, it is  
different, but I thought that when using plain match and %in%,  
matching would be determined by ==.)

Also I did not mean for my first comment to sound like a criticism of  
R for treating NAs inconsistently.  Nonetheless I am still curious  
why the particular choice was made that match (and therefore %in%)  
acts differently from == with respect to NA's and NaN's (with the  
default and the only implemented value of the incomparables  
parameter)?

Thank you,
David


On May 28, 2006, at 1:10 AM, Prof Brian Ripley wrote:

 You start with very general comments, but only use one specific  
 function, match (see ?%in%, a help page entitled `value matching').

 Matching and equality are treated differently.  By definition, NA  
 matches NA and nothing else, and NaN matches NaN and nothing else.   
 In comparisons, these values are not comparable.

 As you will have seen from the help page, match() has the expansion  
 capacity for declaring values non-comparable.  That has not been  
 implemented for a decade and no one has supplied code to implement  
 it, so it seems no want has much need of it.

 I have added notes to the help pages for match and == to say  
 explicitly what matches and what is comparable.  If the *Draft* R  
 Language Definition were ever to be finished it would have such  
 details: it already has a useful commentary.

 On Sat, 27 May 2006, David Soloveichik wrote:

 I am wondering whether there is a well-accepted approach to handling
 missing values (NA's) in a programming language such as R.  For
 example, most functions seem to propagate NA to the output when the
 value of the missing entry could have mattered.  In other words, most
 functions are not willing to take a stand on what the missing value
 was.  However, some functions don't seem to do this.  For example,

  c(1,2,3,NA) %in% c(2,3)
 [1] FALSE  TRUE  TRUE FALSE

 rather than: FALSE  TRUE  TRUE NA


 Also, what is the logic of the following:
  c(1,2,3,NA) %in% c(2,3,NA)
 [1] FALSE  TRUE  TRUE  TRUE

 Why is the last output value TRUE?  Why does R claim that the NA on
 the left hand side of %in% is the same as the NA on the right hand
 side of %in%?

 It does not: it reports that it *matches*.  Please do read the help  
 page bwofre posting, as the posting guide asked you to.

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

 -- 
 Brian D. Ripley,  [EMAIL PROTECTED]
 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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[R] Query: lme output

[Pryseley Assam]

Dear R-Users
   
  I have a problem accessing some values in the output from the summary of an 
lme fit. 
   
  I fit the model below:
   
  ggg - lme (ST~ -1 + as.factor(endp):Z.sas + as.factor(endp), data=dat4a,
  random=~-1 + as.factor(endp) + as.factor(endp):Z.sas|as.factor(trials),
  correlation = corSymm(form=~1|as.factor(trials)/as.factor(id)), 
weights=varIdent(form=~1|endp))
   
  hh - summary(ggg)
  hh
   
  Below is the following part of the output of interest:
   
  Correlation Structure: General
 Formula: ~1 | as.factor(trials)/as.factor(id) 
 Parameter estimate(s):
 Correlation: 
  1
2 0.785
Variance function:
 Structure: Different standard deviations per stratum
 Formula: ~1 | endp 
 Parameter estimates:
   -1 1 
1.000 0.9692405 
   
  I wish to access the value of the correlation (0.785) and the vector of the 
variance function estimates (1,0.969). I know these can be done throught the 
intervals function, but sometimes when the estimated Hessian matrix is not 
positive definite or something like that (i am not quite sure), the intervals 
function delivers an error message. 
   
  Thus, i will like to ask if there is another way to access these values. I 
tried using the following code:
   
  hh$modelStruct$corStruct[1]
  hh$modelStruct$varStruct[1]
   
  Rather the output was:
   
   hh$modelStruct$corStruct[1]
[1] -1.308580
 hh$modelStruct$varStruct[1]
[1] -0.03124255
   
  I presume there is a way to calculate the correlation and variance function 
coefficients using these values. 
   
  Could someone tell me how to access those values (without using the intervals 
function) or better still how to calculate the values from the last output 
values. 
   
  Kind regards
  Pryseley
   
   


-

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Re: [R] multiple comparisons of time series data (Stephan Moratti)

[Stephan Moratti]

To account for the strong serial correlation you 
could try the lme() function of the nlme package. 
There you can apply different covariance 
structures in your linear model such as a 
first-order autoregressive covariance structure (AR1).

example:

model.fit - lme(response ~ condition * time, 
data = time.series.data, random=~1|case, correlation = corCAR1());

This model uses an autoregressive process for 
continous data. The random expression defines the 
intercept for each case (or observation, subject) 
as a random factor. Condition and time would be 
fixed factors in this case. See also help(lme) and help(corClasses).

Hopes that helps,

Stephan




Stephan Moratti, PhD
Centro de Magnetoencefalografía Dr. Perez Modrego
Faculdad de Medicina
Universidad Complutense de Madrid
Pabellón 8
Avda. Complutense, s/n
28040 Madrid
Spain
email: [EMAIL PROTECTED]
Tel.:   +34 91 394 2292
Fax.:   +34 91 394 2294  
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Re: [R] Query: lme output

[Andrew Robinson]

You can try dissecting the output of 

VarCorr(ggg)

but, once again, I can't guarantee that it's what you want because we
can't fit your model, and you have not sent us a simple, reproducible
example.  

Andrew


On Tue, May 30, 2006 at 03:37:19AM -0700, Pryseley Assam wrote:
 Dear R-Users

   I have a problem accessing some values in the output from the summary of an 
 lme fit. 

   I fit the model below:

   ggg - lme (ST~ -1 + as.factor(endp):Z.sas + as.factor(endp), data=dat4a,
   random=~-1 + as.factor(endp) + as.factor(endp):Z.sas|as.factor(trials),
   correlation = corSymm(form=~1|as.factor(trials)/as.factor(id)), 
 weights=varIdent(form=~1|endp))

   hh - summary(ggg)
   hh

   Below is the following part of the output of interest:

   Correlation Structure: General
  Formula: ~1 | as.factor(trials)/as.factor(id) 
  Parameter estimate(s):
  Correlation: 
   1
 2 0.785
 Variance function:
  Structure: Different standard deviations per stratum
  Formula: ~1 | endp 
  Parameter estimates:
-1 1 
 1.000 0.9692405 

   I wish to access the value of the correlation (0.785) and the vector of the 
 variance function estimates (1,0.969). I know these can be done throught the 
 intervals function, but sometimes when the estimated Hessian matrix is not 
 positive definite or something like that (i am not quite sure), the intervals 
 function delivers an error message. 

   Thus, i will like to ask if there is another way to access these values. I 
 tried using the following code:

   hh$modelStruct$corStruct[1]
   hh$modelStruct$varStruct[1]

   Rather the output was:

hh$modelStruct$corStruct[1]
 [1] -1.308580
  hh$modelStruct$varStruct[1]
 [1] -0.03124255

   I presume there is a way to calculate the correlation and variance function 
 coefficients using these values. 

   Could someone tell me how to access those values (without using the 
 intervals function) or better still how to calculate the values from the last 
 output values. 

   Kind regards
   Pryseley


 
   
 -
 
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-- 
Andrew Robinson  
Department of Mathematics and StatisticsTel: +61-3-8344-9763
University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599
Email: [EMAIL PROTECTED] http://www.ms.unimelb.edu.au

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[R] Automate concatenation?

[Robert Lundqvist]

I have this typical problem of joining a number of vectors with similar
names - a1, a2,..., a10 - which should be concatenated into one. Using
c(a1,a2,a3,a4,a5,a6,a,a8,a9,a10) naturally works, but I would like to do
it with less manual input. My attempts to use paste() gives a vector of
the vector names, see below. The question is how to do the the
concatenation? Any suggestions?

paste(a,1:10,sep=)

Robert

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[R] Time rather than dates?

[Robert Lundqvist]

Using strptime() and other functions for dates has been very helpful with
the kind of data I often work with. However, I haven't found out how time
as such should be specified. All my attempts result in time *and* date:

treatment_time-c(01:02:03,02:03:04) # hours:minutes:seconds
time.2-strptime(treatment_time,format=%H:%M:%S)

time.2
[1] 1900-01-01 01:02:03 1900-01-01 02:03:04

Why the 1900-...? I had hoped for some easy conversion from time to
numeric data and possibly back. Assistance would be appreciated.

Robert

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Re: [R] Time rather than dates?

[Prof Brian Ripley]

On Tue, 30 May 2006, Robert Lundqvist wrote:

 Using strptime() and other functions for dates has been very helpful with
 the kind of data I often work with. However, I haven't found out how time
 as such should be specified. All my attempts result in time *and* date:

 treatment_time-c(01:02:03,02:03:04) # hours:minutes:seconds
 time.2-strptime(treatment_time,format=%H:%M:%S)

 time.2
 [1] 1900-01-01 01:02:03 1900-01-01 02:03:04

 Why the 1900-...? I had hoped for some easy conversion from time to
 numeric data and possibly back. Assistance would be appreciated.

You asked to print a datetime object of class POSIXlt, and unspecified 
fields are set to their earliest values.  But printing is only part of the 
story.

What do you actually want from this?

3600*time.2$hour + 60*time.2$min + time.2$sec

gives you the number of seconds since midnight, for example.  See
?DateTimeClasses

-- 
Brian D. Ripley,  [EMAIL PROTECTED]
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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Re: [R] Automate concatenation?

[Dimitris Rizopoulos]

you need get(), look also at FAQ 7.21

a1 - 1:3
a2 - 4:5
a3 - 6:10
a4 - 11:20
a5 - 21:25
#
lapply(paste(a, 1:5, sep = ), get)

I hope it helps.

Best,
Dimitris


Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
 http://www.student.kuleuven.be/~m0390867/dimitris.htm


- Original Message - 
From: Robert Lundqvist [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Tuesday, May 30, 2006 2:07 PM
Subject: [R] Automate concatenation?


I have this typical problem of joining a number of vectors with 
similar
 names - a1, a2,..., a10 - which should be concatenated into one. 
 Using
 c(a1,a2,a3,a4,a5,a6,a,a8,a9,a10) naturally works, but I would like 
 to do
 it with less manual input. My attempts to use paste() gives a vector 
 of
 the vector names, see below. The question is how to do the the
 concatenation? Any suggestions?

 paste(a,1:10,sep=)

 Robert

 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide! 
 http://www.R-project.org/posting-guide.html
 


Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm

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Re: [R] Time rather than dates?

[Gabor Grothendieck]

Try using the times class in the chron package.

 library(chron)
 times(c(01:02:03,02:03:04))
[1] 01:02:03 02:03:04

On 5/30/06, Robert Lundqvist [EMAIL PROTECTED] wrote:
 Using strptime() and other functions for dates has been very helpful with
 the kind of data I often work with. However, I haven't found out how time
 as such should be specified. All my attempts result in time *and* date:

 treatment_time-c(01:02:03,02:03:04) # hours:minutes:seconds
 time.2-strptime(treatment_time,format=%H:%M:%S)

 time.2
 [1] 1900-01-01 01:02:03 1900-01-01 02:03:04

 Why the 1900-...? I had hoped for some easy conversion from time to
 numeric data and possibly back. Assistance would be appreciated.

 Robert

 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html


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Re: [R] Time rather than dates?

[Petr Pikal]

Hi

Which version of R do you use?
Version 2.3.1 beta (2006-05-23 r38179)

 treatment_time-c(01:02:03,02:03:04)
 strptime(treatment_time,format=%H:%M:%S)
[1] 2006-05-30 01:02:03 2006-05-30 02:03:04
 ttt-strptime(treatment_time,format=%H:%M:%S)

For changing format of POSIX variables use format
 format(ttt, %H:%M)
[1] 01:02 02:03

Maybe also consult locale setting.

HTH
Petr



On 30 May 2006 at 14:16, Robert Lundqvist wrote:

To: r-help@stat.math.ethz.ch
From:   Robert Lundqvist [EMAIL PROTECTED]
Date sent:  Tue, 30 May 2006 14:16:29 +0200
Subject:[R] Time rather than dates?
Send reply to:  [EMAIL PROTECTED]
mailto:[EMAIL PROTECTED]
mailto:[EMAIL PROTECTED]

 Using strptime() and other functions for dates has been very helpful
 with the kind of data I often work with. However, I haven't found out
 how time as such should be specified. All my attempts result in time
 *and* date:
 
 treatment_time-c(01:02:03,02:03:04) # hours:minutes:seconds
 time.2-strptime(treatment_time,format=%H:%M:%S)
 
 time.2
 [1] 1900-01-01 01:02:03 1900-01-01 02:03:04
 
 Why the 1900-...? I had hoped for some easy conversion from time to
 numeric data and possibly back. Assistance would be appreciated.
 
 Robert
 
 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide!
 http://www.R-project.org/posting-guide.html

Petr Pikal
[EMAIL PROTECTED]

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Re: [R] Automate concatenation?

[Peter Dalgaard]

Robert Lundqvist [EMAIL PROTECTED] writes:

 I have this typical problem of joining a number of vectors with similar
 names - a1, a2,..., a10 - which should be concatenated into one. Using
 c(a1,a2,a3,a4,a5,a6,a,a8,a9,a10) naturally works, but I would like to do
 it with less manual input. 

And less error-prone (where did the 7 go?)...

 My attempts to use paste() gives a vector of
 the vector names, see below. The question is how to do the the
 concatenation? Any suggestions?
 
 paste(a,1:10,sep=)

I think this should work:

unlist(lapply( paste(a,1:10,sep=), get))

(and of course, the usual sermon applies: You're likely better off
using a list of vectors rather than vectors with similar names.)

-- 
   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
~~ - ([EMAIL PROTECTED])  FAX: (+45) 35327907

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[R] Add text/numbers to x axis?

[Robert Lundqvist]

Is there any simple way to add text to the x axis in R? I have tried to
add simple characters in R without any greater sucess. As an example of
what I want to do is the added C in the following dumb plot (produced
with MacAnova using the option dumb:T):

++--+---+--+---+--+---+--+---++
   1+  : *+
|  :   *  *   |
|  : *  * |
|  :  *  .|
 0.8+  :   * .+
|  : .|
|  :*.|
|  : *   .|
 0.6+  : .+
|  :  *  .|
|  * .|
|  : .|
|   *  : .|
 0.4+  : .+
|* : .|
| *: .|
|  : .|
 0.2+  *   : .+
|   *  : .|
| *  * : .|
|*  *  *   : .|
   0++--+---+--+---+--+---+--.---+---++
-2-1.5 -1-0.5  0 0.5  1  C  1.5  2


My attempts to use locator() and text() in R's plot() has not been working
as good as expected. Never really knows where the added symbols end up...

BTW, anyone who knows how such dumb plots could be achieved in R? This
option is avaliable in S, so a port to R shouldn't be impossible, should
it? Don't know how to do it myself however...

Robert

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Re: [R] Add text/numbers to x axis?

[Petr Pikal]

Hi

maybe something like

 plot(1:10,1:10, axes=F)
 axis(1, at=1:10, labels=c((1:7)/10, c, 9:10))
 axis(2, at=1:10, labels=c(letters[1:5], 6:10))
 box()

HTH
Petr


On 30 May 2006 at 14:45, Robert Lundqvist wrote:

To: r-help@stat.math.ethz.ch
From:   Robert Lundqvist [EMAIL PROTECTED]
Date sent:  Tue, 30 May 2006 14:45:41 +0200
Subject:[R] Add text/numbers to x axis?
Send reply to:  [EMAIL PROTECTED]
mailto:[EMAIL PROTECTED]
mailto:[EMAIL PROTECTED]

 Is there any simple way to add text to the x axis in R? I have tried
 to add simple characters in R without any greater sucess. As an
 example of what I want to do is the added C in the following dumb
 plot (produced with MacAnova using the option dumb:T):
 
 ++--+---+--+---+--+---+--+---++
1+  : *+
 |  :   *  *   | | 
 : *  * | |
  :  *  .|
  0.8+  :   * .+
 |  : .| | 
 :*.| |
  : *   .|
  0.6+  : .+
 |  :  *  .| | 
 * .| |
  : .| |   
*  : .|
  0.4+  : .+
 |* : .| | 
*: .| |
  : .|
  0.2+  *   : .+
 |   *  : .| | 
*  * : .| |*  *
  *   : .|
0++--+---+--+---+--+---+--.---+---++
 -2-1.5 -1-0.5  0 0.5  1  C  1.5  2
 
 
 My attempts to use locator() and text() in R's plot() has not been
 working as good as expected. Never really knows where the added
 symbols end up...
 
 BTW, anyone who knows how such dumb plots could be achieved in R?
 This option is avaliable in S, so a port to R shouldn't be impossible,
 should it? Don't know how to do it myself however...
 
 Robert
 
 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide!
 http://www.R-project.org/posting-guide.html

Petr Pikal
[EMAIL PROTECTED]

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Re: [R] Time rather than dates?

[Prof Brian Ripley]

On Tue, 30 May 2006, Petr Pikal wrote:

 Which version of R do you use?
 Version 2.3.1 beta (2006-05-23 r38179)

 treatment_time-c(01:02:03,02:03:04)
 strptime(treatment_time,format=%H:%M:%S)
 [1] 2006-05-30 01:02:03 2006-05-30 02:03:04

From ?strptime

  If the date string does not specify the date completely, the
  returned answer may be system-specific.  The most common behaviour
  is to assume that unspecified seconds, minutes or hours are zero,
  and a missing year, month or day is the current one.

which explains the different answer.  Neither of you told us your OS.

 ttt-strptime(treatment_time,format=%H:%M:%S)

 For changing format of POSIX variables use format
 format(ttt, %H:%M)
 [1] 01:02 02:03

 Maybe also consult locale setting.

 HTH
 Petr



 On 30 May 2006 at 14:16, Robert Lundqvist wrote:

 To:   r-help@stat.math.ethz.ch
 From: Robert Lundqvist [EMAIL PROTECTED]
 Date sent:Tue, 30 May 2006 14:16:29 +0200
 Subject:  [R] Time rather than dates?
 Send reply to:[EMAIL PROTECTED]
   mailto:[EMAIL PROTECTED]
   mailto:[EMAIL PROTECTED]

 Using strptime() and other functions for dates has been very helpful
 with the kind of data I often work with. However, I haven't found out
 how time as such should be specified. All my attempts result in time
 *and* date:

 treatment_time-c(01:02:03,02:03:04) # hours:minutes:seconds
 time.2-strptime(treatment_time,format=%H:%M:%S)

 time.2
 [1] 1900-01-01 01:02:03 1900-01-01 02:03:04

 Why the 1900-...? I had hoped for some easy conversion from time to
 numeric data and possibly back. Assistance would be appreciated.

 Robert

-- 
Brian D. Ripley,  [EMAIL PROTECTED]
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

__
R-help@stat.math.ethz.ch mailing list
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[R] boosting - second posting

[Kuhn, Max]

The family arg appears to be the problem. Either bernoulli or adaboost
are appropriate for classification problems.

Max

 Perhaps by following the Posting Guide you're likely to get more
helpful
 responses.  You have not shown an example that others can reproduce,
not
 given version information for R or gbm.  The output you showed does
not use
 type=response, either.
  
 Andy
 
   _  
 
 From: r-help-bounces at stat.math.ethz.ch on behalf of stephenc
 Sent: Sat 5/27/2006 4:02 PM
 To: 'R Help'
 Subject: [R] boosting - second posting [Broadcast]
 
 
 
 Hi 
   
 I am using boosting for a classification and prediction problem. 
   
 For some reason it is giving me an outcome that doesn't fall between 0

 and 1 for the predictions.  I have tried type=response but it made
no 
 difference. 
   
 Can anyone see what I am doing wrong? 
   
 Screen output shown below: 
   
   
  boost.model - gbm(as.factor(train$simNuance) ~ ., # formula

 +  data=train,   # dataset 
 +   # +1: monotone increase, 
 +   #  0: no monotone restrictions

 +  distribution=gaussian, # bernoulli, adaboost,
gaussian, 
 +   # poisson, and coxph available

 +  n.trees=3000,# number of trees 
 +  shrinkage=0.005, # shrinkage or learning rate, 
 +   # 0.001 to 0.1 usually work 
 +  interaction.depth=3, # 1: additive model, 2:
two-way 
 interactions, etc. 
 +  bag.fraction = 0.5,  # subsampling fraction, 0.5 is

 probably best 
 +  train.fraction = 0.5,# fraction of data for
training, 
 +   # first train.fraction*N used 
 for training 
 +  n.minobsinnode = 10, # minimum total weight needed
in 
 each node 
 +  cv.folds = 5,# do 5-fold cross-validation 
 +  keep.data=TRUE,  # keep a copy of the dataset 
 with the object 
 +  verbose=FALSE)# print out progress 
  
  best.iter = gbm.perf(boost.model,method=cv) 
  pred = predict.gbm(boost.model, test, best.iter) 
  summary(pred) 
Min. 1st Qu.  MedianMean 3rd Qu.Max. 
  0.4772  1.5140  1.6760  1.5100  1.7190  1.9420
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Re: [R] Add text/numbers to x axis?

[Peter Dalgaard]

Robert Lundqvist [EMAIL PROTECTED] writes:

 Is there any simple way to add text to the x axis in R? I have tried to
 add simple characters in R without any greater sucess. As an example of
 what I want to do is the added C in the following dumb plot (produced
 with MacAnova using the option dumb:T):
 
 ++--+---+--+---+--+---+--+---++
1+  : *+
 |  :   *  *   |
 |  : *  * |
 |  :  *  .|
  0.8+  :   * .+
 |  : .|
 |  :*.|
 |  : *   .|
  0.6+  : .+
 |  :  *  .|
 |  * .|
 |  : .|
 |   *  : .|
  0.4+  : .+
 |* : .|
 | *: .|
 |  : .|
  0.2+  *   : .+
 |   *  : .|
 | *  * : .|
 |*  *  *   : .|
0++--+---+--+---+--+---+--.---+---++
 -2-1.5 -1-0.5  0 0.5  1  C  1.5  2
 
 
 My attempts to use locator() and text() in R's plot() has not been working
 as good as expected. Never really knows where the added symbols end up...

mtext() is your friend.
 
 BTW, anyone who knows how such dumb plots could be achieved in R? This
 option is avaliable in S, so a port to R shouldn't be impossible, should
 it? Don't know how to do it myself however...

Presumably you just need someone to write a device driver for it.
The structure of those is somewhat different from S, so it's not a
straightforward port (even if the S code was publically available).
As I recall it (fortunately, it was decades ago), some cosmetic issues
become tricky due to the high granularity of such line printer plots;
things like alignment of strings and continuation of polylines.
However if someone wants to read up on Bresenham's line drawing
algorithm and all that...

-- 
   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
~~ - ([EMAIL PROTECTED])  FAX: (+45) 35327907

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Re: [R] Add text/numbers to x axis?

[Prof Brian Ripley]

On Tue, 30 May 2006, Robert Lundqvist wrote:

 Is there any simple way to add text to the x axis in R? I have tried to
 add simple characters in R without any greater sucess. As an example of
 what I want to do is the added C in the following dumb plot (produced
 with MacAnova using the option dumb:T):

++--+---+--+---+--+---+--+---++
   1+  : *+
|  :   *  *   |
|  : *  * |
|  :  *  .|
 0.8+  :   * .+
|  : .|
|  :*.|
|  : *   .|
 0.6+  : .+
|  :  *  .|
|  * .|
|  : .|
|   *  : .|
 0.4+  : .+
|* : .|
| *: .|
|  : .|
 0.2+  *   : .+
|   *  : .|
| *  * : .|
|*  *  *   : .|
   0++--+---+--+---+--+---+--.---+---++
-2-1.5 -1-0.5  0 0.5  1  C  1.5  2


 My attempts to use locator() and text() in R's plot() has not been working
 as good as expected. Never really knows where the added symbols end up...

Well, they end up where you request them.  In that particular case I would
use axis() to add a label that matched the tick marks, but mtext() is 
often useful.

 BTW, anyone who knows how such dumb plots could be achieved in R? This
 option is avaliable in S, so a port to R shouldn't be impossible, should
 it? Don't know how to do it myself however...

There is the little matter of having access to the S source code and 
permission to make use of it, as well as knowing how the S graphics device 
model works (which AFAIK has never been publicly documented but it 
believed to be somewhat different from R).

The devil is in the details: notice how the y axis on the plot shown is 
uneven, with a stretch in the middle?  Should such things be allowed (I 
think not)?

-- 
Brian D. Ripley,  [EMAIL PROTECTED]
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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[R] correlating values

[Ahamarshan jn]

  I did a correlation for the values 

-   EX4577  EX4599  EX4566EX4522
WL917   2.53528 0.79077 0.21499  -0.01084
WP429S  -0.192723715
WP819   -1.016997552
WP977   1.378674-0.070710.6250890.4728363
WI205S  -0.24443-1.789526   0.648923-0.775867


by using
round(cor(t(person.data),use=pairwise.complete.obs))

i got the result as

- WL917 WP429S  WP819   WP977   WI205S
WL917   1   NA  NA  0.344   -0.11424
WP429S  NA  NA  NA  NA   NA
WP819   NA  NA  NA  NA   NA
WP977   0.34461 NA  NA  1   .23294
WI205S  -0.11424NA  NA  0.23294 1

i notice that 
for correlation between WP429S x wp429S the value is
given as NA where as it should be 1
same is the case with WP819 x WP819 

Can someone please help me reason out why this is
happening? and any corrective measure that I need to
take so that I can get the true value of 1 for them.

Thank you
- ash#

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Re: [R] boosting - second posting

[Weiwei Shi]

I remember if you use distribution=bernoulli, then you don't have to
as.factor(your_response_variable) either.

Weiwei

On 5/30/06, Kuhn, Max [EMAIL PROTECTED] wrote:

 The family arg appears to be the problem. Either bernoulli or adaboost
 are appropriate for classification problems.

 Max

  Perhaps by following the Posting Guide you're likely to get more
 helpful
  responses.  You have not shown an example that others can reproduce,
 not
  given version information for R or gbm.  The output you showed does
 not use
  type=response, either.
 
  Andy
 
_
 
  From: r-help-bounces at stat.math.ethz.ch on behalf of stephenc
  Sent: Sat 5/27/2006 4:02 PM
  To: 'R Help'
  Subject: [R] boosting - second posting [Broadcast]
 
 
 
  Hi
 
  I am using boosting for a classification and prediction problem.
 
  For some reason it is giving me an outcome that doesn't fall between 0

  and 1 for the predictions.  I have tried type=response but it made
 no
  difference.
 
  Can anyone see what I am doing wrong?
 
  Screen output shown below:
 
 
   boost.model - gbm(as.factor(train$simNuance) ~ ., # formula

  +  data=train,   # dataset
  +   # +1: monotone increase,
  +   #  0: no monotone restrictions

  +  distribution=gaussian, # bernoulli, adaboost,
 gaussian,
  +   # poisson, and coxph available

  +  n.trees=3000,# number of trees
  +  shrinkage=0.005, # shrinkage or learning rate,
  +   # 0.001 to 0.1 usually work
  +  interaction.depth=3, # 1: additive model, 2:
 two-way
  interactions, etc.
  +  bag.fraction = 0.5,  # subsampling fraction, 0.5 is

  probably best
  +  train.fraction = 0.5,# fraction of data for
 training,
  +   # first train.fraction*N used
  for training
  +  n.minobsinnode = 10, # minimum total weight needed
 in
  each node
  +  cv.folds = 5,# do 5-fold cross-validation
  +  keep.data=TRUE,  # keep a copy of the dataset
  with the object
  +  verbose=FALSE)# print out progress
  
   best.iter = gbm.perf(boost.model,method=cv)
   pred = predict.gbm(boost.model, test, best.iter)
   summary(pred)
 Min. 1st Qu.  MedianMean 3rd Qu.Max.
  0.4772  1.5140  1.6760  1.5100  1.7190  1.9420
 --
 LEGAL NOTICE\ Unless expressly stated otherwise, this messag...{{dropped}}

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-- 
Weiwei Shi, Ph.D

Did you always know?
No, I did not. But I believed...
---Matrix III

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Re: [R] Add text/numbers to x axis?

[Gabor Grothendieck]

 BTW, anyone who knows how such dumb plots could be achieved in R? This
 option is avaliable in S, so a port to R shouldn't be impossible, should
 it? Don't know how to do it myself however...

Use gnuplot or MacAnova.  RSiteSearch(MacAnova) has an example.

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Re: [R] correlating values

[Sundar Dorai-Raj]

Ahamarshan jn wrote:
   I did a correlation for the values 
 
 -   EX4577  EX4599  EX4566  EX4522
 WL917 2.53528 0.79077 0.21499  -0.01084
 WP429S-0.192723715
 WP819 -1.016997552
 WP977 1.378674-0.070710.6250890.4728363
 WI205S-0.24443-1.789526   0.648923-0.775867
 
 
 by using
 round(cor(t(person.data),use=pairwise.complete.obs))
 
 i got the result as
 
 - WL917   WP429S  WP819   WP977   WI205S
 WL917 1   NA  NA  0.344   -0.11424
 WP429SNA  NA  NA  NA   NA
 WP819 NA  NA  NA  NA   NA
 WP977 0.34461 NA  NA  1   .23294
 WI205S-0.11424NA  NA  0.23294 1
 
 i notice that 
 for correlation between WP429S x wp429S the value is
 given as NA where as it should be 1
 same is the case with WP819 x WP819 
 
 Can someone please help me reason out why this is
 happening? and any corrective measure that I need to
 take so that I can get the true value of 1 for them.
 
 Thank you
 - ash#
 

Hi, Ash,

It's difficult to tell with the data you provided (maybe just poor email 
formatting), but it looks like both WP429S and WP819 have only one 
value. You cannot obtain a correlation with only one value (see any 
intro stats book for the definition or correlation or just Google it). Try:

x - rnorm(1)
y - rnorm(1)
cor(x, y)
cor(matrix(x, 1, 1))

This is because x and y have zero variance. Forcing the NA to 1 as a 
corrective measure would be incorrect and misleading.

HTH,

--sundar

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[R] when dimensionality is larger than the number of observations?

[Weiwei Shi]

Hi, there:

Can anyone here kindly point some good reference or links on this topic?
Esp. some solutions from BioConductor or R, when dealing with
microarray-like, fat data?

thanks,


-- 
Weiwei Shi, Ph.D

Did you always know?
No, I did not. But I believed...
---Matrix III

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[R] executable file with R

[Romain Lorrilliere]

Hi,

I made an R function, and I want make an executable applet with it. Do 
you know how it is possible?


Thank for your help.


Romain


-- 

Lorrillière Romain

 

UMR 8079 Laboratoire Ecologie, Systématique et Evolution

Bât. 362

Université Paris-Sud

91405 Orsay cedex

France

 

tel : 01 69 15 56 85

fax : 01 69 15 56 96

mobile : 06 81 70 90 70

 

email : [EMAIL PROTECTED]

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[R] average by group...

[r user]

I have a dataframe with 700,000 rows and 2 vectors
(columns): “group” and “score”.

I  wish to calculate a third vector of length 70:
the average score by group.  Even though the avarge
value will repeat, I wish to return the average for
that particular group for each row.

(I know I can do this by calculating each group’s
average and then using the merge command, but as my
calculations get more complex and my data set gets
larger, the merge command seems to be fairly slow.)

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Re: [R] when dimensionality is larger than the number of observations?

[Gabor Grothendieck]

On 5/30/06, Weiwei Shi [EMAIL PROTECTED] wrote:
 Hi, there:

 Can anyone here kindly point some good reference or links on this topic?
 Esp. some solutions from BioConductor or R, when dealing with
 microarray-like, fat data?


In that case there will be an entire subspace of coefficient
vectors that will give the same fitted values.

Lets take 3 rows of the iris data set
and regress column 1 on the rest.   There will
be an entire subspace of coefficients that correspond
to the same (unique) fitted values and we can get
one of those coefficient vectors using the generalized
inverse:

 # test data
 iris3 - iris[c(1, 51, 101),]
 y - iris3[,1]
 y
[1] 5.1 7.0 6.3
 X - model.matrix(~., iris3[,2:5])
 X
(Intercept) Sepal.Width Petal.Length Petal.Width Speciesversicolor
Speciesvirginica
1 1 3.5  1.4 0.2 0
   0
511 3.2  4.7 1.4 1
   0
101   1 3.3  6.0 2.5 0
   1
attr(,assign)
[1] 0 1 2 3 4 4
attr(,contrasts)
attr(,contrasts)$Species
[1] contr.treatment


 library(MASS) # needed for ginv
 coefs - c(ginv(crossprod(X)) %*% crossprod(X, y))
 names(coefs) - colnames(X)
 coefs
  (Intercept)   Sepal.Width  Petal.Length
Petal.Width Speciesversicolor  Speciesvirginica
0.3619361 1.1497417 0.5443438
-0.2405670 0.7372685-0.5207289
 X %*% coefs # fitted values
[,1]
15.1
51   7.0
101  6.3

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[R] position of number at risk in survplot() graphs

[Osman Al-Radi]

Dear R-help

How can one get survplot() to place the number at risk just below the
survival curve as opposed to the default which is just above the x-axis?
I tried the code bellow but the result is not satisfactory as some numbers
are repeated several times at different y coordinates and the position of
the n.risk numbers corresponds to the x-axis tick marks not the survival
curve time of censoring.

n - 20
set.seed(731)
cens - 15*runif(n)
h - .02*exp(2)
dt - -log(runif(n))/h
label(dt) - 'Follow-up Time'
e - ifelse(dt = cens,1,0)
dt - pmin(dt, cens)
units(dt) - Year
S - Surv(dt,e)
km-survfit(S~1)
survplot(km,n.risk=T,conf='none', y.n.risk=unique(summary(km)$surv))

Any suggestion on addressing this problem would be apprecited.

Also, is there a way to add a tick mark to the survival curve at times of
censoring similar to the mark.time=T argument in plot.survplot()?

Thanks

Osman

-- 
Osman O. Al-Radi, MD, MSc, FRCSC
Fellow, Cardiovascular Surgery
The Hospital for Sick Children
University of Toronto, Canada

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Re: [R] average by group...

[Doran, Harold]

?tapply
 

 -Original Message-
 From: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED] On Behalf Of r user
 Sent: Tuesday, May 30, 2006 12:27 PM
 To: rhelp
 Subject: [R] average by group...
 
 I have a dataframe with 700,000 rows and 2 vectors
 (columns): group and score.
 
 I  wish to calculate a third vector of length 70:
 the average score by group.  Even though the avarge value 
 will repeat, I wish to return the average for that particular 
 group for each row.
 
 (I know I can do this by calculating each group's average and 
 then using the merge command, but as my calculations get more 
 complex and my data set gets larger, the merge command seems 
 to be fairly slow.)
 
 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide! 
 http://www.R-project.org/posting-guide.html


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Re: [R] average by group...

[Peter Dalgaard]

Doran, Harold [EMAIL PROTECTED] writes:

 ?tapply

Nope. 

?ave
  
 
  -Original Message-
  From: [EMAIL PROTECTED] 
  [mailto:[EMAIL PROTECTED] On Behalf Of r user
  Sent: Tuesday, May 30, 2006 12:27 PM
  To: rhelp
  Subject: [R] average by group...
  
  I have a dataframe with 700,000 rows and 2 vectors
  (columns): group and score.
  
  I  wish to calculate a third vector of length 70:


  the average score by group.  Even though the avarge value 
  will repeat, I wish to return the average for that particular 
  group for each row.
  
  (I know I can do this by calculating each group's average and 
  then using the merge command, but as my calculations get more 
  complex and my data set gets larger, the merge command seems 
  to be fairly slow.)
  
  __
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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
 
 
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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
~~ - ([EMAIL PROTECTED])  FAX: (+45) 35327907

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Re: [R] combinatorial programming problem

[Spencer Graves]

Hi, Martin:  Thanks much.  I don't know if it solved Kjetil's problem, 
but it helps educate me.  Best Wishes, Spencer Graves

Martin Maechler wrote:
 SpG == Spencer Graves [EMAIL PROTECTED]
 on Sun, 28 May 2006 16:21:53 -0700 writes:
 
 SpGI'm not sure I understand your question, but
 SpG are you asking how to index choose(k, r) objects?
 SpG Almost 3 years ago, I asked a question like this.  Andy
 SpG Liaw referred me to nchoosek(vsn)
 SpG (http://finzi.psych.upenn.edu/R/Rhelp02a/archive/12518.html).
 SpG This produces a matrix of dimension (r, choose(k, r)).
 SpG With this matrix, you could convert an integer between
 SpG 1 and choose(k, r) into an r-vector by table look-up.
 SpG Reading the code for nchoosek might help you further if
 SpG this does not seem appropriate for you.
 
 Note that *if* the above is the answer,
 I'd rather recommend to use  combn() from package combinat,
 since a (slightly improved) version of combn() has been part of R-devel
 (to become R 2.4.0 in October) for a while.  Also, combn() from
 combinat precedes nchoosek() historically and is also faster.
 
 Martin Maechler, ETH Zurich
 
 SpGI found this just now using 'RSiteSearch(all
 SpG subsets of a size)', which produced 102 hits.  Another
 SpG one that looked like it might help you is
 SpG http://finzi.psych.upenn.edu/R/Rhelp02a/archive/1717.html;.
 
 SpGHope this helps, Spencer Graves
 
 SpG Kjetil Brinchmann Halvorsen wrote:
  Hola!
  
  I am programming a class (S3) symarray for storing the
  results of functions symmetric in its k
  arguments. Intended use is for association indices for
  more than two variables, for instance coresistivity
  against antibiotics.
  
  There is one programming problem I haven't solved, making
  an inverse of the index function indx() --- se code
  below. It could for instance return the original k
  indexes in strictly increasing order, to make indx()
  formally invertible.
  
  Any ideas?
  
  Kjetil
  
  
  Code:
  
  
  # Implementing an S3 class for symarrays with array rank
  r for dimension # [k, k, ..., k] with k=r repeated r
  times. We do not store the diagonal.
  
  # Storage requirement is given by {r, k}= choose(k, r) #
  where r=array rank, k=maximum index
  
  symarray - function(data=NA, dims=c(1,1)){ r - dims[1]
  k - dims[2] if(r  k) stop(symarray needs dimension
  larger than array rank) len - choose(k, r) out -
  data[1:len] attr(out, dims) - dims class(out) -
  symarray out }
  
  # Index calculation:
  
  indx - function(inds, k){ r - length(inds) if(r==1)
  return(inds) else { if(inds[1]==1) { return(
  indx(inds[-1]-1, k-1 ) ) } else { return(
  indx(c(inds[1]-1, seq(from=k-r+2, by=1, to=k)), k) +
  indx( inds[-1]-inds[1], k-inds[1] )) } } } # end indx
  
  # Methods for assignment and indexing:
  
  [.symarray - function(x, inds, drop=FALSE){ dims -
  attr(x, dims) k - dims[2] inds - indx(inds, k) res -
  NextMethod([, x) res }
  
  [-.symarray - function(x, inds, value){ dims -
  attr(x, dims) k - dims[2] inds - indx(inds, k) res -
  NextMethod([-, x) res }
  
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[R] Help with adding minutes to time

[Jean-Louis Abitbol]

Dear R Helpers,

I need to read time from a .csv file which is formated as chartime
(09:12:00) below. I need to add one minute (cf chartime2).
Then I need to output the value just as 09:13 without the seconds for
writing a csv file and input in another program.

I get it with the following reproducible example but I can't help
thinking that there must a less clumsy way to do that ! 

Thanks for any input and eventually a pointer to an example of how to
add minutes or seconds to a time without a date which I think makes
POSIX not relevant in this case.

Best regards, Jean-louis

 
library(chron)
chartime-09:12:00
chartime2-00:01:00
chrontime-times(chartime)
chrontime2-times(chartime2)
test-chrontime+chrontime2
test-as.character(test)
test-sub(:00,,test)

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[R] Weighting cluster variables in R?

[Quin Wills]

Are there functions to weight variables for clustering in R? I can't seem to
find anything, so apologies if there is.

 

I am particularly interested in weighting variables (starting with kmeans)
to optimise inter/intra-cluster distances. It seems to me that if certain
variables do show a strong cluster structure, this would be a wise thing to
do.

 

Any advice welcome.

 

Quin

 

 

 


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Re: [R] average by group...

[Doran, Harold]

I didn't know about ave(). What about this, though:

dat - data.frame(score = rnorm(100), group = gl(10,10))
group.score - with(dat, tapply(score, group, mean))
dat$group.score - group.score[as.character(dat$group)] 

Harold

 -Original Message-
 From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On 
 Behalf Of Peter Dalgaard
 Sent: Tuesday, May 30, 2006 1:09 PM
 To: Doran, Harold
 Cc: r user; rhelp
 Subject: Re: [R] average by group...
 
 Doran, Harold [EMAIL PROTECTED] writes:
 
  ?tapply
 
 Nope. 
 
 ?ave
   
  
   -Original Message-
   From: [EMAIL PROTECTED] 
   [mailto:[EMAIL PROTECTED] On Behalf Of r user
   Sent: Tuesday, May 30, 2006 12:27 PM
   To: rhelp
   Subject: [R] average by group...
   
   I have a dataframe with 700,000 rows and 2 vectors
   (columns): group and score.
   
   I  wish to calculate a third vector of length 70:
 
 
   the average score by group.  Even though the avarge value will 
   repeat, I wish to return the average for that particular 
 group for 
   each row.
   
   (I know I can do this by calculating each group's average 
 and then 
   using the merge command, but as my calculations get more 
 complex and 
   my data set gets larger, the merge command seems to be 
 fairly slow.)
   
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  (*) \(*) -- University of Copenhagen   Denmark  Ph:  
 (+45) 35327918
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[R] na.pass

[Bahamonde Natalia]

Hello...

What does na.pass?


x=c(2.4, 2.4, 1.9, 2.5, 2.1)
xNA=replace(x, 3, NA)

p=c(acf(x, type=c(covariance), plot=FALSE)$acf)
pNA=c(acf(xNA, type=c(covariance), na.action=na.pass, plot=FALSE)$acf)

  p
[1]  0.05040 -0.03112  0.00816  0.00224 -0.00448
  pNA
[1]  0.02250 -0.01167  0.00250 -0.00100 -0.00250

In the manual say na.pass returns objet unchanged...

Thanks, Natalia

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[R] max / pmax

[Brian Perron]

Hello R users,

I am relatively new to R and cannot seem to crack a coding problem.  I 
am working with substance abuse data, and I have a variable called 
primary.drug which is considered the drug of choice for each 
subject.   I have just a few missing values on that variable.  Instead 
of using a multiple imputation method like chained equations, I would 
prefer to derive these values from other survey responses.  
Specifically, I have a frequency of use (in days) for each of the major 
drugs, so I would like the missing values to be replaced by that drug 
with the highest level of use.  I am starting with the ifelse and 
max statements, but I know it is wrong:

impute.primary.drug -   ifelse(is.na(primary.drug), max(marijuana, 
crack, cocaine, heroin), primary.drug)

Here are the problems.  First, the max statement (should it be pmax?), 
returns the highest numeric quantity rather than the variable itself.  
In other words, I want to test which drug has the highest value, but 
return the variable name rather than the observed value.   Second, if 
ties are observed, how can I specify the value to be NA?  Or, how can I 
specify one of the values to be randomly selected?   

 Thank in advance for your assistance.

Regards,
Brian

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Re: [R] Help with adding minutes to time

[Sebastian Luque]

Jean-Louis Abitbol [EMAIL PROTECTED] wrote:

 Dear R Helpers,

 I need to read time from a .csv file which is formated as chartime
 (09:12:00) below. I need to add one minute (cf chartime2).
 Then I need to output the value just as 09:13 without the seconds for
 writing a csv file and input in another program.

[...]

The numeric representation of chron objects is expressed in day units, and
there are sensible arithmetic methods:

chrontime - times(09:12:00) + 1/24/60

and then:

paste(hours(chrontime), minutes(chrontime), sep=:)


Cheers,

-- 
Seb

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[R] merging

[Gavin Simpson]

Dear List,

Given,

y - matrix(c(0,1,1,1,0,0,0,4,4), ncol = 3, byrow = TRUE)
rownames(y) - c(a,b,c)
colnames(y) - c(1,2,3)
y
y2 - y[2:3, ]
rownames(y2) - c(x,z)
y2

how can I stop

merge(y, y2, all = TRUE, sort = FALSE)

squishing the extra rows? Ideally I want the same as:

rbind(y, y2)

in this case. This is specific example of situation where two data
matrices have same column variables and all I want is to stick the two
sets of rows together, but I have been using merge for cases such as the
one below, where the second matrix has extra column(s):

y3 - matrix(c(0,1,1,1,0,0,0,4,4,5,6,7), ncol = 4, byrow = TRUE)
rownames(y3) - c(d,e,f)
colnames(y3) - c(1,2,3,4)
y3
merge(y, y3, all = TRUE, sort = FALSE)

We don't know before hand if the columns will match. But I see now that
even this doesn't work as I was expecting/thinking!

So I'm looking for a general way to merge two matrices such that the
number of rows in the merged matrix is nrow(mat1) + nrow(mat2) and the
number of columns in the merged matrix is length(unique(colnames(mat1),
colnames(mat2).

Is there a function in R to do this, or can someone suggest a way to
achieve this? My R version info is at the end.

Just to be clear, for the y, y3 example I want something like this
returned:

  1 2 3 4
a 0 1 1 NA
b 1 0 0 NA
c 0 4 4 NA
d 0 1 1 1
e 0 0 0 4
f 4 5 6 7

and for the y, y2 example, I want something like this returned:

  1 2 3
a 0 1 1
b 1 0 0
c 0 4 4
x 1 0 0
z 0 4 4

Many thanks,

Gav

 version
   _
platform   i686-pc-linux-gnu
arch   i686
os linux-gnu
system i686, linux-gnu
status Patched
major  2
minor  3.0
year   2006
month  05
day03
svn rev37978
language   R
version.string Version 2.3.0 Patched (2006-05-03 r37978)
-- 
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
*  Note new Address, Telephone  Fax numbers from 6th April 2006  *
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
Gavin Simpson
ECRC  ENSIS  [t] +44 (0)20 7679 0522
UCL Department of Geography   [f] +44 (0)20 7679 0565
Pearson Building  [e] gavin.simpsonATNOSPAMucl.ac.uk
Gower Street  [w] http://www.ucl.ac.uk/~ucfagls/cv/
London, UK.   [w] http://www.ucl.ac.uk/~ucfagls/
WC1E 6BT.
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%

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Re: [R] Compiling R to run natively on Windows x64

[roger bos]

I am today where Alastiar was back in August 2005.  I am considering buying
a 64 bit machine with 16GB of memory to run bigger objects in R, which
brougt me down the Linux path.  The problem is that my company's IT staff
doesn't support Linux and this path was getting very bumpy for me, so I am
oscilating back towards 64bit Windows.  I thought the lack of a 64
bit Windows binary was due to there being no open-source gcc compiler for
64bit Windows, but reading Prof. Ripley reply seems to imply that I could
not even buy a compiler and compiler the source code myself because there
are none that treat a long as 64 bits.  Is that the only remaining obstacle
and does that obstacle still exist?

Thanks,

Roger





On 8/19/05, Prof Brian Ripley [EMAIL PROTECTED] wrote:

 On Thu, 18 Aug 2005, Alastair Cooper wrote:

  I am looking at getting a PC preinstalled with Windows XP x64. What I
  want to know is, has anyone successfully compiled a version of R for
  64-bit Windows (Amd64 - not Itanium), and if so did they find any
  performance boost?

 Hmm, where do you get a reliable C99-compatible compiler for 64-bit
 Windows?  We don't know of one, and the R sources are written assuming
 long is 64-bit on a 64-bit platform (and that is not the Win64
 convention) so there would still be a lot of 32-bit restrictions until we
 change that (which is on my TODO list).

 (We don't support building R with VC++, and although there have been a
 number of attempts none has produced a version that passes make check: I
 recall finding VC++ thought -Inf  3, for example.)

 Based on extensive experience on other platforms, I would expect a
 noticeable performance hit for a 64-bit build, but the ability to run
 bigger tasks: this is discussed with data and reasoning in the latest
 R-admin manual (in the R-devel version of R).

 --
 Brian D. Ripley,  [EMAIL PROTECTED]
 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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[R] Piecewise (broken stick) models in R

[Milton Cezar]

Hi there,
  I´m a newbie in R and I´m looking for some advice how to teste Piecewise 
(Broken Stick) models. 
   
  I´have another simple question: how can I compute descriptive statistics for 
data grouped for two or more variables? See below, please.
   
  Year  Site Repetition rainfall
   1980  1 1 ...
   1980  1 2 ...
   1980  1 3 ...
   1980  2 1 ...
   1980  2 2 ...
   1980  2 3 ...
   
   1990  1 1 ...
   1990  1 2 ...
   1990  1 3 ...
   1990  2 1 ...
   1990  2 2 ...
   1990  2 3 ...
   
  I´d like to compute the mean rainfall for each YEAR * SITE combination.
   
  Kind regards,
   
  Miltinho


-

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Re: [R] Compiling R to run natively on Windows x64

[Duncan Murdoch]

On 5/30/2006 2:18 PM, roger bos wrote:
 I am today where Alastiar was back in August 2005.  I am considering buying
 a 64 bit machine with 16GB of memory to run bigger objects in R, which
 brougt me down the Linux path.  The problem is that my company's IT staff
 doesn't support Linux and this path was getting very bumpy for me, so I am
 oscilating back towards 64bit Windows.  I thought the lack of a 64
 bit Windows binary was due to there being no open-source gcc compiler for
 64bit Windows, but reading Prof. Ripley reply seems to imply that I could
 not even buy a compiler and compiler the source code myself because there
 are none that treat a long as 64 bits.  Is that the only remaining obstacle
 and does that obstacle still exist?

I would somewhat seriously suggest that you give your support staff a 
copy of the R source, and get them to work out the details of compiling 
it in Win64.  I'm sure lots of people would appreciate their work.

Duncan Murdoch

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Re: [R] Help with adding minutes to time

[Gabor Grothendieck]

Try this:

 hhmm - function(x) sub(:..$, , format(times(x)))
 y - times(01:02:00, out.format = hhmm) + times(00:01:00)
 y  # or format(y)
[1] 01:03

or

 # hhmm is from above
 x - times(01:02:00)
 hhmm(x + times(00:01:00))
[1] 01:03

On 5/30/06, Jean-Louis Abitbol [EMAIL PROTECTED] wrote:
 Dear R Helpers,

 I need to read time from a .csv file which is formated as chartime
 (09:12:00) below. I need to add one minute (cf chartime2).
 Then I need to output the value just as 09:13 without the seconds for
 writing a csv file and input in another program.

 I get it with the following reproducible example but I can't help
 thinking that there must a less clumsy way to do that !

 Thanks for any input and eventually a pointer to an example of how to
 add minutes or seconds to a time without a date which I think makes
 POSIX not relevant in this case.

 Best regards, Jean-louis


 library(chron)
 chartime-09:12:00
 chartime2-00:01:00
 chrontime-times(chartime)
 chrontime2-times(chartime2)
 test-chrontime+chrontime2
 test-as.character(test)
 test-sub(:00,,test)

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Re: [R] max / pmax

[Tony Plate]

Here's an example of how I think you can do what you want.  Play with 
the definition of the function highest.use() to get random selection of 
multiple maxima.

  drug.names - c(marijuana, crack, cocaine, heroin)
  drugs - factor(drug.names, levels=drug.names)
  drugs
[1] marijuana crack cocaine   heroin
Levels: marijuana crack cocaine heroin
  as.numeric(drugs)
[1] 1 2 3 4
  N - 20
  set.seed(1)
  primary.drug - sample(drugs, N, rep=T)
  primary.drug[sample(1:20, 10)] - NA
  primary.drug
  [1] NA  crack NA  NA  NA  NA  heroin
  [8] cocaine   cocaine   marijuana NA  NA  cocaine   crack
[15] heroinNA  cocaine   heroinNA  NA
Levels: marijuana crack cocaine heroin
  # usage frequencies
  marijuana - sample(1:3, N, rep=T)
  crack - sample(1:3, N, rep=T)
  cocaine - sample(1:3, N, rep=T)
  heroin - sample(1:3, N, rep=T)
  cbind(marijuana, crack, cocaine, heroin)
   marijuana crack cocaine heroin
  [1,] 2 2   2  1
  [2,] 2 3   3  1
  [3,] 2 2   2  2
  [4,] 1 1   2  3
  [5,] 3 1   2  3
  [6,] 3 1   3  3
  [7,] 3 1   3  2
  [8,] 1 2   2  2
  [9,] 3 2   3  3
[10,] 2 2   3  2
[11,] 3 3   2  2
[12,] 2 1   3  2
[13,] 3 2   2  1
[14,] 2 1   1  3
[15,] 2 2   3  2
[16,] 3 1   1  1
[17,] 1 2   3  1
[18,] 2 3   1  2
[19,] 3 1   1  3
[20,] 3 3   1  2
  highest.use - function(x) {y - which(x==max(x, na.rm=T)); if 
(length(y)==1) return(y) else return(NA)}
  apply(cbind(marijuana, crack, cocaine, heroin), 1, highest.use)
  [1] NA NA NA  4 NA NA NA NA NA  3 NA  3  1  4  3  1  3  2 NA NA
  impute.primary.drug - drugs[ifelse(is.na(primary.drug), 
apply(cbind(marijuana, crack, cocaine, heroin), 1, highest.use), 
as.numeric(primary.drug))]
  data.frame(primary.drug, impute.primary.drug)
primary.drug impute.primary.drug
1  NANA
2 crack   crack
3  NANA
4  NA  heroin
5  NANA
6  NANA
7heroin  heroin
8   cocaine cocaine
9   cocaine cocaine
10marijuana   marijuana
11 NANA
12 NA cocaine
13  cocaine cocaine
14crack   crack
15   heroin  heroin
16 NA   marijuana
17  cocaine cocaine
18   heroin  heroin
19 NANA
20 NANA
 


Brian Perron wrote:
 Hello R users,
 
 I am relatively new to R and cannot seem to crack a coding problem.  I 
 am working with substance abuse data, and I have a variable called 
 primary.drug which is considered the drug of choice for each 
 subject.   I have just a few missing values on that variable.  Instead 
 of using a multiple imputation method like chained equations, I would 
 prefer to derive these values from other survey responses.  
 Specifically, I have a frequency of use (in days) for each of the major 
 drugs, so I would like the missing values to be replaced by that drug 
 with the highest level of use.  I am starting with the ifelse and 
 max statements, but I know it is wrong:
 
 impute.primary.drug -   ifelse(is.na(primary.drug), max(marijuana, 
 crack, cocaine, heroin), primary.drug)
 
 Here are the problems.  First, the max statement (should it be pmax?), 
 returns the highest numeric quantity rather than the variable itself.  
 In other words, I want to test which drug has the highest value, but 
 return the variable name rather than the observed value.   Second, if 
 ties are observed, how can I specify the value to be NA?  Or, how can I 
 specify one of the values to be randomly selected?   
 
  Thank in advance for your assistance.
 
 Regards,
 Brian
 
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Re: [R] Piecewise (broken stick) models in R

[jim holtman]

?aggregate

 y
   year site rep  rain
1  19801   1 2.6550866
7  19801   2 3.7212390
13 19801   3 5.7285336
4  19802   1 9.0820779
10 19802   2 2.0168193
16 19802   3 8.9838968
2  19811   1 9.4467527
8  19811   2 6.6079779
14 19811   3 6.2911404
5  19812   1 0.6178627
11 19812   2 2.0597457
17 19812   3 1.7655675
3  19821   1 6.8702285
9  19821   2 3.8410372
15 19821   3 7.6984142
6  19822   1 4.9769924
12 19822   2 7.1761851
18 19822   3 9.9190609
 aggregate(y, list(y$year, y$site), mean)
  Group.1 Group.2 year site rep rain
11980   1 19801   2 4.034953
21981   1 19811   2 7.448624
31982   1 19821   2 6.136560
41980   2 19802   2 6.694265
51981   2 19812   2 1.481059
61982   2 19822   2 7.357413




On 5/30/06, Milton Cezar [EMAIL PROTECTED] wrote:

 Hi there,
 I´m a newbie in R and I´m looking for some advice how to teste Piecewise
 (Broken Stick) models.

 I´have another simple question: how can I compute descriptive statistics
 for data grouped for two or more variables? See below, please.

 Year  Site Repetition rainfall
   1980  1 1 ...
   1980  1 2 ...
   1980  1 3 ...
   1980  2 1 ...
   1980  2 2 ...
   1980  2 3 ...

   1990  1 1 ...
   1990  1 2 ...
   1990  1 3 ...
   1990  2 1 ...
   1990  2 2 ...
   1990  2 3 ...

 I´d like to compute the mean rainfall for each YEAR * SITE combination.

 Kind regards,

 Miltinho


 -

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-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390 (Cell)
+1 513 247 0281 (Home)

What is the problem you are trying to solve?

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Re: [R] Compiling R to run natively on Windows x64

[roger bos]

Duncan,

Thanks for the reply.  I was hoping to hear that other people were
successful and then I would try doing so myself, but no one in the R
community has gotten R to compile under 64 bit Windows, I unfortunately I
not going to be the one to overcome this obstacle since I only took a few
C++ courses 7 years ago--and haven't used it since.  I have a better chance
of learning how to use Linux.

Thanks,

Roger


On 5/30/06, Duncan Murdoch [EMAIL PROTECTED] wrote:

 On 5/30/2006 2:18 PM, roger bos wrote:
  I am today where Alastiar was back in August 2005.  I am considering
 buying
  a 64 bit machine with 16GB of memory to run bigger objects in R, which
  brougt me down the Linux path.  The problem is that my company's IT
 staff
  doesn't support Linux and this path was getting very bumpy for me, so I
 am
  oscilating back towards 64bit Windows.  I thought the lack of a 64
  bit Windows binary was due to there being no open-source gcc compiler
 for
  64bit Windows, but reading Prof. Ripley reply seems to imply that I
 could
  not even buy a compiler and compiler the source code myself because
 there
  are none that treat a long as 64 bits.  Is that the only remaining
 obstacle
  and does that obstacle still exist?

 I would somewhat seriously suggest that you give your support staff a
 copy of the R source, and get them to work out the details of compiling
 it in Win64.  I'm sure lots of people would appreciate their work.

 Duncan Murdoch


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[R] Faster way to zero-pad a data frame...?

[Pete Cap]

Hello List,
 
 I am working on creating periodograms from IP network traffic logs using the 
Fast Fourier Transform.  The FFT requires all the data points to be 
evenly-spaced in the time domain (constant delta-T), so I have a step where I 
zero-pad the data.
 
 Lately I've been wondering if there is a faster way to do this.  Here's what 
I've got:
 
 * data1 is a data frame consisting of a timestamp, in seconds, from the 
beginning of the network log, and the number of network events that fell on 
that timestamp.
 Example:
 time,events
 0,1
 1,30
 5,14
 10,4
 
 *data2 is the zero-padded data frame.  It has length equal to the greatest 
value of time in data2:
 time,events
 1,0
 2,0
 3,0
 4,0
 5,0
 6,0
 7,0
 8,0
 9,0
 10,0
 
 So I run this for loop:
 for(i in 1:length(data1[,1])) {
 data2[data1[i,1],2]-data1[i,2]
 }
 
 Which goes to each row in data1, reads the timestamp, and writes the events 
to the corresponding row in data2.  The result is:
 time,events
 0,1
 1,30
 2,0
 3,0
 4,0
 5,14
 6,0
 7,0
 9,0
 9,0
 10,4
 
 For a 24-hour log (86,400 seconds) this can take a while...Any advice on how 
to speed it up would be appreciated.
 
 Thanks,
 Pete Cap
 

-

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[R] weighted correlation coefficient

[Young-Jin Lee]

Dear R-listers

Is there a R-package that I can use to compute a weighted correlation
coefficient along with its p-values?
I want to be able to assign weights to the data points before computing a
Pearson correlation coefficient.

Thanks in advance.

Young-Jin

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Re: [R] merging

[Sundar Dorai-Raj]

Gavin Simpson wrote:
 Dear List,
 
 Given,
 
 y - matrix(c(0,1,1,1,0,0,0,4,4), ncol = 3, byrow = TRUE)
 rownames(y) - c(a,b,c)
 colnames(y) - c(1,2,3)
 y
 y2 - y[2:3, ]
 rownames(y2) - c(x,z)
 y2
 
 how can I stop
 
 merge(y, y2, all = TRUE, sort = FALSE)
 
 squishing the extra rows? Ideally I want the same as:
 
 rbind(y, y2)
 
 in this case. This is specific example of situation where two data
 matrices have same column variables and all I want is to stick the two
 sets of rows together, but I have been using merge for cases such as the
 one below, where the second matrix has extra column(s):
 
 y3 - matrix(c(0,1,1,1,0,0,0,4,4,5,6,7), ncol = 4, byrow = TRUE)
 rownames(y3) - c(d,e,f)
 colnames(y3) - c(1,2,3,4)
 y3
 merge(y, y3, all = TRUE, sort = FALSE)
 
 We don't know before hand if the columns will match. But I see now that
 even this doesn't work as I was expecting/thinking!
 
 So I'm looking for a general way to merge two matrices such that the
 number of rows in the merged matrix is nrow(mat1) + nrow(mat2) and the
 number of columns in the merged matrix is length(unique(colnames(mat1),
 colnames(mat2).
 
 Is there a function in R to do this, or can someone suggest a way to
 achieve this? My R version info is at the end.
 
 Just to be clear, for the y, y3 example I want something like this
 returned:
 
   1 2 3 4
 a 0 1 1 NA
 b 1 0 0 NA
 c 0 4 4 NA
 d 0 1 1 1
 e 0 0 0 4
 f 4 5 6 7
 
 and for the y, y2 example, I want something like this returned:
 
   1 2 3
 a 0 1 1
 b 1 0 0
 c 0 4 4
 x 1 0 0
 z 0 4 4
 
 Many thanks,
 
 Gav
 
 
version
 
_
 platform   i686-pc-linux-gnu
 arch   i686
 os linux-gnu
 system i686, linux-gnu
 status Patched
 major  2
 minor  3.0
 year   2006
 month  05
 day03
 svn rev37978
 language   R
 version.string Version 2.3.0 Patched (2006-05-03 r37978)


Will this help:

rbind.all - function(...) {
   x - list(...)
   cn - unique(unlist(lapply(x, colnames)))
   for(i in seq(along = x)) {
 if(any(m - !cn %in% colnames(x[[i]]))) {
   na - matrix(NA, nrow(x[[i]]), sum(m))
   dimnames(na) - list(rownames(x[[i]]), cn[m])
   x[[i]] - cbind(x[[i]], na)
 }
   }
   do.call(rbind, x)
}

y - matrix(c(0,1,1,1,0,0,0,4,4), ncol = 3, byrow = TRUE)
rownames(y) - c(a,b,c)
colnames(y) - c(1,2,3)
y2 - y[2:3, 2:3]
rownames(y2) - c(x,z)
y3 - matrix(c(0,1,1,1,0,0,0,4,4,5,6,7), ncol = 4, byrow = TRUE)
rownames(y3) - c(d,e,f)
colnames(y3) - c(1,2,3,4)

rbind.all(y, y2, as.data.frame(y3))

It does very little error-checking, so be careful how you use it.

--sundar

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Re: [R] Install R problem

[Pramod Anugu]

I have copied R-2.3.0.tar.gx and uncompressed the directory R-2.3.0 is
created.
./configure
Make

Typed R in the directory /root/downloads/R-2.3.0/bin/exec
-bash-2.05b# pwd
/root/downloads/R-2.3.0/bin/exec
-bash-2.05b# R
Fatal error: R home directory is not defined

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Re: [R] Faster way to zero-pad a data frame...?

[jim holtman]

How about starting your time from 1 instead of 0 to make indexing earier
(you can always substract one later).  If so:

 x
  time events
11  1
22 30
36 14
4   11  4
 y - data.frame(time=seq(max(x$time)), events=rep(0, max(x$time)))
 y
   time events
1 1  0
2 2  0
3 3  0
4 4  0
5 5  0
6 6  0
7 7  0
8 8  0
9 9  0
10   10  0
11   11  0
 y$events[x$time] - x$events
 y
   time events
1 1  1
2 2 30
3 3  0
4 4  0
5 5  0
6 6 14
7 7  0
8 8  0
9 9  0
10   10  0
11   11  4




On 5/30/06, Pete Cap [EMAIL PROTECTED] wrote:

 Hello List,

 I am working on creating periodograms from IP network traffic logs using
 the Fast Fourier Transform.  The FFT requires all the data points to be
 evenly-spaced in the time domain (constant delta-T), so I have a step where
 I zero-pad the data.

 Lately I've been wondering if there is a faster way to do this.  Here's
 what I've got:

 * data1 is a data frame consisting of a timestamp, in seconds, from the
 beginning of the network log, and the number of network events that fell on
 that timestamp.
 Example:
 time,events
 0,1
 1,30
 5,14
 10,4

 *data2 is the zero-padded data frame.  It has length equal to the greatest
 value of time in data2:
 time,events
 1,0
 2,0
 3,0
 4,0
 5,0
 6,0
 7,0
 8,0
 9,0
 10,0

 So I run this for loop:
 for(i in 1:length(data1[,1])) {
 data2[data1[i,1],2]-data1[i,2]
 }

 Which goes to each row in data1, reads the timestamp, and writes the
 events to the corresponding row in data2.  The result is:
 time,events
 0,1
 1,30
 2,0
 3,0
 4,0
 5,14
 6,0
 7,0
 9,0
 9,0
 10,4

 For a 24-hour log (86,400 seconds) this can take a while...Any advice on
 how to speed it up would be appreciated.

 Thanks,
 Pete Cap


 -

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-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390 (Cell)
+1 513 247 0281 (Home)

What is the problem you are trying to solve?

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[R] En: R: Piecewise (broken stick) models in R Stats for groups

[Milton Cezar]

Data: Tue, 30 May 2006 16:08:26 -0300 (ART)
De: Milton Cezar [EMAIL PROTECTED]
Assunto: R: Piecewise (broken stick) models in R  Stats for groups
Para: [EMAIL PROTECTED]


  Chris Barker and everyone
   
  Thanks for your fast replay.
  Regarding the descreptive Stats, I can compute it for the full dataset, but I 
can´t comput it for groups of that. 
  I used mean ( Rainfall [year==1980] ) without problem. But how can I get the 
mean Rainfall for Year==1980 AND Site=1; Year=1980 AND Site=2 etc.
   
  Regarding my question for Pieacewise Regression model It can really be done 
with liner models functions. In fact, the results are a set of models that work 
bether in different range of an X variable (for  X varying 0 to 0.15 use 
model_1; for 0.16 to 0.46 use model_2... for X0,47 use model_3 etc.
   
  Sorry for my broken english,
   
  Regards a lot,
  Miltinho
   
  From: Barker, Chris [SCIUS] 
  To: 'Milton Cezar' 
  Sent: Tuesday, May 30, 2006 3:47 PM
  Subject: RE: [R] Piecewise (broken stick) models in R
  You can get descriptive statistics with mean(), var() or summary(). 
AS to broken stick, I can only speculate that's a variation on a regression.  
You may need to be more specific in your questions to the list, as I suspect 
its something easily done in the linear models functions.
  Chris Barker 
Associate Director, Biostatistics 
Scios Inc. 
6500 Paseo Padre Parkway 
 Fremont, CA 94555 
Tel 510 248 2439 
Fax 510 248 2451 
  ==
  From:   [EMAIL PROTECTED] On Behalf Of Milton Cezar 
Sent:   Tuesday, May 30, 2006 11:42 AM 
To: r-help@stat.math.ethz.ch 
Subject:[R] Piecewise (broken stick) models in R 
File: ATT17883177.txt  Hi there, 
  I´m a newbie in R and I´m looking for some advice how to teste Piecewise 
(Broken Stick) models. 
   
  I´have another simple question: how can I compute descriptive statistics for 
data grouped for two or more variables? See below, please.
 
  Year  Site Repetition rainfall 
   1980  1 1 ... 
   1980  1 2 ... 
   1980  1 3 ... 
   1980  2 1 ... 
   1980  2 2 ... 
   1980  2 3 ... 
   
   1990  1 1 ... 
   1990  1 2 ... 
   1990  1 3 ... 
   1990  2 1 ... 
   1990  2 2 ... 
   1990  2 3 ... 
   
  I´d like to compute the mean rainfall for each YEAR * SITE combination. 
   
  Kind regards, 
   
  Miltinho 
  
- 
   


-

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Re: [R] merging

[Marc Schwartz (via MN)]

On Tue, 2006-05-30 at 19:08 +0100, Gavin Simpson wrote:
 Dear List,
 
 Given,
 
 y - matrix(c(0,1,1,1,0,0,0,4,4), ncol = 3, byrow = TRUE)
 rownames(y) - c(a,b,c)
 colnames(y) - c(1,2,3)
 y
 y2 - y[2:3, ]
 rownames(y2) - c(x,z)
 y2
 
 how can I stop
 
 merge(y, y2, all = TRUE, sort = FALSE)
 
 squishing the extra rows? Ideally I want the same as:
 
 rbind(y, y2)
 
 in this case. This is specific example of situation where two data
 matrices have same column variables and all I want is to stick the two
 sets of rows together, but I have been using merge for cases such as the
 one below, where the second matrix has extra column(s):
 
 y3 - matrix(c(0,1,1,1,0,0,0,4,4,5,6,7), ncol = 4, byrow = TRUE)
 rownames(y3) - c(d,e,f)
 colnames(y3) - c(1,2,3,4)
 y3
 merge(y, y3, all = TRUE, sort = FALSE)
 
 We don't know before hand if the columns will match. But I see now that
 even this doesn't work as I was expecting/thinking!
 
 So I'm looking for a general way to merge two matrices such that the
 number of rows in the merged matrix is nrow(mat1) + nrow(mat2) and the
 number of columns in the merged matrix is length(unique(colnames(mat1),
 colnames(mat2).
 
 Is there a function in R to do this, or can someone suggest a way to
 achieve this? My R version info is at the end.
 
 Just to be clear, for the y, y3 example I want something like this
 returned:
 
   1 2 3 4
 a 0 1 1 NA
 b 1 0 0 NA
 c 0 4 4 NA
 d 0 1 1 1
 e 0 0 0 4
 f 4 5 6 7
 
 and for the y, y2 example, I want something like this returned:
 
   1 2 3
 a 0 1 1
 b 1 0 0
 c 0 4 4
 x 1 0 0
 z 0 4 4

Gavin,

Here is a possible solution, though not fully tested.

It uses the row.names for the two matrices as part of the 'by'
matching process. This is noted in the Details section in ?merge.

So for y and y2:

 res - merge(y, y2, 
   by = c(row.names, intersect(colnames(y),
  colnames(y2))), 
   all = TRUE)

# Note that the row names are now the first col
 res
  Row.names 1 2 3
1 a 0 1 1
2 b 1 0 0
3 c 0 4 4
4 x 1 0 0
5 z 0 4 4

# Subset res, leaving out the first col
 mat - res[, -1]

# Set the rownames from res
 rownames(mat) - res[, 1]

 mat
  1 2 3
a 0 1 1
b 1 0 0
c 0 4 4
x 1 0 0
z 0 4 4

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Re: [R] merging

[Marc Schwartz (via MN)]

On Tue, 2006-05-30 at 15:38 -0500, Marc Schwartz (via MN) wrote:
 On Tue, 2006-05-30 at 19:08 +0100, Gavin Simpson wrote:
  Dear List,
  
  Given,
  
  y - matrix(c(0,1,1,1,0,0,0,4,4), ncol = 3, byrow = TRUE)
  rownames(y) - c(a,b,c)
  colnames(y) - c(1,2,3)
  y
  y2 - y[2:3, ]
  rownames(y2) - c(x,z)
  y2
  
  how can I stop
  
  merge(y, y2, all = TRUE, sort = FALSE)
  
  squishing the extra rows? Ideally I want the same as:
  
  rbind(y, y2)
  
  in this case. This is specific example of situation where two data
  matrices have same column variables and all I want is to stick the two
  sets of rows together, but I have been using merge for cases such as the
  one below, where the second matrix has extra column(s):
  
  y3 - matrix(c(0,1,1,1,0,0,0,4,4,5,6,7), ncol = 4, byrow = TRUE)
  rownames(y3) - c(d,e,f)
  colnames(y3) - c(1,2,3,4)
  y3
  merge(y, y3, all = TRUE, sort = FALSE)
  
  We don't know before hand if the columns will match. But I see now that
  even this doesn't work as I was expecting/thinking!
  
  So I'm looking for a general way to merge two matrices such that the
  number of rows in the merged matrix is nrow(mat1) + nrow(mat2) and the
  number of columns in the merged matrix is length(unique(colnames(mat1),
  colnames(mat2).
  
  Is there a function in R to do this, or can someone suggest a way to
  achieve this? My R version info is at the end.
  
  Just to be clear, for the y, y3 example I want something like this
  returned:
  
1 2 3 4
  a 0 1 1 NA
  b 1 0 0 NA
  c 0 4 4 NA
  d 0 1 1 1
  e 0 0 0 4
  f 4 5 6 7
  
  and for the y, y2 example, I want something like this returned:
  
1 2 3
  a 0 1 1
  b 1 0 0
  c 0 4 4
  x 1 0 0
  z 0 4 4
 
 Gavin,
 
 Here is a possible solution, though not fully tested.
 
 It uses the row.names for the two matrices as part of the 'by'
 matching process. This is noted in the Details section in ?merge.
 
 So for y and y2:
 
  res - merge(y, y2, 
by = c(row.names, intersect(colnames(y),
   colnames(y2))), 
all = TRUE)
 
 # Note that the row names are now the first col
  res
   Row.names 1 2 3
 1 a 0 1 1
 2 b 1 0 0
 3 c 0 4 4
 4 x 1 0 0
 5 z 0 4 4
 
 # Subset res, leaving out the first col
  mat - res[, -1]
 
 # Set the rownames from res
  rownames(mat) - res[, 1]
 
  mat
   1 2 3
 a 0 1 1
 b 1 0 0
 c 0 4 4
 x 1 0 0
 z 0 4 4

Ack...hit the wrong button. Sorry.  

Must be the long weekendyeah, that's my story and I'm sticking to
it...  ;-)

Here is the solution for y and y3:


 res2 - merge(y, y3, 
by = c(row.names, intersect(colnames(y),
   colnames(y3))), 
all = TRUE)

 res2
  Row.names 1 2 3  4
1 a 0 1 1 NA
2 b 1 0 0 NA
3 c 0 4 4 NA
4 d 0 1 1  1
5 e 0 0 0  4
6 f 4 5 6  7

 mat2 - res2[, -1]

 rownames(mat2) - res2[, 1]
 
 mat2
  1 2 3  4
a 0 1 1 NA
b 1 0 0 NA
c 0 4 4 NA
d 0 1 1  1
e 0 0 0  4
f 4 5 6  7


HTH,

Marc Schwartz

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Re: [R] Faster way to zero-pad a data frame...?

[Gabor Grothendieck]

Try this:

Lines - time,events
 0,1
 1,30
 5,14
 10,4

library(zoo)
data1 - read.zoo(textConnection(Lines), header = TRUE, sep = ,)
data2 - as.ts(data1)
data2[is.na(data2)] - 0 # omit this lines if NAs in extra positions is ok


On 5/30/06, Pete Cap [EMAIL PROTECTED] wrote:
 Hello List,

  I am working on creating periodograms from IP network traffic logs using the 
 Fast Fourier Transform.  The FFT requires all the data points to be 
 evenly-spaced in the time domain (constant delta-T), so I have a step where I 
 zero-pad the data.

  Lately I've been wondering if there is a faster way to do this.  Here's what 
 I've got:

  * data1 is a data frame consisting of a timestamp, in seconds, from the 
 beginning of the network log, and the number of network events that fell on 
 that timestamp.
  Example:
  time,events
  0,1
  1,30
  5,14
  10,4

  *data2 is the zero-padded data frame.  It has length equal to the greatest 
 value of time in data2:
  time,events
  1,0
  2,0
  3,0
  4,0
  5,0
  6,0
  7,0
  8,0
  9,0
  10,0

  So I run this for loop:
  for(i in 1:length(data1[,1])) {
 data2[data1[i,1],2]-data1[i,2]
  }

  Which goes to each row in data1, reads the timestamp, and writes the 
 events to the corresponding row in data2.  The result is:
  time,events
  0,1
  1,30
  2,0
  3,0
  4,0
  5,14
  6,0
  7,0
  9,0
  9,0
  10,4

  For a 24-hour log (86,400 seconds) this can take a while...Any advice on how 
 to speed it up would be appreciated.

  Thanks,
  Pete Cap


 -

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Re: [R] Install R problem

[Seth Falcon]

Pramod Anugu [EMAIL PROTECTED] writes:

 I have copied R-2.3.0.tar.gx and uncompressed the directory R-2.3.0 is
 created.
 ./configure
 Make

 Typed R in the directory /root/downloads/R-2.3.0/bin/exec
 -bash-2.05b# pwd
 /root/downloads/R-2.3.0/bin/exec
 -bash-2.05b# R
 Fatal error: R home directory is not defined

Try:

.../R-2.3.0/bin/R

Or consider make install.

+ seth

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Re: [R] position of number at risk in survplot() graphs

[Frank E Harrell Jr]

Osman Al-Radi wrote:
 Dear R-help
 
 How can one get survplot() to place the number at risk just below the
 survival curve as opposed to the default which is just above the x-axis?
 I tried the code bellow but the result is not satisfactory as some numbers
 are repeated several times at different y coordinates and the position of
 the n.risk numbers corresponds to the x-axis tick marks not the survival
 curve time of censoring.
 
 n - 20
 set.seed(731)
 cens - 15*runif(n)
 h - .02*exp(2)
 dt - -log(runif(n))/h
 label(dt) - 'Follow-up Time'
 e - ifelse(dt = cens,1,0)
 dt - pmin(dt, cens)
 units(dt) - Year
 S - Surv(dt,e)
 km-survfit(S~1)
 survplot(km,n.risk=T,conf='none', y.n.risk=unique(summary(km)$surv))
 
 Any suggestion on addressing this problem would be apprecited.
 
 Also, is there a way to add a tick mark to the survival curve at times of
 censoring similar to the mark.time=T argument in plot.survplot()?
 
 Thanks
 
 Osman
 

Osman,

y.n.risk has to be a scalar and gives the y-coordinate of the bottom 
line of number at risk.  I take it that you want the numbers not all at 
the same height.  This will require a customization of survplot or 
fetching information from the km object and using text( ).

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

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Re: [R] Automate concatenation?

[Paul Roebuck]

On Tue, 30 May 2006, Robert Lundqvist wrote:

 I have this typical problem of joining a number of vectors with similar
 names - a1, a2,..., a10 - which should be concatenated into one. Using
 c(a1,a2,a3,a4,a5,a6,a,a8,a9,a10) naturally works, but I would like to do
 it with less manual input. My attempts to use paste() gives a vector of
 the vector names, see below. The question is how to do the the
 concatenation? Any suggestions?

 paste(a,1:10,sep=)

a1 - c(5, 4)
a2 - 2
a3 - 6:9
cmd - sprintf(c(%s),
   paste(a, 1:3, sep = , collapse = , ))
eval(parse(text = cmd))


--
SIGSIG -- signature too long (core dumped)

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Re: [R] Faster way to zero-pad a data frame...?

[Rolf Turner]

Why not something simple like:

# Toy example:
data1 - data.frame(time=c(0,1,5,10),events=c(1,30,14,4))
data2 - rep(0,11) # Or more generally data2 - rep(0,1+max(data1$time))

# You don't need a for loop! Use the indexing capabilities of R!
data2[data1$time+1] - data1$events # The ``+1'' is to allow for 0-origin.
data2 - ts(data2,start=0)

???

cheers,

Rolf Turner
[EMAIL PROTECTED]

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[R] sib TDT transmission/disequilibrium test

[Farrel Buchinsky]

Does anyone know if the sib TDT has been implemented in R

1. Spielman, R.S., and Ewens, W.J. (1998) A sibship test for linkage in the 
presence of association: the sib transmission/disequilibrium test. Am J Hum 
Genet 62, 450-458


-- 
Farrel Buchinsky, MD
Pediatric Otolaryngologist
Allegheny General Hospital
Pittsburgh, PA

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Re: [R] Automate concatenation?

[Rolf Turner]

Well, I like do.call():

puddy.tat - do.call(c,lapply(paste(a,1:10,sep=),get))

cheers,

Rolf Turner
[EMAIL PROTECTED]

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Re: [R] position of number at risk in survplot() graphs

[Osman Al-Radi]

Prof. Harrell,

Thank you for your response. Potting the n.risk on the curve is probably
only useful for a small data sets with a few censored events. The default
present method is more general and has more applications. For my purpose I
figured out a way of editing the postscript file in WinEdt.

Regards,

Osman



On 5/30/06, Frank E Harrell Jr [EMAIL PROTECTED] wrote:

 Osman Al-Radi wrote:
  Dear R-help
 
  How can one get survplot() to place the number at risk just below the
  survival curve as opposed to the default which is just above the x-axis?
  I tried the code bellow but the result is not satisfactory as some
 numbers
  are repeated several times at different y coordinates and the position
 of
  the n.risk numbers corresponds to the x-axis tick marks not the survival
  curve time of censoring.
 
  n - 20
  set.seed(731)
  cens - 15*runif(n)
  h - .02*exp(2)
  dt - -log(runif(n))/h
  label(dt) - 'Follow-up Time'
  e - ifelse(dt = cens,1,0)
  dt - pmin(dt, cens)
  units(dt) - Year
  S - Surv(dt,e)
  km-survfit(S~1)
  survplot(km,n.risk=T,conf='none', y.n.risk=unique(summary(km)$surv))
 
  Any suggestion on addressing this problem would be apprecited.
 
  Also, is there a way to add a tick mark to the survival curve at times
 of
  censoring similar to the mark.time=T argument in plot.survplot()?
 
  Thanks
 
  Osman
 

 Osman,

 y.n.risk has to be a scalar and gives the y-coordinate of the bottom
 line of number at risk.  I take it that you want the numbers not all at
 the same height.  This will require a customization of survplot or
 fetching information from the km object and using text( ).

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




-- 
Osman O. Al-Radi, MD, MSc, FRCSC
Fellow, Cardiovascular Surgery
The Hospital for Sick Children
University of Toronto, Canada

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[R] changes in RSiteSearch() and http://finzi.psych.upenn.edu/search.html

[Jonathan Baron]

This is about my searchable archive.  The function RSiteSearch()
now searches this archive.  I'm considering the following
changes.  If you have comments, please write me.  Try to avoid
cc'ing the list.

The Rhelp02a directory, which now contains all list mail from
2002 (about 100 MB), is getting larger and larger.  This probably
cannot go on forever, and performance might even improve if it
got smaller now.

I've considered two changes.  One is to start a new archive.
This would break RSiteSearch.  Even if that were fixed, the fix
would spread very slowly.

The second solution, which I plan to implement unless I hear a
better one, is to make this major archive include a maximum
four-year window, so that it would now start in 2003 rather than
2002, then (next year) would start in 2004, and so on.  The main
disadvantage is that references to current message in my archive
would be lost, because the message numbers would change.  (I will
try to save the old archive.)  There aren't very many of these
(244, including replies, in the archive itself).  Although the
archive would start in 2003, it would still be called Rhelp02a,
so that RSiteSearch(), etc., still work.

Maybe someone experienced with hypermail or namazu will tell me
that 100 MB is nothing and I should wait until I get a GB before
getting nervous.

I also plan to change the default number of items per page from
20 to 100.  This will not change in RSiteSearch() until the next
version, but this is not so big a deal.

Jon
-- 
Jonathan Baron, Professor of Psychology, University of Pennsylvania
Home page: http://www.sas.upenn.edu/~baron

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[R] How to choose columns in data.frame by parts of columns' names?

[Guo Wei-Wei]

Dear all,

I have a data.frame which has names as following.
[1] XG1  YG1  XEST YEST
[2] XNOEMP1  XNOEMP2 YNOEMP1  YNOEMP2
[3] XBUS10   XBUS10A XBUS10B  XBUS10C
[4] YBUS10   YBUS10A  YBUS10B  YBUS10C
[5] XOWNBUS  XSELFEST  YOWNBUS  YSELFEST

Those columns have names beginning with X or Y. Each X is paired
by a Y, e.g. XG1 and YG1, but they are not in the order of X Y
X Y  I want to combine X* and Y* like this:

data.new[,G1]  - (data.old[,XG1] + endata.use[,YG1])/2

How to choose columns by parts of names? For example, I can pick out
XG1 and YG1 because they have the common part G1.

Thank you.

Wei-Wei

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Re: [R] TsayData

[Spencer Graves]

  I don't know how to get the TsayData in fSeries.

  However, the data for the first edition of Tsay 2002) Analysis of 
Financial Time Series (Wiley) can be downloaded from 
http://www.gsb.uchicago.edu/fac/ruey.tsay/teaching/fts/;.  For the 
second edition, try 
http://www.gsb.uchicago.edu/fac/ruey.tsay/teaching/fts2/;.

  If you'd like further help, you might include a brief, self-contained 
example, so someone else can try what you tried and get (presumably) the 
same response, as suggested in the posting guide! 
(www.R-project.org/posting-guide.html).  Posts more consistent with that 
style often receive quicker and more helpful answers.

  Hope this helps.
  Spencer Graves

SUMANTA BASAK wrote:
 Hi,
 
 I'm trying to work with TsayData in fSeries package.
 How can i fetch any time series data of this package.
 Please advice.
 
 Thanks,
 Sumanta Basak.
 
 
 Send instant messages to your online friends http://in.messenger.yahoo.com
 
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Re: [R] How to choose columns in data.frame by parts of columns' names?

[Gabor Grothendieck]

On 5/30/06, Guo Wei-Wei [EMAIL PROTECTED] wrote:
 Dear all,

 I have a data.frame which has names as following.
 [1] XG1  YG1  XEST YEST
 [2] XNOEMP1  XNOEMP2 YNOEMP1  YNOEMP2
 [3] XBUS10   XBUS10A XBUS10B  XBUS10C
 [4] YBUS10   YBUS10A  YBUS10B  YBUS10C
 [5] XOWNBUS  XSELFEST  YOWNBUS  YSELFEST

 Those columns have names beginning with X or Y. Each X is paired
 by a Y, e.g. XG1 and YG1, but they are not in the order of X Y
 X Y  I want to combine X* and Y* like this:

 data.new[,G1]  - (data.old[,XG1] + endata.use[,YG1])/2

 How to choose columns by parts of names? For example, I can pick out
 XG1 and YG1 because they have the common part G1.



This gives all columns whose column name contains G1:

data.old[, regexpr(G1, colnames(data.old))  0]

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Re: [R] How to choose columns in data.frame by parts of columns' names?

[Peter Alspach]

Wei-wei

 I have a data.frame which has names as following.
 [1] XG1  YG1  XEST YEST
 [2] XNOEMP1  XNOEMP2 YNOEMP1  YNOEMP2
 [3] XBUS10   XBUS10A XBUS10B  XBUS10C
 [4] YBUS10   YBUS10A  YBUS10B  YBUS10C
 [5] XOWNBUS  XSELFEST  YOWNBUS  YSELFEST

 Those columns have names beginning with X or Y. Each X
 is paired by a Y, e.g. XG1 and YG1, but they are not in
 the order of X Y X Y  I want to combine X* and Y* like this:

 data.new[,G1]  - (data.old[,XG1] + endata.use[,YG1])/2

 How to choose columns by parts of names? For example, I can pick out
 XG1 and YG1 because they have the common part G1.

Not entirely sure what you mean but one approach might be to re-order
the columns so that they are in order.

yourNames
 [1] XG1  YG1  XEST YEST XNOEMP1  XNOEMP2
 [7] YNOEMP1  YNOEMP2  XBUS10   XBUS10A  XBUS10B  XBUS10C
[13] YBUS10   YBUS10A  YBUS10B  YBUS10C  XOWNBUS  XSELFEST
[19] YOWNBUS  YSELFEST
yourNames[order(substring(yourNames,2), substring(yourNames, 1,1))]
 [1] XBUS10   YBUS10   XBUS10A  YBUS10A  XBUS10B  YBUS10B
 [7] XBUS10C  YBUS10C  XEST YEST XG1  YG1
[13] XNOEMP1  YNOEMP1  XNOEMP2  YNOEMP2  XOWNBUS  YOWNBUS
[19] XSELFEST YSELFEST

gives an idea of what I mean ...

Peter Alspach



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Re: [R] na.pass

[Gabor Grothendieck]

Check out the source code to na.pass.  It just returns its first
argument unchanged:

 na.pass
function (object, ...)
object
environment: namespace:stats

On 5/30/06, Bahamonde Natalia [EMAIL PROTECTED] wrote:
 Hello...

 What does na.pass?


x=c(2.4, 2.4, 1.9, 2.5, 2.1)
xNA=replace(x, 3, NA)

p=c(acf(x, type=c(covariance), plot=FALSE)$acf)
pNA=c(acf(xNA, type=c(covariance), na.action=na.pass, plot=FALSE)$acf)

   p
 [1]  0.05040 -0.03112  0.00816  0.00224 -0.00448
   pNA
 [1]  0.02250 -0.01167  0.00250 -0.00100 -0.00250

 In the manual say na.pass returns objet unchanged...

 Thanks, Natalia

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Re: [R] How to choose columns in data.frame by parts of columns' names?

[Guo Wei-Wei]

Thank you. I made a mistake in my previous email. What I mean is:

data.new[,G1]  - (data.old[,XG1] + data.old[,YG1])/2

data.old[, regexpr(G1, colnames(data.old))  0]
is a nice way, but there are about 100 X*s and Y*s. Can I do some
comparision on all those column names and get columns with similar
parts?

2006/5/31, Gabor Grothendieck [EMAIL PROTECTED]:
 On 5/30/06, Guo Wei-Wei [EMAIL PROTECTED] wrote:
  Dear all,
 
  I have a data.frame which has names as following.
  [1] XG1  YG1  XEST YEST
  [2] XNOEMP1  XNOEMP2 YNOEMP1  YNOEMP2
  [3] XBUS10   XBUS10A XBUS10B  XBUS10C
  [4] YBUS10   YBUS10A  YBUS10B  YBUS10C
  [5] XOWNBUS  XSELFEST  YOWNBUS  YSELFEST
 
  Those columns have names beginning with X or Y. Each X is paired
  by a Y, e.g. XG1 and YG1, but they are not in the order of X Y
  X Y  I want to combine X* and Y* like this:
 
  data.new[,G1]  - (data.old[,XG1] + endata.use[,YG1])/2
 
  How to choose columns by parts of names? For example, I can pick out
  XG1 and YG1 because they have the common part G1.
 


 This gives all columns whose column name contains G1:

 data.old[, regexpr(G1, colnames(data.old))  0]


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Re: [R] How to choose columns in data.frame by parts of columns' names?

[Guo Wei-Wei]

Peter,

Thank you,
I made a mistake in my previous email. What I mean is:

data.new[,G1]  - (data.old[,XG1] + data.old[,YG1])/2

Does your way have effects on data? or only have effects on those column names?
I tried on my data and get a list of numbers. Can I rearrange the
order of columns of data.frame by your way?

2006/5/31, Peter Alspach [EMAIL PROTECTED]:

 Wei-wei

 yourNames
  [1] XG1  YG1  XEST YEST XNOEMP1  XNOEMP2
  [7] YNOEMP1  YNOEMP2  XBUS10   XBUS10A  XBUS10B  XBUS10C
 [13] YBUS10   YBUS10A  YBUS10B  YBUS10C  XOWNBUS  XSELFEST
 [19] YOWNBUS  YSELFEST
 yourNames[order(substring(yourNames,2), substring(yourNames, 1,1))]
  [1] XBUS10   YBUS10   XBUS10A  YBUS10A  XBUS10B  YBUS10B
  [7] XBUS10C  YBUS10C  XEST YEST XG1  YG1
 [13] XNOEMP1  YNOEMP1  XNOEMP2  YNOEMP2  XOWNBUS  YOWNBUS
 [19] XSELFEST YSELFEST

 gives an idea of what I mean ...

 Peter Alspach



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Re: [R] How to choose columns in data.frame by parts of columns' names?

[Gabor Grothendieck]

This is not restricted to single matches:

 colnames(iris)
[1] Sepal.Length Sepal.Width  Petal.Length Petal.Width  Species
 regexpr(Sepal, colnames(iris))  0
[1]  TRUE  TRUE FALSE FALSE FALSE

On 5/30/06, Guo Wei-Wei [EMAIL PROTECTED] wrote:
 Thank you. I made a mistake in my previous email. What I mean is:

 data.new[,G1]  - (data.old[,XG1] + data.old[,YG1])/2

 data.old[, regexpr(G1, colnames(data.old))  0]
 is a nice way, but there are about 100 X*s and Y*s. Can I do some
 comparision on all those column names and get columns with similar
 parts?

 2006/5/31, Gabor Grothendieck [EMAIL PROTECTED]:
  On 5/30/06, Guo Wei-Wei [EMAIL PROTECTED] wrote:
   Dear all,
  
   I have a data.frame which has names as following.
   [1] XG1  YG1  XEST YEST
   [2] XNOEMP1  XNOEMP2 YNOEMP1  YNOEMP2
   [3] XBUS10   XBUS10A XBUS10B  XBUS10C
   [4] YBUS10   YBUS10A  YBUS10B  YBUS10C
   [5] XOWNBUS  XSELFEST  YOWNBUS  YSELFEST
  
   Those columns have names beginning with X or Y. Each X is paired
   by a Y, e.g. XG1 and YG1, but they are not in the order of X Y
   X Y  I want to combine X* and Y* like this:
  
   data.new[,G1]  - (data.old[,XG1] + endata.use[,YG1])/2
  
   How to choose columns by parts of names? For example, I can pick out
   XG1 and YG1 because they have the common part G1.
  
 
 
  This gives all columns whose column name contains G1:
 
  data.old[, regexpr(G1, colnames(data.old))  0]
 

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Re: [R] How to choose columns in data.frame by parts of columns' names?

[Guo Wei-Wei]

Gabor and Peter,

Thank you. Both of you give me excellent ways. I have a further problem.

How can I get the common parts of column names as column names in a
new data.frame? For example, I combines data of XG1 and YG1 in
data.old and get a new column in data.new named G1. Can It be done
automaticlly?

 data.new[,G1]  - (data.old[,XG1] + data.old[,YG1])/2


2006/5/31, Gabor Grothendieck [EMAIL PROTECTED]:
 This is not restricted to single matches:

  colnames(iris)
 [1] Sepal.Length Sepal.Width  Petal.Length Petal.Width  Species
  regexpr(Sepal, colnames(iris))  0
 [1]  TRUE  TRUE FALSE FALSE FALSE


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