date:20121121

HI,
Your question is not very clear to me.

dat1-data.frame(x=c(1,1,2,2,2,3,4,4),y=c(23,45,87,46,78,12,87,79))
If I select randomly 80% of data:
 dat1[sample(nrow(dat1),0.8*nrow(dat1)),]
#  x  y
#4 2 46
#6 3 12
#7 4 87
#3 2 87
#2 1 45
#8 4 79

In this case, it is a mix of all the records that was in the original data.  
So, if I understand correctly, then you want to get the original data, instead 
of the 80% data (as you want all the records).  But, suppose if the sample is:
 set.seed(244)
 dat2-dat1[sample(nrow(dat1),0.8*nrow(dat1)),]
 dat2
#  x  y
#2 1 45
#7 4 87
#8 4 79
#1 1 23
#5 2 78
#4 2 46

You wanted to eliminate the records 3 from the original dataset.  Is that what 
you meant?
If that is the case:
dat1[dat1$x%in%dat2$x,]
#  x  y
#1 1 23
#2 1 45
#3 2 87
#4 2 46
#5 2 78
#7 4 87
#8 4 79

A.K.





- Original Message -
From: Madhu Ganganapalli mganganapa...@upstreamsoftware.com
To: r-help@r-project.org r-help@r-project.org
Cc: 
Sent: Wednesday, November 21, 2012 4:10 AM
Subject: Re: [R] FW: Select a random subset of rows out of matrix

Hi,
  This is Madhu and I have a following doubt please give a solution...

**i have the following data frame
from this i want to select a 80% of data randomly in such a way that
if the selected records are 1 and then we have to get the all records 
corresponding to 1
similarly for 2 also and soon. help me

    data-data.frame(x=c(1,1,2,2,2,3,4,4,4),y=c(23,45,87,46,78,12,87,79));**


Regards,
Madhu.

    [[alternative HTML version deleted]]

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Re: [R] FW: Select a random subset of rows out of matrix


Hello,

It's not, in general, possible to fullfill the two constraints, to both 
select 80% of the rows and to select all x %in% sampled x. Maybe the 
following is a way close enough.


set.seed(244)
s - sample(unique(dat$x), length(unique(dat$x))*0.8, prob = table(dat$x))
dat[dat$x %in% s,]


Hope this helps,

Rui Barradas
Em 21-11-2012 13:25, arun escreveu:

HI,
Your question is not very clear to me.

dat1-data.frame(x=c(1,1,2,2,2,3,4,4),y=c(23,45,87,46,78,12,87,79))
If I select randomly 80% of data:
  dat1[sample(nrow(dat1),0.8*nrow(dat1)),]
#  x  y
#4 2 46
#6 3 12
#7 4 87
#3 2 87
#2 1 45
#8 4 79

In this case, it is a mix of all the records that was in the original data.  
So, if I understand correctly, then you want to get the original data, instead 
of the 80% data (as you want all the records).  But, suppose if the sample is:
  set.seed(244)
  dat2-dat1[sample(nrow(dat1),0.8*nrow(dat1)),]
  dat2
#  x  y
#2 1 45
#7 4 87
#8 4 79
#1 1 23
#5 2 78
#4 2 46

You wanted to eliminate the records 3 from the original dataset.  Is that what 
you meant?
If that is the case:
dat1[dat1$x%in%dat2$x,]
#  x  y
#1 1 23
#2 1 45
#3 2 87
#4 2 46
#5 2 78
#7 4 87
#8 4 79

A.K.





- Original Message -
From: Madhu Ganganapalli mganganapa...@upstreamsoftware.com
To: r-help@r-project.org r-help@r-project.org
Cc:
Sent: Wednesday, November 21, 2012 4:10 AM
Subject: Re: [R] FW: Select a random subset of rows out of matrix

Hi,
   This is Madhu and I have a following doubt please give a solution...

**i have the following data frame
from this i want to select a 80% of data randomly in such a way that
if the selected records are 1 and then we have to get the all records 
corresponding to 1
similarly for 2 also and soon. help me

 data-data.frame(x=c(1,1,2,2,2,3,4,4,4),y=c(23,45,87,46,78,12,87,79));**


Regards,
Madhu.

 [[alternative HTML version deleted]]

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Re: [R] Sourcing files with Umlaut in path no longer works

2012-11-21 Thread peter dalgaard


On Nov 21, 2012, at 12:49 , Ulrike Grömping wrote:

 Dear helpeRs,
 
 on my new machine - Windows 7 64Bit, R 2.15-2, I can no longer source files 
 for which the path contains the o-Umlaut (ö). As a historical burden, my 
 username is Grömping and contains that Umlaut (I wouldn't have chosen it now, 
 but decided to keep it for easy transfer). I have had the difficulty that I 
 couldn't use the R CMD tools on files with Umlauts in the path (and solved it 
 by having a dedicated directory without umlaut in the path for that purpose), 
 but so far (Windows XP and earlier R versions) sourcing from within R worked 
 fine.
 
 The issue is that the o-Umlaut (ö) is always replaced by a capital A with 
 tilde above together with a paragraph symbol (Ã¶). Apparently, the c3b6  used 
 for the o-Umlaut in some encoding is translated back to the Ã¶. Can I somehow 
 fix this? Or is it a bug?

Without a reproducible example, it is hard to tell where things are going 
wrong, but the symptoms are those of utf-8 codes being sent to something that 
expects latin1 (or one of the 8th bit ASCII variations, anyway).

 
 Best regards,
 Ulrike
 
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-- 
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk  Priv: pda...@gmail.com

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Re: [R] Sourcing files with Umlaut in path no longer works

2012-11-21 Thread Duncan Murdoch


On 21/11/2012 6:49 AM, Ulrike Grömping wrote:

Dear helpeRs,

on my new machine - Windows 7 64Bit, R 2.15-2, I can no longer source
files for which the path contains the o-Umlaut (ö). As a historical
burden, my username is Grömping and contains that Umlaut (I wouldn't
have chosen it now, but decided to keep it for easy transfer). I have
had the difficulty that I couldn't use the R CMD tools on files with
Umlauts in the path (and solved it by having a dedicated directory
without umlaut in the path for that purpose), but so far (Windows XP and
earlier R versions) sourcing from within R worked fine.

The issue is that the o-Umlaut (ö) is always replaced by a capital A
with tilde above together with a paragraph symbol (Ã¶). Apparently, the
c3b6  used for the o-Umlaut in some encoding is translated back to the
Ã¶. Can I somehow fix this? Or is it a bug?



I can't reproduce this.  The file d:/temp/Grömping/test.R contains

cat(it worked!\n)

I am on Windows 7 64bit, and I see the following in both 32 and 64 bit R:

 source(d:/temp/Grömping/test.R)
it worked!

If I look at Encoding(d:/temp/Grömping/test.R) I get latin1.  If I run

iconv(d:/temp/Grömping/test.R, latin1, UTF-8)

I get a proper UTF-8 string, but if I run

iconv(d:/temp/Grömping/test.R, latin1, utf8)

I get a string with encoding marked as unknown, and displayed as

[1] d:/temp/GrÃ¶mping/test.R

So perhaps you or we have used an unofficial name of the UTF-8 encoding 
in some conversion.  We need to know exactly what you did to pursue this.


Duncan Murdoch


Duncan Murdoch

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[R] Create BATCH file

2012-11-21 Thread R_Antony

Hi,

I have a .R file written many functions into that. My requirement what is,
i need  to create a batch file for this.
No idea, how to create it. Tried it many other ways, but no result !. 
Could anyone please help me out ?

Thanks.
Antony.




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[R] Statistical significance in robust estimation rlm()

2012-11-21 Thread fxen3k

Hi there,

I used the rlm() function for doing a robust estimation based on
M-estimates.
Obviously, you only get the estimate, standard error and t- value by
implementing this rlm() function.
So, how can I say if a coefficient is statistical significant without the
presence of a p-value?

Thanks in advance!





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[R] [lattice] print only legend

2012-11-21 Thread AnjaM

Is it possible to plot (and save) only the legend of a lattice plot? Of
course I could make the axes and axis labels transparent and use an empty
panel function, but additionally to being a very dirty solution, there
would be still a lot of free space on the plot, while I would like to save
only the legend as a separate plot without any white space around it. Is
there a (nice) way to do this?



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Re: [R] lattice density plot: add vertical lines at groupwise medians for all panels

2012-11-21 Thread Anja Mirenska

Dear José

Thanks for your reply.

I've tried out your suggestion, but this code plots the overall median for
each group rather than plotting the medians for the individual panels. What
I am looking for is a way to calculate the medians per group for each
panel. Sorry if I didn't state it clearly before.

Best wishes

Anja


2012/11/21 Jose Iparraguirre jose.iparragui...@ageuk.org.uk

 Dear AnjaM

 There may be better ways of dealing with this, but the following works:


 densityplot(~gcsescore | factor(score), groups=gender, data=Chem97,
 auto.key=TRUE,  plot.points=FALSE, ref=TRUE,
 panel=function(x,...){
   panel.densityplot(x,...)
   median.values - tapply(Chem97$gcsescore, Chem97$gender,
 median)
   panel.abline(v=median.values, col=c(blue,red))
 panel.grid(v=2)}
 )


 Regards,


 José Iparraguirre
 Chief Economist
 Age UK


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
 On Behalf Of AnjaM
 Sent: 20 November 2012 14:13
 To: r-help@r-project.org
 Subject: [R] lattice density plot: add vertical lines at groupwise medians
 for all panels

 Suppose you have the following code:

 ## Start code##

 data(Chem97, package=mlmRev)

 densityplot(~gcsescore | factor(score), groups=gender, data=Chem97,
 auto.key=TRUE,  plot.points=FALSE, ref=TRUE,
 panel=function(x,...){
   panel.densityplot(x,...)
   median.values - median(x)
   panel.abline(v=median.values)}
 )

 ##  End code##

 For some reason I don't understand, this adds only the median for one group
 (in this case for gender == M) into each panel. How do I calculate and add
 the medians for both groups in the right group colour?

 This is what I tried:

 ## Start code##

 densityplot(~gcsescore | factor(score), groups=gender, data=Chem97,
 auto.key=TRUE,  plot.points=FALSE, ref=TRUE,
 panel=function(x,..., groups){
   panel.densityplot(x,..., groups)
   median.values - tapply(x, groups, median)
   panel.abline(v=median.values)}
 )

 ##  End code##

 However, this gives an error.  What's going wrong and how can I solve this
 problem?



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[R] dúvidas com matriz de correlação e covariancia

2012-11-21 Thread alanarocha


Bom dia eu chamo-me Ana, estou a tentar fazer matriz de correlação e
covariancia para comparar 4 variaveis e saber quais são as mehores...
fiz o segundo codigo mas devo tar a fazer alguma confusão para nao me 
aparecer o output correcto:

cor(dados[,2:5],method=c(pearson))
cor_with_p_test(dados[2:5])
ct - cor.test(dados[,2:5])
ct$p
ct$v

#matriz de covariancias
cov(dados[,2:5], na.rm=TRUE)
mando o arquivo de dados em anexo.
Aguardando uma resposta, desde já agradeço.
eSubescrevo-m Ana Rocha

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[R] creat an interactive graph

2012-11-21 Thread Henry Smith

Hello!

Does anyone get any idea how to generate a following graph by using R?

http://www.fastcodesign.com/multisite_files/codesign/imagecache/inline-large/post-inline/inline-north-carolina-gay-rights.jpg

Any information or hint will be highly appreciated.

Kind regards,
Henry

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Re: [R] Stepwise analysis with fixed variables

2012-11-21 Thread Einat

These are my questions: 

1. For example, if this is my code: 

RegModel =
lm(glucose~sex+BMI+height+weight+education+ses,weight=w_without_non_response)
  summary(RegModel) 
step(RegModel, direction =backward,scope=list(lower=?,upper=?)) 

and I want the sex and height variables to be fixed, but the rest of the
variables to go into the backward analysis, how should I write the scope
function?
   
2.How can I add an alpha level to the step function as a criterion for the
backward regression analysis?
   
Thank you :) 



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[R] size of the side panel in gvisMotionChart

2012-11-21 Thread xavier abulker

Hello,

Is there an option parameter to increase the size of the bottom right side 
panel in gvisMotionChart?

My data labels are too long for this panel (for example Call butterfly spread 
@ 2400/2600/2800) and the names are always cut even if I increase the width 
and height.

Regards
Xavier

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[R] Creating a frequency table for binomial varaible

Hello, 
I have simulated 30  observations from a binomial(5,0.1) distribution. 
Now I need to make frequency table( that means I need to tally how many 0's
, 1's  2's... 5's)
I know that the simple  R function table() will do this, but I am afraid
that some times I may get zero frequency for some particular values (for
example in the above there are 5-0's 10-1's , 14-2's, 10-3's , 11-4's but no
any 5's )

So I want to make by frequecy table ( as a date frame) as
value   freq
0   5
1  10
2  14
3  10
4   11
5   0


How can I create such a table?
Forgive me if this is a very basic question. I am new to R.

Thank you very much. 





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Re: [R] Stepwise analysis with fixed variables

2012-11-21 Thread Einat

These are my questions:

1. For example, if this is my code: 

RegModel =
lm(glucose~sex+BMI+height+weight+education+ses,weight=w_without_non_response)
 summary(RegModel) 
step(RegModel, direction =backward,scope=list(lower=?,upper=?)) 

and I want the sex and height variables to be fixed, but the rest of the
variables to go into the backward analysis, how should I write the scope
function?
 
2.How can I add an alpha level to the step function as a criterion for the
backward regression analysis?
 
Thank you :)



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Re: [R] Ben Bolker's '‘emdbook’ Package , rbetabinom

I understood that from the source code. Currently I am referring the
suggested articles. 
Thank you very much for the off-list answer and this reply. 




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Re: [R] Controlling the number of interactions of a lme

2012-11-21 Thread cleberchaves

Is it, bbolker!

Thank you very, very much!



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[R] Weighted least squares

2012-11-21 Thread rbowman16

Hi everyone, 

I admit I am a bit of an R novice, and I was hoping someone could help me
with this error message:

Warning message:
In lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
  extra arguments weigths are just disregarded.


My equation is:


 lm( Y ~ X1 + X2 + X3, weigths = seq(0.1, 1, by = 0.1))



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Re: [R] Creating a frequency table for binomial varaible

As I have to to this in a simulation study for 1000 such binomial variables,
I have created a R function as below..
Am I doing in the correct way? or is there any other simplest ways?

#using a user defined function to create a frequency distribution
create.freq.table- function(x){
  values-seq(0,5,by=1)
  level-seq(0,6,by=1)
  freq.cut-cut(x,breaks=level,right=FALSE)
  int.freq.table-table(freq.cut)
  int.dataframe-data.frame(int.freq.table)
  final.freq.table-data.frame(cbind(values, Freq=int.dataframe[,2]))
  return(final.freq.table)
}




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Re: [R] Creating a frequency table for binomial varaible


Hello,

Try

table(x)  # or table(dat$value)

Hope this helps,

Rui Barradas
Em 21-11-2012 13:19, arun4 escreveu:

Hello,
I have simulated 30  observations from a binomial(5,0.1) distribution.
Now I need to make frequency table( that means I need to tally how many 0's
, 1's  2's... 5's)
I know that the simple  R function table() will do this, but I am afraid
that some times I may get zero frequency for some particular values (for
example in the above there are 5-0's 10-1's , 14-2's, 10-3's , 11-4's but no
any 5's )

So I want to make by frequecy table ( as a date frame) as
value   freq
0   5
1  10
2  14
3  10
4   11
5   0


How can I create such a table?
Forgive me if this is a very basic question. I am new to R.

Thank you very much.





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Re: [R] dúvidas com matriz de correlação e covariancia

Olá

É melhor escreveres em inglês, esta lista é anglófona.
As for your file, r-help doesn't like attachments, use ?dput instead:

dput( head(dados, 30) )  # paste the output of this in a post

Hope this helps,

Rui Barradas
Em 21-11-2012 11:03, alanaro...@sapo.pt escreveu:
 Bom dia eu chamo-me Ana, estou a tentar fazer matriz de correlação e 
 covariancia para comparar 4 variaveis e saber quais são as mehores...
 fiz o segundo codigo mas devo tar a fazer alguma confusão para nao me 
 aparecer o output correcto:
 cor(dados[,2:5],method=c(pearson))
 cor_with_p_test(dados[2:5])
 ct - cor.test(dados[,2:5])
 ct$p
 ct$v

 #matriz de covariancias
 cov(dados[,2:5], na.rm=TRUE)
 mando o arquivo de dados em anexo.
 Aguardando uma resposta, desde já agradeço.
 eSubescrevo-m Ana Rocha



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Re: [R] Weighted least squares

2012-11-21 Thread Martyn Byng

Hi,

That message usually means that an unknown argument has been supplied to
the function.

In this case you have spelt weights incorrectly.

Best wishes

Martyn

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of rbowman16
Sent: 21 November 2012 14:34
To: r-help@r-project.org
Subject: [R] Weighted least squares

Hi everyone, 

I admit I am a bit of an R novice, and I was hoping someone could help
me with this error message:

Warning message:
In lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
  extra arguments weigths are just disregarded.


My equation is:


 lm( Y ~ X1 + X2 + X3, weigths = seq(0.1, 1, by = 0.1))



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Re: [R] Creating a frequency table for binomial varaible

2012-11-21 Thread William Dunlap

 I know that the simple  R function table() will do this, but I am afraid
 that some times I may get zero frequency for some particular values 

Make a factor out of your data, specifying all the levels you want counts
for, and pass that factor to table().  E.g.,

  x - rep(0:6, c(5,2,0,3,0,4,0))
  table(x)
   x
   0 1 3 5 
   5 2 3 4
  table(factor(x, levels=0:10))
   
0  1  2  3  4  5  6  7  8  9 10 
5  2  0  3  0  4  0  0  0  0  0

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf
 Of arun4
 Sent: Wednesday, November 21, 2012 5:20 AM
 To: r-help@r-project.org
 Subject: [R] Creating a frequency table for binomial varaible
 
 Hello,
 I have simulated 30  observations from a binomial(5,0.1) distribution.
 Now I need to make frequency table( that means I need to tally how many 0's
 , 1's  2's... 5's)
 I know that the simple  R function table() will do this, but I am afraid
 that some times I may get zero frequency for some particular values (for
 example in the above there are 5-0's 10-1's , 14-2's, 10-3's , 11-4's but no
 any 5's )
 
 So I want to make by frequecy table ( as a date frame) as
 value   freq
 0   5
 1  10
 2  14
 3  10
 4   11
 5   0
 
 
 How can I create such a table?
 Forgive me if this is a very basic question. I am new to R.
 
 Thank you very much.
 
 
 
 
 
 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Creating-a-frequency-
 table-for-binomial-varaible-tp4650286.html
 Sent from the R help mailing list archive at Nabble.com.
 
 __
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Re: [R] Stepwise analysis with fixed variables

On Nov 21, 2012, at 6:41 AM, Einat einatgra...@gmail.com wrote:

 These are my questions: 
 
 1. For example, if this is my code: 
 
 RegModel =
 lm(glucose~sex+BMI+height+weight+education+ses,weight=w_without_non_response)
 summary(RegModel) 
 step(RegModel, direction =backward,scope=list(lower=?,upper=?)) 
 
 and I want the sex and height variables to be fixed, but the rest of the
 variables to go into the backward analysis, how should I write the scope
 function?
 
 2.How can I add an alpha level to the step function as a criterion for the
 backward regression analysis?
 
 Thank you :) 



First, I strongly suggest that you search on the problems associated with using 
stepwise regression and alternative approaches. This subject has been discussed 
ad infinitum on this list and is not a reasonable approach to covariate 
selection.

To your first question, which Uwe replied to and which I thought was pretty 
clear:

  scope = list(upper = ~ sex + BMI + height + weight + education + ses, 
   lower = ~ sex + height)

It is described in the Details section of ?step and there is an example of this 
in ?stepAIC in VR's MASS package, which is a default part of R and is linked 
in the See Also section of ?step.

To your second question, you can't. It uses AIC and this has also been 
discussed frequently on this list. You might look at Frank's fastbw() function 
in his 'rms' package on CRAN.

Regards,

Marc Schwartz

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Re: [R] Stepwise analysis with fixed variables

2012-11-21 Thread PIKAL Petr

Hi

I am not at all an expert in step. From the help page I inferred that

step(lm1, scope=list(lower=~Catholic))

keeps Catholic in model.

So something like

step(RegModel, direction =backward,scope=list(lower=~sex+height))

shall leave those two in a model.

AFAIK there is no parameter alpha in a step function.

Regards
Petr

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of Einat
 Sent: Wednesday, November 21, 2012 1:41 PM
 To: r-help@r-project.org
 Subject: Re: [R] Stepwise analysis with fixed variables
 
 These are my questions:
 
 1. For example, if this is my code:
 
 RegModel =
 lm(glucose~sex+BMI+height+weight+education+ses,weight=w_without_non_res
 ponse)
   summary(RegModel)
 step(RegModel, direction =backward,scope=list(lower=?,upper=?))
 
 and I want the sex and height variables to be fixed, but the rest of
 the variables to go into the backward analysis, how should I write the
 scope function?
 
 2.How can I add an alpha level to the step function as a criterion for
 the backward regression analysis?
 
 Thank you :)
 
 
 
 --
 View this message in context: http://r.789695.n4.nabble.com/Stepwise-
 analysis-with-fixed-variables-tp4650015p4650283.html
 Sent from the R help mailing list archive at Nabble.com.
 
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Re: [R] creat an interactive graph

2012-11-21 Thread R. Michael Weylandt

I'd use ggplot2 and combine geom_bar() and coord_polar()

Google the LearningR blog and find the consultant's chart entry.

Michael

On Wednesday, November 21, 2012, Henry Smith wrote:

 Hello!

 Does anyone get any idea how to generate a following graph by using R?


 http://www.fastcodesign.com/multisite_files/codesign/imagecache/inline-large/post-inline/inline-north-carolina-gay-rights.jpg

 Any information or hint will be highly appreciated.

 Kind regards,
 Henry

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[R] lists as matrix cells ?

2012-11-21 Thread Asis Hallab

Dear R experts,

since more or less half a year I am using R.

In many of my computations I construct huge matrices. Often I do so using
'cbind' on named lists:

do.call( 'cbind',
  list(
Column_A=list(Row_one=1.0, Row_two=2.0, Row_three=3.0),
Column_B=list(Row_one=4.0, Row_two=5.0, Row_three=6.0)
  )
)

# Returns:

  Column_A Column_B
Row_one   14
Row_two   25
Row_three 36

In some cases I even construct matrices with lists as cell content:

do.call( 'cbind',
  list(
Column_A=list(Row_one=list(1.0, 2.0), Row_two=list(2.0, 3.0),
Row_three=list(3.0, 4.0)),
Column_B=list(Row_one=list(4.0, 5.0), Row_two=list(5.0, 6.0),
Row_three=list(6.0, 7.0))
  )
)

# Returns:
  Column_A Column_B
Row_one   List,2   List,2
Row_two   List,2   List,2
Row_three List,2   List,2

Interestingly I seem not to be able to initialize a 3*2 matrix and
subsequently fill its cells with lists in order to produce the latter
example:

m - matrix( nrow=3, ncol=2,
  dimnames=list(
c(Row_one, Row_two, Row_three),
c(Column_A, Column_B)
  )
)
# Returns:
  Column_A Column_B
Row_one NA   NA
Row_two NA   NA
Row_three   NA   NA

# The following expressions produce the same error:
m[ Row_one, Column_A ]   - list(1.0, 2.0)
m[[ Row_one, Column_A ]] - list(1.0, 2.0)
m[ 1,1 ]   - list(1.0, 2.0)
m[[ 1,1 ]] - list(1.0, 2.0)

# Error returned by each of the above expressions:
*number of items to replace is not a multiple of replacement length*

*So, now my questions:*
*1)* What am I doing wrong? How to get the above to work?

*2)* Or am I misusing matrices in R? Is it just by coincidence ( bug ) that
my 'cbind' or 'rbind' calls can generate matrices with lists as cell
content, while I seem not to be able to do so by direct assignment (as in
above example expressions)?

Any ideas, comments and help will be much appreciated!
Kind regards!
Josef

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[R] Scaling values 0-255 - -1 , 1 - how can this be done?

2012-11-21 Thread Brian Feeny


I have a dataframe in which I have values 0-255, I wish to transpose them such 
that:

if value   127.5 value = 1
if value  127.5 value = -1

I did something similar using the binarize function of the biclust package, 
this transforms my dataframe to 0 and 1 values, but I wish
to use -1 and 1 and looking for a way in R to do this.

Brian

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Re: [R] Creating a frequency table for binomial varaible

Thank you A.K
Btw in which package count() is available? 



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[R] installing Rmpi on centos 6 with mpich

2012-11-21 Thread Ricardo Román Brenes

Hello everyone

im trying to install Rmpi library on centos 6. I have already installed
mpich2 1.4

[root@localhost ~]# R --version
 R version 2.15.1 (2012-06-22) -- Roasted Marshmallows
 Copyright (C) 2012 The R Foundation for Statistical Computing
 ISBN 3-900051-07-0
 Platform: x86_64-redhat-linux-gnu (64-bit)



[root@localhost ~]# mpich2version
 MPICH2 Version: 1.4.1p1
 MPICH2 Release date: Thu Sep  1 13:53:02 CDT 2011
 MPICH2 Device: ch3:nemesis
 MPICH2 configure:
 MPICH2 CC: gcc-O2
 MPICH2 CXX: c++   -O2
 MPICH2 F77: gfortran   -O2
 MPICH2 FC: f95   -O2




when i issue the install order i get this output:


 install.packages(Rmpi)
 Installing package(s) into /usr/lib64/R/library
 (as lib is unspecified)
 trying URL 'http://cran.rstudio.com/src/contrib/Rmpi_0.6-1.tar.gz'
 Content type 'application/x-gzip' length 92977 bytes (90 Kb)
 opened URL
 ==
 downloaded 90 Kb
 * installing *source* package Rmpi ...
 ** package Rmpi successfully unpacked and MD5 sums checked
 checking for gcc... gcc -m64 -std=gnu99
 checking whether the C compiler works... yes
 checking for C compiler default output file name... a.out
 checking for suffix of executables...
 checking whether we are cross compiling... no
 checking for suffix of object files... o
 checking whether we are using the GNU C compiler... yes
 checking whether gcc -m64 -std=gnu99 accepts -g... yes
 checking for gcc -m64 -std=gnu99 option to accept ISO C89... none needed
 I am here /usr/local and it is OpenMPI
 Trying to find mpi.h ...
 Found in /usr/local/include
 Trying to find libmpi.so or libmpich.a ...
 checking for main in -lmpi... no
 libmpi not found. exiting...
 ERROR: configuration failed for package Rmpi
 * removing /usr/lib64/R/library/Rmpi
 The downloaded source packages are in
 /tmp/RtmpBvTTqc/downloaded_packages
 Updating HTML index of packages in '.Library'
 Making packages.html  ... done
 Warning message:
 In install.packages(Rmpi) :
   installation of package Rmpi had non-zero exit status



I have tried with different configure.args but the result is the same, i
cant seem to install it...

Is there someone who has been through the same problem? or that knows a
guide i can follow or something?

Thanks in advance

Ricardo

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Re: [R] Creating a frequency table for binomial varaible

Thank you Bill Dunlap . This seems very simple. 



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[R] Help with loop

2012-11-21 Thread Hefri

Hi,

I have used R for some time, but managed to avoid writing  loops. But this
time I am afraid there is no way around it. 

I have a dataframe with time and salinity (see below). I would like to
extract the time intervals where salinity changes by less than 0.05. So
using the values below this would mean that a subset was made from
2003-07-19 to 2003-07-24, where the change in salinity exceeded 0.05, and
then stating again from 2003-07-24. In other words, I want to make subset of
the time intervals where the one variable stays stabile within a certain set
range. 

 head (D, 10)
   Salt   time
1  35.65114 2003-07-19
2  35.64226 2003-07-20
3  35.62411 2003-07-21
4  35.62473 2003-07-22
5  35.65893 2003-07-23
6  35.70140 2003-07-24
7  35.62157 2003-07-25
8  35.64122 2003-07-26
9  35.63515 2003-07-27
10 35.63798 2003-07-28

Is this possible? Can some please help with how I need to write the loop to
achieve this?

Many thanks, Helen




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[R] installing Rmpi on centos 6 with mpich

2012-11-21 Thread Ricardo Román Brenes

Hello everyone

im trying to install Rmpi library on centos 6. I have already installed
mpich2 1.4

[root@localhost ~]# R --version
 R version 2.15.1 (2012-06-22) -- Roasted Marshmallows
 Copyright (C) 2012 The R Foundation for Statistical Computing
 ISBN 3-900051-07-0
 Platform: x86_64-redhat-linux-gnu (64-bit)



[root@localhost ~]# mpich2version
 MPICH2 Version: 1.4.1p1
 MPICH2 Release date: Thu Sep  1 13:53:02 CDT 2011
 MPICH2 Device: ch3:nemesis
 MPICH2 configure:
 MPICH2 CC: gcc-O2
 MPICH2 CXX: c++   -O2
 MPICH2 F77: gfortran   -O2
 MPICH2 FC: f95   -O2




when i issue the install order i get this output:


 install.packages(Rmpi)
 Installing package(s) into /usr/lib64/R/library
 (as lib is unspecified)
 trying URL 'http://cran.rstudio.com/src/contrib/Rmpi_0.6-1.tar.gz'
 Content type 'application/x-gzip' length 92977 bytes (90 Kb)
 opened URL
 ==
 downloaded 90 Kb
 * installing *source* package Rmpi ...
 ** package Rmpi successfully unpacked and MD5 sums checked
 checking for gcc... gcc -m64 -std=gnu99
 checking whether the C compiler works... yes
 checking for C compiler default output file name... a.out
 checking for suffix of executables...
 checking whether we are cross compiling... no
 checking for suffix of object files... o
 checking whether we are using the GNU C compiler... yes
 checking whether gcc -m64 -std=gnu99 accepts -g... yes
 checking for gcc -m64 -std=gnu99 option to accept ISO C89... none needed
 I am here /usr/local and it is OpenMPI
 Trying to find mpi.h ...
 Found in /usr/local/include
 Trying to find libmpi.so or libmpich.a ...
 checking for main in -lmpi... no
 libmpi not found. exiting...
 ERROR: configuration failed for package Rmpi
 * removing /usr/lib64/R/library/Rmpi
 The downloaded source packages are in
 /tmp/RtmpBvTTqc/downloaded_packages
 Updating HTML index of packages in '.Library'
 Making packages.html  ... done
 Warning message:
 In install.packages(Rmpi) :
   installation of package Rmpi had non-zero exit status



I have tried with different configure.args but the result is the same, i
cant seem to install it...

Is there someone who has been through the same problem? or that knows a
guide i can follow or something?

Thanks in advance

Ricardo

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[R] help with if statement

2012-11-21 Thread york8866

Hi all,

I had a dataset A like:

TIME  DV
0  0
1   10
520
24  30
36   80
48  60
72 15

I would like to add 24 to those values higher than 24 in the TIME column. 

I did the following:

If (A$TIME=24) {
A$TIME - A$TIME+24}

It did not work.  How should I do it?

Thanks,



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[R] Finding a max

2012-11-21 Thread Ignacio Martinez

My data looks like this:

X Y1(X) Y2(X)

i want to find the values of x that maximize Y1 and Y2.
Right now I'm getting the answer but I would like to know if there is a
more efficient/elegant way of doing this.

This code reproduces what I'm doing:
[code]
prop3-structure(list(effort = c(0, 0.008989899, 0.017979798, 0.026969697,
  0.035959596, 0.044949495, 0.053939394,
0.062929293, 0.071919192,
  0.080909091, 0.08989899, 0.09889,
0.107878788, 0.116868687,
  0.125858586, 0.134848485, 0.143838384,
0.152828283, 0.161818182,
  0.170808081, 0.17979798, 0.188787879,
0.19778, 0.206767677,
  0.215757576, 0.224747475, 0.233737374,
0.242727273, 0.251717172,
  0.260707071, 0.26969697, 0.278686869,
0.287676768, 0.29667,
  0.305656566, 0.314646465, 0.323636364,
0.332626263, 0.341616162,
  0.350606061, 0.35959596, 0.368585859,
0.377575758, 0.386565657,
  0.39556, 0.404545455, 0.413535354,
0.422525253, 0.431515152,
  0.440505051, 0.44949495, 0.458484849,
0.467474748, 0.476464647,
  0.485454546, 0.49445, 0.503434344,
0.512424243, 0.521414142,
  0.530404041, 0.53939394, 0.548383839,
0.557373738, 0.566363637,
  0.575353536, 0.584343435, 0.59334,
0.602323233, 0.611313132,
  0.620303031, 0.62929293, 0.638282829,
0.647272728, 0.656262627,
  0.665252526, 0.674242425, 0.683232324,
0.69223, 0.701212122,
  0.710202021, 0.71919192, 0.728181819,
0.737171718, 0.746161617,
  0.755151516, 0.764141415, 0.773131314,
0.782121213, 0.79112,
  0.800101011, 0.80909091, 0.818080809,
0.827070708, 0.836060607,
  0.845050506, 0.854040405, 0.863030304,
0.872020203, 0.881010102,
  0.89001), Low = c(7118.22889879,
7198.74588723, 7202.19756567,
7205.63654441,
7209.06211889, 7212.47354473, 7215.87003521, 7219.25075859,
7222.61483517,
7225.96133418, 7229.28927041, 7232.59760064, 7235.88521975,
7239.15095652,
7242.39356918, 7245.61174055, 7248.8040728, 7251.96908185,
7255.10519126,
7258.21072566, 7261.28390362, 7264.32282997, 7267.32548746,
7270.28972767,
7273.21326125, 7276.0936472, 7278.92828132, 7281.7143836,
7284.44898447,
7287.12890981, 7289.75076469, 7292.31091545, 7294.80547025,
7297.23025772,
7299.5808036, 7301.85230504, 7304.03960253, 7306.13714891,
7308.13897531,
7310.03865366, 7311.82925527, 7313.50330516, 7315.05273141,
7316.46880922,
7317.74209876, 7318.86237619, 7319.81855692, 7320.59861012,
7321.18946333,
7321.57689588, 7321.74541961, 7321.678145, 7321.35663099,
7320.76071577,
7319.86832617, 7318.65526223, 7317.09495336, 7315.15818179,
7312.81276812,
7310.02321309, 7306.7502885, 7302.95056882, 7298.57589369,
7293.57274947,
7287.8817, 7281.43583878, 7274.16127505, 7265.97457404,
7256.78217653,
7246.47872816, 7234.94528475, 7222.04719437, 7207.63158811,
7191.52439555,
7173.52677949, 7153.41085818, 7130.91454853, 7105.73531868,
7077.52257998,
7045.86837049, 7010.29587987, 6970.24522727, 6925.05571843,
6873.94355449,
6815.97361626, 6750.02346243, 6674.73699741, 6588.46429282,
6489.18264441,
6374.39189638, 6240.97402156, 6085.00235313, 5901.47881048,
5683.96641688,
5424.06674216, 5110.66298423, 4728.80081891, 4257.99514316,
3669.60082983,
2922.6078062), High = c(7118.22889879, 7198.73069287,

   7202.13567036, 7205.49469474, 7208.80521654, 7212.06454229,
7215.26982532,

   7218.41805599, 7221.50605116, 7224.53044288, 7227.48766626,
7230.37394647,

   7233.18528464, 7235.91744284, 7238.56592778, 7241.12597328,
7243.59252142,

   7245.96020213, 7248.22331114, 7250.3757862, 7252.41118136,
7254.32263905,

   7256.1028599, 7257.74407005, 7259.23798555,

Re: [R] Scaling values 0-255 - -1 , 1 - how can this be done?

On Nov 21, 2012, at 9:32 AM, Brian Feeny bfe...@me.com wrote:

 
 I have a dataframe in which I have values 0-255, I wish to transpose them 
 such that:
 
 if value   127.5 value = 1
 if value  127.5 value = -1
 
 I did something similar using the binarize function of the biclust package, 
 this transforms my dataframe to 0 and 1 values, but I wish
 to use -1 and 1 and looking for a way in R to do this.
 
 Brian

See ?ifelse

  ifelse(value  127.5, 1, -1)

You might want to think about what happens when value == 127.5 ...

Regards,

Marc Schwartz

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Re: [R] Scaling values 0-255 - -1 , 1 - how can this be done?

2012-11-21 Thread Sarah Goslee

fakedata - data.frame(matrix(sample(1:255, 50, replace=TRUE), ncol=5))
ifelse(fakedata  127.5, 1, -1)

Sarah

On Wed, Nov 21, 2012 at 10:32 AM, Brian Feeny bfe...@me.com wrote:

 I have a dataframe in which I have values 0-255, I wish to transpose them 
 such that:

 if value   127.5 value = 1
 if value  127.5 value = -1

 I did something similar using the binarize function of the biclust package, 
 this transforms my dataframe to 0 and 1 values, but I wish
 to use -1 and 1 and looking for a way in R to do this.

 Brian



--
Sarah Goslee
http://www.functionaldiversity.org

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

Re: [R] lists as matrix cells ?

2012-11-21 Thread Sarah Goslee

A matrix may only contain one data type. By not specifying when you
created m, it was filled with logical values of NA.

A logical value can't hold a list.

You can see that with
str(m)
which returns:
 str(m)
 logi [1:3, 1:2] NA NA NA NA NA NA
 - attr(*, dimnames)=List of 2
  ..$ : chr [1:3] Row_one Row_two Row_three
  ..$ : chr [1:2] Column_A Column_B

If you were to instead specify that the elements of m should be list objects:


m - matrix(list(), nrow=3, ncol=2,
  dimnames=list(
c(Row_one, Row_two, Row_three),
c(Column_A, Column_B)
  )
)

(and check with str(m) again)

then you can replace those list objects with other list objects
m[[ 1,1 ]] - list(1.0, 2.0)


Please also note that [ and [[ are not the same thing.

I'm not sure this is the most efficient solution for your problem, but
since I'm not really sure what your problem is I solved the question
asked rather than the underlying problem.

Sarah

On Wed, Nov 21, 2012 at 11:11 AM, Asis Hallab asis.hal...@gmail.com wrote:
 Dear R experts,

 since more or less half a year I am using R.

 In many of my computations I construct huge matrices. Often I do so using
 'cbind' on named lists:

 do.call( 'cbind',
   list(
 Column_A=list(Row_one=1.0, Row_two=2.0, Row_three=3.0),
 Column_B=list(Row_one=4.0, Row_two=5.0, Row_three=6.0)
   )
 )

 # Returns:

   Column_A Column_B
 Row_one   14
 Row_two   25
 Row_three 36

 In some cases I even construct matrices with lists as cell content:

 do.call( 'cbind',
   list(
 Column_A=list(Row_one=list(1.0, 2.0), Row_two=list(2.0, 3.0),
 Row_three=list(3.0, 4.0)),
 Column_B=list(Row_one=list(4.0, 5.0), Row_two=list(5.0, 6.0),
 Row_three=list(6.0, 7.0))
   )
 )

 # Returns:
   Column_A Column_B
 Row_one   List,2   List,2
 Row_two   List,2   List,2
 Row_three List,2   List,2

 Interestingly I seem not to be able to initialize a 3*2 matrix and
 subsequently fill its cells with lists in order to produce the latter
 example:

 m - matrix( nrow=3, ncol=2,
   dimnames=list(
 c(Row_one, Row_two, Row_three),
 c(Column_A, Column_B)
   )
 )
 # Returns:
   Column_A Column_B
 Row_one NA   NA
 Row_two NA   NA
 Row_three   NA   NA

 # The following expressions produce the same error:
 m[ Row_one, Column_A ]   - list(1.0, 2.0)
 m[[ Row_one, Column_A ]] - list(1.0, 2.0)
 m[ 1,1 ]   - list(1.0, 2.0)
 m[[ 1,1 ]] - list(1.0, 2.0)

 # Error returned by each of the above expressions:
 *number of items to replace is not a multiple of replacement length*

 *So, now my questions:*
 *1)* What am I doing wrong? How to get the above to work?

 *2)* Or am I misusing matrices in R? Is it just by coincidence ( bug ) that
 my 'cbind' or 'rbind' calls can generate matrices with lists as cell
 content, while I seem not to be able to do so by direct assignment (as in
 above example expressions)?

 Any ideas, comments and help will be much appreciated!
 Kind regards!
 Josef


--
Sarah Goslee
http://www.functionaldiversity.org

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] help with if statement

2012-11-21 Thread Sarah Goslee

Use ifelse().

ifelse(A$TIME = 24, A$TIME + 24, A$TIME)

Please in the future use dput() to provide your data, and explain what
did not work means.

Sarah

On Wed, Nov 21, 2012 at 12:05 PM, york8866 yu_y...@hotmail.com wrote:
 Hi all,

 I had a dataset A like:

 TIME  DV
 0  0
 1   10
 520
 24  30
 36   80
 48  60
 72 15

 I would like to add 24 to those values higher than 24 in the TIME column.

 I did the following:

 If (A$TIME=24) {
 A$TIME - A$TIME+24}

 It did not work.  How should I do it?

 Thanks,



--
Sarah Goslee
http://www.functionaldiversity.org

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

Re: [R] Create BATCH file

2012-11-21 Thread Suzen, Mehmet

Try this:
http://stat.ethz.ch/R-manual/R-patched/library/utils/html/BATCH.html

On Wed, Nov 21, 2012 at 11:58 AM, R_Antony antony.akk...@ge.com wrote:
 Hi,

 I have a .R file written many functions into that. My requirement what is,
 i need  to create a batch file for this.
 No idea, how to create it. Tried it many other ways, but no result !.
 Could anyone please help me out ?

 Thanks.
 Antony.




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 View this message in context: 
 http://r.789695.n4.nabble.com/Create-BATCH-file-tp4650277.html
 Sent from the R help mailing list archive at Nabble.com.

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

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Re: [R] help with if statement

2012-11-21 Thread Rainer Schuermann

Does 

A$TIME - ifelse( A$TIME = 24, A$TIME + 24, A$TIME )

what you want?

Rgds,
Rainer


On Wednesday 21 November 2012 09:05:39 york8866 wrote:
 Hi all,
 
 I had a dataset A like:
 
 TIME  DV
 0  0
 1   10
 520
 24  30
 36   80
 48  60
 72 15
 
 I would like to add 24 to those values higher than 24 in the TIME column. 
 
 I did the following:
 
 If (A$TIME=24) {
 A$TIME - A$TIME+24}
 
 It did not work.  How should I do it?
 
 Thanks,
 
 
 
 --
 View this message in context: 
 http://r.789695.n4.nabble.com/help-with-if-statement-tp4650315.html
 Sent from the R help mailing list archive at Nabble.com.
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.

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

Re: [R] installing Rmpi on centos 6 with mpich


On Nov 21, 2012, at 10:40 AM, Ricardo Román Brenes rro...@cenat.ac.cr wrote:

 Hello everyone
 
 im trying to install Rmpi library on centos 6. I have already installed
 mpich2 1.4
 
 [root@localhost ~]# R --version
 R version 2.15.1 (2012-06-22) -- Roasted Marshmallows
 Copyright (C) 2012 The R Foundation for Statistical Computing
 ISBN 3-900051-07-0
 Platform: x86_64-redhat-linux-gnu (64-bit)
 
 
 
 [root@localhost ~]# mpich2version
 MPICH2 Version: 1.4.1p1
 MPICH2 Release date: Thu Sep  1 13:53:02 CDT 2011
 MPICH2 Device: ch3:nemesis
 MPICH2 configure:
 MPICH2 CC: gcc-O2
 MPICH2 CXX: c++   -O2
 MPICH2 F77: gfortran   -O2
 MPICH2 FC: f95   -O2
 
 
 
 
 when i issue the install order i get this output:
 
 
 install.packages(Rmpi)
 Installing package(s) into ‘/usr/lib64/R/library’
 (as ‘lib’ is unspecified)
 trying URL 'http://cran.rstudio.com/src/contrib/Rmpi_0.6-1.tar.gz'
 Content type 'application/x-gzip' length 92977 bytes (90 Kb)
 opened URL
 ==
 downloaded 90 Kb
 * installing *source* package ‘Rmpi’ ...
 ** package ‘Rmpi’ successfully unpacked and MD5 sums checked
 checking for gcc... gcc -m64 -std=gnu99
 checking whether the C compiler works... yes
 checking for C compiler default output file name... a.out
 checking for suffix of executables...
 checking whether we are cross compiling... no
 checking for suffix of object files... o
 checking whether we are using the GNU C compiler... yes
 checking whether gcc -m64 -std=gnu99 accepts -g... yes
 checking for gcc -m64 -std=gnu99 option to accept ISO C89... none needed
 I am here /usr/local and it is OpenMPI
 Trying to find mpi.h ...
 Found in /usr/local/include
 Trying to find libmpi.so or libmpich.a ...
 checking for main in -lmpi... no
 libmpi not found. exiting...
 ERROR: configuration failed for package ‘Rmpi’
 * removing ‘/usr/lib64/R/library/Rmpi’
 The downloaded source packages are in
 ‘/tmp/RtmpBvTTqc/downloaded_packages’
 Updating HTML index of packages in '.Library'
 Making packages.html  ... done
 Warning message:
 In install.packages(Rmpi) :
  installation of package ‘Rmpi’ had non-zero exit status
 
 
 
 I have tried with different configure.args but the result is the same, i
 cant seem to install it...
 
 Is there someone who has been through the same problem? or that knows a
 guide i can follow or something?
 
 Thanks in advance
 
 Ricardo


There is a page here:

  http://www.stats.uwo.ca/faculty/yu/Rmpi/

which is linked from the CRAN page for the package. On the above page, there is 
an Installation for Linux link, which is likely to be helpful.

Regards,

Marc Schwartz

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

Re: [R] installing Rmpi on centos 6 with mpich

2012-11-21 Thread Ricardo Román Brenes

i've followed that guide already to no success

On Wed, Nov 21, 2012 at 11:34 AM, Marc Schwartz marc_schwa...@me.comwrote:


 On Nov 21, 2012, at 10:40 AM, Ricardo Román Brenes rro...@cenat.ac.cr
 wrote:

  Hello everyone
 
  im trying to install Rmpi library on centos 6. I have already installed
  mpich2 1.4
 
  [root@localhost ~]# R --version
  R version 2.15.1 (2012-06-22) -- Roasted Marshmallows
  Copyright (C) 2012 The R Foundation for Statistical Computing
  ISBN 3-900051-07-0
  Platform: x86_64-redhat-linux-gnu (64-bit)
 
 
 
  [root@localhost ~]# mpich2version
  MPICH2 Version: 1.4.1p1
  MPICH2 Release date: Thu Sep  1 13:53:02 CDT 2011
  MPICH2 Device: ch3:nemesis
  MPICH2 configure:
  MPICH2 CC: gcc-O2
  MPICH2 CXX: c++   -O2
  MPICH2 F77: gfortran   -O2
  MPICH2 FC: f95   -O2
 
 
 
 
  when i issue the install order i get this output:
 
 
  install.packages(Rmpi)
  Installing package(s) into /usr/lib64/R/library
  (as lib is unspecified)
  trying URL 'http://cran.rstudio.com/src/contrib/Rmpi_0.6-1.tar.gz'
  Content type 'application/x-gzip' length 92977 bytes (90 Kb)
  opened URL
  ==
  downloaded 90 Kb
  * installing *source* package Rmpi ...
  ** package Rmpi successfully unpacked and MD5 sums checked
  checking for gcc... gcc -m64 -std=gnu99
  checking whether the C compiler works... yes
  checking for C compiler default output file name... a.out
  checking for suffix of executables...
  checking whether we are cross compiling... no
  checking for suffix of object files... o
  checking whether we are using the GNU C compiler... yes
  checking whether gcc -m64 -std=gnu99 accepts -g... yes
  checking for gcc -m64 -std=gnu99 option to accept ISO C89... none needed
  I am here /usr/local and it is OpenMPI
  Trying to find mpi.h ...
  Found in /usr/local/include
  Trying to find libmpi.so or libmpich.a ...
  checking for main in -lmpi... no
  libmpi not found. exiting...
  ERROR: configuration failed for package Rmpi
  * removing /usr/lib64/R/library/Rmpi
  The downloaded source packages are in
  /tmp/RtmpBvTTqc/downloaded_packages
  Updating HTML index of packages in '.Library'
  Making packages.html  ... done
  Warning message:
  In install.packages(Rmpi) :
   installation of package Rmpi had non-zero exit status
 
 
 
  I have tried with different configure.args but the result is the same, i
  cant seem to install it...
 
  Is there someone who has been through the same problem? or that knows a
  guide i can follow or something?
 
  Thanks in advance
 
  Ricardo


 There is a page here:

   http://www.stats.uwo.ca/faculty/yu/Rmpi/

 which is linked from the CRAN page for the package. On the above page,
 there is an Installation for Linux link, which is likely to be helpful.

 Regards,

 Marc Schwartz



[[alternative HTML version deleted]]

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

Re: [R] Help with loop


Hello,

If I understand it well, this might avoid a loop.

dat - read.table(text=
 Salt   time
1  35.65114 2003-07-19
2  35.64226 2003-07-20
3  35.62411 2003-07-21
4  35.62473 2003-07-22
5  35.65893 2003-07-23
6  35.70140 2003-07-24
7  35.62157 2003-07-25
8  35.64122 2003-07-26
9  35.63515 2003-07-27
10 35.63798 2003-07-28
, header = TRUE, stringsAsFactors = FALSE)

dat$time - as.Date(dat$time)

change - cumsum(c(FALSE, abs(diff(dat$Salt))  0.05))
split(dat, change)


Hope this helps,

Rui Barradas
Em 21-11-2012 16:39, Hefri escreveu:

Hi,

I have used R for some time, but managed to avoid writing  loops. But this
time I am afraid there is no way around it.

I have a dataframe with time and salinity (see below). I would like to
extract the time intervals where salinity changes by less than 0.05. So
using the values below this would mean that a subset was made from
2003-07-19 to 2003-07-24, where the change in salinity exceeded 0.05, and
then stating again from 2003-07-24. In other words, I want to make subset of
the time intervals where the one variable stays stabile within a certain set
range.


head (D, 10)

Salt   time
1  35.65114 2003-07-19
2  35.64226 2003-07-20
3  35.62411 2003-07-21
4  35.62473 2003-07-22
5  35.65893 2003-07-23
6  35.70140 2003-07-24
7  35.62157 2003-07-25
8  35.64122 2003-07-26
9  35.63515 2003-07-27
10 35.63798 2003-07-28

Is this possible? Can some please help with how I need to write the loop to
achieve this?

Many thanks, Helen




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Re: [R] installing Rmpi on centos 6 with mpich


On Nov 21, 2012, at 11:36 AM, Ricardo Román Brenes rro...@cenat.ac.cr wrote:

 i've followed that guide already to no success


That installation guide shows the installation taking place from the CLI 
outside of R, using R CMD INSTALL ..., not using install.packages() from within 
an R session, which is what you are showing below.

If you have tried that and failed, then I would recommend contacting the 
package maintainer for assistance or perhaps posting to R-SIG-HPC:

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

Marc


 
 On Wed, Nov 21, 2012 at 11:34 AM, Marc Schwartz marc_schwa...@me.com wrote:
 
 On Nov 21, 2012, at 10:40 AM, Ricardo Román Brenes rro...@cenat.ac.cr wrote:
 
  Hello everyone
 
  im trying to install Rmpi library on centos 6. I have already installed
  mpich2 1.4
 
  [root@localhost ~]# R --version
  R version 2.15.1 (2012-06-22) -- Roasted Marshmallows
  Copyright (C) 2012 The R Foundation for Statistical Computing
  ISBN 3-900051-07-0
  Platform: x86_64-redhat-linux-gnu (64-bit)
 
 
 
  [root@localhost ~]# mpich2version
  MPICH2 Version: 1.4.1p1
  MPICH2 Release date: Thu Sep  1 13:53:02 CDT 2011
  MPICH2 Device: ch3:nemesis
  MPICH2 configure:
  MPICH2 CC: gcc-O2
  MPICH2 CXX: c++   -O2
  MPICH2 F77: gfortran   -O2
  MPICH2 FC: f95   -O2
 
 
 
 
  when i issue the install order i get this output:
 
 
  install.packages(Rmpi)
  Installing package(s) into /usr/lib64/R/library
  (as lib is unspecified)
  trying URL 'http://cran.rstudio.com/src/contrib/Rmpi_0.6-1.tar.gz'
  Content type 'application/x-gzip' length 92977 bytes (90 Kb)
  opened URL
  ==
  downloaded 90 Kb
  * installing *source* package Rmpi ...
  ** package Rmpi successfully unpacked and MD5 sums checked
  checking for gcc... gcc -m64 -std=gnu99
  checking whether the C compiler works... yes
  checking for C compiler default output file name... a.out
  checking for suffix of executables...
  checking whether we are cross compiling... no
  checking for suffix of object files... o
  checking whether we are using the GNU C compiler... yes
  checking whether gcc -m64 -std=gnu99 accepts -g... yes
  checking for gcc -m64 -std=gnu99 option to accept ISO C89... none needed
  I am here /usr/local and it is OpenMPI
  Trying to find mpi.h ...
  Found in /usr/local/include
  Trying to find libmpi.so or libmpich.a ...
  checking for main in -lmpi... no
  libmpi not found. exiting...
  ERROR: configuration failed for package Rmpi
  * removing /usr/lib64/R/library/Rmpi
  The downloaded source packages are in
  /tmp/RtmpBvTTqc/downloaded_packages
  Updating HTML index of packages in '.Library'
  Making packages.html  ... done
  Warning message:
  In install.packages(Rmpi) :
   installation of package Rmpi had non-zero exit status
 
 
 
  I have tried with different configure.args but the result is the same, i
  cant seem to install it...
 
  Is there someone who has been through the same problem? or that knows a
  guide i can follow or something?
 
  Thanks in advance
 
  Ricardo
 
 
 There is a page here:
 
   http://www.stats.uwo.ca/faculty/yu/Rmpi/
 
 which is linked from the CRAN page for the package. On the above page, there 
 is an Installation for Linux link, which is likely to be helpful.
 
 Regards,
 
 Marc Schwartz
 
 


[[alternative HTML version deleted]]

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

Re: [R] Scaling values 0-255 - -1 , 1 - how can this be done?

2012-11-21 Thread Rainer Schuermann

x - as.data.frame( matrix( 0:255, nrow = 16 ) )
ifelse( x  127.5, 1, -1 )

Is that what you want?

Rgds,
Rainer


On Wednesday 21 November 2012 10:32:49 Brian Feeny wrote:
 
 I have a dataframe in which I have values 0-255, I wish to transpose them 
 such that:
 
 if value   127.5 value = 1
 if value  127.5 value = -1
 
 I did something similar using the binarize function of the biclust package, 
 this transforms my dataframe to 0 and 1 values, but I wish
 to use -1 and 1 and looking for a way in R to do this.
 
 Brian
 
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[R] Help: is there a R Package for L1-regression (not regression with L1-penalty)

2012-11-21 Thread Chee Chen

Dear All,
Is there a R package for L1-regression (meaning, optimize the sum of absolute 
deviations, NOT TO BE UNDERSTOOD as regression with L1-penalty) ?
Any information will be appreciated.
Regards,
Chee
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Re: [R] Finding a max


Hello,

You're complicating what is simple:


prop3$effort[which.max(prop3$Low)]  # First maximum of Low
prop3$effort[which.max(prop3$High)] # Ditto, of High

which.max(prop3$Low)   # Row number that maximizes Low
which.max(prop3$High)  # Row number that maximizes High


Hope this helps,

Rui Barradas

Em 21-11-2012 17:10, Ignacio Martinez escreveu:

My data looks like this:

X Y1(X) Y2(X)

i want to find the values of x that maximize Y1 and Y2.
Right now I'm getting the answer but I would like to know if there is a
more efficient/elegant way of doing this.

This code reproduces what I'm doing:
[code]
prop3-structure(list(effort = c(0, 0.008989899, 0.017979798, 0.026969697,
   0.035959596, 0.044949495, 0.053939394,
0.062929293, 0.071919192,
   0.080909091, 0.08989899, 0.09889,
0.107878788, 0.116868687,
   0.125858586, 0.134848485, 0.143838384,
0.152828283, 0.161818182,
   0.170808081, 0.17979798, 0.188787879,
0.19778, 0.206767677,
   0.215757576, 0.224747475, 0.233737374,
0.242727273, 0.251717172,
   0.260707071, 0.26969697, 0.278686869,
0.287676768, 0.29667,
   0.305656566, 0.314646465, 0.323636364,
0.332626263, 0.341616162,
   0.350606061, 0.35959596, 0.368585859,
0.377575758, 0.386565657,
   0.39556, 0.404545455, 0.413535354,
0.422525253, 0.431515152,
   0.440505051, 0.44949495, 0.458484849,
0.467474748, 0.476464647,
   0.485454546, 0.49445, 0.503434344,
0.512424243, 0.521414142,
   0.530404041, 0.53939394, 0.548383839,
0.557373738, 0.566363637,
   0.575353536, 0.584343435, 0.59334,
0.602323233, 0.611313132,
   0.620303031, 0.62929293, 0.638282829,
0.647272728, 0.656262627,
   0.665252526, 0.674242425, 0.683232324,
0.69223, 0.701212122,
   0.710202021, 0.71919192, 0.728181819,
0.737171718, 0.746161617,
   0.755151516, 0.764141415, 0.773131314,
0.782121213, 0.79112,
   0.800101011, 0.80909091, 0.818080809,
0.827070708, 0.836060607,
   0.845050506, 0.854040405, 0.863030304,
0.872020203, 0.881010102,
   0.89001), Low = c(7118.22889879,
7198.74588723, 7202.19756567,
 7205.63654441,
7209.06211889, 7212.47354473, 7215.87003521, 7219.25075859,
 7222.61483517,
7225.96133418, 7229.28927041, 7232.59760064, 7235.88521975,
 7239.15095652,
7242.39356918, 7245.61174055, 7248.8040728, 7251.96908185,
 7255.10519126,
7258.21072566, 7261.28390362, 7264.32282997, 7267.32548746,
 7270.28972767,
7273.21326125, 7276.0936472, 7278.92828132, 7281.7143836,
 7284.44898447,
7287.12890981, 7289.75076469, 7292.31091545, 7294.80547025,
 7297.23025772,
7299.5808036, 7301.85230504, 7304.03960253, 7306.13714891,
 7308.13897531,
7310.03865366, 7311.82925527, 7313.50330516, 7315.05273141,
 7316.46880922,
7317.74209876, 7318.86237619, 7319.81855692, 7320.59861012,
 7321.18946333,
7321.57689588, 7321.74541961, 7321.678145, 7321.35663099,
 7320.76071577,
7319.86832617, 7318.65526223, 7317.09495336, 7315.15818179,
 7312.81276812,
7310.02321309, 7306.7502885, 7302.95056882, 7298.57589369,
 7293.57274947,
7287.8817, 7281.43583878, 7274.16127505, 7265.97457404,
 7256.78217653,
7246.47872816, 7234.94528475, 7222.04719437, 7207.63158811,
 7191.52439555,
7173.52677949, 7153.41085818, 7130.91454853, 7105.73531868,
 7077.52257998,
7045.86837049, 7010.29587987, 6970.24522727, 6925.05571843,
 6873.94355449,
6815.97361626, 6750.02346243, 6674.73699741, 6588.46429282,
 6489.18264441,
6374.39189638, 6240.97402156, 6085.00235313, 5901.47881048,
 5683.96641688,
5424.06674216, 5110.66298423, 4728.80081891, 4257.99514316,
 3669.60082983,
2922.6078062), High = c(7118.22889879, 7198.73069287,

Re: [R] Finding a max

2012-11-21 Thread Bert Gunter

On Wed, Nov 21, 2012 at 9:51 AM, Rui Barradas ruipbarra...@sapo.pt wrote:
 Hello,

 You're complicating what is simple:

Fortune?

(Well, it's a profound truism that we all should live by -- but I
leave it to others to judge whether it meets Fortunes criteria).

-- Bert


 prop3$effort[which.max(prop3$Low)]  # First maximum of Low
 prop3$effort[which.max(prop3$High)] # Ditto, of High

 which.max(prop3$Low)   # Row number that maximizes Low
 which.max(prop3$High)  # Row number that maximizes High


 Hope this helps,

 Rui Barradas

 Em 21-11-2012 17:10, Ignacio Martinez escreveu:

 My data looks like this:

 X Y1(X) Y2(X)

 i want to find the values of x that maximize Y1 and Y2.
 Right now I'm getting the answer but I would like to know if there is a
 more efficient/elegant way of doing this.

 This code reproduces what I'm doing:
 [code]
 prop3-structure(list(effort = c(0, 0.008989899, 0.017979798, 0.026969697,
0.035959596, 0.044949495, 0.053939394,
 0.062929293, 0.071919192,
0.080909091, 0.08989899, 0.09889,
 0.107878788, 0.116868687,
0.125858586, 0.134848485, 0.143838384,
 0.152828283, 0.161818182,
0.170808081, 0.17979798, 0.188787879,
 0.19778, 0.206767677,
0.215757576, 0.224747475, 0.233737374,
 0.242727273, 0.251717172,
0.260707071, 0.26969697, 0.278686869,
 0.287676768, 0.29667,
0.305656566, 0.314646465, 0.323636364,
 0.332626263, 0.341616162,
0.350606061, 0.35959596, 0.368585859,
 0.377575758, 0.386565657,
0.39556, 0.404545455, 0.413535354,
 0.422525253, 0.431515152,
0.440505051, 0.44949495, 0.458484849,
 0.467474748, 0.476464647,
0.485454546, 0.49445, 0.503434344,
 0.512424243, 0.521414142,
0.530404041, 0.53939394, 0.548383839,
 0.557373738, 0.566363637,
0.575353536, 0.584343435, 0.59334,
 0.602323233, 0.611313132,
0.620303031, 0.62929293, 0.638282829,
 0.647272728, 0.656262627,
0.665252526, 0.674242425, 0.683232324,
 0.69223, 0.701212122,
0.710202021, 0.71919192, 0.728181819,
 0.737171718, 0.746161617,
0.755151516, 0.764141415, 0.773131314,
 0.782121213, 0.79112,
0.800101011, 0.80909091, 0.818080809,
 0.827070708, 0.836060607,
0.845050506, 0.854040405, 0.863030304,
 0.872020203, 0.881010102,
0.89001), Low = c(7118.22889879,
 7198.74588723, 7202.19756567,
  7205.63654441,
 7209.06211889, 7212.47354473, 7215.87003521, 7219.25075859,
  7222.61483517,
 7225.96133418, 7229.28927041, 7232.59760064, 7235.88521975,
  7239.15095652,
 7242.39356918, 7245.61174055, 7248.8040728, 7251.96908185,
  7255.10519126,
 7258.21072566, 7261.28390362, 7264.32282997, 7267.32548746,
  7270.28972767,
 7273.21326125, 7276.0936472, 7278.92828132, 7281.7143836,
  7284.44898447,
 7287.12890981, 7289.75076469, 7292.31091545, 7294.80547025,
  7297.23025772,
 7299.5808036, 7301.85230504, 7304.03960253, 7306.13714891,
  7308.13897531,
 7310.03865366, 7311.82925527, 7313.50330516, 7315.05273141,
  7316.46880922,
 7317.74209876, 7318.86237619, 7319.81855692, 7320.59861012,
  7321.18946333,
 7321.57689588, 7321.74541961, 7321.678145, 7321.35663099,
  7320.76071577,
 7319.86832617, 7318.65526223, 7317.09495336, 7315.15818179,
  7312.81276812,
 7310.02321309, 7306.7502885, 7302.95056882, 7298.57589369,
  7293.57274947,
 7287.8817, 7281.43583878, 7274.16127505, 7265.97457404,
  7256.78217653,
 7246.47872816, 7234.94528475, 7222.04719437, 7207.63158811,
  7191.52439555,
 7173.52677949, 7153.41085818, 7130.91454853, 7105.73531868,
  7077.52257998,
 7045.86837049, 7010.29587987, 6970.24522727, 6925.05571843,
  6873.94355449,
 6815.97361626, 6750.02346243, 6674.73699741, 6588.46429282,
  6489.18264441,

Re: [R] Scaling values 0-255 - -1 , 1 - how can this be done?

HI,
You could also use:
set.seed(5)
 x1-data.frame(matrix(sample(0:255,80,replace=TRUE),ncol=10))
library(car)
 do.call(cbind,lapply(x1,function(x) x-recode(x,0:127.5=-1;127.6:255=1)))
# X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
#[1,] -1  1 -1 -1 -1  1  1 -1  1  -1
#[2,]  1 -1  1 -1 -1 -1  1 -1  1   1
#[3,]  1 -1  1 -1 -1  1 -1 -1  1  -1
#[4,] -1 -1  1  1 -1  1  1 -1 -1   1
#[5,] -1 -1  1 -1  1  1  1  1  1  -1
#[6,]  1  1  1  1  1  1  1 -1  1   1
#[7,]  1 -1 -1 -1 -1  1  1  1  1  -1
#[8,]  1 -1 -1 -1 -1 -1 -1  1  1  -1
A.K.




- Original Message -
From: Brian Feeny bfe...@me.com
To: r-help@r-project.org
Cc: 
Sent: Wednesday, November 21, 2012 10:32 AM
Subject: [R] Scaling values 0-255 - -1 , 1 - how can this be done?


I have a dataframe in which I have values 0-255, I wish to transpose them such 
that:

if value   127.5 value = 1
if value  127.5 value = -1

I did something similar using the binarize function of the biclust package, 
this transforms my dataframe to 0 and 1 values, but I wish
to use -1 and 1 and looking for a way in R to do this.

Brian

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


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

Re: [R] Create BATCH file

2012-11-21 Thread FJ M


C:\Program Files\R\R-2.15.1\bin\x64\R.exe CMD BATCH
C:\Users\Frank\Documents\R\Projects\Current_Yield\Divs.txt 
C:\Users\Frank\Documents\R\Projects\Current_Yield\Divs.out
Divs.txt contains my R code and the output goes to Divs.out. I always check 
Divs.out first.
If you want to write objects to divs.out, I seem to need to use the print 
command, not just have the object name:
ndivs-nrow(Divs)
print(ndivs)
Divs_Per_Year-4
print(Divs_Per_Year)
Let me know if you have any questions.
Thanks, 
Frank


 Date: Wed, 21 Nov 2012 02:58:53 -0800
 From: antony.akk...@ge.com
 To: r-help@r-project.org
 Subject: [R] Create BATCH file

 Hi,

 I have a .R file written many functions into that. My requirement what is,
 i need to create a batch file for this.
 No idea, how to create it. Tried it many other ways, but no result !.
 Could anyone please help me out ?

 Thanks.
 Antony.




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Re: [R] Sourcing files with Umlaut in path no longer works

2012-11-21 Thread Ulrike Grömping


Duncan and Peter, thank you very much!
Actually, I made a mistake with the path, due to my inexperience with 
Windows 7 (user instead of users). The R warning then confused me, 
because I was told that 'C:/user/grÃ¶mping/documents/publicat/...' 
wasn't found.

In fact, once the path was correct, source worked without problems.
Thus, only the warning message did not properly handle the encoding.

Best regards,
Ulrike Grömping

Am 21.11.2012 15:59, schrieb Duncan Murdoch:

On 21/11/2012 6:49 AM, Ulrike Grömping wrote:

Dear helpeRs,

on my new machine - Windows 7 64Bit, R 2.15-2, I can no longer source
files for which the path contains the o-Umlaut (ö). As a historical
burden, my username is Grömping and contains that Umlaut (I wouldn't
have chosen it now, but decided to keep it for easy transfer). I have
had the difficulty that I couldn't use the R CMD tools on files with
Umlauts in the path (and solved it by having a dedicated directory
without umlaut in the path for that purpose), but so far (Windows XP and
earlier R versions) sourcing from within R worked fine.

The issue is that the o-Umlaut (ö) is always replaced by a capital A
with tilde above together with a paragraph symbol (Ã¶). Apparently, the
c3b6  used for the o-Umlaut in some encoding is translated back to the
Ã¶. Can I somehow fix this? Or is it a bug?



I can't reproduce this.  The file d:/temp/Grömping/test.R contains

cat(it worked!\n)

I am on Windows 7 64bit, and I see the following in both 32 and 64 bit R:

 source(d:/temp/Grömping/test.R)
it worked!

If I look at Encoding(d:/temp/Grömping/test.R) I get latin1. If I run

iconv(d:/temp/Grömping/test.R, latin1, UTF-8)

I get a proper UTF-8 string, but if I run

iconv(d:/temp/Grömping/test.R, latin1, utf8)

I get a string with encoding marked as unknown, and displayed as

[1] d:/temp/GrÃ¶mping/test.R

So perhaps you or we have used an unofficial name of the UTF-8 
encoding in some conversion.  We need to know exactly what you did to 
pursue this.


Duncan Murdoch


Duncan Murdoch


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Re: [R] GEE - order of data?

2012-11-21 Thread JoAnn Alvarez

Hi Anna,

In the geeglm help file, it states Data are assumed to be sorted so that
observations on a cluster are contiguous rows for all entities in the
formula.

I'm not sure if you are asking how to sort data, or how your data should be
sorted. If your data come from a data frame called dat, it could be done in
this way:

dat - dat[order(dat$Route), ]
geeglm(Pass~Distance, id=Route, corstr=ar1)

I think you only need to sort by the id variable. If you need to sort by two
variables, you can add them to the order function: 
dat[order(dat$Var1, dat$Var2), ]

I would think you may want to include the day or time as model covariates,
but it would depend on the nature of the problem. 

Hope this is helpful,
JoAnn



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Re: [R] Remove Column from matrix

2012-11-21 Thread JoAnn Alvarez

Hi frespider,

I think the problem is first that you are referring to column names that you
haven't yet defined. To add the column names you can use the dimnames
argument of the matrix function.

Asse - matrix(0,nrow=5,ncol=length(namVar), dimnames = list(NULL, namVar)) 

JoAnn



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Re: [R] help with if statement



On Nov 21, 2012, at 9:05 AM, york8866 wrote:


Hi all,

I had a dataset A like:

TIME  DV
0  0
1   10
520
24  30
36   80
48  60
72 15

I would like to add 24 to those values higher than 24 in the TIME  
column.


I did the following:

If (A$TIME=24) {
A$TIME - A$TIME+24}

It did not work.  How should I do it?


You should study the help pages for if and ifelse:

?if
?ifelse

--

David Winsemius, MD
Alameda, CA, USA

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Re: [R] Spider Graph

2012-11-21 Thread Britt Aronovich

Hi,

Is the stars command in the base package or do I need to download?
I am looking to make star/radar/spider charts.

Thanks!
Britt

Britt Aronovich
Marketing Analyst
BAM (Brooklyn Academy of Music)
...

Peter Jay Sharp Building
30 Lafayette Ave.
Brooklyn, NY 11217-01486
...

P: 718.724.8038
E: baronov...@bam.orgmailto:baronov...@bam.org

BAM.org
Facebook.com/BAMstage
Twitter.com/BAM_Brooklyn


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[R] Modify Y axis

2012-11-21 Thread Elli

How you can change the Y-axis intervals in a graph?

Use and ylim command but this command only changes the minimum and maximum
values of the axis, I want to do is take the axis values from 10 to 10
for example
How I can do?

thanks



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[R] Remove Column from matrix

2012-11-21 Thread frespider

Hi,,
Can I get help with this. I need to remove a column form the matrix if that
specific column has his all entry zero, here the code I wrote but it is not
working can you help me please

namVar -
c(TrlgWSST,TrlgWSSE,TrlgWSSR,SSdiff,TrlgWMSE,TrlgWR2,TrlgWR2adj,TrlgSSE,TrlgMSE,TrlgR2,TrlgR2adj,TrSSE,TrMSE,TrR2,TrR2adj,rdf2,
TelgSSE,TelgMSE,TelgR2,TelgR2adj,TelgWSSE,TelgWMSE
,TelgWR2,TelgWR2adj,TeSSE,TeMSE,TeR2,TeR2adj,edf2,Runtime)
Asse - matrix(0,nrow=5,ncol=length(namVar))
Asse[,1:3]-2
Asse[2,4] -2000
Asse[,5:10]-3
Asse[,11:20]-5
Asse[,21:30]-12
if(all(Asse[,SSdiff]==0)==TRUE){
Asse - Asse[,-which(colnames(Asse)%in%SSdiff)]}






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[R] histogram help

2012-11-21 Thread Rosario Scandurra

Hi,

I want to generate an histogram and plot on the y axis the percentage of a
categorical variable and on the x axis a nominal variable. I want to move
the origin to have 2 categories below 0. Hope somebody could help me.
Thanks.

Best,
-- 
Rosario Ivano Scandurra

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[R] Bayesian cluster analysis - R functions

2012-11-21 Thread KitKat

I want to try Bayesian cluster analysis. Someone suggested using package
mcclust. Is there a website that says how to install mcclust or another
appropriate Bayesian package? Including the appropriate R functions that I
can follow?

I am trying to get probability of membership for each individual I am trying
to cluster

Thank you!



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[R] plotting 1000 simulations, error message: plot.new has not been called yet

2012-11-21 Thread Maximilian Lklweryc

Hi,
I know this is not a mailing list for r, but I posted my question on
several help pages and did not get any help. I really don't know how
to solve my problem, maybe you could help me?

want to simulate stock paths. I have simulated 1000 paths with 22
trading days (1 starting value). Now I want to include it into my
presentation, but animated, so I need the png files.

I want to create 1000 png files, starting with the first stock path,
then the second and so on.

So I start with the first path, add a second to the plot, add the
third and so on, so at the end I have a plot with 1000 simulations,
here is my code:

for(i in 1:1000){
#jpeg(paste(1000s,i,.png,sep=))
plot(c(1:23),matrix[,1],type=l,ylim=c(17,24))
lines(c(1:23),matrix[,i],type=l,col=i)
#dev.off()
}

Here is the problem, that each additional part disappears when the
loop gets to the next value, so I tried:

plot(0,0 , xlim=c(1,23),ylim=c(17,24),xlab=,ylab=)
for(i in 1:1000){
jpeg(paste(1000s,i,.png,sep=))
lines(c(1:23),matrix[,i],type=l,col=i)
dev.off()
}

(I know this is not a working example, but my problem is just a
logical one with the loop) I get the following error message when I
the last code: plot.new has not been called yet.

The matrix has 1000 columns and 23 row entries, this should be 1000
simulations of stock pathes for 22 trading days.

How can I change that the error does not appear anymore? Thanks!

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Re: [R] cluster analysis in R

2012-11-21 Thread KitKat

Thank you for replying! 
I made a new post asking if there are any websites or files on how to
download package mclust (or other Bayesian cluster analysis packages) and
the appropriate R functions? Sorry I don't know how this forum works yet



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[R] printing difftime summary

2012-11-21 Thread Sam Steingold

Hi,
I have a vector of difftime objects and I want to see its summary.
Alas:
--8---cut here---start-8---
 summary(infl$delay)
  LengthClass Mode 
 9008386 difftime  numeric 
--8---cut here---end---8---
this is almost completely useless.
I can use as.numeric:
--8---cut here---start-8---
 s - summary(as.numeric(infl$delay))
 dput(s)
structure(c(0.5, 1027, 5969, 29870, 28970, 603100), .Names = c(Min., 
1st Qu., Median, Mean, 3rd Qu., Max.), class = c(summaryDefault, 
table))
 s
Min.  1st Qu.   Median Mean  3rd Qu. Max. 
 0.5   1027.0   5969.0  29870.0  28970.0 603100.0 
--8---cut here---end---8---
but the printed representation is very unreadable: the fact that
603100.0 is almost exactly 7 days is not obvious.
Okay, maybe as.difftime will help?
--8---cut here---start-8---
 as.difftime(s,units=secs)
Time differences in secs
Min.  1st Qu.   Median Mean  3rd Qu. Max. 
 0.5   1027.0   5969.0  29870.0  28970.0 603100.0 
 as.difftime(s/3600,units=hours)
Time differences in hours
Min.  1st Qu.   Median Mean  3rd Qu. Max. 
1.39e-04 2.852778e-01 1.658056e+00 8.297222e+00 8.047222e+00 1.675278e+02 
--8---cut here---end---8---
nope; still unreadable.

What I really want to see _printed_ is something likes this:
--8---cut here---start-8---
 sapply(s,difftime2string)
   Min. 1st Qu.  MedianMean 3rd Qu.Max. 
500.00 ms 17.12 min 99.48 min  8.30 hrs  8.05 hrs 6.98 days 
--8---cut here---end---8---
except that the quotes are not needed in the printed output.
Here I wrote:
--8---cut here---start-8---
difftime2string - function (x) {
  if (x  1) return(sprintf(%.2f ms,x*1000))
  if (x  100) return(sprintf(%.2f sec,x))
  if (x  6000) return(sprintf(%.2f min,x/60))
  if (x  108000) return(sprintf(%.2f hrs,x/3600))
  if (x  400*24*3600) return(sprintf(%.2f days,x/(24*3600)))
  sprintf(%.2f years,x/(365.25*24*3600))
}
--8---cut here---end---8---

So, what is The Right R Way to print a summary of difftime objects?
Thanks!
-- 
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
http://www.childpsy.net/ http://openvotingconsortium.org
http://memri.org http://camera.org http://mideasttruth.com http://pmw.org.il
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Re: [R] side by side boxplots

2012-11-21 Thread JoAnn Alvarez

Hi Charlie,

I'm not sure what you mean by opposite direction. It would be helpful if
you include a simple, self-contained dataset here to illustrate your
problem. 

For side-by-side boxplots, you can use tplot.

JoAnn



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[R] remote connection to an Oracle database - using RODBC - RMySQL..?

2012-11-21 Thread Raffaello Vardavas


Dear users,

I can access an database oracle database using sql developer. This was done by 
importing an xml file that contains the login details - username, password and 
specifies that it uses the KERBEROS_AUTHENTICATION.

I'm trying to connect R -  so that it can access this database - do sql queries 
and convert the resulting tables into dataframes. 

I am a novice in SQL and database access - but a friend provided me with the 
following approach:


library(DBI)
library(RMySQL)
drvr-dbDriver(MySQL) #Or another driver, say from the RODBC package?
acon-dbConnect(drvr, user=ENTER_USERID, dbname=ENTER_NAME, 
host=ENTER_HOST,
port=1521,password=NULL) #password maybe non-null?
cmds-dbSendQuery(acon,statement=YOUR SQL QUERY HERE)
yourdata-fetch(cmds, n=-1) #Collects all rows and columns of data requested 
query.

I have provided this info changing the relevant info in the dbConnect command 
and provided the password. However this doesn't work. I suspect because in this 
command there is not specification of the encryption of the password (i.e., 
KERBEROS_AUTHENTICATION)


When I look at the details of the connection in SQL developer - what is 
specified is the follow:

connection name, username, password (that I cannot see), hostname, port and the 
SID.

Note that although the password here cannot be seen - 
I believe it is computed by the longer password displaced the the xml file I 
use to set up the connection with sql developer using the 
KERBEROS_AUTHENTICATION.

Any ideas on how to proceed.

Please help.

Thank you.

Raff.
  
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Re: [R] Scaling values 0-255 - -1 , 1 - how can this be done?



On Nov 21, 2012, at 7:32 AM, Brian Feeny wrote:



I have a dataframe in which I have values 0-255, I wish to transpose  
them such that:


if value   127.5 value = 1
if value  127.5 value = -1



c(-1, 1)[ 1+(value  127.5) ]

I suspect most will find this less intuitive than `ifelse`, but I find  
it useful when picking colors or other values from a vector,  
expecially when the argument is built with findInterval, e.g.


c(-1, 1)[ findInterval(value, c(0, 127.5, 255) ) ] Which generalizes  
much more compactly to multiple intervals than does ifelse.



I did something similar using the binarize function of the biclust  
package, this transforms my dataframe to 0 and 1 values, but I wish

to use -1 and 1 and looking for a way in R to do this.

--

David Winsemius, MD
Alameda, CA, USA

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Re: [R] histogram help

2012-11-21 Thread John Kane

i think we need some sample data.  

I would not have thought that you could calculate the percentage of a 
categorical data so perhaps a simple example of what you are doing would help.

John Kane
Kingston ON Canada


 -Original Message-
 From: rosario.scandu...@gmail.com
 Sent: Wed, 21 Nov 2012 19:15:53 +0100
 To: r-help@r-project.org
 Subject: [R] histogram help
 
 Hi,
 
 I want to generate an histogram and plot on the y axis the percentage of
 a
 categorical variable and on the x axis a nominal variable. I want to move
 the origin to have 2 categories below 0. Hope somebody could help me.
 Thanks.
 
 Best,
 --
 Rosario Ivano Scandurra
 
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Re: [R] Modify Y axis

2012-11-21 Thread jim holtman

?axis



On Wed, Nov 21, 2012 at 12:44 PM, Elli ellilti_...@hotmail.com wrote:
 How you can change the Y-axis intervals in a graph?

 Use and ylim command but this command only changes the minimum and maximum
 values of the axis, I want to do is take the axis values from 10 to 10
 for example
 How I can do?

 thanks



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-- 
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Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

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Re: [R] Spider Graph



On Nov 21, 2012, at 9:09 AM, Britt Aronovich wrote:


Hi,

Is the stars command in the base package


No.

or do I need to download?

No

I am looking to make star/radar/spider charts.


The `stars` function is in the graphics package which is loaded by  
default. You should ahve been able to determine this from the console  
by typing:


?stars

--
David Winsemius, MD
Alameda, CA, USA

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Re: [R] Modify Y axis



On Nov 21, 2012, at 9:44 AM, Elli wrote:


How you can change the Y-axis intervals in a graph?

Use and ylim command but this command only changes the minimum and  
maximum
values of the axis, I want to do is take the axis values  
from 10 to 10

for example
How I can do?


?axis

You will probably need to suppress the default axis annotation. In  
base graphics this is done with yaxt=n, which should be remembered  
as y-axis type is 'none'.

--

David Winsemius, MD
Alameda, CA, USA

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Re: [R] Spider Graph

2012-11-21 Thread John Kane

is this link of any use? 
http://www.inside-r.org/packages/cran/fmsb/docs/radarchart

John Kane
Kingston ON Canada


 -Original Message-
 From: baronov...@bam.org
 Sent: Wed, 21 Nov 2012 12:09:38 -0500
 To: r-help@r-project.org
 Subject: Re: [R] Spider Graph
 
 Hi,
 
 Is the stars command in the base package or do I need to download?
 I am looking to make star/radar/spider charts.
 
 Thanks!
 Britt
 
 Britt Aronovich
 Marketing Analyst
 BAM (Brooklyn Academy of Music)
 ...
 
 Peter Jay Sharp Building
 30 Lafayette Ave.
 Brooklyn, NY 11217-01486
 ...
 
 P: 718.724.8038
 E: baronov...@bam.orgmailto:baronov...@bam.org
 
 BAM.org
 Facebook.com/BAMstage
 Twitter.com/BAM_Brooklyn
 
 
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Re: [R] Scaling values 0-255 - -1 , 1 - how can this be done?

2012-11-21 Thread S Ellison

  Subject: Re: [R] Scaling values 0-255 - -1 , 1 - how can 
 this be done?

#Also by taking advantage of numerical interpretation of booleans:
x- sample(255)  #toy data

2*( x  127.5 )  - 1

#S Ellison

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Re: [R] cluster analysis in R

2012-11-21 Thread Brian Feeny



http://cran.r-project.org/web/views/Cluster.html

might be a good start

Brian

On Nov 21, 2012, at 1:36 PM, KitKat wrote:

 Thank you for replying! 
 I made a new post asking if there are any websites or files on how to
 download package mclust (or other Bayesian cluster analysis packages) and
 the appropriate R functions? Sorry I don't know how this forum works yet
 
 
 
 --
 View this message in context: 
 http://r.789695.n4.nabble.com/cluster-analysis-in-R-tp4649635p4650341.html
 Sent from the R help mailing list archive at Nabble.com.
 
 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.

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Re: [R] remote connection to an Oracle database - using RODBC - RMySQL..?


On Nov 21, 2012, at 1:52 PM, Raffaello Vardavas r_varda...@hotmail.com wrote:

 
 Dear users,
 
 I can access an database oracle database using sql developer. This was done 
 by importing an xml file that contains the login details - username, password 
 and specifies that it uses the KERBEROS_AUTHENTICATION.
 
 I'm trying to connect R -  so that it can access this database - do sql 
 queries and convert the resulting tables into dataframes. 
 
 I am a novice in SQL and database access - but a friend provided me with the 
 following approach:
 
 
 library(DBI)
 library(RMySQL)
 drvr-dbDriver(MySQL) #Or another driver, say from the RODBC package?
 acon-dbConnect(drvr, user=ENTER_USERID, dbname=ENTER_NAME, 
 host=ENTER_HOST,
port=1521,password=NULL) #password maybe non-null?
 cmds-dbSendQuery(acon,statement=YOUR SQL QUERY HERE)
 yourdata-fetch(cmds, n=-1) #Collects all rows and columns of data requested 
 query.
 
 I have provided this info changing the relevant info in the dbConnect command 
 and provided the password. However this doesn't work. I suspect because in 
 this command there is not specification of the encryption of the password 
 (i.e., KERBEROS_AUTHENTICATION)
 
 
 When I look at the details of the connection in SQL developer - what is 
 specified is the follow:
 
 connection name, username, password (that I cannot see), hostname, port and 
 the SID.
 
 Note that although the password here cannot be seen - 
 I believe it is computed by the longer password displaced the the xml file I 
 use to set up the connection with sql developer using the 
 KERBEROS_AUTHENTICATION.
 
 Any ideas on how to proceed.
 
 Please help.
 
 Thank you.
 
 Raff.


Several comments:

1. Future posts on this subject should be made to R-SIG-DB, not here. More info:

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

2. Why would you expect to use an R package and driver for MySQL when 
attempting to access an Oracle server?

3. There is a good starting point on this subject generally in the R Data 
Import/Export manual:

  
http://cran.r-project.org/doc/manuals/r-release/R-data.html#Relational-databases

4. I would recommend using RODBC, which is what I use. You will of course need 
to have an ODBC driver for Oracle installed on your system and properly 
configured. You may need to get that from Oracle or other parties depending 
upon your OS which is unstated here. You may also need to get assistance with 
that process from your SysAdmin or DBAdmin.

5. If you use RODBC, there is additional, quite good information in the package 
vignette, which is accessible by using:

  vignette(RODBC)

post package installation.

6. I don't have any experience using Kerberos authentication on my Oracle 
server here, so you may have to follow up on the R-SIG-DB list on that point. A 
search of the archives did not reveal anything material on that point.

7. Alternatives to RODBC would include ROracle and RJDBC via CRAN.

Regards,

Marc Schwartz

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Re: [R] Remove Column from matrix


Hello,

Three things:
1. You don't need an explicit comparison to TRUE,

if(all(Asse[,SSdiff]==0)){

will do the same.
2. Your matrix Asse doesn't have colnames, try to see the output of

colnames(Asse)

You forgot to assign colnames(Asse) - namVar.
3. Even if it did, SSdiff is the 4th column, which you have set to 2000 
so the if statement would return FALSE.


Hope this helps,

Rui Barradas
Em 21-11-2012 17:52, frespider escreveu:

Hi,,
Can I get help with this. I need to remove a column form the matrix if that
specific column has his all entry zero, here the code I wrote but it is not
working can you help me please

namVar -
c(TrlgWSST,TrlgWSSE,TrlgWSSR,SSdiff,TrlgWMSE,TrlgWR2,TrlgWR2adj,TrlgSSE,TrlgMSE,TrlgR2,TrlgR2adj,TrSSE,TrMSE,TrR2,TrR2adj,rdf2,
TelgSSE,TelgMSE,TelgR2,TelgR2adj,TelgWSSE,TelgWMSE
,TelgWR2,TelgWR2adj,TeSSE,TeMSE,TeR2,TeR2adj,edf2,Runtime)
Asse - matrix(0,nrow=5,ncol=length(namVar))
Asse[,1:3]-2
Asse[2,4] -2000
Asse[,5:10]-3
Asse[,11:20]-5
Asse[,21:30]-12
if(all(Asse[,SSdiff]==0)==TRUE){
Asse - Asse[,-which(colnames(Asse)%in%SSdiff)]}
 






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[R] Using doMC to run parallel SVM grid search?

2012-11-21 Thread Brian Feeny

Has anyone used doMC to speed up an SVM grid search?  I am considering doing 
like so:

library(doMC)
registerDoMC()
foreach (i=0:3) %dopar% {
tuned_part1 - tune.svm(label~., data = trainset, gamma = 10^(-10:-6), 
cost = 10^(-1:1))
tuned_part2 - tune.svm(label~., data = trainset, gamma = 10^(-5:0),
 cost = 10^(-1:1))
tuned_part3 - tune.svm(label~., data = trainset, gamma = 10^(1:-5),
 cost = 10^(-1:1))
tuned_part4 - tune.svm(label~., data = trainset, gamma = 10^(5:10),
cost = 10^(-1:1))
}


I have a Quad Core processor, so if I understand correctly the above could 
split that up across the cores.

My goal would be a coarse grid search, not sure if the above parameters are 
good for that, it just seemed like 
some good starting points.

I would just manually look at each of the resulting files, although it would be 
cool if it resulted in an instance variable
being set of the best values. 

Has anyone used doMC for something like this?  Is there a better library to 
potentially use than doMC for doing 
something like splitting up an SVM grid search over multiple cores?

Brian

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Re: [R] Help: is there a R Package for L1-regression (not regression with L1-penalty)

2012-11-21 Thread cberry

Chee Chen chee.c...@yahoo.com writes:

 Dear All,
 Is there a R package for L1-regression (meaning, optimize the sum of
 absolute deviations, NOT TO BE UNDERSTOOD as regression with
 L1-penalty) ?

Yes.

 Any information will be appreciated.

See 

http://cran.cnr.berkeley.edu/web/views/Robust.html


HTH,

Chuck

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[R] Regression: standardized coefficients CI

2012-11-21 Thread Torvon

I run 9 WLS regressions in R, with 7 predictors each.

What I want to do now is compare:
(1) The strength of predictors within each model (assuming all predictors
are significant). That is, I want to say whether x1 is stronger than x2,
and also say whether it is significantly stronger. I compare strength by
simply comparing standardized beta weights, correct? How do I compare if
one predictor is significantly stronger than the others? I thought about
comparing confidence intervals, but if I understand correctly the
confidence intervals are calculated from the unstandardized beta weights,
which in this case would not help me, correct?
(2) The strength of the same predictor over different models. I want to say
whether x1 affects y1 - y9 equally strong or not. How would I do this?

I hope that I provided all information that is needed.
Thank you
T

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Re: [R] Integration in R

2012-11-21 Thread Rolf Turner


On 21/11/12 22:26, Rehena Sultana wrote:

Dear R - Experts,

I am trying to integrate lognormal distribution (mu = -0.3 and sigma2 = 
0.00041.. ) based on the some hypothetical data. But I am getting 0 as the 
result. I have checked that my R-code is correct as code is giving me result 
for some other data.

As I understand, when I am integrating some pdf with in the range of (0, Inf), 
I should not get 0. How to handle this kind of problem?

Please help. Looking forward for your reply.


Reproducible example?

When I do

integrate(function(x){dlnorm(x,meanlog=-0.3,sdlog=sqrt(0.00041))},0,Inf)

I get

1 with absolute error  3.5e-06

No problema.

Why do you want to integrate it anyhow?  You know the answer is 1.

Or if you want the integral from 0 to x for some x  infinity, just use
plnorm().  (???)

cheers,

Rolf Turner

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Re: [R] Regression: standardized coefficients CI

2012-11-21 Thread Bert Gunter

1. This is a statistics, not an R, question. Post on a statistics
list, like stats.stackexchange.com

Also...

On Wed, Nov 21, 2012 at 12:39 PM, Torvon tor...@gmail.com wrote:
I run 9 WLS regressions in R, with 7 predictors each.

What I want to do now is compare:
(1) The strength of predictors within each model (assuming all predictors
are significant). That is, I want to say whether x1 is stronger than x2,
and also say whether it is **significantly stronger.**

-- I have no idea what this means, though perhaps it is defined
somewhere and in some way that I am not familiar with. When you post
to a stats list, I suggest you provide a reference so the folks there
know what you mean by this.

-- Bert

I compare strength by
simply comparing standardized beta weights, correct? How do I compare if
one predictor is significantly stronger than the others? I thought about
comparing confidence intervals, but if I understand correctly the
confidence intervals are calculated from the unstandardized beta weights,
which in this case would not help me, correct?
(2) The strength of the same predictor over different models. I want to say
whether x1 affects y1 - y9 equally strong or not. How would I do this?

I hope that I provided all information that is needed.
Thank you
T

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Bert Gunter
Genentech Nonclinical Biostatistics

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

Re: [R] [lattice] print only legend


On Nov 21, 2012, at 1:45 AM, AnjaM wrote:

 Is it possible to plot (and save) only the legend of a lattice plot? Of
 course I could make the axes and axis labels transparent and use an empty
 panel function, but additionally to being a very dirty solution, there
 would be still a lot of free space on the plot, while I would like to save
 only the legend as a separate plot without any white space around it. Is
 there a (nice) way to do this?
 

If you read the help page:

? xyplot

... it should not be too hard to find the section that names the grid function 
that draws a key. Unsurprisingly, it is `draw.key`.

-- 
David Winsemius, MD
Alameda, CA, USA

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Re: [R] Integration in R

2012-11-21 Thread William Dunlap

But if you use a smaller sdlog value then integrate does get it wrong because 
it does not find the delta-like function hidden somewhere between 0 and 
infinity.
   integrate(function(x){dlnorm(x,meanlog=-0.3,sdlog=0.00041)},0,Inf)
  0 with absolute error  0
   integrate(function(x){dlnorm(x,meanlog=-0.3,sdlog=0.00041)},0,1)
  0 with absolute error  0
Tell it where the bulk of the mass is and it works
   integrate(function(x)dlnorm(x,-0.3,0.00041), 0.738, 0.743, 
subdivisions=10^3)
  1 with absolute error  4.4e-05

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf
 Of Rolf Turner
 Sent: Wednesday, November 21, 2012 12:57 PM
 To: Rehena Sultana
 Cc: r-help@r-project.org
 Subject: Re: [R] Integration in R
 
 On 21/11/12 22:26, Rehena Sultana wrote:
  Dear R - Experts,
 
  I am trying to integrate lognormal distribution (mu = -0.3 and sigma2 = 
  0.00041.. )
 based on the some hypothetical data. But I am getting 0 as the result. I have 
 checked
 that my R-code is correct as code is giving me result for some other data.
 
  As I understand, when I am integrating some pdf with in the range of (0, 
  Inf), I should
 not get 0. How to handle this kind of problem?
 
  Please help. Looking forward for your reply.
 
 Reproducible example?
 
 When I do
 
 integrate(function(x){dlnorm(x,meanlog=-0.3,sdlog=sqrt(0.00041))},0,Inf)
 
 I get
 
  1 with absolute error  3.5e-06
 
 No problema.
 
 Why do you want to integrate it anyhow?  You know the answer is 1.
 
 Or if you want the integral from 0 to x for some x  infinity, just use
 plnorm().  (???)
 
  cheers,
 
  Rolf Turner
 
 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.

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Re: [R] Finding a max

2012-11-21 Thread Ignacio Martinez

Thanks a lot!


On Wed, Nov 21, 2012 at 12:51 PM, Rui Barradas ruipbarra...@sapo.pt wrote:

 Hello,

 You're complicating what is simple:


 prop3$effort[which.max(prop3$**Low)]  # First maximum of Low
 prop3$effort[which.max(prop3$**High)] # Ditto, of High

 which.max(prop3$Low)   # Row number that maximizes Low
 which.max(prop3$High)  # Row number that maximizes High


 Hope this helps,

 Rui Barradas

 Em 21-11-2012 17:10, Ignacio Martinez escreveu:

 My data looks like this:

 X Y1(X) Y2(X)

 i want to find the values of x that maximize Y1 and Y2.
 Right now I'm getting the answer but I would like to know if there is a
 more efficient/elegant way of doing this.

 This code reproduces what I'm doing:
 [code]
 prop3-structure(list(effort = c(0, 0.008989899, 0.017979798, 0.026969697,
0.035959596, 0.044949495, 0.053939394,
 0.062929293, 0.071919192,
0.080909091, 0.08989899, 0.09889,
 0.107878788, 0.116868687,
0.125858586, 0.134848485, 0.143838384,
 0.152828283, 0.161818182,
0.170808081, 0.17979798, 0.188787879,
 0.19778, 0.206767677,
0.215757576, 0.224747475, 0.233737374,
 0.242727273, 0.251717172,
0.260707071, 0.26969697, 0.278686869,
 0.287676768, 0.29667,
0.305656566, 0.314646465, 0.323636364,
 0.332626263, 0.341616162,
0.350606061, 0.35959596, 0.368585859,
 0.377575758, 0.386565657,
0.39556, 0.404545455, 0.413535354,
 0.422525253, 0.431515152,
0.440505051, 0.44949495, 0.458484849,
 0.467474748, 0.476464647,
0.485454546, 0.49445, 0.503434344,
 0.512424243, 0.521414142,
0.530404041, 0.53939394, 0.548383839,
 0.557373738, 0.566363637,
0.575353536, 0.584343435, 0.59334,
 0.602323233, 0.611313132,
0.620303031, 0.62929293, 0.638282829,
 0.647272728, 0.656262627,
0.665252526, 0.674242425, 0.683232324,
 0.69223, 0.701212122,
0.710202021, 0.71919192, 0.728181819,
 0.737171718, 0.746161617,
0.755151516, 0.764141415, 0.773131314,
 0.782121213, 0.79112,
0.800101011, 0.80909091, 0.818080809,
 0.827070708, 0.836060607,
0.845050506, 0.854040405, 0.863030304,
 0.872020203, 0.881010102,
0.89001), Low = c(7118.22889879,
 7198.74588723, 7202.19756567,
  7205.63654441,
 7209.06211889, 7212.47354473, 7215.87003521, 7219.25075859,
  7222.61483517,
 7225.96133418, 7229.28927041, 7232.59760064, 7235.88521975,
  7239.15095652,
 7242.39356918, 7245.61174055, 7248.8040728, 7251.96908185,
  7255.10519126,
 7258.21072566, 7261.28390362, 7264.32282997, 7267.32548746,
  7270.28972767,
 7273.21326125, 7276.0936472, 7278.92828132, 7281.7143836,
  7284.44898447,
 7287.12890981, 7289.75076469, 7292.31091545, 7294.80547025,
  7297.23025772,
 7299.5808036, 7301.85230504, 7304.03960253, 7306.13714891,
  7308.13897531,
 7310.03865366, 7311.82925527, 7313.50330516, 7315.05273141,
  7316.46880922,
 7317.74209876, 7318.86237619, 7319.81855692, 7320.59861012,
  7321.18946333,
 7321.57689588, 7321.74541961, 7321.678145, 7321.35663099,
  7320.76071577,
 7319.86832617, 7318.65526223, 7317.09495336, 7315.15818179,
  7312.81276812,
 7310.02321309, 7306.7502885, 7302.95056882, 7298.57589369,
  7293.57274947,
 7287.8817, 7281.43583878, 7274.16127505, 7265.97457404,
  7256.78217653,
 7246.47872816, 7234.94528475, 7222.04719437, 7207.63158811,
  7191.52439555,
 7173.52677949, 7153.41085818, 7130.91454853, 7105.73531868,
  7077.52257998,
 7045.86837049, 7010.29587987, 6970.24522727, 6925.05571843,
  6873.94355449,
 6815.97361626, 6750.02346243, 6674.73699741, 6588.46429282,
  6489.18264441,
 6374.39189638, 6240.97402156, 6085.00235313, 5901.47881048,
  5683.96641688,

Re: [R] update fit (removing insignificant variables)

HI,

I am not sure about the ?update() method.
You could try this:

set.seed(232)
mat1-matrix(sample(1:100,80,replace=TRUE),ncol=8) #with 8 columns
dat1-data.frame(mat1)
 names(dat1)[1]-Y
 fit-lm(Y~.,data=dat1)
res-coef(summary(fit))

 res
# Estimate Std. Error t value   Pr(|t|)
#(Intercept) 143.763081344 21.9902865  6.53757200 0.02260698
#X2   -0.009030461  0.1103611 -0.08182650 0.94223654
#X3   -1.844809745  0.2655513 -6.94709326 0.02009766
#X4    1.436647351  0.2540660  5.65462216 0.02988003
#X5   -0.005406684  0.2906822 -0.01859998 0.98684896
#X6   -0.050101169  0.1486149 -0.33712071 0.76811703
#X7    0.563284987  0.1204945  4.67477901 0.04284010
#X8   -1.901918462  0.2697166 -7.05154460 0.01952387

Wtd-row.names(res[res[,Pr(|t|)]0.05,])
 Wtd1-Wtd[!grepl(\\(.*\\),Wtd)]
fit1-lm(as.formula(paste(Y~,paste(Wtd1,collapse=+))),data=dat1)
 coef(summary(fit1))
#   Estimate  Std. Error   t value Pr(|t|)
#(Intercept) 146.8921451 10.87053736  13.51287 3.975711e-05
#X3   -1.9002023  0.10667909 -17.81232 1.023637e-05
#X4    1.4421062  0.09790384  14.72982 2.606788e-05
#X7    0.5726324  0.05396720  10.61075 1.285606e-04
#X8   -1.9560052  0.14204298 -13.77052 3.625118e-05

A.K.

From: farnoosh sheikhi farnoosh...@yahoo.com
To: arun smartpink...@yahoo.com 
Sent: Wednesday, November 21, 2012 2:47 PM
Subject: update fit (removing insignificant variables)


Hi there,

I have a question for removing insignificant predictors from fit.
The output looks like below but with 170 predictors.
Lab_1                                                              0.037928 *  
Lab_2                                                              0.030423 *  
Lab_4                                                              0.616122    
Lab_7                                                              0.101278 
Lab_8                                                              2.44e-06 ***
Lab_12                                                             0.003181 ** 
Lab_14                                                             0.096809 

I need to update my model with significant ones.
I know if I want to remove one, I use the following code, but I don't know how 
to do that for 70 insignificant predictors.
fit2-update(fit, .~. -Lab_7)



Thanks a lot.



Best,Farnoosh Sheikhi 

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Re: [R] Remove Column from matrix

2012-11-21 Thread frespider

Hi,

I edited the code sorry,

I forgot the line before

Can you have look again please?

Thanks 



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Re: [R] Remove Column from matrix

Hi,

In this particular example, you may not be able to remove any columns:
because:
namVar - 
c(TrlgWSST,TrlgWSSE,TrlgWSSR,SSdiff,TrlgWMSE,TrlgWR2,TrlgWR2adj,TrlgSSE,TrlgMSE,TrlgR2,TrlgR2adj,TrSSE,TrMSE,TrR2,TrR2adj,rdf2,
 
TelgSSE,TelgMSE,TelgR2,TelgR2adj,TelgWSSE,TelgWMSE 
,TelgWR2,TelgWR2adj,TeSSE,TeMSE,TeR2,TeR2adj,edf2,Runtime) Asse 
- matrix(0,nrow=5,ncol=length(namVar), dimnames = list(NULL, namVar)) #JoAnn's 
suggestion
Asse[,1:3]-2 
Asse[2,4] -2000 
Asse[,5:10]-3 
Asse[,11:20]-5 
Asse[,21:30]-12 
Asse[,1:5]
# TrlgWSST TrlgWSSE TrlgWSSR SSdiff TrlgWMSE
#[1,]    2    2    2  0    3
#[2,]    2    2    2   2000    3
#[3,]    2    2    2  0    3
#[4,]    2    2    2  0    3
#[5,]    2    2    2  0    3

So, if you are not assigning :
#Asse[2,4] -2000 
Then, Asse[,4] will be all zeros.

Asse[,!colSums(abs(Asse))==0] #will remove the columns which are all zeros
#or
Asse[,!apply(Asse,2,function(x) all(x==0))]
#will get the results you wanted.
A.K.



- Original Message -
From: frespider frespi...@hotmail.com
To: r-help@r-project.org
Cc: 
Sent: Wednesday, November 21, 2012 12:52 PM
Subject: [R] Remove Column from matrix

Hi,,
Can I get help with this. I need to remove a column form the matrix if that
specific column has his all entry zero, here the code I wrote but it is not
working can you help me please

namVar -
c(TrlgWSST,TrlgWSSE,TrlgWSSR,SSdiff,TrlgWMSE,TrlgWR2,TrlgWR2adj,TrlgSSE,TrlgMSE,TrlgR2,TrlgR2adj,TrSSE,TrMSE,TrR2,TrR2adj,rdf2,
TelgSSE,TelgMSE,TelgR2,TelgR2adj,TelgWSSE,TelgWMSE
,TelgWR2,TelgWR2adj,TeSSE,TeMSE,TeR2,TeR2adj,edf2,Runtime)
Asse - matrix(0,nrow=5,ncol=length(namVar))
Asse[,1:3]-2
Asse[2,4] -2000
Asse[,5:10]-3
Asse[,11:20]-5
Asse[,21:30]-12
if(all(Asse[,SSdiff]==0)==TRUE){
Asse - Asse[,-which(colnames(Asse)%in%SSdiff)]}
            





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[R] lmer model specification

2012-11-21 Thread Roy

I am running version 2.15.2 64 bit version on 64 bit Windows 7. I have a 
data set with the following structure:

Fixed Effect: locationFact
Random Effects: datefact, timefact nested in datefact, interactions of 
datefact and timefact with locationFact

I fit the model with the latest version of lme4.

The formula is:   Thick2 ~ locationFact + (1 | datefact) + (1 | 
datefact/timefact) +  (1 | locationFact:datefact) + (1 | 
datefact/locationFact:timefact)


Other elements of  output object are:

Linear mixed model fit by REML

Random effects:
 GroupsName Variance Std.Dev.
 locationFact:timefact:datefact  (Intercept)  34.614   5.8833
 timefact:datefact(Intercept)  96.795 9.8385
 locationFact:datefact (Intercept)  56.375   7.5083
 datefact   (Intercept) 20.341   
4.5101
 datefact   (Intercept) 20.338   
4.5098
 datefact   (Intercept) 20.340   
4.5100

 Residual 447.252  21.1483
I tested this model using rmel with another software package and found 
that the datefact variance is the sum of the 3 datefact variance 
estimates above. Is there a way to specify the model so I do not get the 
3 datefact estimates? The same applies to the datefact BLUPs. I have to 
add them to get the actual BLUP.


Thank you,

Roy Robertson

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Re: [R] Creating a frequency table for binomial varaible