[R] MANOVA usage

[Aalim Weljie]

Hello,

I had a couple questions about  manova modeling in R.

I have calculated a manova model, and generated a summary.manova output
using both the Wilks test and Pillai test.

The output is essentially the same, except that the Wilks lambda = 1 -
Pillai. Is this normal? (The output from both is appended below.)

My other question is about the use of MANOVA. If I have one variable which
has a higher F-stat value (~60) than the MANOVA fstat (~20), but the other
15 variables all have lower ANOVA F-stat values (~5-10), am I still okay in
using the MANOVA? (If I remove the one highly significant variable, the
MANOVA f-stat is still larger than the other univariate F-stats).


Thanks very much for your help,

Aalim

Output from Manova tests:

 summary.manova(fit)
  Df  Pillai approx F num Df den DfPr(F)
response  1  0.9725  19.8967 16  9 4.193e-05 ***
Residuals 24
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
 summary.manova(fit, test=Wilks)
  Df   Wilks approx F num Df den DfPr(F)
response 1  0.0275  19.8967 16  9 4.193e-05 ***
Residuals 24
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

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Re: [R] applying lm on an array of observations with common design matrix

[Prof Brian Ripley]

On Thu, 22 Feb 2007, Petr Klasterecky wrote:

 Ranjan Maitra napsal(a):
 On Sun, 18 Feb 2007 07:46:56 + (GMT) Prof Brian Ripley [EMAIL 
 PROTECTED] wrote:

 On Sat, 17 Feb 2007, Ranjan Maitra wrote:

 Dear list,

 I have a 4-dimensional array Y of dimension 330 x 67 x 35 x 51. I have a
 design matrix X of dimension 330 x 4. I want to fit a linear regression
 of each

 lm( Y[, i, j, k] ~ X). for each i, j, k.

 Can I do it in one shot without a loop?
 Yes.

 YY - YY
 dim(YY) - c(330, 67*35*51)
 fit - lm(YY ~ X)

 Actually, I am also interested in getting the p-values of some of the
 coefficients -- lets say the coefficient corresponding to the second
 column of the design matrix. Can the same be done using array-based
 operations?
 Use lapply(summary(fit), function(x) coef(x)[3,4])  (since there is a
 intercept, you want the third coefficient).

 In this context, can one also get the variance-covariance matrix of the 
 coefficients?

 Sure:

 lapply(summary(fit), function(x) {$(x,cov.unscaled)})

But that is not the variance-covariance matrix (and it is an unusual way 
to write x$cov.unscaled)!

 Add indexing if you do not want the whole matrix. You can extract
 whatever you want, just take a look at ?summary.lm, section Value.

It is unclear to me what the questioner expects: the estimated 
coefficients for different responses are independent.  For a list of 
matrices applying to each response one could mimic vcov.lm and do

lapply(summary(fit, corr=FALSE),
function(so) so$sigma^2 * so$cov.unscaled)

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

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Re: [R] write fixed format

[Gregor Gorjanc]

YIHSU CHEN yihsu.chen at ucmerced.edu writes:
 Dear R users;
 
 Is there a function in R that I can put text with proper alignments in 
 a fixed format.  For instance, if I have three fields: A, B and C, where 
 both A and C are text with 3 characters and left alignment; B is a 
 numeric one with 2 decimals and 3 integer space digits.   How can I put 
 the following row in a file? (note there is a space between a and 2, and 
 after b.) (A=aaa, B=23.11 and C=bb)
 aaa 23.11bb 

Check write.fwf() in gdata package. It might help you, but you will have to 
create new column to have B and C together i.e. without any space between.

Gregor

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[R] how to install a package in R on a linux machine?

[gallon li]

I downloaded the tar.gz file from r-project website (and saved it in a local
directory) and wish to use the package in R.

But I am not sure how to use the install.packages command. I tried a few
times and still couldn't figure out the correct way to install this package.

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[R] Spatial error model estimation

[Dong GUO]

Greetings to the list,

I was trying to estimate spatial error model in R, somehow I got the
message below. Would you please help me with it? Many thanks in
advance.

Error in solve.default(asyvar, tol = tol.solve) :
system is computationally singular: reciprocal condition number = 
5.6964e-18


Regards,
Dong

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Re: [R] 3F2 hypergeometric function

[Robin Hankin]

Hello Joe


On 21 Feb 2007, at 16:22, Lucke, Joseph F wrote:

 Does anyone have code for the 3F2 hypergeometric function? I am  
 looking
 for code similar to the 2F1 hypergeometric function implemented as
 hyperg_2F1 in the GSL package. TIA.  ---Joe



The GSL library does not have a hyperg_3F2, so neither does the gsl  
package.

BUT the Davies package does have a hyperg() function that is written in
such a way that it would be a cinch to convert it from 2F1 to 3F2.

Let me know how you get on


Robin





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--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton
European Way, Southampton SO14 3ZH, UK
  tel  023-8059-7743

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[R] Move y label in xyplot

[Ola Caster]

Is there any way to make xyplot draw the y label to the right of the graph
instead of to the left? Any help appreciated.

Regards,
M.Sc. Ola Caster
Uppsala University, Sweden

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Re: [R] how to install a package in R on a linux machine?

[Gabor Csardi]

The easiest is perhaps to do 

install.packages(packagename)

this downloads the package and installs it into the default R package
library on your machine. If you want to install it to a different 
directory use the 'lib' argument of 'install.packages'.

If you don't want to download the package again but want to use the 
downloaded one, use the following command:

install.packages(repos=NULL, pkgs=the.file.you've.downloaded)

You can also install R packages from the command line, like this:

R CMD INSTALL -l lib.directectory downloaded.package.file

Gabor

On Thu, Feb 22, 2007 at 04:44:25PM +0800, gallon li wrote:
 I downloaded the tar.gz file from r-project website (and saved it in a local
 directory) and wish to use the package in R.
 
 But I am not sure how to use the install.packages command. I tried a few
 times and still couldn't figure out the correct way to install this package.
 
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Csardi Gabor [EMAIL PROTECTED]MTA RMKI, ELTE TTK

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Re: [R] MANOVA usage

[Prof Brian Ripley]

On Thu, 22 Feb 2007, Aalim Weljie wrote:

 Hello,

 I had a couple questions about  manova modeling in R.

 I have calculated a manova model, and generated a summary.manova output
 using both the Wilks test and Pillai test.

 The output is essentially the same, except that the Wilks lambda = 1 -
 Pillai. Is this normal? (The output from both is appended below.)

For a 1-df test, yes: all the statistics give the same test (as 
?summary.manova does say).

 My other question is about the use of MANOVA. If I have one variable which
 has a higher F-stat value (~60) than the MANOVA fstat (~20), but the other
 15 variables all have lower ANOVA F-stat values (~5-10), am I still okay in
 using the MANOVA? (If I remove the one highly significant variable, the
 MANOVA f-stat is still larger than the other univariate F-stats).

OK for what purpose?  Using (M)ANOVA for variable selection has many 
issues.

 Thanks very much for your help,

 Aalim

 Output from Manova tests:

 summary.manova(fit)
  Df  Pillai approx F num Df den DfPr(F)
 response  1  0.9725  19.8967 16  9 4.193e-05 ***
 Residuals 24
 ---
 Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
 summary.manova(fit, test=Wilks)
  Df   Wilks approx F num Df den DfPr(F)
 response 1  0.0275  19.8967 16  9 4.193e-05 ***
 Residuals 24
 ---
 Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

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


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

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[R] Accessing the class of an object with two elements.

[Shubha Vishwanath Karanth]

Hi R,

 

Here's my question about accessing the class of an object.

 

I have an object dat which can take any two of the classes, (dates
times) or (chron dates times). Note that the classes have two
elements within it. I want to read these classes in such a way that 

 

v=class(dat) # let class(dat)= dates times

If(class(dat)==v) k=1 else k=0

 

The problem is I can't read the above class. The error which I get for
the above if statement is as follows:

 

Warning message:

the condition has length  1 and only the first element will be used in:
if (class(index(intra)) == v) k = 1 

 

 

How should I proceed with this? Any ideas? I tried with readline to read
the class and access it in the 'if' statement...But doesn't work :-(

 

Thanks in advance,

Shubha


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[R] Sorting rows of a binary matrix

[Serguei Kaniovski]

Hallo,

The command:

x - 3
mat - as.matrix(expand.grid(rep(list(0:1), x)))

generates a matrix with 2^x columns containing the binary representations
of the decimals from 0 to (2^x-1), here from 0 to 7. But the rows are not
sorted in this order.

How can sort the rows the ascending order of the decimals they represent,
preferably without a function which converts binaries to decimals (which I
have)? Alternatively, generate a matrix that has the rows sorted that way?

Thanks,
Serguei
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Re: [R] Sorting rows of a binary matrix

[David Barron]

I'm sure there are more elegant ways, but this should work:

 ix-order(mat[,1],mat[,2],mat[,3])
 ix
[1] 1 5 3 7 2 6 4 8
 mat[ix,]
  Var1 Var2 Var3
1000
5001
3010
7011
2100
6101
4110
8111


On 22/02/07, Serguei Kaniovski [EMAIL PROTECTED] wrote:

 Hallo,

 The command:

 x - 3
 mat - as.matrix(expand.grid(rep(list(0:1), x)))

 generates a matrix with 2^x columns containing the binary representations
 of the decimals from 0 to (2^x-1), here from 0 to 7. But the rows are not
 sorted in this order.

 How can sort the rows the ascending order of the decimals they represent,
 preferably without a function which converts binaries to decimals (which I
 have)? Alternatively, generate a matrix that has the rows sorted that way?

 Thanks,
 Serguei
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-- 
=
David Barron
Said Business School
University of Oxford
Park End Street
Oxford OX1 1HP

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Re: [R] Trying to get an apply to work with a list in applying names totables

[John Kane]

--- Marc Schwartz [EMAIL PROTECTED] wrote:

 I might suggest an alternative, since you seem to be
 creating the
 underlying data set from scratch.

Thanks Marc.  I see what you're suggesting but I am
not creating the data from scratch.  The data base
represented by cc  is an SPSS file that I have
inherited.  

Still I am having to do enough mucking about with the
SPSS file anyway that your approach might be the best
way to go if I do this again. I wish I had asked this
question a week ago!

 Create the data frame with the requisite data
 structures to start with
 and then perform the table operations:
 
 
 # First create your vectors as factors. See ?factor
 aa - factor(c(2,2,1,1,2), levels = 1:2, labels =
 c(yes, no))
 bb - factor(c(5,6,6,7,4), levels = 4:7, labels =
 letters[1:4])
 
 
 # Now create your data frame using the names you
 want for each column
 cc - data.frame(abby = aa, billy = bb)
 
 
 Now run the table on each column:
 
  lapply(cc, table)
 $abby
 
 yes  no 
   2   3 
 
 $billy
 
 a b c d 
 1 1 2 1 
 
 
 See ?lapply as well. Note that a data frame is a
 list:
 
  is.list(cc)
 [1] TRUE
 
  is.data.frame(cc)
 [1] TRUE
 
 
  as.list(cc)
 $abby
 [1] no  no  yes yes no 
 Levels: yes no
 
 $billy
 [1] b c c d a
 Levels: a b c d
 
 
 HTH,
 
 Marc Schwartz
 
 
 On Wed, 2007-02-21 at 17:15 +0100, ONKELINX, Thierry
 wrote:
  John,
  
  Two things. You don't need to pout the cc variable
 in the apply. Use
  instead something like this.
  
  apply(cc, 2, fn1, y = mylist)
  
  But this still doesn't solve your problem. You'll
 need to rewrite your
  function like this.
  
   fn2 - function(x, y, i){
  +   tt - table(x[, i])
  +   names(tt) - y[[i]]
  +   return(tt)
  + }
   sapply(1:ncol(cc), fn2, x = cc, y = mylist)
  [[1]]
  yes  no 
2   3 
  
  [[2]]
  a b c d 
  1 1 2 1 
  
  Cheers,
  
  Thierry
 
  -Oorspronkelijk bericht-
  Van: [EMAIL PROTECTED]
  [mailto:[EMAIL PROTECTED] Namens
 John Kane
  Verzonden: woensdag 21 februari 2007 16:47
  Aan: R R-help
  Onderwerp: [R] Trying to get an apply to work with
 a list in applying
  names totables
  
  I am trying to use apply and a  list to supply
 names
  to a set of tables I want to generate. Below is an
  example that I hope mimics the larger original
  problem.
  
  EXAMPLE
  
  aa - c( 2,2,1,1,2)
  bb - c(5,6,6,7,4)
  aan - c(yes, no)
  bbn - c(a, b, c, d)
  mynames - c(abby, billy)
  mylist - list(aan, bbn);   names(mylist) -
 mynames
  
  cc - data.frame(aa,bb)
  fn1 - function(x,y) {tt - table(x); names(tt)-
  mylist[[y]]}
  jj -apply(cc, 2, fn1(cc,mylist))
  
  RESULT:  
  Error in fn1(cc, mylist) : invalid subscript type
  
  To be honest I didn't expect it to work since that
  fin1(cc  looks recursive but oh well...
  
  Can anyone offer a solution or some advice here. 
 It
  would be greatly appreciated
 
 


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Re: [R] Spatial error model estimation

[Roger Bivand]

On Thu, 22 Feb 2007, Dong GUO wrote:

 Greetings to the list,
 
 I was trying to estimate spatial error model in R, somehow I got the
 message below. Would you please help me with it? Many thanks in
 advance.
 
 Error in solve.default(asyvar, tol = tol.solve) :
   system is computationally singular: reciprocal condition number = 
 5.6964e-18

(This refers to function errorsarlm() in package spdep)

Please see ?errorsarlm, the problem is explained there as follows:

tol.solve: the tolerance for detecting linear dependencies in the
  columns of matrices to be inverted - passed to 'solve()'
  (default=1.0e-10). This may be used if necessary to extract
  coefficient standard errors (for instance lowering to 1e-12),
  but errors in 'solve()' may constitute indications of poorly
  scaled variables: if the variables have scales differing much
  from the autoregressive coefficient, the values in this
  matrix may be very different in scale, and inverting such a
  matrix is analytically possible by definition, but
  numerically unstable; rescaling the RHS variables alleviates
  this better than setting tol.solve to a very small value.



 
 
 Regards,
 Dong
 
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-- 
Roger Bivand
Economic Geography Section, Department of Economics, Norwegian School of
Economics and Business Administration, Helleveien 30, N-5045 Bergen,
Norway. voice: +47 55 95 93 55; fax +47 55 95 95 43
e-mail: [EMAIL PROTECTED]

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[R] daisy function in cluster- coerced NAs

[Roger Humphry]

I am currently using the function daisy in package cluster to create a
dissimilarity matrix because my multivariate dataset contain missing
data and variables of various types including factors, symmetric and
asymmetric binary and quantitative. This is a step prior to using pco
within ecodist.

 

There is a warning which comes twice 

NAs introduced by coercion

 

I've used the option:

type = list(asymm = c(4:30,32:42), symm= c(31,43))

to indicate which are binary variables and their symmetry.

 

The remaining variables I have left alone since if they are numerical
they are already in the data frame as int or num and if they are
factors they are already in the dataframe as Factor.

 

Can anyone advise me: a) does this coercion matter? b) If it does
matter, how can I track down where/how it is happening?

 

Thank you in advance,

Roger Humphry

 

 

 Roger Humphry, PhD.

 EPI, SCRI, 

 Dundee, DD2 5DA

 01382 560042 (Internal Ext: 2528)

 

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

SCRI, Invergowrie, Dundee, DD2 5DA.  
The Scottish Crop Research Institute is a charitable company limited by 
guarantee. 
Registered in Scotland No: SC 29367.
Recognised by the Inland Revenue as a Scottish Charity No: SC 006662.


DISCLAIMER:
\ 
\ This email is from the Scottish Crop Re...{{dropped}}

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Re: [R] random uniform sample of points on an ellipsoid (e.g. WGS84)

[Roger Bivand]

On 21 Feb 2007, Russell Senior wrote:

 
 I am interested in making a random sample from a uniform distribution
 of points over the surface of the earth, using the WGS84 ellipsoid as
 a model for the earth.  I know how to do this for a sphere, but would
 like to do better.  I can supply random numbers, want latitude
 longitude pairs out.
 
 Can anyone point me at a solution?  Thanks very much.
 

http://www.csit.fsu.edu/~burkardt/f_src/random_data/random_data.html

looks promising, untried.



 
 

-- 
Roger Bivand
Economic Geography Section, Department of Economics, Norwegian School of
Economics and Business Administration, Helleveien 30, N-5045 Bergen,
Norway. voice: +47 55 95 93 55; fax +47 55 95 95 43
e-mail: [EMAIL PROTECTED]

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[R] investigating interactions with mixed models

[R. Baker]

I'm investigating a number of dependent variables using mixed models, e.g.

data.lmer45 = lmer(ampStopB ~ (type + stress + MorD)^3 + (1|speaker) + 
(1|word), data=data)

The p-values for some of the 2-way and 3-way interactions are significant 
at a 0.05 level and I have been trying to find out how to understand the 
exact nature of the interactions. Does anyone know if it is possible to run 
post-hoc tests on mixed model (lmer) objects? I have read about TukeyHSD 
but it seems that this can only be run on anova (aov) objects.

Any suggestions would be gratefully appreciated!

Rachel Baker

-- 
--
PhD student
Dept of Linguistics
Sidgwick Avenue
University of Cambridge  
Cambridge

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[R] residuals and glm

[Martin Olivier]

Hi all,

I have some problems to compute the residuals from a glm model with
binomial distribution.

Suppose I have the following result :
resfit-glm(y~x1+x2,weights=we,family=binomial(link=logit))

Now I would like to obtain the residuals .
the command residuals(resfit) and the vector resfit$residuals give 
different
results, and they do not correspond to residuals E(yi)-yi (that is 
resfit$fitted-resfit$y)

so, I would like to know what formula  is applied to compute these 
residuals.
And moreover,  if I want to compute the standardized residuals, what is the
right command from the glmfit result.
Is it 
(resfit$fitted-resfit$y)/sqrt(1/resfit$prior*resfit$fitt*(1-resfit$fitt)) ??


Thanks for your help,
Olivier.

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Re: [R] Sorting rows of a binary matrix

[Adrian Dusa]

Hello Serguei,

Is this what you need?

myfunc - function(x) {
create - function(idx) {
rep.int(c(rep.int(0,2^(idx-1)), rep.int(1,2^(idx-1))),
2^x/2^idx)
}
sapply(rev(seq(x)), create)
}

 myfunc(3)
 [,1] [,2] [,3]
[1,]000
[2,]001
[3,]010
[4,]011
[5,]100
[6,]101
[7,]110
[8,]111

For numerical values only, this is faster than expand.grid().
Alternatively (for multiple values in separate varaibles), you could use the 
function createMatrix() in package QCA.

HTH,
Adrian

On Thursday 22 February 2007 12:50, Serguei Kaniovski wrote:
 Hallo,

 The command:

 x - 3
 mat - as.matrix(expand.grid(rep(list(0:1), x)))

 generates a matrix with 2^x columns containing the binary representations
 of the decimals from 0 to (2^x-1), here from 0 to 7. But the rows are not
 sorted in this order.

 How can sort the rows the ascending order of the decimals they represent,
 preferably without a function which converts binaries to decimals (which I
 have)? Alternatively, generate a matrix that has the rows sorted that way?

 Thanks,
 Serguei
   [[alternative HTML version deleted]]

-- 
Adrian Dusa
Romanian Social Data Archive
1, Schitu Magureanu Bd
050025 Bucharest sector 5
Romania
Tel./Fax: +40 21 3126618 \
  +40 21 3120210 / int.101

__
R-help@stat.math.ethz.ch 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] residuals and glm

[David Barron]

You really need to look at ?glm and ?residuals.glm.

resfit$residuals are the *working* residuals, which are not typically
very useful in themselves.  Far better to use the extractor function.
This enables you to obtain a number of different types of residuals,
but the default (and therefore the type you have obtained) are
deviance residuals.  You can also specify other types, such as pearson
residuals.

Hope this helps.
David

On 22/02/07, Martin Olivier [EMAIL PROTECTED] wrote:
 Hi all,

 I have some problems to compute the residuals from a glm model with
 binomial distribution.

 Suppose I have the following result :
 resfit-glm(y~x1+x2,weights=we,family=binomial(link=logit))

 Now I would like to obtain the residuals .
 the command residuals(resfit) and the vector resfit$residuals give
 different
 results, and they do not correspond to residuals E(yi)-yi (that is
 resfit$fitted-resfit$y)

 so, I would like to know what formula  is applied to compute these
 residuals.
 And moreover,  if I want to compute the standardized residuals, what is the
 right command from the glmfit result.
 Is it
 (resfit$fitted-resfit$y)/sqrt(1/resfit$prior*resfit$fitt*(1-resfit$fitt)) ??


 Thanks for your help,
 Olivier.

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



-- 
=
David Barron
Said Business School
University of Oxford
Park End Street
Oxford OX1 1HP

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Re: [R] residuals and glm

[Prof Brian Ripley]

On Thu, 22 Feb 2007, Martin Olivier wrote:

 I have some problems to compute the residuals from a glm model with 
 binomial distribution.

 Suppose I have the following result :
 resfit-glm(y~x1+x2,weights=we,family=binomial(link=logit))

 Now I would like to obtain the residuals . the command residuals(resfit) 
 and the vector resfit$residuals give different results, and they do not 
 correspond to residuals E(yi)-yi (that is resfit$fitted-resfit$y)

?residuals.glm should help you, but note that residuals are always 
conventionally (observed - fitted), not as you give.

?glm tells you what resfit$residuals is, and it seems to be not what you 
think it is.

 so, I would like to know what formula is applied to compute these 
 residuals. And moreover, if I want to compute the standardized 
 residuals, what is the right command from the glmfit result. Is it 
 (resfit$fitted-resfit$y)/sqrt(1/resfit$prior*resfit$fitt*(1-resfit$fitt)) 
 ??

See ?rstandard and print(rstandard.glm).

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

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

[R] Combining tapply() and cor.test()?

[Sergey Goriatchev]

 Hello, fellow R-users.

Let me describe the setup first. I have a data.frame, a sample of
which is reported below:

   Company.Name  Periods  Returns   MFR.Factor
350 Wartsila Oyj A  1996-07-31  6.82 0.02
351Custodia Holding AG  1996-07-31  4.15-0.02
352   Wartsila Oyj   1996-07-31  7.73 0.09
353   GEA Group AG   1996-07-3110.12 0.04
354LEGRAND ORD 1996-07-31 -7.46-0.20
355 Mayr-Melnhof Karton AG 1996-07-31 4.71-0.05
356GEVAERT NPV  1996-08-30  NA  NA
357NOKIA K FMA2.50  1996-08-30 7.65 0.03
358   Altadis S.A. 1996-08-30 7.65 0.55
359   Metrovacesa S.A. 1996-08-30 4.55-0.17
360   Oce N.V.  1996-08-309.43 0.23

The variable Periods is a date object, shows the month.
Variables Returns and MFR.Factor are numeric.
For each month the number of Returns and MFR.Factors varies, sometimes
it is 350, sometimes 320 etc.

What I need is to use cor.test(Returns, MFR.Factor,...) for each
month, and produce a dataframe with columns: Period, cor.estimate,
p.value.

The simplest way would be with tapply() using variable Period as a
factor, but tapply() only applies FUN to just one cell.

What is the most painless way to achieve my objective?

Thank you in advance for your help!

Best,
Sergey

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

Re: [R] residuals and glm

[Frank E Harrell Jr]

David Barron wrote:
 You really need to look at ?glm and ?residuals.glm.
 
 resfit$residuals are the *working* residuals, which are not typically
 very useful in themselves.  Far better to use the extractor function.
 This enables you to obtain a number of different types of residuals,
 but the default (and therefore the type you have obtained) are
 deviance residuals.  You can also specify other types, such as pearson
 residuals.
 
 Hope this helps.
 David

And in the Design package's lrm and residuals.lrm function (called by 
resid(fit)) you can get partial residuals that when smoothed estimate 
covariate effects.  I do prefer direct modeling instead of looking at 
any of the residuals though, e.g., adding nonlinear or interaction terms 
to logistic models.  Usually I find that residuals in simple binary 
logistic models are most useful in checking for overly influential 
observations.

Frank Harrell

 
 On 22/02/07, Martin Olivier [EMAIL PROTECTED] wrote:
 Hi all,

 I have some problems to compute the residuals from a glm model with
 binomial distribution.

 Suppose I have the following result :
 resfit-glm(y~x1+x2,weights=we,family=binomial(link=logit))

 Now I would like to obtain the residuals .
 the command residuals(resfit) and the vector resfit$residuals give
 different
 results, and they do not correspond to residuals E(yi)-yi (that is
 resfit$fitted-resfit$y)

 so, I would like to know what formula  is applied to compute these
 residuals.
 And moreover,  if I want to compute the standardized residuals, what is the
 right command from the glmfit result.
 Is it
 (resfit$fitted-resfit$y)/sqrt(1/resfit$prior*resfit$fitt*(1-resfit$fitt)) ??


 Thanks for your help,
 Olivier.

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

 
 


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

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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] Sorting rows of a binary matrix

[Adrian Dusa]

And, for multiple bases:

myfunc - function(cols, bases) {
   create - function(idx) {
rep.int(c(sapply(seq_len(bases)-1, function(x)
rep.int(x, bases^(idx-1, bases^cols/bases^idx)
}
sapply(rev(seq_len(cols)), create)
}

# For 3 columns in base 2
myfunc(3, 2)

# For 3 columns in base 3
myfunc(3, 3)

hth,
Adrian


On Thursday 22 February 2007 15:00, Adrian Dusa wrote:
 Hello Serguei,

 Is this what you need?

 myfunc - function(x) {
 create - function(idx) {
 rep.int(c(rep.int(0,2^(idx-1)), rep.int(1,2^(idx-1))),
 2^x/2^idx)
 }
 sapply(rev(seq(x)), create)
 }

  myfunc(3)

  [,1] [,2] [,3]
 [1,]000
 [2,]001
 [3,]010
 [4,]011
 [5,]100
 [6,]101
 [7,]110
 [8,]111

 For numerical values only, this is faster than expand.grid().
 Alternatively (for multiple values in separate varaibles), you could use
 the function createMatrix() in package QCA.

 HTH,
 Adrian

 On Thursday 22 February 2007 12:50, Serguei Kaniovski wrote:
  Hallo,
 
  The command:
 
  x - 3
  mat - as.matrix(expand.grid(rep(list(0:1), x)))
 
  generates a matrix with 2^x columns containing the binary representations
  of the decimals from 0 to (2^x-1), here from 0 to 7. But the rows are not
  sorted in this order.
 
  How can sort the rows the ascending order of the decimals they represent,
  preferably without a function which converts binaries to decimals (which
  I have)? Alternatively, generate a matrix that has the rows sorted that
  way?
 
  Thanks,
  Serguei
  [[alternative HTML version deleted]]

-- 
Adrian Dusa
Romanian Social Data Archive
1, Schitu Magureanu Bd
050025 Bucharest sector 5
Romania
Tel./Fax: +40 21 3126618 \
  +40 21 3120210 / int.101

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

Re: [R] Combining tapply() and cor.test()?

[jim holtman]

?by

On 2/22/07, Sergey Goriatchev [EMAIL PROTECTED] wrote:

 Hello, fellow R-users.

 Let me describe the setup first. I have a data.frame, a sample of
 which is reported below:

   Company.Name  Periods  Returns   MFR.Factor
 350 Wartsila Oyj A  1996-07-31  6.82 0.02
 351Custodia Holding AG  1996-07-31  4.15-0.02
 352   Wartsila Oyj   1996-07-31  7.73 0.09
 353   GEA Group AG   1996-07-3110.12 0.04
 354LEGRAND ORD 1996-07-31 -7.46-0.20
 355 Mayr-Melnhof Karton AG 1996-07-31 4.71-0.05
 356GEVAERT NPV  1996-08-30  NA  NA
 357NOKIA K FMA2.50  1996-08-30 7.65 0.03
 358   Altadis S.A. 1996-08-30 7.65 0.55
 359   Metrovacesa S.A. 1996-08-30 4.55-0.17
 360   Oce N.V.  1996-08-309.43 0.23

 The variable Periods is a date object, shows the month.
 Variables Returns and MFR.Factor are numeric.
 For each month the number of Returns and MFR.Factors varies, sometimes
 it is 350, sometimes 320 etc.

 What I need is to use cor.test(Returns, MFR.Factor,...) for each
 month, and produce a dataframe with columns: Period, cor.estimate,
 p.value.

 The simplest way would be with tapply() using variable Period as a
 factor, but tapply() only applies FUN to just one cell.

 What is the most painless way to achieve my objective?

 Thank you in advance for your help!

 Best,
 Sergey

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




-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem you are trying to solve?

[[alternative HTML version deleted]]

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

[R] Error in solve.default

[Grant, Mark D.]

I am trying to run the following function (a hierarchical bayes linear
model) and receive the error in solve.default.  The function was
originally written for an older version of SPlus.  Can anyone give me some
insights into where the problem is?

Thanks

R 2.4.1 on MAC OSX 2mb ram

Mark Grant
[EMAIL PROTECTED]

 attach(Aspirin.frame)
 hblm(Diff ~ 1, s = SE)
Error in solve.default(R, rinv) : 'a' is 0-diml

 traceback()
6: .Call(La_dgesv, a, b, tol, PACKAGE = base)
5: solve.default(R, rinv)
4: solve(R, rinv)
3: summary.blm(fit)
2: eb.calc(rho[i], X, Y, s.e., df.se, corrs, prior, ...)
1: hblm(Diff ~ 1, s = SE)


 hblm
function(formula, s.e., df.se = Inf, corrs = F, prior = NULL, fast.calc = F,
...)
{
# hblm()
# main program to create hblm object
call - match.call()
call.ab - abbrev.hblm.call(call)
taumin - (sum(1/s.e.^2))^-0.5
if(is.null(prior$error)) {
prior$error$df - 1
prior$error$sd - taumin
}
s.e..old - s.e.
if(max(s.e.) == Inf)
s.e.[s.e. == Inf] - taumin * 1
wts - s.e.^(-2)  # changed 1.10.2007 DMR df.se
- 1/mean(1/df.se)
if(df.se == Inf) {
if(is.null(prior$tau))
prior$tau - taumin * sqrt(length(wts))
}
fit - lm(formula, weights = wts, qr = T, x = T, y = T, singular =
T,
...)
X - fit$x
Y - fit$y
Terms - fit$terms
rss - sum(wts * fit$residuals^2)
levs - hat(fit$qr)
tau.rss - (max(0, rss - fit$df.residual)/sum((1 - levs) *
wts))^0.5
log.r - if(tau.rss  0) log(tau.rss) else log(taumin)
maxval - max1d(l.p.d.log.r, start = log.r, step =
taumin/exp(log.r), x
 = X, y = Y, s.e. = s.e., df.se = df.se, prior = prior,
...)
log.r - maxval$x
se.lr - (2 * maxval$h)^(-0.5)
g.h.v - gauss.hermite.vals(log.r, se.lr, fast.calc)
rho - exp(g.h.v$x)
r - length(rho)
k - nrow(X)
p - ncol(X)
tau - sig - lpd - array(0, dim = r)
coefs - array(0, dim = c(r, p, 3), dimnames = list(NULL,
dimnames(X)[[
2]], c(Mean, S.D., Prob  0)))
cov.c - array(0, dim = c(r, p, p), dimnames =
append(dimnames(X)[c(2,
2)], list(NULL), 0))
means - array(0, dim = c(r, k, 3), dimnames = list(NULL,
names(Y), c(
Post.Mn, Post.SD, Prob  0)))
if(corrs)
cov.m - array(0, dim = c(r, k, k), dimnames = list(NULL,
names(
Y), names(Y)))
for(i in 1:r) {
fit - eb.calc(rho[i], X, Y, s.e., df.se, corrs, prior,
...)
tau[i] - fit$tau
sig[i] - fit$sigma
lpd[i] - fit$lpd
coefs[i,  ,  ] - fit$coefs
cov.c[i,  ,  ] - fit$cov.c
means[i,  ,  ] - fit$means
if(corrs)
cov.m[i,  ,  ] - fit$cov.means
}
prob - (exp(lpd) * g.h.v$w)/sum(exp(lpd) * g.h.v$w)
if(fit$df.post == Inf) {
K1 - K2 - 1
}
else {
K1 - sqrt(fit$df.post/2) * exp(lgamma((fit$df.post -
1)/2) -
lgamma(fit$df.post/2))
K2 - fit$df.post/(fit$df.post - 2)
}
tausq.m - K2 * prob %*% tau^2
tau.m - K1 * prob %*% tau
tau.sd - sqrt(tausq.m - tau.m^2)
sigsq.m - K2 * prob %*% sig^2
sig.m - K1 * prob %*% sig
sig.sd - sqrt(sigsq.m - sig.m^2)
coefs.m - prob %ip% coefs
cov.c.m - K2 * (prob %ip% cov.c)
coef.d - coefs[,  , 1] - matrix(coefs.m[, 1], nrow = r, ncol = p,

byrow = T)
coef.v - array(0, dim = dim(cov.c.m), dimnames =
dimnames(cov.c.m))
for(i in 1:r)
coef.v - coef.v + prob[i] * outer(coef.d[i,  ], coef.d[i,
 ])
coefs.m[, 2] - sqrt(diag(cov.c.m) + diag(coef.v))
shrink - matrix(0, nrow = k, ncol = 6, dimnames = list(names(Y),
c(Y,
Prior Mn, (Y-Prior)/SE, Post.Mn, Post.SD, Prob 
0)))
fitted - X %*% coefs.m[, 1]
residuals - (Y - fitted)/s.e.
shrink[, 1:3] - matrix(c(Y, fitted, residuals), ncol = 3)
shrink[, 4:6] - prob %ip% means
means.d - means[,  , 1] - matrix(shrink[, 4], nrow = r, ncol = k,

byrow = T)
means.v - prob %*% means.d^2
shrink[, 5] - sqrt(K2 * (prob %*% means[,  , 2]^2) + means.v)
corr.m.m - if(!corrs) NULL else {
covm - matrix(0, nrow = k, ncol = k, dimnames =
list(names(Y),
names(Y)))
for(i in 1:r) {
covd - means.d[i,  ] %o% means.d[i,  ]
covm - covm + prob[i] * (covd + K2 * cov.m[i,  ,
])
}
covm/(shrink[, 5] %o% shrink[, 5])
}
trace - list(rho = rho, tau = tau, sigma = sig, lpd = lpd, prob =
prob,

Re: [R] Combining tapply() and cor.test()?

[Dimitris Rizopoulos]

one approach is the following:

dat - data.frame(
Period = as.Date(rep(c(1996-07-31, 1996-08-31, 1996-09-30), 
each = 15)),
Returns = rnorm(45),
MFR.Factor = runif(45)
)

###

do.call(rbind, lapply(split(dat[c(Returns, MFR.Factor)], 
dat$Period),
function (x) {
cr - cor.test(x$Returns, x$MFR.Factor, method = spearman)
c(estimate = cr$estimate, p.value = cr$p.value)
}))


I hope it helps.

Best,
Dimitris


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

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


- Original Message - 
From: Sergey Goriatchev [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Thursday, February 22, 2007 2:35 PM
Subject: [R] Combining tapply() and cor.test()?


 Hello, fellow R-users.

 Let me describe the setup first. I have a data.frame, a sample of
 which is reported below:

   Company.Name  Periods  Returns   MFR.Factor
 350 Wartsila Oyj A  1996-07-31  6.82 0.02
 351Custodia Holding AG  1996-07-31  4.15-0.02
 352   Wartsila Oyj   1996-07-31  7.73 0.09
 353   GEA Group AG   1996-07-3110.12 0.04
 354LEGRAND ORD 1996-07-31 -7.46-0.20
 355 Mayr-Melnhof Karton AG 1996-07-31 4.71-0.05
 356GEVAERT NPV  1996-08-30  NA  NA
 357NOKIA K FMA2.50  1996-08-30 7.65 0.03
 358   Altadis S.A. 1996-08-30 7.65 0.55
 359   Metrovacesa S.A. 1996-08-30 4.55-0.17
 360   Oce N.V.  1996-08-309.43 0.23

 The variable Periods is a date object, shows the month.
 Variables Returns and MFR.Factor are numeric.
 For each month the number of Returns and MFR.Factors varies, 
 sometimes
 it is 350, sometimes 320 etc.

 What I need is to use cor.test(Returns, MFR.Factor,...) for each
 month, and produce a dataframe with columns: Period, 
 cor.estimate,
 p.value.

 The simplest way would be with tapply() using variable Period as a
 factor, but tapply() only applies FUN to just one cell.

 What is the most painless way to achieve my objective?

 Thank you in advance for your help!

 Best,
 Sergey

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


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

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

Re: [R] random uniform sample of points on an ellipsoid (e.g. WG

[Ted Harding]

On 22-Feb-07 Roger Bivand wrote:
 On 21 Feb 2007, Russell Senior wrote:
 
 
 I am interested in making a random sample from a uniform distribution
 of points over the surface of the earth, using the WGS84 ellipsoid as
 a model for the earth.  I know how to do this for a sphere, but would
 like to do better.  I can supply random numbers, want latitude
 longitude pairs out.
 
 Can anyone point me at a solution?  Thanks very much.
 
 
 http://www.csit.fsu.edu/~burkardt/f_src/random_data/random_data.html
 
 looks promising, untried.

Hmmm ... That page didn't seem to be directly useful, since
on my understanding of the code (and comments) listed under
subroutine uniform_on_ellipsoid_map(dim_num, n, a, r, seed, x)
UNIFORM_ON_ELLIPSOID_MAP maps uniform points onto an ellipsoid.
in

http://www.csit.fsu.edu/~burkardt/f_src/random_data/random_data.f90

it takes points uniformly distributed on a sphere and then
linearly transforms these onto an ellipsoid. This will not
give unform density over the surface of the ellipsoid: indeed
the example graph they show of points on an ellipse generated
in this way clearly appear to be more dense at the ends of
the ellipse, and less dense on its sides. See:

http://www.csit.fsu.edu/~burkardt/f_src/random_data/
uniform_on_ellipsoid_map.png
[all one line]

Indeed, if I understand their method correctly, in the case
of a horizontal ellipse it is equivalent (modulo rotating
the result) to distributing the points uniformly over a circle,
and then stretching the circle sideways. This will preserve
the vertical distribution (so at the two ends of the major axis
it has the same density as on the circle) but diluting the
horizontal distribution (so that at the two ends of the minor
axis the density isless than on the circle).

I did have a notion about this, but sat on it expecting that
someone would come up with a slick solution -- which hasn't
happened yet.

For the application you have in hand, uniform distribution
over a sphere is a fairly close approximation to uniform
distriobution over the ellipspoid -- but not quite.

But a rejection method, applied to points uniform on the sphere,
can give you points uniform on the ellipsoid and, because of
the close approximation of the sphere to the ellipsoid, you
would not be rejecting many points.

The outline strategy I had in mind (I haven't worked out details)
is based on the following.

Consider a point X0 on the sphere, at radial distance r0 from
the centre of the sphere (same as the centre of the ellipsoid).
Let the radius through that point meet the ellipsoid at a point
X1, at radial distance R1.

Let dS0 be an element of area at X0 on the sphere, which projects
radially onto an element of area dS1 on the ellipsoid. You want
all elements dS1 of equal size to be equally likely to receive
a random point.

Let the angle between the tangent plane to the ellipsoid at X1,
and the tangent plane to the sphere at X0, be phi.

The the ratio of areas dS1/dS0 is R(X0), say, where

  R(X0) = dS1/dS0 = r1^2/(r0^2 * cos(phi))

and the smaller this ratio, the less likely you want a point
u.d. on the sphere to give rise to a point on the ellipsoid.

Now define a rejection probability P(X0) by

  P(X0) = R(X0)/sup[R(X)]

taking the supremum over X on the sphere. Then sample points X0
unformly on the sphere, rejecting each on with probability
P(X0), and continue sampling until you have the number of
points that you need.

Maybe someone has a better idea ... (or code for the above!)

Ted.


E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 22-Feb-07   Time: 14:10:13
-- XFMail --

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[R] Updating or installing R packages on Windows Vista

[Christopher Albert]

Hi,

Windows Vista includes additional security mechanisms (User Access Control)
whose defaults make it difficult to install or update R packages.
To avoid these problems you need to go to Computer- Program Files
Right click on the R directory and select properties. Now select the
security tab.
Give your user ( which is the use R whose priviledges R runs under) Full
Control to the R directory.
This should solve the install/update issues.

Keep up the good work.

Chris Albert

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

Re: [R] Updating or installing R packages on Windows Vista

[Charles Annis, P.E.]

Or you can right-click on the R icon and choose Run as administrator.  That
way you won't alter the security settings and forget to re-set them. After
the packages are installed R will load in the usual way by clicking on the
icon.

Charles Annis, P.E.

[EMAIL PROTECTED]
phone: 561-352-9699
eFax:  614-455-3265
http://www.StatisticalEngineering.com
 

-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Christopher Albert
Sent: Wednesday, February 21, 2007 10:23 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Updating or installing R packages on Windows Vista

Hi,

Windows Vista includes additional security mechanisms (User Access Control)
whose defaults make it difficult to install or update R packages.
To avoid these problems you need to go to Computer- Program Files
Right click on the R directory and select properties. Now select the
security tab.
Give your user ( which is the use R whose priviledges R runs under) Full
Control to the R directory.
This should solve the install/update issues.

Keep up the good work.

Chris Albert

[[alternative HTML version deleted]]

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Re: [R] how much performance penalty does this incur, scalar as a vector of one element?

[Luke Tierney]

I think the short answer is not much.

Longer answer: In an interpreted framework with double precision
floating point scalars there is little chance of avoiding fresh
allocations for each scalar; given that, the overhead associated with
length checks can be made negligible.  (That isn't to say it currently
is--it may or may not be, but you asked about design.)  Systems that
support integer scalars often represent them as immediate values within
pointers by sacrificing one or two bits of precision in the integers,
but that doesn't work for double precision floats except possibly on
64-bit systems.  Though even there it would be possible to use an
efficient internal representation of vectors of length one without
changing the concept that everything is a vector.

As we think about compilation there are opportunities to produce more
efficient code if values can be assumed to be scalars, but that can be
accomplished by adding a declaration mechanism.  So again the answer
in terms of efficiency cost is not much.

The APL view of everything as an array, with zero-dimensional arrays
being scalars and higher-dimensional arrays being real entities rather
than decorated vectors, is in many ways conceptually cleaner and might
in hindsight have been a better choice for that reason, but efficiency
isn't really a consideration.

Best,

luke

On Wed, 21 Feb 2007, Jason Liao wrote:


 I have been comparing R with other languages and systems. One peculiar 
 feature of R is there is no scalar. Instead, it is just a vector of length 
 one. I wondered how much performance penalty this deign cause, particular in 
 situations with many scalars in a program. Thanks.



 Jason Liao, http://www.geocities.com/jg_liao
 Associate Professor of Biostatistics
 Drexel University School of Public Health
 245 N. 15th Street, Mail Stop 660
 Philadelphia, PA 19102-1192
 phone 215-762-3934





 
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 Check out Tonight's Picks on Yahoo! TV.

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-- 
Luke Tierney
Chair, Statistics and Actuarial Science
Ralph E. Wareham Professor of Mathematical Sciences
University of Iowa  Phone: 319-335-3386
Department of Statistics andFax:   319-335-3017
Actuarial Science
241 Schaeffer Hall  email:  [EMAIL PROTECTED]
Iowa City, IA 52242 WWW:  http://www.stat.uiowa.edu

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[R] List filtration

[Johannes Graumann]

Hello R-ologists,

Imagine you have a list list like so:

list
[[1]]
[1] IPI00776145.1 IPI00776187.1

[[2]]
[1] Something IPI00807764.1 IPI00807887.1

[[3]]
[1] IPI00807764.1

[[4]]
[1] Somethingelse

What I need to achieve is a filtered list list2 like so:

list2
[[1]]
[1] IPI00776145.1

[[2]]
[1] IPI00807764.1

[[3]]
[1] IPI00807764.1

So: 
- if sublist-entry 1 start with ^IPI make it the list-entry.
- otherwise chose the first ^IPI sublist-entry present.
- delete the list-entry if not ^IPI sublist-entry present.

Can anybody nudge me towards an elegant solution without looping - I have
LOTS of entries to process ...

Thanks for your Teachings,

Joh

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[R] several Filled.contour plots on the same device...

[Alexis . berg]

hello -

a question about filled.contour plots, for which i haven't found a
response in previous posts - sorry if already treated.

i'd like to draw several filled.contour plots (that is, maps) on the same
device (a postscript file, actually). I know about layout(matrix) ,
split.screen or par(mfrow) : it works well for simple plots, but with
filled.contour plots, i get several pages instead of one page divided into
several cells.
I'd really like to get these maps directly on one graphs, without having
to process them afterwards.
Does anyone know something about that ?

Thank you for your help,

Alexis Berg

Ingénieur de recherche
LOCEAN (IPSL) - Paris
[EMAIL PROTECTED]

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[R] Package for Screen Scrapers?

[Edward Bolanger]

Hi All,

I was doing clustering on some genes, and wanted to verify the clustering 
results using another website.

Essentially, I upload two files (in Step 1  Step 3) at this website: 
http://db.yeastgenome.org/cgi-bin/GO/goTermFinder 
The website then outputs a graph, which I save. 

Are there any package that might help me with the screen scraping? Ideally, I 
would be calling the screen scraper from within the R code.

many thanks!
 
-
Have a burning question? Go to Yahoo! Answers and get answers from real people 
who know.
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[R] Diagnostic Tests: Jarque-Bera Test / RAMSEY

[Simon P. Kempf]

Hello R-Users,

 

The following questions are not R-technical, but more of general statistical
nature. 

 

1. NORMALITY 

I built a normal linear regression model and now I want to check for the
residual normality assumption. If I check the distribution graphically and
look at the descriptive characteristics (skewness and kurtosis are below 1),
I would confirm that the residuals are normally distributed. 

 

 basicStats(IQR.in.mi02.nw.tdv.mix$residuals)

round.ans..digits...6.

nobs  19316.00

NAs   0.00

Minimum  -0.639527

Maximum   0.693383

1. Quartile  -0.083753

3. Quartile   0.06

Mean  0.00

Median0.004641

Sum   0.00

SE Mean   0.001047

LCL Mean -0.002053

UCL Mean  0.002053

Variance  0.021186

Stdev 0.145554

Skewness -0.164821

Kurtosis  0.937282

 

However, when I use the jarque.bera.test(), the assumption of normality is
rejected.

 

 jarque.bera.test(IQR.in.mi02.nw.tdv.mix$residuals)

 

Jarque Bera Test

 

data:  IQR.in.mi02.nw.tdv.mix$residuals 

X-squared = 795.1296, df = 2, p-value  2.2e-16

 

Therefore, I am wondering how good are the diagnostic test for the normality
assumption? Do you they work for large sample as well? If not, what other
diagnostic measures for normality exist for large samples? 

 

In many statistic textbooks it is mentioned that for large samples the
significance tests are independent from the distributions of residuals. Why
is that?

 

2. RAMSEY TEST

I made several scatterplots (residuals ~ predictors). Here everything looks
fine but when I apply the RAMSEY resettest(), the assumption of linearity is
rejected. My model does not show any signs of multicollinearity and all
independent variables have strong linear relationship with the dependent
variable. Therefore, I am wondering how well the RAMSEY resettest works in
general. What other tests exist?

 

Again, I would like to apologize for the asking no specific R-technical
questions. But I would really appreciate any help.

 

Thanks,

 

Simon

 


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[R] Cross-tabulations next to each other

[Charilaos Skiadas]

I have the following relatively simple problem. Say we have three  
factors, and we want to create a cross-tabulation against each of the  
other two:

x - factor(rbinom(5, 1, 1/2))
y - factor(rbinom(5, 1, 1/2))
z - factor(rbinom(5, 1, 1/2))
table(x,y)
table(x,z)

This looks like:

y
x   0 1
   0 2 0
   1 1 2

z
x   0 1
   0 1 1
   1 2 1

I would like to get (surely this will look a mess in non-monospaced  
fonts):

yz
x   0 1  0 1
   0 2 0  1 1
   1 1 2  2 1

Or something along those lines. Then I would like to convert this to  
a LaTeX table, in the obvious sort of way.

I couldn't find an answer with a quick look through the  
documentation. Are these two things already done, before I try to  
roll my own?

Haris Skiadas
Department of Mathematics and Computer Science
Hanover College

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[R] R CMD CHECK question

[Joerg van den Hoff]

hi,

I have two private packages, the first (`pkc') depending on the second one
(`roiutils'). The source code and DESCRIPTION files describes the dependency
as it should be ('Imports', `require'), at least I think so.

now, running

R CMD CHECK pkc 

yields the following output in which I have inserted my
questions (lines starting with --):

* checking for working latex ... OK
* using log directory '/Users/vdh/rfiles/Rlibrary/.check/pkc.Rcheck'
* using R version 2.4.0 (2006-10-03)
* checking for file 'pkc/DESCRIPTION' ... OK
* this is package 'pkc' version '1.1'
* checking package dependencies ... OK
* checking if this is a source package ... OK
* checking whether package 'pkc' can be installed ... WARNING
Found the following significant warnings:
   missing link(s):  readroi readroi readroi figure readroi conv3exmodel 
readroi
   missing link(s):  figure readroi

-- there _are_ links to the mentioned functions (from `roiutils') in the
-- manpages of `pkc'. after installing the libs, the help system works just
-- fine. why is it, that CHECK is complaining? can one selectively switch of
-- this warning? or how have I to specify the links to tell CHECK that
-- everything is OK?

* checking package directory ... OK
* checking for portable file names ... OK
* checking for sufficient/correct file permissions ... OK
* checking DESCRIPTION meta-information ... OK
* checking top-level files ... OK
* checking index information ... OK
* checking package subdirectories ... OK
* checking R files for syntax errors ... OK
* checking R files for non-ASCII characters ... OK
* checking whether the package can be loaded ... OK
* checking whether the package can be loaded with stated dependencies ... OK
* checking whether the name space can be loaded with stated dependencies ... OK
* checking S3 generic/method consistency ... OK
* checking replacement functions ... OK
* checking foreign function calls ... OK
* checking R code for possible problems ... OK
* checking Rd files ... WARNING
Rd files with unknown sections:
  /Users/vdh/rfiles/Rlibrary/pkc/man/fitdemo.Rd: example

See the chapter 'Writing R documentation files' in manual 'Writing R
Extensions'.
* checking Rd cross-references ... WARNING
Missing link(s) in documentation object 'compfit.Rd':
  readroi readroi readroi figure readroi conv3exmodel readroi

Missing link(s) in documentation object 'exp3fit.Rd':
  figure readroi

-- this seems the same problem as above, right?





any hints appreciated,

joerg

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[R] Crosstabbing multiple response data

[Michael Wexler]

Using R version 2.4.1 (2006-12-18) on Windows, I have a dataset which resembles 
this:

idatt1att2att3
1110
2100
3011
4111

ratings - data.frame(id = c(1,2,3,4), att1 = c(1,1,0,1), att2 = c(1,0,0,1), 
att3 = c(0,1,1,1))

I would like to get a cross tab of counts of co-ocurrence, which might resemble 
this:

att1att2att3
att1 2   1
att222
att312

with the hope of understanding, at least pairwise, what things hang together. 
  (Yes, there are much, much better ways to do this statistically including 
clustering and binary corrected correlation, but the audience I am working with 
asked for this version for a specific reason.)  

(Later on, I would also like to convert to percentages of the total unique pop, 
so the final version of the table would be


att1att2att3

att1 50%   25%

att250%50%

att325%50%


But I can do this in excel if I can get the first table out.)

I have tried the reshape library, but could not get anything resembling this 
(both on its own, as well as feeding in to table()).  (I have also played with 
transposing and using some comments from this list from 2002 and 2004, but the 
questioners appear to assume more knowledge than I have in use of R; the 
example in the posting guide was also more complex than I was ready for, I'm 
afraid.)

Sample of some of my efforts:
library(reshape)
melt(ratings,id=c(id))

ds1 - melt(ratings,id=c(id))
table(ds1$variable, ds1$variable) # returns only rowcounts, 3 along diagonal
xtabs(formula = value ~ ds1$variable + ds1$variable , data=ds1) # returns only 
a single row of collapsed counts, appears to not allow 1 variable in multiple 
uses

I suspect I am close, so any nudges in the right direction would be helpful.

Thanks much, Michael

PS: www.rseek.org is very impressive, I heartily encourage its use.


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Re: [R] List filtration

[Dimitris Rizopoulos]

try this:

lis. - lapply(lis, function(x) if (length(ind - grep(^IPI, x))) 
x[ind[1]] else NULL)
lis.[!sapply(lis., is.null)]


I hope it helps.

Best,
Dimitris


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

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


- Original Message - 
From: Johannes Graumann [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Thursday, February 22, 2007 3:33 PM
Subject: [R] List filtration


 Hello R-ologists,

 Imagine you have a list list like so:

list
 [[1]]
 [1] IPI00776145.1 IPI00776187.1

 [[2]]
 [1] Something IPI00807764.1 IPI00807887.1

 [[3]]
 [1] IPI00807764.1

 [[4]]
 [1] Somethingelse

 What I need to achieve is a filtered list list2 like so:

list2
 [[1]]
 [1] IPI00776145.1

 [[2]]
 [1] IPI00807764.1

 [[3]]
 [1] IPI00807764.1

 So:
 - if sublist-entry 1 start with ^IPI make it the list-entry.
 - otherwise chose the first ^IPI sublist-entry present.
 - delete the list-entry if not ^IPI sublist-entry present.

 Can anybody nudge me towards an elegant solution without looping - I 
 have
 LOTS of entries to process ...

 Thanks for your Teachings,

 Joh

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


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

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Re: [R] List filtration

[Rajarshi Guha]

On Thu, 2007-02-22 at 15:33 +0100, Johannes Graumann wrote:
 Hello R-ologists,
 
[snip]
 
 So: 
 - if sublist-entry 1 start with ^IPI make it the list-entry.
 - otherwise chose the first ^IPI sublist-entry present.
 - delete the list-entry if not ^IPI sublist-entry present.

One way to do it would be:

l - list(c(IPI00776145.1, IPI00776187.1),
  c(Something, IPI00807764.1, IPI00807887.1),
  c(IPI00807764.1),
  c(Somethingelse))

f - function(x) {
  r - grep(^IPI, x, value=TRUE)
  if (length(r)  0) return(r[1])
  else return(NA)
}

l2 - unlist(lapply(l, f))
l2 - l2[!is.na(l2)]

But I'm sure that more elegant solutions will be posted

---
Rajarshi Guha [EMAIL PROTECTED]
GPG Fingerprint: 0CCA 8EE2 2EEB 25E2 AB04 06F7 1BB9 E634 9B87 56EE
---
Writing software is more fun than working.

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[R] how to show date with this subset

[Alfonso Sammassimo]

Dear List,

Thankyou to Jim and Marc for their help on my previous question.

I have a data frame of five columns, the first being a list of dates and the 
other four columns are numeric values. I wanted to list the days where all 4 
columns of values are less than in the previous row. I used the following 
which works fine, except it doesnt show the dates for each row (the values 
from column 1).

differences - apply(x, 2, diff)
all.lower.diffs - subset(differences, apply(differences, 1, 
function(x){all(x0)}

I tried using the following loop instead, but it would only apply to the 
first column of every row(warning condition has length1 and only first 
element will be used):

for ( i in 2:length(x[,1]))  if (x[i,]x[i-1,]) {print(x[i,])}

How can I resolve with either method? Any help much appreciated.

Regards,
Alf Sammassimo.

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[R] Debugging S Plus

[cjkogan111]

Does anyone know the basic commands to debug in s plus. I know how to debug
in r, but I'm having trouble finding similar tools. Mostly what I want to
know is what is the equivalent of the debug() command, and what do I use to
move to the next line, and get the values of different variables. 
Thanks!
-cjkogan111
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View this message in context: 
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Sent from the R help mailing list archive at Nabble.com.

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Re: [R] List filtration

[Johannes Graumann]

Thanks for your help!

Joh

Dimitris Rizopoulos wrote:

 try this:
 
 lis. - lapply(lis, function(x) if (length(ind - grep(^IPI, x)))
 x[ind[1]] else NULL)
 lis.[!sapply(lis., is.null)]
 
 
 I hope it helps.
 
 Best,
 Dimitris
 
 
 Dimitris Rizopoulos
 Ph.D. Student
 Biostatistical Centre
 School of Public Health
 Catholic University of Leuven
 
 Address: Kapucijnenvoer 35, Leuven, Belgium
 Tel: +32/(0)16/336899
 Fax: +32/(0)16/337015
 Web: http://med.kuleuven.be/biostat/
  http://www.student.kuleuven.be/~m0390867/dimitris.htm
 
 
 - Original Message -
 From: Johannes Graumann [EMAIL PROTECTED]
 To: r-help@stat.math.ethz.ch
 Sent: Thursday, February 22, 2007 3:33 PM
 Subject: [R] List filtration
 
 
 Hello R-ologists,

 Imagine you have a list list like so:

list
 [[1]]
 [1] IPI00776145.1 IPI00776187.1

 [[2]]
 [1] Something IPI00807764.1 IPI00807887.1

 [[3]]
 [1] IPI00807764.1

 [[4]]
 [1] Somethingelse

 What I need to achieve is a filtered list list2 like so:

list2
 [[1]]
 [1] IPI00776145.1

 [[2]]
 [1] IPI00807764.1

 [[3]]
 [1] IPI00807764.1

 So:
 - if sublist-entry 1 start with ^IPI make it the list-entry.
 - otherwise chose the first ^IPI sublist-entry present.
 - delete the list-entry if not ^IPI sublist-entry present.

 Can anybody nudge me towards an elegant solution without looping - I
 have
 LOTS of entries to process ...

 Thanks for your Teachings,

 Joh

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

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Re: [R] Cross-tabulations next to each other

[Dimitris Rizopoulos]

maybe cbind() is close to what you're looking for, e.g.,

tb1 - table(x, y)
tb2 - table(x, z)

cbind(tb1, tb2)


Best,
Dimitris


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

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


- Original Message - 
From: Charilaos Skiadas [EMAIL PROTECTED]
To: R-Mailingliste r-help@stat.math.ethz.ch
Sent: Thursday, February 22, 2007 4:01 PM
Subject: [R] Cross-tabulations next to each other


I have the following relatively simple problem. Say we have three
 factors, and we want to create a cross-tabulation against each of 
 the
 other two:

 x - factor(rbinom(5, 1, 1/2))
 y - factor(rbinom(5, 1, 1/2))
 z - factor(rbinom(5, 1, 1/2))
 table(x,y)
 table(x,z)

 This looks like:

y
 x   0 1
   0 2 0
   1 1 2

z
 x   0 1
   0 1 1
   1 2 1

 I would like to get (surely this will look a mess in non-monospaced
 fonts):

yz
 x   0 1  0 1
   0 2 0  1 1
   1 1 2  2 1

 Or something along those lines. Then I would like to convert this to
 a LaTeX table, in the obvious sort of way.

 I couldn't find an answer with a quick look through the
 documentation. Are these two things already done, before I try to
 roll my own?

 Haris Skiadas
 Department of Mathematics and Computer Science
 Hanover College

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

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Re: [R] Cross-tabulations next to each other

[Charilaos Skiadas]

Hi Dimitris,

On Feb 22, 2007, at 10:27 AM, Dimitris Rizopoulos wrote:

 maybe cbind() is close to what you're looking for, e.g.,

 tb1 - table(x, y)
 tb2 - table(x, z)

 cbind(tb1, tb2)

Yes, that was my first thought too, and it does place the values  
where I want them, but it completely destroys the names, which I'd  
like to keep, i.e. it doesn't treat it as a table any more. The  
resulting LaTeX table I would like to have a very top row, with  
multicolumn titles, one for each factor, then a second row with the  
levels for each factor, and then below those the data. I could I  
guess add that stuff separately, but I was hoping someone had already  
done that part.

 Best,
 Dimitris

Haris Skiadas
Department of Mathematics and Computer Science
Hanover College

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Re: [R] how to show date with this subset

[Prof Brian Ripley]

On Fri, 23 Feb 2007, Alfonso Sammassimo wrote:

 Dear List,

 Thankyou to Jim and Marc for their help on my previous question.

 I have a data frame of five columns, the first being a list of dates and the
 other four columns are numeric values. I wanted to list the days where all 4
 columns of values are less than in the previous row. I used the following
 which works fine, except it doesnt show the dates for each row (the values
 from column 1).

 differences - apply(x, 2, diff)

Please don't use apply() columnwise on data frames: it turns them into 
matrices. Here you could use

tmp - lapply(x[2:5], diff)
ind - do.call(pmax, tmp)  0 
x[c(FALSE, ind), ]

 all.lower.diffs - subset(differences, apply(differences, 1,
 function(x){all(x0)}

 I tried using the following loop instead, but it would only apply to the
 first column of every row(warning condition has length1 and only first
 element will be used):

 for ( i in 2:length(x[,1]))  if (x[i,]x[i-1,]) {print(x[i,])}
^all(^)

 How can I resolve with either method? Any help much appreciated.

 Regards,
 Alf Sammassimo.

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

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

Re: [R] Crosstabbing multiple response data

[Gabor Grothendieck]

Try this:

tab - crossprod(as.matrix(ratings[,-1]))
tab - tab - diag(diag(tab))
tab

tab / nrow(ratings)


On 2/22/07, Michael Wexler [EMAIL PROTECTED] wrote:
 Using R version 2.4.1 (2006-12-18) on Windows, I have a dataset which 
 resembles this:

 idatt1att2att3
 1110
 2100
 3011
 4111

 ratings - data.frame(id = c(1,2,3,4), att1 = c(1,1,0,1), att2 = c(1,0,0,1), 
 att3 = c(0,1,1,1))

 I would like to get a cross tab of counts of co-ocurrence, which might 
 resemble this:

att1att2att3
 att1 2   1
 att222
 att312

 with the hope of understanding, at least pairwise, what things hang 
 together.   (Yes, there are much, much better ways to do this statistically 
 including clustering and binary corrected correlation, but the audience I am 
 working with asked for this version for a specific reason.)

 (Later on, I would also like to convert to percentages of the total unique 
 pop, so the final version of the table would be


att1att2att3

 att1 50%   25%

 att250%50%

 att325%50%


 But I can do this in excel if I can get the first table out.)

 I have tried the reshape library, but could not get anything resembling this 
 (both on its own, as well as feeding in to table()).  (I have also played 
 with transposing and using some comments from this list from 2002 and 2004, 
 but the questioners appear to assume more knowledge than I have in use of R; 
 the example in the posting guide was also more complex than I was ready for, 
 I'm afraid.)

 Sample of some of my efforts:
 library(reshape)
 melt(ratings,id=c(id))

 ds1 - melt(ratings,id=c(id))
 table(ds1$variable, ds1$variable) # returns only rowcounts, 3 along diagonal
 xtabs(formula = value ~ ds1$variable + ds1$variable , data=ds1) # returns 
 only a single row of collapsed counts, appears to not allow 1 variable in 
 multiple uses

 I suspect I am close, so any nudges in the right direction would be helpful.

 Thanks much, Michael

 PS: www.rseek.org is very impressive, I heartily encourage its use.


[[alternative HTML version deleted]]

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[R] adjacency matrix?

[Jeff D. Hamann]

I'm curious to know if it's possible to easily generate a grid (lattice)
and obtain the adjacency matrix. For example, I would like to display a
3x3 (or 10x10) lattice then then generate the 10 x 10 adjacency matrix

 matrix( 1:9, 3,3, byrow=TRUE )
 [,1] [,2] [,3]
[1,]123
[2,]456
[3,]789


and then use the mathgraph(?) package to generate the adjacency matrix? I've
been able to generate a simple graph as well as some basic graphs like the
one above but have two concatenate lots of simple graphs together,


gr - c( mathgraph( ~ 1 / c(2,4) ),
mathgraph( ~ 2 / c(1,3,5) ),
mathgraph( ~ 3 / c(2,6) ),
mathgraph( ~ 4 / c(1,5,7) ),
mathgraph( ~ 5 / c(2,4,6,8) ),
mathgraph( ~ 6 / c(3,5,9) ),
mathgraph( ~ 7 / c(4,8) ),
mathgraph( ~ 8 / c(5,7,9) ),
mathgraph( ~ 9 / c(6,8) ) )

and then

plot( gr )

and

adjamat( gr )

which yields a correct adjacency matrix. Since the names of the nodes are
the values in the elements, is there are easier way to accomplish this?

Thanks,
Jeff.


-- 
Jeff D. Hamann
Forest Informatics, Inc.
PO Box 1421
Corvallis, Oregon 97339-1421
phone 541-754-1428
[EMAIL PROTECTED]
www.forestinformatics.com

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Re: [R] Accessing the class of an object with two elements.

[Vincent Goulet]

Le Jeudi 22 Février 2007 05:37, Shubha Vishwanath Karanth a écrit :
 Hi R,

 Here's my question about accessing the class of an object.

 I have an object dat which can take any two of the classes, (dates
 times) or (chron dates times). Note that the classes have two
 elements within it. I want to read these classes in such a way that

 v=class(dat) # let class(dat)= dates times

 If(class(dat)==v) k=1 else k=0

 The problem is I can't read the above class. The error which I get for
 the above if statement is as follows:

 Warning message:

 the condition has length  1 and only the first element will be used in:
 if (class(index(intra)) == v) k = 1

 How should I proceed with this? Any ideas? I tried with readline to read
 the class and access it in the 'if' statement...But doesn't work :-(

Well, given your code the condition in the 'if' statement will return c(TRUE, 
TRUE), hence the warning. Executing your code piece by piece would tell you 
that.

That said, you probably rather want to use inherit() for such purposes.

HTH

-- 
  Vincent Goulet, Associate Professor
  École d'actuariat
  Université Laval, Québec 
  [EMAIL PROTECTED]   http://vgoulet.act.ulaval.ca

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Diagnostic Tests: Jarque-Bera Test / RAMSEY

[John C Frain]

I suspect that your data is non-normal.  You might try the diagnostics in
the nortest package and refer to the text Thode (2002), Testing for
Normality, Marcel Decker, quoted in the references to that package.  A QQ
diagram might help to reveal the problems with your data.

John Frain

On 22/02/07, Simon P. Kempf [EMAIL PROTECTED] wrote:

 Hello R-Users,



 The following questions are not R-technical, but more of general
 statistical
 nature.



 1. NORMALITY

 I built a normal linear regression model and now I want to check for the
 residual normality assumption. If I check the distribution graphically and
 look at the descriptive characteristics (skewness and kurtosis are below
 1),
 I would confirm that the residuals are normally distributed.



  basicStats(IQR.in.mi02.nw.tdv.mix$residuals)

 round.ans..digits...6.

 nobs  19316.00

 NAs   0.00

 Minimum  -0.639527

 Maximum   0.693383

 1. Quartile  -0.083753

 3. Quartile   0.06

 Mean  0.00

 Median0.004641

 Sum   0.00

 SE Mean   0.001047

 LCL Mean -0.002053

 UCL Mean  0.002053

 Variance  0.021186

 Stdev 0.145554

 Skewness -0.164821

 Kurtosis  0.937282



 However, when I use the jarque.bera.test(), the assumption of normality is
 rejected.



  jarque.bera.test(IQR.in.mi02.nw.tdv.mix$residuals)



 Jarque Bera Test



 data:  IQR.in.mi02.nw.tdv.mix$residuals

 X-squared = 795.1296, df = 2, p-value  2.2e-16



 Therefore, I am wondering how good are the diagnostic test for the
 normality
 assumption? Do you they work for large sample as well? If not, what other
 diagnostic measures for normality exist for large samples?



 In many statistic textbooks it is mentioned that for large samples the
 significance tests are independent from the distributions of residuals.
 Why
 is that?



 2. RAMSEY TEST

 I made several scatterplots (residuals ~ predictors). Here everything
 looks
 fine but when I apply the RAMSEY resettest(), the assumption of linearity
 is
 rejected. My model does not show any signs of multicollinearity and all
 independent variables have strong linear relationship with the dependent
 variable. Therefore, I am wondering how well the RAMSEY resettest works in
 general. What other tests exist?



 Again, I would like to apologize for the asking no specific R-technical
 questions. But I would really appreciate any help.



 Thanks,



 Simon




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-- 
John C Frain
Trinity College Dublin
Dublin 2
Ireland
www.tcd.ie/Economics/staff/frainj/home.html
mailto:[EMAIL PROTECTED]
mailto:[EMAIL PROTECTED]

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

Re: [R] several Filled.contour plots on the same device...

[Greg Snow]

The problem is that filled.contour uses the layout function internally which 
messes up any other use of layout, split.screen, or mfrow.  One alternative is 
to use the levelplot function from the lattice package, or you could use 
filled.contour to make several full page plots to a pdf file, then use an 
external utility like pdfpages to combine them onto a single page.

Hope this helps,

-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
 
 

 -Original Message-
 From: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED] On Behalf Of 
 [EMAIL PROTECTED]
 Sent: Thursday, February 22, 2007 4:15 AM
 To: r-help@stat.math.ethz.ch
 Subject: [R] several Filled.contour plots on the same device...
 
 
 
 hello -
 
 a question about filled.contour plots, for which i haven't 
 found a response in previous posts - sorry if already treated.
 
 i'd like to draw several filled.contour plots (that is, maps) 
 on the same device (a postscript file, actually). I know 
 about layout(matrix) , split.screen or par(mfrow) : it works 
 well for simple plots, but with filled.contour plots, i get 
 several pages instead of one page divided into several cells.
 I'd really like to get these maps directly on one graphs, 
 without having to process them afterwards.
 Does anyone know something about that ?
 
 Thank you for your help,
 
 Alexis Berg
 
 Ingénieur de recherche
 LOCEAN (IPSL) - Paris
 [EMAIL PROTECTED]
 
 __
 R-help@stat.math.ethz.ch mailing list
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 http://www.R-project.org/posting-guide.html
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Re: [R] Sorting rows of a binary matrix

[Charles C. Berry]

On Thu, 22 Feb 2007, Serguei Kaniovski wrote:


 Hallo,

 The command:

 x - 3
 mat - as.matrix(expand.grid(rep(list(0:1), x)))

 generates a matrix with 2^x columns containing the binary representations
 of the decimals from 0 to (2^x-1), here from 0 to 7. But the rows are not
 sorted in this order.

 How can sort the rows the ascending order of the decimals they represent,
 preferably without a function which converts binaries to decimals (which I
 have)? Alternatively, generate a matrix that has the rows sorted that way?

The alternative:

mat - as.matrix(expand.grid(rep(list(0:1), x))[ , x:1 ]  )



 Thanks,
 Serguei
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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.


Charles C. Berry(858) 534-2098
  Dept of Family/Preventive Medicine
E mailto:[EMAIL PROTECTED]   UC San Diego
http://biostat.ucsd.edu/~cberry/ La Jolla, San Diego 92093-0901

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[R] S Plus Debugging

[cjkogan111]

Hello,
I am trying to debug in S+. I know how to debug in r, but the commands in S+
seem to be different. I am just looking for a command that allows me to
debug the function, and then commands that allow me to step through and find
the values of different variables.
Thanks!
-cjkogan111
-- 
View this message in context: 
http://www.nabble.com/S-Plus-Debugging-tf3274285.html#a9105226
Sent from the R help mailing list archive at Nabble.com.

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[R] Help using Sweave with wireframe or cloud

[Matthieu Cornec]

Hi,


I used the sweave package to get a 3D plot

echo=F,fig=F=
wireframe(volcano, shade = TRUE,
  aspect = c(61/87, 0.4),
  light.source = c(10,0,10))
@

but it gives an error message for the pdf file of this picture.

Any one could help ?

Thanks in advance,

Matthieu

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Re: [R] Crosstabbing multiple response data

[Charles C. Berry]

 res - crossprod( as.matrix( ratings[ , -1] ) )
 diag(res) - 
 print(res, quote=F)
  att1 att2 att3
att1  21
att2 2 2
att3 12
 
 res2 - crossprod(as.matrix( ratings[ , -1])) * 100 / nrow( ratings )
 res2[] - paste( res2, %, sep= )
 diag(res2) - 
 print(res2, quote=F)
  att1 att2 att3
att1  50%  25%
att2 50%   50%
att3 25%  50%


Be sure to bone up on format and sprintf before taking this into 
production.

On Thu, 22 Feb 2007, Michael Wexler wrote:

 Using R version 2.4.1 (2006-12-18) on Windows, I have a dataset which 
 resembles this:

 idatt1att2att3
 1110
 2100
 3011
 4111

 ratings - data.frame(id = c(1,2,3,4), att1 = c(1,1,0,1), att2 = c(1,0,0,1), 
 att3 = c(0,1,1,1))

 I would like to get a cross tab of counts of co-ocurrence, which might 
 resemble this:

att1att2att3
 att1 2   1
 att222
 att312

 with the hope of understanding, at least pairwise, what things hang 
 together.   (Yes, there are much, much better ways to do this statistically 
 including clustering and binary corrected correlation, but the audience I am 
 working with asked for this version for a specific reason.)

 (Later on, I would also like to convert to percentages of the total unique 
 pop, so the final version of the table would be


att1att2att3

 att1 50%   25%

 att250%50%

 att325%50%


 But I can do this in excel if I can get the first table out.)

 I have tried the reshape library, but could not get anything resembling this 
 (both on its own, as well as feeding in to table()).  (I have also played 
 with transposing and using some comments from this list from 2002 and 2004, 
 but the questioners appear to assume more knowledge than I have in use of R; 
 the example in the posting guide was also more complex than I was ready for, 
 I'm afraid.)

 Sample of some of my efforts:
 library(reshape)
 melt(ratings,id=c(id))

 ds1 - melt(ratings,id=c(id))
 table(ds1$variable, ds1$variable) # returns only rowcounts, 3 along diagonal
 xtabs(formula = value ~ ds1$variable + ds1$variable , data=ds1) # returns 
 only a single row of collapsed counts, appears to not allow 1 variable in 
 multiple uses

 I suspect I am close, so any nudges in the right direction would be helpful.

 Thanks much, Michael

 PS: www.rseek.org is very impressive, I heartily encourage its use.


   [[alternative HTML version deleted]]

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


Charles C. Berry(858) 534-2098
  Dept of Family/Preventive Medicine
E mailto:[EMAIL PROTECTED]   UC San Diego
http://biostat.ucsd.edu/~cberry/ La Jolla, San Diego 92093-0901

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R-help@stat.math.ethz.ch mailing list
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Re: [R] Help using Sweave with wireframe or cloud

[Kevin E. Thorpe]

Matthieu Cornec wrote:
 Hi,
 
 
 I used the sweave package to get a 3D plot
 
 echo=F,fig=F=
 wireframe(volcano, shade = TRUE,
   aspect = c(61/87, 0.4),
   light.source = c(10,0,10))
 @
 
 but it gives an error message for the pdf file of this picture.
 
 Any one could help ?
 

Try this.

echo=FALSE,fig=TRUE=
library(lattice)
print(wireframe(volcano, shade = TRUE, aspect = c(61/87, 0.4),
light.source = c(10,0,10)))
@


 Thanks in advance,
 
 Matthieu
 


-- 
Kevin E. Thorpe
Biostatistician/Trialist, Knowledge Translation Program
Assistant Professor, Department of Public Health Sciences
Faculty of Medicine, University of Toronto
email: [EMAIL PROTECTED]  Tel: 416.864.5776  Fax: 416.864.6057

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[R] How to print a double quote

[Bos, Roger]

Can anyone tell me how to get R to include a double quote in the middle
of a character string?  

For example, the following code is close:

 fnd-Open fnd 'test'
cat(fnd)
Open fnd 'test'


But instead of Open fnd 'test' I need: Open fnd test.  Difference
seems minor, but I am writing batch files for another program to read in
and it has to have the double quotes to work.  

Thanks in advance for any help or ideas,

Roger

** * 
This message is for the named person's use only. It may 
contain confidential, proprietary or legally privileged 
information. No right to confidential or privileged treatment 
of this message is waived or lost by any error in 
transmission. If you have received this message in error, 
please immediately notify the sender by e-mail, 
delete the message and all copies from your system and destroy 
any hard copies. You must not, directly or indirectly, use, 
disclose, distribute, print or copy any part of this message 
if you are not the intended recipient.

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Re: [R] How to print a double quote

[Tony Plate]

  cat('Open fnd test\n')
Open fnd test
  cat(Open fnd \test\\n)
Open fnd test
 

Bos, Roger wrote:
 Can anyone tell me how to get R to include a double quote in the middle
 of a character string?  
 
 For example, the following code is close:
 
 fnd-Open fnd 'test'
cat(fnd)
 Open fnd 'test'
 
 But instead of Open fnd 'test' I need: Open fnd test.  Difference
 seems minor, but I am writing batch files for another program to read in
 and it has to have the double quotes to work.  
 
 Thanks in advance for any help or ideas,
 
 Roger
 
 ** * 
 This message is for the named person's use only. It may 
 contain confidential, proprietary or legally privileged 
 information. No right to confidential or privileged treatment 
 of this message is waived or lost by any error in 
 transmission. If you have received this message in error, 
 please immediately notify the sender by e-mail, 
 delete the message and all copies from your system and destroy 
 any hard copies. You must not, directly or indirectly, use, 
 disclose, distribute, print or copy any part of this message 
 if you are not the intended recipient.
 
 __
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Re: [R] How to print a double quote

[Ranjan Maitra]

try 

 cat(Open fnd \test\)

which is the same as for C.

HTH.

Ranjan

On Thu, 22 Feb 2007 14:09:26 -0500 Bos, Roger [EMAIL PROTECTED] wrote:

 Can anyone tell me how to get R to include a double quote in the middle
 of a character string?  
 
 For example, the following code is close:
 
  fnd-Open fnd 'test'
 cat(fnd)
 Open fnd 'test'
 
 
 But instead of Open fnd 'test' I need: Open fnd test.  Difference
 seems minor, but I am writing batch files for another program to read in
 and it has to have the double quotes to work.  
 
 Thanks in advance for any help or ideas,
 
 Roger
 
 ** * 
 This message is for the named person's use only. It may 
 contain confidential, proprietary or legally privileged 
 information. No right to confidential or privileged treatment 
 of this message is waived or lost by any error in 
 transmission. If you have received this message in error, 
 please immediately notify the sender by e-mail, 
 delete the message and all copies from your system and destroy 
 any hard copies. You must not, directly or indirectly, use, 
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 if you are not the intended recipient.
 
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Re: [R] applying lm on an array of observations with common design matrix

[Ranjan Maitra]

On Thu, 22 Feb 2007 08:17:38 + (GMT) Prof Brian Ripley [EMAIL PROTECTED] 
wrote:

 On Thu, 22 Feb 2007, Petr Klasterecky wrote:
 
  Ranjan Maitra napsal(a):
  On Sun, 18 Feb 2007 07:46:56 + (GMT) Prof Brian Ripley [EMAIL 
  PROTECTED] wrote:
 
  On Sat, 17 Feb 2007, Ranjan Maitra wrote:
 
  Dear list,
 
  I have a 4-dimensional array Y of dimension 330 x 67 x 35 x 51. I have a
  design matrix X of dimension 330 x 4. I want to fit a linear regression
  of each
 
  lm( Y[, i, j, k] ~ X). for each i, j, k.
 
  Can I do it in one shot without a loop?
  Yes.
 
  YY - YY
  dim(YY) - c(330, 67*35*51)
  fit - lm(YY ~ X)
 
  Actually, I am also interested in getting the p-values of some of the
  coefficients -- lets say the coefficient corresponding to the second
  column of the design matrix. Can the same be done using array-based
  operations?
  Use lapply(summary(fit), function(x) coef(x)[3,4])  (since there is a
  intercept, you want the third coefficient).
 
  In this context, can one also get the variance-covariance matrix of the 
  coefficients?
 
  Sure:
 
  lapply(summary(fit), function(x) {$(x,cov.unscaled)})
 
 But that is not the variance-covariance matrix (and it is an unusual way 
 to write x$cov.unscaled)!
 
  Add indexing if you do not want the whole matrix. You can extract
  whatever you want, just take a look at ?summary.lm, section Value.
 
 It is unclear to me what the questioner expects: the estimated 
 coefficients for different responses are independent.  For a list of 
 matrices applying to each response one could mimic vcov.lm and do
 
 lapply(summary(fit, corr=FALSE),
 function(so) so$sigma^2 * so$cov.unscaled)

Thanks! Actually, I am really looking to compare the coefficients (let us say 
second and the third) beta2 - beta4 = 0 for each regression. Basically, get the 
two-sided p-value for the test statistic for each regression. 

One way of doing that is to get the dispersion matrix of each regression and 
then to compute the t-statistic and the p-value. That is the genesis of the 
question above. Is there a better way?

Many thanks and best wishes,
Ranjan

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Re: [R] How to print a double quote

[Roger Bivand]

On Thu, 22 Feb 2007, Bos, Roger wrote:

 Can anyone tell me how to get R to include a double quote in the middle
 of a character string?  

FAQ 7.37 Why does backslash behave strangely inside strings?

 
 For example, the following code is close:
 
  fnd-Open fnd 'test'
 cat(fnd)
 Open fnd 'test'
 
 
 But instead of Open fnd 'test' I need: Open fnd test.  Difference
 seems minor, but I am writing batch files for another program to read in
 and it has to have the double quotes to work.  
 
 Thanks in advance for any help or ideas,
 
 Roger
 
 ** * 
 This message is for the named person's use only. It may 
 contain confidential, proprietary or legally privileged 
 information. No right to confidential or privileged treatment 
 of this message is waived or lost by any error in 
 transmission. If you have received this message in error, 
 please immediately notify the sender by e-mail, 
 delete the message and all copies from your system and destroy 
 any hard copies. You must not, directly or indirectly, use, 
 disclose, distribute, print or copy any part of this message 
 if you are not the intended recipient.
 
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-- 
Roger Bivand
Economic Geography Section, Department of Economics, Norwegian School of
Economics and Business Administration, Helleveien 30, N-5045 Bergen,
Norway. voice: +47 55 95 93 55; fax +47 55 95 95 43
e-mail: [EMAIL PROTECTED]

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[R] Principal Component Analysis explained variance

[Milton Cezar Ribeiro]

Hi there,

How can I know the explaned variance of a PC axis generated by prcomp()?

Kind regards,

miltinho
Brazil

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Re: [R] Principal Component Analysis explained variance

[Chuck Cleland]

Milton Cezar Ribeiro wrote:
 Hi there,
 
 How can I know the explaned variance of a PC axis generated by prcomp()?

  From the standard deviations of each component, you could do something
like this maybe:

prcomp(USArrests, scale = TRUE)$sdev^2 / ncol(USArrests)

[1] 0.62006039 0.24744129 0.08914080 0.04335752

 Kind regards,
 
 miltinho
 Brazil
 
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Re: [R] how to install a package in R on a linux machine?

[Scionforbai]

 I tried a few times and still couldn't figure out the correct way
 to install this package.

Help us to help you, Gallon. Which error comes out?

 install.packages(packagename)
 this downloads the package and installs it into the default R package
 library on your machine.

Of course, on a normal installation R need to be run by root to do
this, in order to write in a system directory like /usr/lib/.

 If you want to install it to a different
 directory use the 'lib' argument of 'install.packages'.

Again, the user running R must have writing access to the directory
specified by the 'lib' option. This option allows to install R
packages in a directory in your /home, typically.

Maybe you want to try (dependencies=TRUE) to automagically download
and install dependencies.

And of course you downloaded the appropriate package for you R version.

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[R] question about boxplot

[Smith, Phil \(CDC/CCID/NCIRD\)]

Here is a question from an old guy:
 
I want to use the boxplot function as follows:
 
boxplot( p.prop ~ R  + bins )
 
This command makes nice boxplot for the factor R crossed with the factor 
bins.
 
I am having alot of trouble getting control of the labels on the X axis. I want 
to control it more by specifying what the labels are, controling the 'size' of 
those labels (by using cex), and then control the rotation of the character 
strings of those labels (by using srt or crt).
 
There is a names argument to boxplot, but I haven't had much luck controlling 
what it prints, the size of the font, and the character rotation.
 
Can somebody enlighten me on how to do this?
 
Please respond to my work email address: [EMAIL PROTECTED]
 
Many thanks!
Phil Smith
Centers for Disease COntrol and Prevention

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[R] problem with weights on lmer function

[Ronaldo Reis Junior]

Hi,

I try to make a model using lmer, but the weigths is not accept.

m1-lmer(ocup/total~tempo+(tempo|estacao),family=binomial,weights=total)

Erro em lmer(ocup/total ~ tempo + (tempo | estacao), family = binomial,  : 
object `weights' of incorrect type

I dont understand why this error, with glm this work. the total object is a 
vector.

Any idea?

Thanks
Ronaldo
-- 
God is subtle, but he is not malicious.
-- Albert Einstein
--
 Prof. Ronaldo Reis Júnior
|  .''`. UNIMONTES/Depto. Biologia Geral/Lab. Ecologia Evolutiva
| : :'  : Campus Universitário Prof. Darcy Ribeiro, Vila Mauricéia
| `. `'` CP: 126, CEP: 39401-089, Montes Claros - MG - Brasil
|   `- Fone: (38) 3229-8190 | [EMAIL PROTECTED] | [EMAIL PROTECTED]
| ICQ#: 5692561 | LinuxUser#: 205366

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[R] JGR launcher for linux

[Ronaldo Reis Junior]

Hi,

anybody have a JGR launcher for linux? Maybe a script that launch JGR directly 
without open R then library(JGR) and JGR().

Thanks
Ronaldo
-- 
Deflector shields just came on, Captain.
--
 Prof. Ronaldo Reis Júnior
|  .''`. UNIMONTES/Depto. Biologia Geral/Lab. Ecologia Evolutiva
| : :'  : Campus Universitário Prof. Darcy Ribeiro, Vila Mauricéia
| `. `'` CP: 126, CEP: 39401-089, Montes Claros - MG - Brasil
|   `- Fone: (38) 3229-8190 | [EMAIL PROTECTED] | [EMAIL PROTECTED]
| ICQ#: 5692561 | LinuxUser#: 205366

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[R] confidence intervals

[Bart Joosen]

Hi,

I'm having trouble with the confidence interval of the nls function.
I did my home work, and searched acros the support list until I came up with 
following solution of Peter Dalgaard:

example(predict.nls) 
se.fit - sqrt(apply(attr(predict(fm,list(Time = 
tt)),gradient),1, 
function(x) sum(vcov(fm)*outer(x,x
matplot(tt, predict(fm,list(Time = tt))+
   outer(se.fit,qnorm(c(.5, .025,.975))),type=l)
points(demand ~ Time, data = BOD) 
One slight issue is that it doesn't work if newdata is omitted, 
but then you can easily get the gradient from fm$m$gradient() 

I tried this with my own data:
Data - data.frame(Temp=rep(c(25,40),each=3), Mnd = c(1:3),Degr = 
c(0.057,0.077,0.108,0.148,0.198,0.223))
model - nls(Degr~exp((A/(Temp)+log(Mnd))*B),Data,start=list(A=-10,B=1))
Months - c(1,2,3,6,9,12,24,48)
se.fit - sqrt(apply(attr(predict(model,list(Temp = 
25,Mnd=Months)),gradient),1, function(x) sum(vcov(fm)*outer(x,x

But unfortunately I get an error ( Error in apply(attr(predict(model, list(Temp 
= 25, Mnd = Months)), gradient),  : 
dim(X) must have a positive length)

When I try using the gradient of the model instead of using the new data then 
there is no problem:
se.fit - sqrt(apply(model$m$gradient(),1, function(x) 
sum(vcov(model)*outer(x,x
matplot(Data$Mnd, predict(model,list(Temp = 
Data$Temp,Mnd=Data$Mnd))+outer(se.fit,qnorm(c(.5, 025,.975))),type=l)

But how about calculating confidence intervals of new data? How do I get an 
gradient for these values?

I'm using Windows XP, R 2.4.1.


Thanks

Bart
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[R] R equivalent of the VAR function in S+Finmetrics

[Leeds, Mark \(IED\)]

I was looking at the systemfit package and it seems like I could use it
to solve OLS systems (
which is essentially what VARs are ) but the lag length would have to be
known beforehand, I think. Does anyone know
if there is an equivalent of the VAR function in Eric Zivot's
S+Finmetrics package where the lag length can be selected based on
some kind of criterion such as BIC for example. Thanks for any
idea/suggestions.


This is not an offer (or solicitation of an offer) to buy/se...{{dropped}}

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Re: [R] JGR launcher for linux

[Liaw, Andy]

Isn't it right in front of you?  I get:

 JGR()
Starting JGR ...
(You can use /usr/local/lib64/R/library/JGR/cont/run to start JGR directly)
 ^^^

Andy

From: Ronaldo Reis Junior
 
 Hi,
 
 anybody have a JGR launcher for linux? Maybe a script that 
 launch JGR directly without open R then library(JGR) and JGR().
 
 Thanks
 Ronaldo
 --
 Deflector shields just came on, Captain.
 --
  Prof. Ronaldo Reis Júnior
 |  .''`. UNIMONTES/Depto. Biologia Geral/Lab. Ecologia Evolutiva
 | : :'  : Campus Universitário Prof. Darcy Ribeiro, Vila Mauricéia `. 
 | `'` CP: 126, CEP: 39401-089, Montes Claros - MG - Brasil
 |   `- Fone: (38) 3229-8190 | [EMAIL PROTECTED] | 
 | [EMAIL PROTECTED]
 | ICQ#: 5692561 | LinuxUser#: 205366
 
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Re: [R] question about boxplot

[Michael Kubovy]

On Feb 22, 2007, at 2:56 PM, Smith, Phil ((CDC/CCID/NCIRD)) wrote:

 boxplot( p.prop ~ R  + bins )

 This command makes nice boxplot for the factor R crossed with the  
 factor bins.

 I am having alot of trouble getting control of the labels on the X  
 axis. I want to control it more by specifying what the labels are,  
 controling the 'size' of those labels (by using cex), and then  
 control the rotation of the character strings of those labels (by  
 using srt or crt).

 There is a names argument to boxplot, but I haven't had much luck  
 controlling what it prints, the size of the font, and the character  
 rotation.

Would you consider an easy way out---an alternative with reasonable  
defaults?

data(ToothGrowth)
bwplot(dose ~ len | supp, ToothGrowth)
_
Professor Michael Kubovy
University of Virginia
Department of Psychology
USPS: P.O.Box 400400Charlottesville, VA 22904-4400
Parcels:Room 102Gilmer Hall
 McCormick RoadCharlottesville, VA 22903
Office:B011+1-434-982-4729
Lab:B019+1-434-982-4751
Fax:+1-434-982-4766
WWW:http://www.people.virginia.edu/~mk9y/

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[R] is a time series regression model a causal forecasting model?

[sj]

I have a semantics question, I am reading Bowerman and O'Connell and they
state that forecasting models fall into two categories, univariate and
causal. My question is whether a time series regression model that relates
the time series of interest to functions of time such as the day of the week
or month of the year could be considered a causal model?

thanks,

spencer

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Re: [R] how to install a package in R on a linux machine?

[Roberto Perdisci]

Hi,
  try this:

$sudo R CMD INSTALL downloaded.package.tar.gz

If you don't use 'sudo' (or do not have privileges to do so), you need
to either become root (with su) or ask the administrator of the
machine you are using to install the package for you

regards,
Roberto


On 2/22/07, Gabor Csardi [EMAIL PROTECTED] wrote:
 The easiest is perhaps to do

 install.packages(packagename)

 this downloads the package and installs it into the default R package
 library on your machine. If you want to install it to a different
 directory use the 'lib' argument of 'install.packages'.

 If you don't want to download the package again but want to use the
 downloaded one, use the following command:

 install.packages(repos=NULL, pkgs=the.file.you've.downloaded)

 You can also install R packages from the command line, like this:

 R CMD INSTALL -l lib.directectory downloaded.package.file

 Gabor

 On Thu, Feb 22, 2007 at 04:44:25PM +0800, gallon li wrote:
  I downloaded the tar.gz file from r-project website (and saved it in a local
  directory) and wish to use the package in R.
 
  But I am not sure how to use the install.packages command. I tried a few
  times and still couldn't figure out the correct way to install this package.
 
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 --
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[R] A question regarding cutree

[Jun Ding]

Hi Everyone, 

I am doing hierarchical clustering analysis and have a
question regarding cutree. 

I am doing things like this:

hc - hclust(dist(X))
a - cutree(hc, k=2)

Basically a is a vector containing the assignments
of 1 or 2 for each sample. May I know how cutree
decides to assign 1 and 2's to each sample (in other
words, how clusters 1 and 2 are decided)? I am having
the feeling that Sample 1 will be always assigned to
Cluster 1, but I am not sure about this. 

Thank you!

Best,
Jun

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Re: [R] investigating interactions with mixed models

[Andrew Robinson]

Hello Rachel,

I don't think that there is any infrastructure for these procedures on
lmer objects, yet.  If you are willing to use lme instead, then the
multcomp package seems to provide post-hoc tests.  It is worth noting
that there is some doubt as to the validity of the reference
distributions for tests of fixed effects in the presence of random
effects. 

http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-are-p_002dvalues-not-displayed-when-using-lmer_0028_0029_003f

Cheers

Andrew

On Thu, Feb 22, 2007 at 12:32:44PM +, R. Baker wrote:
 I'm investigating a number of dependent variables using mixed models, e.g.
 
 data.lmer45 = lmer(ampStopB ~ (type + stress + MorD)^3 + (1|speaker) + 
 (1|word), data=data)
 
 The p-values for some of the 2-way and 3-way interactions are significant 
 at a 0.05 level and I have been trying to find out how to understand the 
 exact nature of the interactions. Does anyone know if it is possible to run 
 post-hoc tests on mixed model (lmer) objects? I have read about TukeyHSD 
 but it seems that this can only be run on anova (aov) objects.
 
 Any suggestions would be gratefully appreciated!
 
 Rachel Baker
 
 -- 
 --
 PhD student
 Dept of Linguistics
 Sidgwick Avenue
 University of Cambridge  
 Cambridge
 
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University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599
http://www.ms.unimelb.edu.au/~andrewpr
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Re: [R] problem with weights on lmer function

[Andrew Robinson]

Hi Ronaldo,

I suggest that you send us a small, well-documented, code example that
we can reproduce.  It certainly looks as though there is a problem,
but given this information it's hard to know what it is!

Cheers

Andrew

On Thu, Feb 22, 2007 at 06:22:03PM -0200, Ronaldo Reis Junior wrote:
 Hi,
 
 I try to make a model using lmer, but the weigths is not accept.
 
 m1-lmer(ocup/total~tempo+(tempo|estacao),family=binomial,weights=total)
 
 Erro em lmer(ocup/total ~ tempo + (tempo | estacao), family = binomial,  : 
   object `weights' of incorrect type
 
 I dont understand why this error, with glm this work. the total object is a 
 vector.
 
 Any idea?
 
 Thanks
 Ronaldo
 -- 
 God is subtle, but he is not malicious.
   -- Albert Einstein
 --
  Prof. Ronaldo Reis J?nior
 |  .''`. UNIMONTES/Depto. Biologia Geral/Lab. Ecologia Evolutiva
 | : :'  : Campus Universit?rio Prof. Darcy Ribeiro, Vila Mauric?ia
 | `. `'` CP: 126, CEP: 39401-089, Montes Claros - MG - Brasil
 |   `- Fone: (38) 3229-8190 | [EMAIL PROTECTED] | [EMAIL PROTECTED]
 | ICQ#: 5692561 | LinuxUser#: 205366
 
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-- 
Andrew Robinson  
Department of Mathematics and StatisticsTel: +61-3-8344-9763
University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599
http://www.ms.unimelb.edu.au/~andrewpr
http://blogs.mbs.edu/fishing-in-the-bay/

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Re: [R] investigating interactions with mixed models

[Bert Gunter]

?interaction.plot

Should help you. This works on the data, not the model. A 3-way interaction
just means that the 2-way interaction differs among the various levels of
the 3rd factor. Clever use of trellis plots (?xyplot -- especially
?panel.linejoin -- gives greater flexibility, but it requires that a steeper
learning curve be climbed).

In general, the presence of interactions is just another manifestation of
the response varying nonlinearly in the factors (**not** in the parameters,
of course -- it's a linear model after all). This is essentially always the
case, it's just a question of whether the signal/noise ratio (which depends
on sample size) is large enough to see it via P-values. So by all means look
at the plots and try to understand and interpret what's going on; but by no
means assume that p-values above and below a threshhold of .05 are a clear
guide to determining this. As usual, statistical significance and scientific
relevance are not equivalent, and the degree of overlap between the two is
often difficult to judge.

Cheers,
Bert Gunter
Genentech Nonclinical Statistics
South San Francisco, CA 94404
650-467-7374


-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Andrew Robinson
Sent: Thursday, February 22, 2007 2:32 PM
To: R. Baker
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] investigating interactions with mixed models

Hello Rachel,

I don't think that there is any infrastructure for these procedures on
lmer objects, yet.  If you are willing to use lme instead, then the
multcomp package seems to provide post-hoc tests.  It is worth noting
that there is some doubt as to the validity of the reference
distributions for tests of fixed effects in the presence of random
effects. 

http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-are-p_002dvalues-not-displa
yed-when-using-lmer_0028_0029_003f

Cheers

Andrew

On Thu, Feb 22, 2007 at 12:32:44PM +, R. Baker wrote:
 I'm investigating a number of dependent variables using mixed models, e.g.
 
 data.lmer45 = lmer(ampStopB ~ (type + stress + MorD)^3 + (1|speaker) + 
 (1|word), data=data)
 
 The p-values for some of the 2-way and 3-way interactions are significant 
 at a 0.05 level and I have been trying to find out how to understand the 
 exact nature of the interactions. Does anyone know if it is possible to
run 
 post-hoc tests on mixed model (lmer) objects? I have read about TukeyHSD 
 but it seems that this can only be run on anova (aov) objects.
 
 Any suggestions would be gratefully appreciated!
 
 Rachel Baker
 
 -- 
 --
 PhD student
 Dept of Linguistics
 Sidgwick Avenue
 University of Cambridge  
 Cambridge
 
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-- 
Andrew Robinson  
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University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599
http://www.ms.unimelb.edu.au/~andrewpr
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Re: [R] Chi-Square test

[Wensui Liu]

?pchisq

On 2/21/07, Mohsen Jafarikia [EMAIL PROTECTED] wrote:
 Hello all,

 I am doing a Likelihood Ratio (LR) test in my simulation and I have a vector
 LR values (each with 1 degree of freedom) at the end of my simulation.

 Can anybody tell me how I can write a 'R' code which gives me the p-value
 for each of those LR values.
 Thanks

 [[alternative HTML version deleted]]

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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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-- 
WenSui Liu
A lousy statistician who happens to know a little programming
(http://spaces.msn.com/statcompute/blog)

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Re: [R] manova discriminant functions: Addendum

[Adam D. I. Kramer]

Three weeks later, I have almost completely solved my problem (quoted below;
about within-subjects manova, and discriminant function analysis to
compliment a manova analysis). So for anyone who was secretly hoping someone
would respond to me:

* manova does not handle within-subjects variables in a manner which is very
   intuitive for those not in the field of statistics or mathematics.  The
   anova.mlm functions allow this sort of analysis, but for applied
   statisticians (read: social scientists), dealing with inner vs outer
   projections is a bit opaque.

* However, a bit of knowledge on what repeated measures MANOVA really does
   is very helpful. (No, really, it helps to know what you're trying to do!)
   The general idea is that you create contrast columns for the
   within-subjects factors, such that the contrast columns are orthogonal.
   Then, you submit these to a general MANOVA with between-subject predictors
   to test the within-between interactions.

* For within-subjects main effects, it's easier (for me) to just compute the
   hypothesis matrix myself. With orthogonal contrast factors, it's pretty
   simple: each row of the hypothesis matrix is the product of the column of
   means for the contrast columns times the mean for that row's contrast,
   times the number of observations. The error matrix can be taken directly
   from the fit for the interaction (see point 2); then, H %*% ginv(E) gives
   the effect matrix for the within subjects main effect, whose eigenvalues
   can be evaluated with the appropriate degrees of freedom. I wrote a
   function to do this, if anyone is interested.

* The current implementation of the Pillai() function uses Pillai's (1954)
   approximation of an F value for a given V, despite the much more robust
   and (seemingly) uncontroversial method of estimating F put forth by Muller
   in 1998. This is, of course, what SPSS and SAS do (unless you require SAS
   to do exact tests, which is still a misnomer), but it doesn't seem to me
   like R ought to do things poorly just to give the same results as SPSS.

* Estimating several within-subjects effects basically amounts to providing
   the relevant contrast matrices to the interaction analyses, and then
   testing subsets relative to the overall residual matrix and the right
   number of df.

* To get discriminant functions and variable loadings, use the lda()
   function. The fact that these are provided by manova() analogues in other
   software packages was just a red herring!

* However, this also does not work for discovering within-subjects
   discriminant functions. And that, I'm afraid, is where I'm now stuck...any
   suggestions on an lda() analogue for within-subject factors? An example:

data - data.frame(treat=factor(rep(c(Control,Treatment),3)),
time1=c(1,5,1,5,1,5),
time2=c(2,8,1,7,2,6),
time3=c(3,9,3,9,2,10))

data.poly - data[,2:4] %*% contr.poly(3)

Now, discriminant functions on predicting the difference between Control and
Treatment, using the three timepoints is easy: Convert them to orthogonal
contrast variables, and then see how to discriminate treatment conditions
based on the linear and quadratic effects of time:

lda(data$treat ~ data.poly)

...but my question is how to do an lda predicting whether an observation
came from the time1, time2, or time3 column.

Thanks,
Adam

On Mon, 5 Feb 2007, Adam D. I. Kramer wrote:

 As an addendum to my earlier post, I am having another difficulty with
 getting manova() to behave as I would like: when I specify contrasts for my
 independent variable(s), I am unsure of how to test them. This is a
 contrived example of both of my questions:

 Assume three alertness measurements, alert1 alert2 alert3, a
 within-subjects variable measuring their alertness at three timepoints,
 minutes after taking the drug (alert1), one hour (alert2), and two hours
 (alert3). The between-subjects variable is dosage, with dose==0 when
 subjects have had no drug, dose==1 when they have had a single dose, dose==2
 when they have had a double dose.

 My intuition says to do the following:

 alert - cbind(alert1,alert2,alert3) %*% contr.poly(3)
 contrasts(dose) - matrix(c(2,-1,-1,0,1,-1),3,2)
 m - manova(alert ~ dose)

 ...what I want is two main effects (dose and alert) and one interaction
 (dose by alert), but also main effect and interaction for the two
 individual contrasts for dose. For the main effect for alert, and all of the
 dose*alert interactions, I need the discriminant function loadings of my two
 alertness contrasts in order to interpret the manner in which alertness is
 varying (e.g., is it varying in a linear or quadratic way).

 m2 - manova (alert ~ dose)
 summary(m2)
 ...gives me a test for the dose * alertness interaction. Good! But I can't
 seem to find the contrasts I asked for for dose. In univariate ANOVA, I
 usually just call summary.lm() which gives me t-test coefficients for each
 level 

[R] Wrinting integers in a matrix faile

[Mohsen Jafarikia]

Hello everyone,
I am using the following program to get the p-value of some numbers
(column 'LR' of the data.dat file). I want to write the 1st and 2nd
column of the output file (data.out) as an integer while the program
change them to double. Could anybody please tell me how I can write
the code which writes the values of the first two columns as integer?
Thanks


library ('MASS')
MP-read.table(file='data.dat')
names(MP)-c('B','R','S','L','LR','Q')
a-as.matrix((1-pchisq(MP$LR, df=1)))
b-cbind(MP$B,MP$R,a,MP$S,MP$L,MP$LR,MP$Q)
write.matrix(b,'data.out')

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

[R] mixture of 2 normals - starting values

[apjaworski]

Hi,

I have a problem of estimating a mixture of two normal distributions.  I
need to find the starting points automatically, since this is a part of a
larger piece of image processing code.

I found the mix2normal1 function in VGAM package that mentions a method of
finding starting values for mu1 and mu2 but refers the reader to a book by
Everitt and Hand.  Unfortunately, I do not have an easy access to this
book.  Could anybody point me to a description of the method that I could
use?

Any help will be appreciated.  Thanks in advance,

Andy

__
Andy Jaworski
518-1-01
Process Laboratory
3M Corporate Research Laboratory
-
E-mail: [EMAIL PROTECTED]
Tel:  (651) 733-6092
Fax:  (651) 736-3122

__
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Re: [R] mixture of 2 normals - starting values

[Xiaohui]

Hi,

Try MCLUST package. You can use the hierarchical clustering to find the 
starting values of your EM.

Xiaohui

[EMAIL PROTECTED] wrote:
 Hi,

 I have a problem of estimating a mixture of two normal distributions.  I
 need to find the starting points automatically, since this is a part of a
 larger piece of image processing code.

 I found the mix2normal1 function in VGAM package that mentions a method of
 finding starting values for mu1 and mu2 but refers the reader to a book by
 Everitt and Hand.  Unfortunately, I do not have an easy access to this
 book.  Could anybody point me to a description of the method that I could
 use?

 Any help will be appreciated.  Thanks in advance,

 Andy

 __
 Andy Jaworski
 518-1-01
 Process Laboratory
 3M Corporate Research Laboratory
 -
 E-mail: [EMAIL PROTECTED]
 Tel:  (651) 733-6092
 Fax:  (651) 736-3122

 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Extracting a subset from a dataframe

[Augusto.Sanabria]

Good day everyone,

Can anyone suggest an effective method to solve
the following problem:

I have 2 dataframes D1 and D2 as follows:

D1:
dates   ws   wc pwc
 2005-10-19:12:00  10.8  80  81
 2005-10-20:12:00  12.3   5  15
 2005-10-21:15:00  12.3   3  15
 2005-10-22:15:00  11.3  13  95
 2005-10-23:12:00  12.3  13   2
 2005-10-24:15:00  10.3   2  95
 2005-10-25:15:00  10.3   2   2

D2:
dates   ws   wc  pwc
 2005-02-02:15:00  17.5   5  96
 2005-02-19:15:00  20.1  15  97
 2005-02-20:18:00  16.5  95  95
 2005-03-03:18:00  10.3  95  95
 2005-03-04:00:00  13.4  13  95
 2005-10-22:15:00  11.3  13  95
 2005-10-25:15:00  10.3   2   2

I want to create another dataframe made up
of the values of dataframe1 which are not common
with dataframe2, ie. newD = D1 - (D1 intersection D2)

that is,
newD: 
   datesws   wc  pwc
 2005-10-19:12:00  10.8  80  81
 2005-10-20:12:00  12.3   5  15
 2005-10-21:15:00  12.3   3  15
 2005-10-23:12:00  12.3  13   2
 2005-10-24:15:00  10.3   2  95
 
Thanks for any help you can provide,

Augusto


Augusto Sanabria. MSc, PhD.
Mathematical Modeller
Risk Research Group
Geospatial  Earth Monitoring Division
Geoscience Australia (www.ga.gov.au)
Cnr. Jerrabomberra Av.  Hindmarsh Dr.
Symonston ACT 2601
Ph. (02) 6249-9155

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Extracting a subset from a dataframe

[Xiaohui]

Try: D1[setdiff(D1$dates,D2$dates) , ]

Xiaohui

[EMAIL PROTECTED] wrote:
 Good day everyone,

 Can anyone suggest an effective method to solve
 the following problem:

 I have 2 dataframes D1 and D2 as follows:

 D1:
 dates   ws   wc pwc
  2005-10-19:12:00  10.8  80  81
  2005-10-20:12:00  12.3   5  15
  2005-10-21:15:00  12.3   3  15
  2005-10-22:15:00  11.3  13  95
  2005-10-23:12:00  12.3  13   2
  2005-10-24:15:00  10.3   2  95
  2005-10-25:15:00  10.3   2   2

 D2:
 dates   ws   wc  pwc
  2005-02-02:15:00  17.5   5  96
  2005-02-19:15:00  20.1  15  97
  2005-02-20:18:00  16.5  95  95
  2005-03-03:18:00  10.3  95  95
  2005-03-04:00:00  13.4  13  95
  2005-10-22:15:00  11.3  13  95
  2005-10-25:15:00  10.3   2   2

 I want to create another dataframe made up
 of the values of dataframe1 which are not common
 with dataframe2, ie. newD = D1 - (D1 intersection D2)

 that is,
 newD: 
datesws   wc  pwc
  2005-10-19:12:00  10.8  80  81
  2005-10-20:12:00  12.3   5  15
  2005-10-21:15:00  12.3   3  15
  2005-10-23:12:00  12.3  13   2
  2005-10-24:15:00  10.3   2  95
  
 Thanks for any help you can provide,

 Augusto

 
 Augusto Sanabria. MSc, PhD.
 Mathematical Modeller
 Risk Research Group
 Geospatial  Earth Monitoring Division
 Geoscience Australia (www.ga.gov.au)
 Cnr. Jerrabomberra Av.  Hindmarsh Dr.
 Symonston ACT 2601
 Ph. (02) 6249-9155

 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Extracting a subset from a dataframe

[Frede Aakmann Tøgersen]

Augusto

cnd - D1$dates %in% D2$dates
D1[!cnd,]

should do it.
 
Med venlig hilsen / Regards

Frede Aakmann Tøgersen
Forsker / Scientist



 AARHUS UNIVERSITET / UNIVERSITY OF AARHUS  
Det Jordbrugsvidenskabelige Fakultet / Faculty of Agricultural Sciences 
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Fra: [EMAIL PROTECTED] på vegne af [EMAIL PROTECTED]
Sendt: fr 23-02-2007 07:26
Til: R-help@stat.math.ethz.ch
Emne: [R] Extracting a subset from a dataframe



Good day everyone,

Can anyone suggest an effective method to solve
the following problem:

I have 2 dataframes D1 and D2 as follows:

D1:
dates   ws   wc pwc
 2005-10-19:12:00  10.8  80  81
 2005-10-20:12:00  12.3   5  15
 2005-10-21:15:00  12.3   3  15
 2005-10-22:15:00  11.3  13  95
 2005-10-23:12:00  12.3  13   2
 2005-10-24:15:00  10.3   2  95
 2005-10-25:15:00  10.3   2   2

D2:
dates   ws   wc  pwc
 2005-02-02:15:00  17.5   5  96
 2005-02-19:15:00  20.1  15  97
 2005-02-20:18:00  16.5  95  95
 2005-03-03:18:00  10.3  95  95
 2005-03-04:00:00  13.4  13  95
 2005-10-22:15:00  11.3  13  95
 2005-10-25:15:00  10.3   2   2

I want to create another dataframe made up
of the values of dataframe1 which are not common
with dataframe2, ie. newD = D1 - (D1 intersection D2)

that is,
newD:
   datesws   wc  pwc
 2005-10-19:12:00  10.8  80  81
 2005-10-20:12:00  12.3   5  15
 2005-10-21:15:00  12.3   3  15
 2005-10-23:12:00  12.3  13   2
 2005-10-24:15:00  10.3   2  95

Thanks for any help you can provide,

Augusto


Augusto Sanabria. MSc, PhD.
Mathematical Modeller
Risk Research Group
Geospatial  Earth Monitoring Division
Geoscience Australia (www.ga.gov.au)
Cnr. Jerrabomberra Av.  Hindmarsh Dr.
Symonston ACT 2601
Ph. (02) 6249-9155

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

__
R-help@stat.math.ethz.ch 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.