date:20090112

Try this:

X$third - rep(X$second[which(!duplicated(X$first))], table(X$first))

On Mon, Jan 12, 2009 at 5:57 AM, Heston Capital hestoncapi...@gmail.comwrote:

 I am trying to write some code where the factor references its
 previous value, but can't find a solution searching through the
 archive.

  X
  first second
 1 A  1
 2 A  2
 3 B  3
 4 B  4
 5 B  5
 6 C  6
 7 C  7

 I need a third column, in pseudo code-
  If value of first=previous value of first:
third=previous value of third
  else third = second

 So the third column would look like:
 0
 0
 3
 3
 3
 6
 6


 Thanks!

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[R] Loading workspaces from the command line

Hi,

Is there any way to load workspaces (e.g. stuff from save.image) from
the command line? I'm on Linux, and would find this very helpful.

I'm guessing this functionality can be duplicated with a skillful bash
script to rename the particular file to .RData (and then back once R
terminates), but I'm wondering if there's a better way.

Zhou Fang

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Re: [R] Loading workspaces from the command line

See ?load

On Mon, Jan 12, 2009 at 10:12 AM, Zhou Fang zhou.zf...@gmail.com wrote:

 Hi,

 Is there any way to load workspaces (e.g. stuff from save.image) from
 the command line? I'm on Linux, and would find this very helpful.

 I'm guessing this functionality can be duplicated with a skillful bash
 script to rename the particular file to .RData (and then back once R
 terminates), but I'm wondering if there's a better way.

 Zhou Fang

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




-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O

[[alternative HTML version deleted]]

__
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Re: [R] Loading workspaces from the command line

That's not really what I meant by 'command line'. I meant, well,
loading from e.g. a bash shell, not from within an interactive R
session itself.

Thanks anyways,

Zhou

(Possibly this email was sent twice. Apologies)

On Mon, Jan 12, 2009 at 12:15 PM, Henrique Dallazuanna www...@gmail.com wrote:
 See ?load

 On Mon, Jan 12, 2009 at 10:12 AM, Zhou Fang zhou.zf...@gmail.com wrote:

 Hi,

 Is there any way to load workspaces (e.g. stuff from save.image) from
 the command line? I'm on Linux, and would find this very helpful.

 I'm guessing this functionality can be duplicated with a skillful bash
 script to rename the particular file to .RData (and then back once R
 terminates), but I'm wondering if there's a better way.

 Zhou Fang

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



 --
 Henrique Dallazuanna
 Curitiba-Paraná-Brasil
 25° 25' 40 S 49° 16' 22 O


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Re: [R] Extraction from an output

2009-01-12 Thread Chuck Cleland

On 1/12/2009 6:42 AM, robert-mcfad...@o2.pl wrote:
 Hello,
 Would you tell my how to extract a result from a test - it's justified 
 because I need to run this test many times. Here is an  example from authors' 
 test:
 
 library(coin)
 lungtumor - data.frame(dose = rep(c(0, 1, 2), c(40, 50, 48)), tumor = 
 c(rep(c(0, 1), c(38, 2)), rep(c(0, 1), c(43, 7)), rep(c(0, 1), c(33, 15
 ca.test-independence_test(tumor ~ dose, data = lungtumor, teststat = quad)
 ca.test
 
 Asymptotic General Independence Test
 
 data:  tumor by dose 
 chi-squared = 10.6381, df = 1, p-value = 0.001108 
 
 How to use ca.test and extract p-value and chi-squared.

 pvalue(ca.test)
[1] 0.001107836

 statistic(ca.test)
[1] 10.63806

  See the Value subsection on the help page for independence_test().

 Robert 
 
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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. 

-- 
Chuck Cleland, Ph.D.
NDRI, Inc. (www.ndri.org)
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
tel: (732) 512-0171 (M, W, F)
fax: (917) 438-0894

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[R] anova() or aov()?

2009-01-12 Thread jass


Dear all,

I have a simple (I think) question that is troubling me lately:

Is there any main difference between anova() command and aov() command when 
performing an ANOVA in Experimental design 
analyses?

Thank you for your time,

Ismini

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Re: [R] anova() or aov()?

2009-01-12 Thread Chuck Cleland

On 1/12/2009 8:57 AM, j...@in.gr wrote:
 Dear all,
 
 I have a simple (I think) question that is troubling me lately:
 
 Is there any main difference between anova() command and aov() command when 
 performing an ANOVA in Experimental design 
 analyses?

  The main difference is that aov() *fits* a single model and anova()
*summarizes* a single model or compares two or more models.

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

-- 
Chuck Cleland, Ph.D.
NDRI, Inc. (www.ndri.org)
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
tel: (732) 512-0171 (M, W, F)
fax: (917) 438-0894

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Loading workspaces from the command line


See if this material is helpful:

http://cran.r-project.org/doc/manuals/R-intro.html#Invoking-R-from-the-command-line

--  
David Winsemius


On Jan 12, 2009, at 7:24 AM, Zhou Fang wrote:


That's not really what I meant by 'command line'. I meant, well,
loading from e.g. a bash shell, not from within an interactive R
session itself.

Thanks anyways,

Zhou

(Possibly this email was sent twice. Apologies)

On Mon, Jan 12, 2009 at 12:15 PM, Henrique Dallazuanna www...@gmail.com 
 wrote:

See ?load

On Mon, Jan 12, 2009 at 10:12 AM, Zhou Fang zhou.zf...@gmail.com  
wrote:


Hi,

Is there any way to load workspaces (e.g. stuff from save.image)  
from

the command line? I'm on Linux, and would find this very helpful.

I'm guessing this functionality can be duplicated with a skillful  
bash

script to rename the particular file to .RData (and then back once R
terminates), but I'm wondering if there's a better way.

Zhou Fang

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




--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O



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Re: [R] How to get solution of following polynomial?

2009-01-12 Thread Ravi Varadhan

Hi Hans,

I actaually meant the polynom package (not polynomial, which was a
typo).  I am curious as to the main differences between polynom and
PolynomF.  

Ron - by vectorizing, I mean that the function fn() can take a vector as an
input and return the function values at all the points in the vector.
sapply is an easy way to do this.  

Ravi.



---

Ravi Varadhan, Ph.D.

Assistant Professor, The Center on Aging and Health

Division of Geriatric Medicine and Gerontology 

Johns Hopkins University

Ph: (410) 502-2619

Fax: (410) 614-9625

Email: rvarad...@jhmi.edu

Webpage:  http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html

 





-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Hans W. Borchers
Sent: Monday, January 12, 2009 4:30 AM
To: r-h...@stat.math.ethz.ch
Subject: Re: [R] How to get solution of following polynomial?

RON70 ron_michael70 at yahoo.com writes:

 
 
 Hi Ravi, Thanks for this reply. However I could not understand meaning 
 of vectorizing the function. Can you please be little bit elaborate on
that?
 Secondly the package polynomial is not available in CRAN it seems. 
 What is the alternate package?
 
 Thanks,


Ravi means 'PolynomF' which is an improved version of the old polynomial
package.

You do not need to recreate the polynomial from points. Instead, calculate
the exact polynomial:

library(PolynomF)
z - polynom()

p11 - 1 - A1[1,1]*z - A2[1,1]*z^2 - A3[1,1]*z^3 - A4[1,1]*z^4
# ...
p - p11*p22 - p12*p21

There is probably a shorter way to generate these four polynoms p11, ...,
p22.
Anyway, the result is

p
# 1 - 1.18*x + 0.2777*x^2 - 0.2941*x^3 - 0.1004*x^4 + 0.3664*x^5 -
# 0.0636*x^6 + 0.062*x^7 - 0.068*x^8

solve(p)
# [1] -1.365976+0.00i -0.737852-1.639581i -0.737852+1.639581i
# [4] -0.012071-1.287727i -0.012071+1.287727i  1.00+0.00i
# [7]  1.388794-0.281841i  1.388794+0.281841i

and the real solutions are 1.0 and -1.365976 !

Regards,  Hans Werner


 Ravi Varadhan wrote:
  
  Hi,
  
  You can use the polynomial package to solve your problem.
  
  The key step is to find the exact polynomial representation of fn(). 
  Noting that it is a 8-th degree polynomial, we can get its exact 
  form using the poly.calc() function.  Once we have that, it is a 
  simple matter of finding the roots using the solve() function.
  
  require(polynomial)
  
  a - c(-0.07, 0.17)
  b - c(1, -4)
  cc - matrix(c(0.24, 0.00, -0.08, -0.31), 2) d - matrix(c(0, 0, 
  -0.13, -0.37), 2) e - matrix(c(0.2, 0, -0.06, -0.34), 2)
  A1 - diag(2) + a %*% t(b) + cc
  A2 - -cc + d
  A3 - -d + e
  A4 - -e
  
  # I am vectorizing your function
  fn - function(z)
 {
  sapply(z, function(z) {
  y - diag(2) - A1*z - A2*z^2 - A3*z^3 - A4*z^4
  det(y)
  })
 }
  
  
  x - seq(-5, 5, length=9) # note we need 9 points to exactly 
  determine a 8-th degree polynomial y - fn(x)
  
  p - poly.calc(x, y)  # uses Lagrange interpolation to determine 
  polynomial form p
  1 - 1.18*x + 0.2777*x^2 - 0.2941*x^3 - 0.1004*x^4 + 0.3664*x^5 -
  0.0636*x^6 + 0.062*x^7 - 0.068*x^8
  
  # plot showing that p is the exact polynomial representation of 
  fn(z) pfunc - as.function(p)
  x1 -seq(-5, 5, length=100)
  plot(x1, fn(x1),type=l)
  lines(x1, pfunc(x1), col=2, lty=2)   
  
  solve(p)  # gives you the roots (some are, of course, complex)
  
  
  Hope this helps,
  Ravi.
  
  
  
  Ravi Varadhan, Ph.D.
  Assistant Professor,
  Division of Geriatric Medicine and Gerontology School of Medicine 
  Johns Hopkins University
  
  Ph. (410) 502-2619
  email: rvaradhan at jhmi.edu
 

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[R] assign a list using expression?

2009-01-12 Thread Skotara


Dear R-users,

I would like to assign elements to a list in the following manner:
mylist - list(a = a, b = b, c = c)

To do this I tried
myexpr - expression(a = a, b = b, c = c)
mylist - list( eval(myexpr) )

It ends up by overwriting a when b is assigned and b when c is assigned. 
Additionally the element of the list does not have a name.

Could you tell me why this is the case?
Thank you very much in advance!

Best regards,
Nils

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Re: [R] Loading workspaces from the command line

Ok, looks like I can do what I want with --args, commandArgs() and an
appropiate .First.

Thanks,

Zhou

On Mon, Jan 12, 2009 at 2:27 PM, David Winsemius dwinsem...@comcast.net wrote:
 See if this material is helpful:

 http://cran.r-project.org/doc/manuals/R-intro.html#Invoking-R-from-the-command-line

 -- David Winsemius

 On Jan 12, 2009, at 7:24 AM, Zhou Fang wrote:

 That's not really what I meant by 'command line'. I meant, well,
 loading from e.g. a bash shell, not from within an interactive R
 session itself.

 Thanks anyways,

 Zhou

 (Possibly this email was sent twice. Apologies)

 On Mon, Jan 12, 2009 at 12:15 PM, Henrique Dallazuanna www...@gmail.com
 wrote:

 See ?load

 On Mon, Jan 12, 2009 at 10:12 AM, Zhou Fang zhou.zf...@gmail.com wrote:

 Hi,

 Is there any way to load workspaces (e.g. stuff from save.image) from
 the command line? I'm on Linux, and would find this very helpful.

 I'm guessing this functionality can be duplicated with a skillful bash
 script to rename the particular file to .RData (and then back once R
 terminates), but I'm wondering if there's a better way.

 Zhou Fang

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



 --
 Henrique Dallazuanna
 Curitiba-Paraná-Brasil
 25° 25' 40 S 49° 16' 22 O


 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
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[R] Problems with impute function from Hmisc library

2009-01-12 Thread Erich Studerus

 

Hello

 

I would like to use the transcan and impute functions from Hmisc library for
single impution.

 

library(Hmisc)

m1-data.frame(x1=rnorm(20),x2=rnorm(20),x3=rnorm(20))

m1[c(2,4),1]-NA

t1-transcan(~x1+x2+x3,data=m1,imputed=T)

impute(t1)

 

Fehler in as.environment(pos) : kein Eintrag names NA im Suchpfad

 

Which means: Error in as.environment(pos): no entry called 'NA' in search
path.

 

What am I doing wrong?

 

Regards,

 

Erich


[[alternative HTML version deleted]]

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Re: [R] Loading workspaces from the command line

Another possibility is to have a separate directory
for each project and place an .RData file in each.
Now just cd to whatever directory corresponds to the
project you wish to work on and start R normally.
No code is needed.

On Mon, Jan 12, 2009 at 10:04 AM, Zhou Fang zhou.zf...@gmail.com wrote:
 Ok, looks like I can do what I want with --args, commandArgs() and an
 appropiate .First.

 Thanks,

 Zhou

 On Mon, Jan 12, 2009 at 2:27 PM, David Winsemius dwinsem...@comcast.net 
 wrote:
 See if this material is helpful:

 http://cran.r-project.org/doc/manuals/R-intro.html#Invoking-R-from-the-command-line

 -- David Winsemius

 On Jan 12, 2009, at 7:24 AM, Zhou Fang wrote:

 That's not really what I meant by 'command line'. I meant, well,
 loading from e.g. a bash shell, not from within an interactive R
 session itself.

 Thanks anyways,

 Zhou

 (Possibly this email was sent twice. Apologies)

 On Mon, Jan 12, 2009 at 12:15 PM, Henrique Dallazuanna www...@gmail.com
 wrote:

 See ?load

 On Mon, Jan 12, 2009 at 10:12 AM, Zhou Fang zhou.zf...@gmail.com wrote:

 Hi,

 Is there any way to load workspaces (e.g. stuff from save.image) from
 the command line? I'm on Linux, and would find this very helpful.

 I'm guessing this functionality can be duplicated with a skillful bash
 script to rename the particular file to .RData (and then back once R
 terminates), but I'm wondering if there's a better way.

 Zhou Fang

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



 --
 Henrique Dallazuanna
 Curitiba-Paraná-Brasil
 25° 25' 40 S 49° 16' 22 O


 __
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Re: [R] assign a list using expression?

2009-01-12 Thread Patrick Burns


I think this is glossing over 9.7 of 'The R Inferno'.
You aren't telling us what you really want to achieve.
It seems hard for me to believe that the approach
you are taking is going to be the easiest route to
whatever that is.

Patrick Burns
patr...@burns-stat.com
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of The R Inferno and A Guide for the Unwilling S User)

Skotara wrote:

Dear R-users,

I would like to assign elements to a list in the following manner:
mylist - list(a = a, b = b, c = c)

To do this I tried
myexpr - expression(a = a, b = b, c = c)
mylist - list( eval(myexpr) )

It ends up by overwriting a when b is assigned and b when c is 
assigned. Additionally the element of the list does not have a name.

Could you tell me why this is the case?
Thank you very much in advance!

Best regards,
Nils

__
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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] assign a list using expression?

The purpose of this is not clear but depending on what a and b are you
might be able to use a data frame (which is a list):

 a - 1:2; b - 3:4
 data.frame(a, b)
  a b
1 1 3
2 2 4

On Mon, Jan 12, 2009 at 9:42 AM, Skotara nils.skot...@uni-hamburg.de wrote:
 Dear R-users,

 I would like to assign elements to a list in the following manner:
 mylist - list(a = a, b = b, c = c)

 To do this I tried
 myexpr - expression(a = a, b = b, c = c)
 mylist - list( eval(myexpr) )

 It ends up by overwriting a when b is assigned and b when c is assigned.
 Additionally the element of the list does not have a name.
 Could you tell me why this is the case?
 Thank you very much in advance!

 Best regards,
 Nils

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


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Re: [R] Loading workspaces from the command line

Well, that isn't ideal for my purposes. (A little context - basically
I have a script that I'm running for a lot of simulations, which is
kinda buggy, and what I'm doing is I'm having the script periodically
save whatever it has done so far to an automatically named file. Then
if something odd happens in between two saves, I can run forward from
a previously saved point to find the problem and figure out why it
happened, and also I won't risk losing everything if something
catastrophic happens.)

Anyways, if anyone's interested, in .Rprofile

.First - function(){
  if (rev(commandArgs())[2] == ld){
load(rev(commandArgs())[1], .GlobalEnv)
  }
}

Then e.g.

alias Rload='R --arg ld'

or make a bash script with

gnome-terminal --command R --args ld $1

and set some Open With options, and you'll be about to open R
workspaces from Nautilus etc by point and click.

Zhou


On Mon, Jan 12, 2009 at 3:14 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
 Another possibility is to have a separate directory
 for each project and place an .RData file in each.
 Now just cd to whatever directory corresponds to the
 project you wish to work on and start R normally.
 No code is needed.

 On Mon, Jan 12, 2009 at 10:04 AM, Zhou Fang zhou.zf...@gmail.com wrote:
 Ok, looks like I can do what I want with --args, commandArgs() and an
 appropiate .First.

 Thanks,

 Zhou

 On Mon, Jan 12, 2009 at 2:27 PM, David Winsemius dwinsem...@comcast.net 
 wrote:
 See if this material is helpful:

 http://cran.r-project.org/doc/manuals/R-intro.html#Invoking-R-from-the-command-line

 -- David Winsemius

 On Jan 12, 2009, at 7:24 AM, Zhou Fang wrote:

 That's not really what I meant by 'command line'. I meant, well,
 loading from e.g. a bash shell, not from within an interactive R
 session itself.

 Thanks anyways,

 Zhou

 (Possibly this email was sent twice. Apologies)

 On Mon, Jan 12, 2009 at 12:15 PM, Henrique Dallazuanna www...@gmail.com
 wrote:

 See ?load

 On Mon, Jan 12, 2009 at 10:12 AM, Zhou Fang zhou.zf...@gmail.com wrote:

 Hi,

 Is there any way to load workspaces (e.g. stuff from save.image) from
 the command line? I'm on Linux, and would find this very helpful.

 I'm guessing this functionality can be duplicated with a skillful bash
 script to rename the particular file to .RData (and then back once R
 terminates), but I'm wondering if there's a better way.

 Zhou Fang

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Re: [R] connecting boxplots

2009-01-12 Thread Michael A. Miller

 johnhj jhar...@web.de wrote:

 Can you also describe me how to describe the standard
 deviation of the boxplots/matrices ?
 
Try tapply:

 x -read.table(file=test.txt)   
 x$group - rep(1:8, each=5)
 boxplot(V3~gruppe, data=x) 


with(x, tapply(V3, gruppe, sd))


Mike

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Re: [R] for loop and if problem

2009-01-12 Thread Sake


I have one final question...
How can I save a CSV ifile with ; separation in stead of , separation?
I know the write.csv(file=filename.csv) an that you can use sep=; when
you open a .csv file, but that doesn't work with the write.csv command.
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[R] Error in svd(S) : infinite or missing values in 'x'

2009-01-12 Thread André Dias

Hi everyone.

I was running correspondence analysis in R with the package 'ca' and I
got a error message that I could not solve.

 summary(ca(gui))
Error in svd(S) : infinite or missing values in 'x'

#where gui is my dat table as follows:

0   90  1   1
0   90  0   0
0   90  5   1
0   95  1   0
0   0   0   0
85  0   0   0
50  0   0   0
90  0   0   0
75  0   0   0
75  0   0   0
0   0   10  25
0   0   5   5
0   0   50  10
0   0   10  20
0   0   1   10
0   0   50  1
0   0   10  5
0   0   20  0
0   0   10  50
0   0   5   5

As far I could undestand, this error should appear due to missing
data, but this is not the case of my table.

I will apreciate any help to undertand what I am doing wrong.

Thanks!
Andre

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[R] problems with download.file() from ftp?

2009-01-12 Thread zack holden


Dear list,
 
I am trying to download a text file from an ftp site using download.file().
 
I used the following code:
url - 
ftp://ftp.wcc.nrcs.usda.gov/data/snow/snow_course/table/history/idaho/13e19.txt;
dest - c:/test/downloadtest.txt
download.file(url, dest)
I get this error message, indicating that R is unable to open the connection:
trying URL 
'ftp://ftp.wcc.nrcs.usda.gov/data/snow/snow_course/table/history/idaho/13e19.txt'
Error in download.file(url, dest) : 
  cannot open URL 
'ftp://ftp.wcc.nrcs.usda.gov/data/snow/snow_course/table/history/idaho/13e19.txt'
 
I'm sure there is a simple solution that allows R to open this connection, but 
after reading the url help file for an hour, I can't seem to find the 
solution. Is anyone willing to help me solve this problem? 
 
thanks in advance for any advice,
 
Zack

 
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[R] (no subject)

2009-01-12 Thread Mélanie-Louise Leblanc

hi,

I'm looking for R codes that can do modified Pearson correlation that corrects
for spatially correlated variables.

thanks,

Mélanie-L. Le Blanc

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[R] Working with duplicated rows

2009-01-12 Thread Steve Murray


Dear all,



I have a dataframe of 3 columns, consisting of 'longitude', 'latitude'
and a corresponding 'value'. Where identical 'longitude' and 'latitude'
pairs occur more than once, I want their corresponding 'value' to be
summed and the 'pair' to only appear once.



For example:



long   lat   value

10 20  5

6  2   3

27-3   9

10 20  10

4 -1   0

6  2   9





would be converted to something like:



long   lat   value


10 20  15


6  2   12


27-3   9


4 -1   0






...as rows 1 and 4, 2 and 6 respectively are matched with respect to
the 'long' and 'lat' columns. Their values in column 3 are then summed
and reported as one row in the new dataframe.



How would I go about coding this in R?



Many thanks,



Steve


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Re: [R] assign a list using expression?

2009-01-12 Thread Skotara


Thank you Patrick and Gabor!
Sorry, I think I have not explainend it well.
The purpose is as follows:
   names - letters[1:3]
   values - data.frame(a = 1:3, b = 4:6, c = 7:9)
With more complicated objects similar to 'names' and 'values' I wrote 
the following line to assign the elements of the list:
   mycommand - parse(text = paste(names,  = values[\, names, \], 
sep=) )

However,
   list(eval(mycommand))
does not do what I want.
whereas
   list(a = values[a], b = values[b], c = values[c])
does.

I can not tell why...
I try to understand, what expression and eval do. I know that many times 
there are other ways to achieve the same goal.
So here, too. But I think there should be a reason why it does not work 
that way.


Best regards!
Nils

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Re: [R] for loop and if problem

Try using:

write.table(..., sep=;)

write.csv just calls write.table

On Mon, Jan 12, 2009 at 6:38 AM, Sake tlep.nav.e...@hccnet.nl wrote:

 I have one final question...
 How can I save a CSV ifile with ; separation in stead of , separation?
 I know the write.csv(file=filename.csv) an that you can use sep=; when
 you open a .csv file, but that doesn't work with the write.csv command.
 --
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 http://www.nabble.com/for-loop-and-if-problem-tp21312449p21412888.html
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-- 
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Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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Re: [R] assign a list using expression?

Try using do.call:

do.call(list, as.list(mycommand))

On Mon, Jan 12, 2009 at 1:58 PM, Skotara nils.skot...@uni-hamburg.dewrote:

 Thank you Patrick and Gabor!
 Sorry, I think I have not explainend it well.
 The purpose is as follows:
   names - letters[1:3]
   values - data.frame(a = 1:3, b = 4:6, c = 7:9)
 With more complicated objects similar to 'names' and 'values' I wrote the
 following line to assign the elements of the list:
   mycommand - parse(text = paste(names,  = values[\, names, \],
 sep=) )
 However,
   list(eval(mycommand))
 does not do what I want.
 whereas
   list(a = values[a], b = values[b], c = values[c])
 does.

 I can not tell why...
 I try to understand, what expression and eval do. I know that many times
 there are other ways to achieve the same goal.
 So here, too. But I think there should be a reason why it does not work
 that way.


 Best regards!
 Nils

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-- 
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25° 25' 40 S 49° 16' 22 O

[[alternative HTML version deleted]]

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Re: [R] Working with duplicated rows

Try this:

aggregate(dt$value, list(long = dt$long, lat = dt$lat), FUN = sum)

On Mon, Jan 12, 2009 at 1:52 PM, Steve Murray smurray...@hotmail.comwrote:


 Dear all,



 I have a dataframe of 3 columns, consisting of 'longitude', 'latitude'
 and a corresponding 'value'. Where identical 'longitude' and 'latitude'
 pairs occur more than once, I want their corresponding 'value' to be
 summed and the 'pair' to only appear once.



 For example:



 long   lat   value

 10 20  5

 6  2   3

 27-3   9

 10 20  10

 4 -1   0

 6  2   9





 would be converted to something like:



 long   lat   value


 10 20  15


 6  2   12


 27-3   9


 4 -1   0






 ...as rows 1 and 4, 2 and 6 respectively are matched with respect to
 the 'long' and 'lat' columns. Their values in column 3 are then summed
 and reported as one row in the new dataframe.



 How would I go about coding this in R?



 Many thanks,



 Steve


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-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O

[[alternative HTML version deleted]]

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Re: [R] assign a list using expression?

This still isn't clear.  In your post, values is already
a list with the required names and values in it so the
whole exercise is pointless -- you are starting
out with the answer.

Just guessing, but maybe your setup is a set of variables
in your workspace and a vector of their names with the
output being a named list of them:

a - 1:2; b - 1:3
nms - c(a, b)
sapply(nms, get, simplify = FALSE)


On Mon, Jan 12, 2009 at 10:58 AM, Skotara nils.skot...@uni-hamburg.de wrote:
 Thank you Patrick and Gabor!
 Sorry, I think I have not explainend it well.
 The purpose is as follows:
   names - letters[1:3]
   values - data.frame(a = 1:3, b = 4:6, c = 7:9)
 With more complicated objects similar to 'names' and 'values' I wrote the
 following line to assign the elements of the list:
   mycommand - parse(text = paste(names,  = values[\, names, \],
 sep=) )
 However,
   list(eval(mycommand))
 does not do what I want.
 whereas
   list(a = values[a], b = values[b], c = values[c])
 does.

 I can not tell why...
 I try to understand, what expression and eval do. I know that many times
 there are other ways to achieve the same goal.
 So here, too. But I think there should be a reason why it does not work that
 way.

 Best regards!
 Nils


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Re: [R] for loop and if problem

write.csv does exactly what you would expect ... creates a *Comma*  
Separated Values file. If you don't want a comma separated value  
format then use write.table with sep=;


You can still name it whatever.csv.

Or you if you also intend commas for decimal points,  use write.csv2  
as described in the help page:


write.csv2 uses a comma for the decimal point and a semicolon for the  
separator, the Excel convention for CSV files in some Western European  
locales.



--
David Winsemius


On Jan 12, 2009, at 6:38 AM, Sake wrote:



I have one final question...
How can I save a CSV ifile with ; separation in stead of , separation?
I know the write.csv(file=filename.csv) an that you can use  
sep=; when
you open a .csv file, but that doesn't work with the write.csv  
command.

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Re: [R] Working with duplicated rows

2009-01-12 Thread Dimitris Rizopoulos


try aggregate(), e.g.,

dat - read.table(textConnection(long   lat   value

10 20  5

6  2   3

27-3   9

10 20  10

4 -1   0

6  2   9), header = TRUE)
closeAllConnections()


aggregate(dat[value], list(Long = dat$long, Lat = dat$lat), sum)


I hope it helps.

Best,
Dimitris


Steve Murray wrote:

Dear all,



I have a dataframe of 3 columns, consisting of 'longitude', 'latitude'
and a corresponding 'value'. Where identical 'longitude' and 'latitude'
pairs occur more than once, I want their corresponding 'value' to be
summed and the 'pair' to only appear once.



For example:



long   lat   value

10 20  5

6  2   3

27-3   9

10 20  10

4 -1   0

6  2   9





would be converted to something like:



long   lat   value


10 20  15


6  2   12


27-3   9


4 -1   0






...as rows 1 and 4, 2 and 6 respectively are matched with respect to
the 'long' and 'lat' columns. Their values in column 3 are then summed
and reported as one row in the new dataframe.



How would I go about coding this in R?



Many thanks,



Steve


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--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus Medical Center

Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014

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Re: [R] (no subject)

2009-01-12 Thread Ben Bolker

Mélanie-Louise Leblanc melanie-louise.leblanc.1 at ulaval.ca writes:

 
 hi,
 
 I'm looking for R codes that can do modified Pearson correlation that corrects
 for spatially correlated variables.
 
 thanks,
 
 Mélanie-L. Le Blanc

  I think you may need to find the answer to the statistical/subject
area question first (i.e., how can I calculate/make inferences
on a measure of association among observations that may also be
correlated due to spatial association?) and then search to see
whether R implements such methods. The only two thoughts that
spring to mind are (1) partial Mantel tests (RSiteSearch(partial mantel)
and read up on the associated literature) and (2) trying this
question out on the r-sig-geo mailing list.

  good luck
Ben Bolker

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Re: [R] Interval censored Data in survreg() with zero values!

2009-01-12 Thread Geraldine Henningsen


Hello again,

I studied your suggestion but still I disagree.  You wrote:

From the way you wrote the problem I assumed 
that there is some number of n looks at the subject and then you count them 
up.

But this is not the case. My data is clearly continuous quantities and no 
discrete choices. I know nothing about the underlying choice process, the only 
thing I know is the final share of one of three regimes. So sorry for the bad 
description of the problem.
So I stick with my censored data model. Still the hint about the p-values is 
very helpful because I actually ran into this problem. So thank you for the 
hint.

Best, Geraldine 



Terry Therneau schrieb:
 Apologies -- you are being more subtle than I thought.  Nevertheless, I think 
 that the censoring language isn't quite right.

   You are thinking of a hierarchical model:
   
 z ~ N(Xb, sigma), where Xb is the linear predictor, whatever covariates 
 you 
 think belong in the model.  Whether the distribution should be Gaussian or 
 somthing else depends not on the overall distribution of z, but on 
 distribution 
 of (z | Xb).  We could have a skewed predictor leading to skewed z, even if 
 the 
 distribution about any given expectation is symmetric.
 
 y = F(z) is what you observe.  The classic tobin model is y= max(0,z), 
 which 
 does lead to censored data. 
 
 In your case y_i = Binomial(n_i, p_i = H(z)).  Note a binomial is k heads 
 out of n tries with a coin of probability p, a Bernouli is a binomial 
 restricted to a single coin flip.  From the way you wrote the problem I 
 assumed 
 that there is some number of n looks at the subject and then you count them 
 up.  Note that var(y) = n p (1-p)
 
 H describes how the probability changes with z.  In biology we very 
 rarely 
 use H(z)= max(min(z,1),0) because it gives a hard threshold, and the 
 probability 
 of nearly anything doesn't go all the way to zero or one.  
 
 If H were as above and 
   var(y) = constant and
   n is sufficiently large so that Binomial dist is approx Gaussian and
   var(y |p)  var(z| Xb)

 then your y will fit a censored Gaussian.  Since at least the second is 
 false, 
 it doesn't.  

A censored model may still be an ok first cut at fitting the data, but I 
 would be suspicious of variance estimates and particularly of any p-values.  
 The 
 bootstrap could help that.

   Terry T.





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Re: [R] merge table rows (\multirow)

2009-01-12 Thread Dieter Menne




Felipe Carrillo wrote:
 
 I am trying to merge the 'Month' column using \multirow. For example for
 the column 'Week' I want July to be merged into one row(weeks 27,28,29,30)
 and so on for the following weeks.
 Below, I am creating a PDF using Sweave, MikTex,R-2.8.1 and windows XP to
 show an example.
 
 

.. Example removed and partially regenerated below

Another try; had a bad hour this morning, too many typos. 
This comes closer to what you want, but not exactly. 
You might try to manually call format.df which is implicitly 
called in latex, and massage that.
Again, I do not know if this is easier with xtable.

If you want to post latex samples, it is not really
necessary to include it in Sweave. The example
below works fine for testing.

Dieter


library(Hmisc)
sampDat - Month Week Estpassage MedFL
July 27665   34
July 28223235
July 29924135
July 3028464   35
Aug  3141049   35
Aug 32 82216   35
Aug 33 230411  35
Aug 34 358541  35
Sept 35747839  35
Sept 36459682  36
Sept 37609567  36
Sept 38979475  36
Sept 39837189  36
DF - read.table(textConnection(sampDat), header = TRUE)
row.names(DF)=DF$Week
latex(DF[,-2],label=tab:hola,longtable=FALSE,caption='Sample table.',
  rowname=,
  rgroup=unique(DF$Month),n.rgroup=table(DF$Month))

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[R] R2WinBUGS stopping execution

2009-01-12 Thread Richard . Cotton

Apologies if this isn't acceptable for the general help list.

I'm running OpenBUGS model via the R2WinBUGS package interface, under 
Windows.  Is it possible to terminate running models, short of using the 
Windows Task Manager to forcibly exit the program?

Regards,
Richie.

Mathematical Sciences Unit
HSL



ATTENTION:

This message contains privileged and confidential inform...{{dropped:20}}

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Re: [R] Working with duplicated rows

2009-01-12 Thread Steve Murray


Thanks - that's great!

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[R] Determining variance components of classed covariates

2009-01-12 Thread Stephen Montgomery

Hi -

I am interested in solving variance components for the data below with
respect to the response variable, Expression within R.

However, the covariates aren't independent and they also have a class
(of which the total variance explained by covariates in that class I am
most interested in).

Very naively, I have tried to look at each individual covariates
variance like this

 lm-lmer(Expression ~ 1 + (1|rs11834524) + (1|rs7074431),
data=input_new)
 lm
Linear mixed-effects model fit by REML 
Formula: Expression ~ 1 + (1 | rs11834524) + (1 | rs7074431) 
   Data: input 
   AIC   BIC logLik MLdeviance REMLdeviance
 108.4 116.5 -51.22  102.5102.4
Random effects:
 Groups NameVariance Std.Dev.
 rs11834524 (Intercept) 0.485538 0.69681 
 rs7074431  (Intercept) 0.013720 0.11713 
 Residual   0.128853 0.35896 
number of obs: 109, groups: rs11834524, 3; rs7074431, 3

Fixed effects:
Estimate Std. Error t value
(Intercept)   9.9524 0.4098   24.29

My assumption is that this is telling me that rs11834524 explains
0.485538 of the variance and rs7074431 explains 0.013720 of the variance
in Expression when looked at independently.  

However, I would like to know how to write a model where I know how much
of the total variance (in Expression) is described by covariates
rs11834524, rs1682421, rs13383869 and rs9457141 (call it class A) and
covariates rs9459617, rs7074431, rs12450785, rs592724 (call it class B).
Assuming an additive model within the class.  The caveats are that there
is missing data and again that there may be correlation between all the
covariates.

Such that a theoretical result may be that
Class A: Explains 60% of the total variance in expression (response)
Class B: Explains 10% of the total variance in expression

Thanks for the help!  I am sorry I am R challenged here...I really
appreciate the guidance!

Stephen


 dump(input_new, file=stdout()) 
input_new -
structure(list(Individual = structure(1:109, .Label = c(NA06984, 
NA06985, NA06986, NA06989, NA06993, NA06994, NA07000, 
NA07022, NA07037, NA07045, NA07051, NA07055, NA07056, 
NA07345, NA07346, NA07347, NA07357, NA07435, NA11829, 
NA11830, NA11831, NA11832, NA11839, NA11840, NA11843, 
NA11881, NA11882, NA11892, NA11893, NA11894, NA11917, 
NA11918, NA11919, NA11920, NA11930, NA11931, NA11992, 
NA11993, NA11994, NA11995, NA12003, NA12005, NA12006, 
NA12043, NA12044, NA12056, NA12057, NA12144, NA12145, 
NA12146, NA12154, NA12155, NA12156, NA12234, NA12239, 
NA12248, NA12249, NA12264, NA12272, NA12273, NA12274, 
NA12275, NA12282, NA12283, NA12286, NA12287, NA12340, 
NA12341, NA12342, NA12343, NA12347, NA12348, NA12383, 
NA12399, NA12400, NA12414, NA12489, NA12546, NA12716, 
NA12718, NA12748, NA12749, NA12750, NA12751, NA12760, 
NA12761, NA12762, NA12763, NA12775, NA12776, NA12777, 
NA12778, NA12812, NA12813, NA12814, NA12815, NA12827, 
NA12828, NA12829, NA12830, NA12842, NA12843, NA12872, 
NA12873, NA12874, NA12875, NA12889, NA12891, NA12892
), class = factor), Expression = c(9.46026823453575, 10.0788903323991,

9.20330296497174, 10.038741467793, 9.33092349416463, 11.0273957217919, 
10.5498875891745, 9.81137299592747, 11.2023261987976, 9.90559354069027, 
10.1524696609679, 10.3171767665993, 9.02155519577685, 9.84917871051438, 
10.658877473136, 9.88895551011107, 8.62335008726357, 9.21529114100886, 
10.7896248923916, 10.1302992505869, 8.64584282787018, 9.56057795866654, 
9.89810004078774, 10.2557482141576, 8.95588077688637, 9.56452454115857, 
9.26525135092154, 10.5438780642797, 9.8468571349548, 10.7416169225352, 
10.5623721612979, 10.6565276881443, 9.67758493445612, 9.75385553511462, 
8.997797236767, 11.0106882086179, 10.362578597992, 9.2745507212906, 
10.7453355016181, 9.75998268015348, 9.45003620116962, 10.055504292376, 
10.7072220720564, 10.0934686444392, 10.0472832129727, 10.1185615033486, 
10.3340911031131, 9.70618910683157, 10.5953304905529, 10.4246307909547, 
9.91463202635336, 10.249081562168, 10.9252022586474, 10.295544143525, 
11.4838109797985, 10.5286570234792, 9.78692800868132, 10.0397050809162, 
9.27914623343747, 10.37600233389, 9.27341681588134, 9.40195375611303, 
10.8979822929135, 9.0398977389, 10.3911745662505, 10.4345408213054, 
9.8548491618724, 10.1897729275437, 10.288149609, 8.9656977165014, 
9.81595398472166, 10.1856794532084, 9.3763789479684, 10.1712420020647, 
10.2964594680427, 10.3515965292101, 8.94492585275159, 11.2529257614993, 
9.25146912450726, 10.1904309237525, 10.7490591053023, 10.3883924463568, 
10.097023765247, 10.0824730785217, 10.0828512817661, 10.6371064852226, 
10.5831044752098, 10.4484786486601, 8.50264408341596, 10.3468670812262, 
9.46061433005316, 8.90027436167269, 9.73630671555279, 9.4022408144, 
10.3220768104446, 8.55132985773453, 10.1678182524815, 10.6145417864386, 
10.4169948161073, 10.0253039670548, 10.2568017077865, 10.5045847076951, 
9.75993936712448, 8.7092895909, 10.674414794, 10.8640943324257, 
10.4295384371541, 10.1987862649656,

Re: [R] problems with download.file() from ftp?

2009-01-12 Thread Prof Brian Ripley


First, try the example in the test file (tests/internet.R)

read.table(ftp://ftp.stats.ox.ac.uk/pub/datasets/csb/ch11b.dat;)

or

download.file(ftp://ftp.stats.ox.ac.uk/pub/datasets/csb/ch11b.dat;, 
test


That is known to work on many, many R systems.

I get the same error as you on that URL, but it looks to me that the 
site required PASV ftp, something R's download.file does not support. 

From the detailed logs (options(internet.info=0)



230-
230-Welcome To The
230- Natural Resources Conservation Service - USDA
230- NATIONAL WATER and CLIMATE CENTER
230-
230-
230-The Local Date and Time is Mon Jan 12 08:30:25 2009
230-Current Directory is /
230-Information Contact is: i...@wcc.nrcs.usda.gov
230----
230-Note that data files herein are at most updated once an hour.
230-
230 Guest login ok, access restrictions apply.

--
19: In download.file(url, dest) :
---

--
20: In download.file(url, dest) : Got 230
21: In download.file(url, dest) : Invalid answer to PASV

Since you are on Windows, using --internet2 (see the rw-FAQ) might 
work.


On Mon, 12 Jan 2009, zack holden wrote:



Dear list,

I am trying to download a text file from an ftp site using download.file().

I used the following code:
url - 
ftp://ftp.wcc.nrcs.usda.gov/data/snow/snow_course/table/history/idaho/13e19.txt;
dest - c:/test/downloadtest.txt
download.file(url, dest)
I get this error message, indicating that R is unable to open the connection:
trying URL 
'ftp://ftp.wcc.nrcs.usda.gov/data/snow/snow_course/table/history/idaho/13e19.txt'
Error in download.file(url, dest) :
 cannot open URL 
'ftp://ftp.wcc.nrcs.usda.gov/data/snow/snow_course/table/history/idaho/13e19.txt'

I'm sure there is a simple solution that allows R to open this connection, but after 
reading the url help file for an hour, I can't seem to find the solution. Is 
anyone willing to help me solve this problem?

thanks in advance for any advice,

Zack


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

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

Re: [R] problems with download.file() from ftp?

2009-01-12 Thread Tony Breyal

I ran your script on Windows Vista Ultimate, SP1 and it worked fine:

## R start...
 url - 
 'ftp://ftp.wcc.nrcs.usda.gov/data/snow/snow_course/table/history/idaho/13e19.txt'
 dest - //PFO-SBS001/Redirected/tonyb/Desktop/test/downloadtest.txt
 download.file(url, dest)
trying URL 'ftp://ftp.wcc.nrcs.usda.gov/data/snow/snow_course/table/
history/idaho/13e19.txt'
ftp data connection made, file length 5026 bytes
opened URL
downloaded 5026 bytes
## R end.


 sessionInfo()
R version 2.8.1 (2008-12-22)
i386-pc-mingw32

locale:
LC_COLLATE=English_United Kingdom.1252;LC_CTYPE=English_United Kingdom.
1252;LC_MONETARY=English_United Kingdom.
1252;LC_NUMERIC=C;LC_TIME=English_United Kingdom.1252

attached base packages:
[1] stats graphics  grDevices utils datasets  methods
base



On 12 Jan, 14:51, zack holden zack_hol...@hotmail.com wrote:
 Dear list,

 I am trying to download a text file from an ftp site using download.file().

 I used the following code:
 url - 
 ftp://ftp.wcc.nrcs.usda.gov/data/snow/snow_course/table/history/idaho...;
 dest - c:/test/downloadtest.txt
 download.file(url, dest)
 I get this error message, indicating that R is unable to open the connection:
 trying URL 
 'ftp://ftp.wcc.nrcs.usda.gov/data/snow/snow_course/table/history/idaho...
 Error in download.file(url, dest) :
   cannot open URL 
 'ftp://ftp.wcc.nrcs.usda.gov/data/snow/snow_course/table/history/idaho...

 I'm sure there is a simple solution that allows R to open this connection, 
 but after reading the url help file for an hour, I can't seem to find the 
 solution. Is anyone willing to help me solve this problem?

 thanks in advance for any advice,

 Zack

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

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

Re: [R] Working with duplicated rows


One of undoubtedly many ways:

# kind of a pain to delete the blank rows. Try to give us full example  
next time?


 txt- long   lat   value
+ 10 20  5
+ 6  2   3
+ 27-3   9
+ 10 20  10
+ 4 -1   0
+ 6  2   9
+ 
 DF2 - read.table(textConnection(txt), header=TRUE)

 DF3 - data.frame(with(DF2,list(lon=long, lat=lat,  
value=ave(value,long, lat, FUN=sum


 DF3
  lon lat value
1  10  2015
2   6   212
3  27  -3 9
4  10  2015
5   4  -1 0
6   6   212

If you want the column name to be long, the fix is obvious.

--
David Winsemius
On Jan 12, 2009, at 10:52 AM, Steve Murray wrote:



Dear all,



I have a dataframe of 3 columns, consisting of 'longitude', 'latitude'
and a corresponding 'value'. Where identical 'longitude' and  
'latitude'

pairs occur more than once, I want their corresponding 'value' to be
summed and the 'pair' to only appear once.



For example:



long   lat   value

10 20  5

6  2   3

27-3   9

10 20  10

4 -1   0

6  2   9





would be converted to something like:



long   lat   value


10 20  15


6  2   12


27-3   9


4 -1   0






...as rows 1 and 4, 2 and 6 respectively are matched with respect to
the 'long' and 'lat' columns. Their values in column 3 are then summed
and reported as one row in the new dataframe.



How would I go about coding this in R?



Many thanks,



Steve


_
Are you a PC?  Upload your PC story and show the world

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] lm: how are polynomial functions interpreted?

2009-01-12 Thread cgw


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

[R] Help with storage of each matrix generated in a loop

2009-01-12 Thread rafamoral


I need to store each matrix generated in a loop.

I've been working with the CUSUM algorithm and I've been trying to implement
it in R.
What I need to do with my dataset is to create 1000 randomized datasets and
cumulative sum them all and store all of those randomized CUSUMed datasets
for further analysis and creation of the simulation envelope in the CUSUM
chart. But I can't manage to store all of them, the script I've written only
stores the last randomized matrix in the variable I called cusumA.

Here's the script I've written:

mat - matrix(data=rep(c(1,2,3,4,5), 16), nrow=16, ncol=5)
# The matrix that will be sampled
A - matrix(data=0, nrow=16, ncol=5)
# The variable that will store the sampled matrix
for(i in 1:1000) {
# I want to do it 1000 times
for(j in 1:nrow(mat)) {
# The number of rows to be sampled)
A[j,] - sample(mat[j,])
# The sample itself - I want to do it 1000 times and then...
cusumA - cumsum(data.frame(A))
# cumulative sum it, and store the 1000 randomized matrices
}}

I've already tried to store it in a null matrix but it will only store the
first column of each matrix, so all I got was 1000 first columns. The code I
used to do it was

mat - matrix(data=rep(c(1,2,3,4,5), 16), nrow=16, ncol=5)
A - matrix(data=0, nrow=16, ncol=5)
cusumA - matrix()
for(i in 1:1000) {
for(j in 1:nrow(mat)) {
A[j,] - sample(mat[j,])
cusumA[i] - cumsum(data.frame(A))
}}

and I even got 50+ warnings...
Any help'd be really appreciated!
-- 
View this message in context: 
http://www.nabble.com/Help-with-storage-of-each-matrix-generated-in-a-loop-tp21418758p21418758.html
Sent from the R help mailing list archive at Nabble.com.

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

Re: [R] Determining variance components of classed covariates

You might want to try the
https://stat.ethz.ch/mailman/listinfo/r-sig-mixed-models
list next time for mixed model questions.

At any rate the Variance column figures are variances, not percentages.

We can use anova with REML=FALSE to make comparisons among
models.  Below we find that removing the rs7074431 term makes very
little difference so we can drop it but removing the rs11834524 term
makes a big difference.  Thus modx0 can be used.

 modxx - lmer(Expression ~ 1 + (1|rs11834524) + (1|rs7074431), input_new, 
 REML = FALSE)
 modx0 - lmer(Expression ~ 1 + (1|rs11834524), input_new, REML = FALSE)
 mod0x - lmer(Expression ~ 1 + (1|rs7074431), input_new, REML = FALSE)
 anova(modxx, modx0)
Data: input_new
Models:
modx0: Expression ~ 1 + (1 | rs11834524)
modxx: Expression ~ 1 + (1 | rs11834524) + (1 | rs7074431)
  Df AIC BIC  logLik  Chisq Chi Df Pr(Chisq)
modx0  3 111.386 119.460 -52.693
modxx  4 110.288 121.053 -51.144 3.0986  10.07836 .
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
 anova(modxx, mod0x)
Data: input_new
Models:
mod0x: Expression ~ 1 + (1 | rs7074431)
modxx: Expression ~ 1 + (1 | rs11834524) + (1 | rs7074431)
  Df  AIC  BIC   logLik  Chisq Chi Df Pr(Chisq)
mod0x  3  206.652  214.726 -100.326
modxx  4  110.288  121.053  -51.144 98.365  1   2.2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

On Mon, Jan 12, 2009 at 11:36 AM, Stephen Montgomery s...@sanger.ac.uk wrote:
 Hi -

 I am interested in solving variance components for the data below with
 respect to the response variable, Expression within R.

 However, the covariates aren't independent and they also have a class
 (of which the total variance explained by covariates in that class I am
 most interested in).

 Very naively, I have tried to look at each individual covariates
 variance like this

 lm-lmer(Expression ~ 1 + (1|rs11834524) + (1|rs7074431),
 data=input_new)
 lm
 Linear mixed-effects model fit by REML
 Formula: Expression ~ 1 + (1 | rs11834524) + (1 | rs7074431)
   Data: input
   AIC   BIC logLik MLdeviance REMLdeviance
  108.4 116.5 -51.22  102.5102.4
 Random effects:
  Groups NameVariance Std.Dev.
  rs11834524 (Intercept) 0.485538 0.69681
  rs7074431  (Intercept) 0.013720 0.11713
  Residual   0.128853 0.35896
 number of obs: 109, groups: rs11834524, 3; rs7074431, 3

 Fixed effects:
Estimate Std. Error t value
 (Intercept)   9.9524 0.4098   24.29

 My assumption is that this is telling me that rs11834524 explains
 0.485538 of the variance and rs7074431 explains 0.013720 of the variance
 in Expression when looked at independently.

 However, I would like to know how to write a model where I know how much
 of the total variance (in Expression) is described by covariates
 rs11834524, rs1682421, rs13383869 and rs9457141 (call it class A) and
 covariates rs9459617, rs7074431, rs12450785, rs592724 (call it class B).
 Assuming an additive model within the class.  The caveats are that there
 is missing data and again that there may be correlation between all the
 covariates.

 Such that a theoretical result may be that
 Class A: Explains 60% of the total variance in expression (response)
 Class B: Explains 10% of the total variance in expression

 Thanks for the help!  I am sorry I am R challenged here...I really
 appreciate the guidance!

 Stephen


 dump(input_new, file=stdout())
 input_new -
 structure(list(Individual = structure(1:109, .Label = c(NA06984,
 NA06985, NA06986, NA06989, NA06993, NA06994, NA07000,
 NA07022, NA07037, NA07045, NA07051, NA07055, NA07056,
 NA07345, NA07346, NA07347, NA07357, NA07435, NA11829,
 NA11830, NA11831, NA11832, NA11839, NA11840, NA11843,
 NA11881, NA11882, NA11892, NA11893, NA11894, NA11917,
 NA11918, NA11919, NA11920, NA11930, NA11931, NA11992,
 NA11993, NA11994, NA11995, NA12003, NA12005, NA12006,
 NA12043, NA12044, NA12056, NA12057, NA12144, NA12145,
 NA12146, NA12154, NA12155, NA12156, NA12234, NA12239,
 NA12248, NA12249, NA12264, NA12272, NA12273, NA12274,
 NA12275, NA12282, NA12283, NA12286, NA12287, NA12340,
 NA12341, NA12342, NA12343, NA12347, NA12348, NA12383,
 NA12399, NA12400, NA12414, NA12489, NA12546, NA12716,
 NA12718, NA12748, NA12749, NA12750, NA12751, NA12760,
 NA12761, NA12762, NA12763, NA12775, NA12776, NA12777,
 NA12778, NA12812, NA12813, NA12814, NA12815, NA12827,
 NA12828, NA12829, NA12830, NA12842, NA12843, NA12872,
 NA12873, NA12874, NA12875, NA12889, NA12891, NA12892
 ), class = factor), Expression = c(9.46026823453575, 10.0788903323991,

 9.20330296497174, 10.038741467793, 9.33092349416463, 11.0273957217919,
 10.5498875891745, 9.81137299592747, 11.2023261987976, 9.90559354069027,
 10.1524696609679, 10.3171767665993, 9.02155519577685, 9.84917871051438,
 10.658877473136, 9.88895551011107, 8.62335008726357, 9.21529114100886,
 10.7896248923916, 10.1302992505869, 8.64584282787018, 9.56057795866654,
 9.89810004078774, 10.2557482141576,

Re: [R] Help with storage of each matrix generated in a loop

If you are satisfied with the structure of cusumA and just want 999  
more randome realizations of the same, then try creating an empty list  
to hold the 1000 dataframes you are creating and then accumulate  
sequentially to the list.


 mat - matrix(data=rep(c(1,2,3,4,5), 16), nrow=16, ncol=5)
 # The matrix that will be sampled
 A - matrix(data=0, nrow=16, ncol=5); cusumA - list()
## ^new empty list^
 # The variable that will store the sampled matrix
 for(i in 1:1000) {
+ # I want to do it 1000 times
+ for(j in 1:nrow(mat)) {
+ # The number of rows to be sampled)
+ A[j,] - sample(mat[j,])
+ # The sample itself - I want to do it 1000 times and then...
+ cusumA[[i]] - cumsum(data.frame(A))
# index 
+ # cumulative sum it, and store the 1000 randomized matrices
+ }}

 str(cusumA)
List of 1000
 $ :'data.frame':   16 obs. of  5 variables:
  ..$ X1: num [1:16] 5 9 10 14 16 21 22 25 26 30 ...
  ..$ X2: num [1:16] 3 4 6 7 10 11 15 17 20 22 ...
  ..$ X3: num [1:16] 2 4 7 12 17 20 25 30 35 40 ...
  ..$ X4: num [1:16] 4 7 12 15 16 20 23 24 28 31 ...
  ..$ X5: num [1:16] 1 6 10 12 16 18 20 24 26 27 ...
 $ :'data.frame':   16 obs. of  5 variables:
  ..$ X1: num [1:16] 4 5 7 9 14 16 17 22 27 28 ...
  ..$ X2: num [1:16] 3 6 10 11 13 17 20 21 23 28 ...
  ..$ X3: num [1:16] 5 7 10 13 14 19 23 26 27 29 ...
  ..$ X4: num [1:16] 2 7 8 12 15 18 20 24 28 31 ...
  ..$ X5: num [1:16] 1 5 10 15 19 20 25 27 30 34 ...
snip long output (no warnings)

--
David Winsemius


On Jan 12, 2009, at 12:07 PM, rafamoral wrote:



I need to store each matrix generated in a loop.

I've been working with the CUSUM algorithm and I've been trying to  
implement

it in R.
What I need to do with my dataset is to create 1000 randomized  
datasets and
cumulative sum them all and store all of those randomized CUSUMed  
datasets
for further analysis and creation of the simulation envelope in the  
CUSUM
chart. But I can't manage to store all of them, the script I've  
written only

stores the last randomized matrix in the variable I called cusumA.

Here's the script I've written:

mat - matrix(data=rep(c(1,2,3,4,5), 16), nrow=16, ncol=5)
# The matrix that will be sampled
A - matrix(data=0, nrow=16, ncol=5)
# The variable that will store the sampled matrix
for(i in 1:1000) {
# I want to do it 1000 times
for(j in 1:nrow(mat)) {
# The number of rows to be sampled)
A[j,] - sample(mat[j,])
# The sample itself - I want to do it 1000 times and then...
cusumA - cumsum(data.frame(A))
# cumulative sum it, and store the 1000 randomized matrices
}}

I've already tried to store it in a null matrix but it will only  
store the
first column of each matrix, so all I got was 1000 first columns.  
The code I

used to do it was

mat - matrix(data=rep(c(1,2,3,4,5), 16), nrow=16, ncol=5)
A - matrix(data=0, nrow=16, ncol=5)
cusumA - matrix()
for(i in 1:1000) {
for(j in 1:nrow(mat)) {
A[j,] - sample(mat[j,])
cusumA[i] - cumsum(data.frame(A))
}}

and I even got 50+ warnings...
Any help'd be really appreciated!
--
View this message in context: 
http://www.nabble.com/Help-with-storage-of-each-matrix-generated-in-a-loop-tp21418758p21418758.html
Sent from the R help mailing list archive at Nabble.com.

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


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Re: [R] lm: how are polynomial functions interpreted?

2009-01-12 Thread Charles C. Berry


On Mon, 12 Jan 2009, c...@witthoft.com wrote:

[nothing deleted]

matplot(1:100, lm(rnorm(100)~poly(1:100,4),x=T)$x ) # for example



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Ahem!


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

..^

Charles C. Berry(858) 534-2098
Dept of Family/Preventive Medicine
E mailto:cbe...@tajo.ucsd.edu   UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901

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Re: [R] R2WinBUGS stopping execution

2009-01-12 Thread Ben Bolker

 Richard.Cotton at hsl.gov.uk writes:

 
 Apologies if this isn't acceptable for the general help list.
 
 I'm running OpenBUGS model via the R2WinBUGS package interface, under 
 Windows.  Is it possible to terminate running models, short of using the 
 Windows Task Manager to forcibly exit the program?
 

  Not that I know of (i.e., probably not).
  
  Ben Bolker

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[R] Two-way repeated measures anova with lme

2009-01-12 Thread Erich Studerus

Dear R-Users,

 

I'm trying to set up a repeated measures anova with two within subjects
factors. I tried it by 3 different anova functions: aov, Anova (from car
package) and lme (from nlme package). I managed to get the same results with
aov and Anova, but the results that I get from lme are slightly different
and I don't figure out why. I guess I did not set up the error structure
correctly. Here's an example, that I analysed with the 3 different anova
functions:

 

set.seed(10)

df-data.frame(Time=as.factor(rep(paste('t',1:3,sep=''),each=36)),Cond=as.fa
ctor(paste('c',1:4,sep='')),Subj=as.factor(rep(paste('s',1:9,sep=''),each=4)
),y=rnorm(108))

 

summary(aov(y ~ (Time*Cond) + Error(Subj/(Time*Cond)), data = df))

 

library(nlme)

anova(lme(y~Time*Cond,random=~1|Subj,data=df))

 

df2-cast(df,Subj~Time+Cond)[-1]

mod-lm(cbind(t1_c1,t1_c2,t1_c3,t1_c4,t2_c1,t2_c2,t2_c3,t2_c4,t3_c1,t3_c2,t3
_c3,t3_c4)~1,df2)

idata=data.frame(Time=gl(3,4),Cond=gl(4,1))

summary(Anova(mod,idata=idata,idesign=~Time*Cond),multivariate=F)

 

 

I would highly appreciate if anyone could explain why the results are
slightly different. I probably spent hours to find an answer on the
internet. I would like to use the lme function, because I need to do
multiple comparisons with ghlt function.

 

Regards,

 

Erich

 


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[R] help on nested mixed effects ANOVA

2009-01-12 Thread Mukta Chakraborty

Hello,
I am trying to run a mixed effects nested ANOVA but none of my codes
are giving me any meaningful results and I am not sure what I am doing
wrong. I am a new user on R and would appreciate some help.
The experimental design is that I have some frogs that have been
exposed to three acoustic Treatments and I am measuring neural
activity (egr), in 12 brain regions. Some frogs also called different
calls than others so I am testing for effect of Call as a fixed
effect.
I have Treatment (Trt), Brain Region (Region), and Call (Call) as
fixed effects and I am modeling Subject (Subj) as a random effect with
brain region nested within subject. I am using lme4 package and I am
not sure if that is the correct one to use.
I also have some missing values as NA as part of the dataset. I have
also set Subj, Region, and Call as factors using as.factor.
I have run the model which I am sure is wrong:
model1-anova(lme(egr ~ Trt* Region, random=~1|Subj/Region,na.action=na.omit))

The output I am getting is:
   numDF   denDFF-value   p-value
 Min.   : 1.00   Min.   : 32   Min.   :  2.891   Min.   :0.000e+00
 1st Qu.: 1.75   1st Qu.:197   1st Qu.: 12.120   1st Qu.:0.000e+00
 Median : 6.50   Median :252   Median : 20.275   Median :1.146e-05
 Mean   : 9.00   Mean   :197   Mean   : 87.353   Mean   :1.383e-05
 3rd Qu.:13.75   3rd Qu.:252   3rd Qu.: 95.508   3rd Qu.:2.530e-05
 Max.   :22.00   Max.   :252   Max.   :305.972   Max.   :3.240e-05

Any help on what I am doing wrong on the coding would be very helpful.
Thanks a lot.
Mukta
-- 
Department of Biology
Wilson Hall, Rm. 316
CB#3280,
University of North Carolina at Chapel Hill
Chapel Hill, NC 27599-3280.
Ph: (919) 962 1445

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[R] RMySQL crashes R

2009-01-12 Thread auburneconomics


This is similar to another current post about RMySQL crashing R, except on
the other post it crashed on connection to the database.  I can successfully
connect, and even can do dbListTables().  But, if I send a query or anything
to actually see the data, R crashes to desktop.  

I have tried this on two computers, one Vista, one XP.  Both crashed.  I am
running 2.8.1, and am using the correct version of DBI and RMySQL.  They are
both loaded with the library() function beforehand.  Versions of MySQL that
it was tried are 5.1 and 6.0a.  

I have searched on the internet and haven't found any answers, and would be
grateful for any help,

Nathan Forczyk
Auburn University
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View this message in context: 
http://www.nabble.com/RMySQL-crashes-R-tp21419908p21419908.html
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[R] merge table rows (\multirow)

2009-01-12 Thread Felipe Carrillo

Hi:
This is what my table should look like:

Month Week   EstpassageMedFL
July-27456634
-28256835
-29328736
-30462337
Aug--31863237
   --32423638
   --33763239
   --34598440
Sept-35125840
-36423541
-37257342
-38423-43
I don't want to reshape my dataset, I basically want a latex table showing only 
one value on the first column instead of repeating the  month over and over. I 
know this can be done with \multirow if the table is created manually. Since 
this table is created dynamically on the fly I don't know how to use \multirow 
like that. Hope is clear. Thanks


--- On Mon, 1/12/09, Dieter Menne dieter.me...@menne-biomed.de wrote:

 From: Dieter Menne dieter.me...@menne-biomed.de
 Subject: Re: [R] merge table rows (\multirow)
 To: r-h...@stat.math.ethz.ch
 Date: Monday, January 12, 2009, 12:41 AM
 Felipe Carrillo mazatlanmexico at yahoo.com
 writes:
 
  I need help merging rows.
  I am trying to merge the 'Month' column using
 \multirow. For example for the
 column 'Week' I want July to be
  merged into one row(weeks 27,28,29,30) and so on for
 the following weeks.
  Below, I am creating a PDF using Sweave,
 MikTex,R-2.8.1 and windows XP to show
 an example.
 
 
 I don't fully understand how the table should look
 like, mainly
 how the Week display should look like. I suggest that you
 try to
 not do the main formatting in latex/xtable, but rather use
 function reshape or package reshape (below). The following
 should give you a starter, I know it is not fully what you
 want.
 
 Dieter
 
 
 sampDat - Month Week Estpassage MedFL
 July 27   665 34
 July 28   2232   35
 July 29   9241   35
 July 30   28464   35
 Aug  31   41049   35
 Aug 32   82216   35
 Aug 33   230411  35
 Aug 34   358541  35
 Sept 35   747839  35
 Sept 36   459682  36
 Sept 37   609567  36
 Sept 38   979475  36
 Sept 39   837189  36
 DF - read.table(textConnection(sampDat), header = TRUE)
 
 library(reshape)
 DFm = melt(DF,id=c(Month,Week))
 cast(DFm,variable~Month+Week)

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[R] rpart with interval censored data crashes R

2009-01-12 Thread Terry Therneau


 Thank you for the input on rpart -- I just saw the message today.
 
 1. You are right, it should not crash.  Why it crashes rpart is simply that I 
(the author) never ever tried using interval censored data in the call.  Real 
users try the most amazing things
  I'll fix it in my local version, but putting in a no no no message.  My 
local version and the R version, maintained by Brian, have drifted quite far 
apart however.
  
  2. Rpart deals with right censored data using the same trick as Cox models, 
by 
thinking of it as observation of a Poisson process; number of events seen over 
a 
given time window.  The fact that the number is always 0 or 1 doesn't hinder 
the 
mathematical trick, which is based in counting process theory.
  BUT - the trick only works for right censored data.
  
  Using the mid points of your intervals is the only approach that comes 
readily 
to mind.
  
Terry Therneau

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Re: [R] RMySQL crashes R

2009-01-12 Thread Duncan Murdoch


On 1/12/2009 1:02 PM, auburneconomics wrote:

This is similar to another current post about RMySQL crashing R, except on
the other post it crashed on connection to the database.  I can successfully
connect, and even can do dbListTables().  But, if I send a query or anything
to actually see the data, R crashes to desktop.  


I have tried this on two computers, one Vista, one XP.  Both crashed.  I am
running 2.8.1, and am using the correct version of DBI and RMySQL.  They are
both loaded with the library() function beforehand.  Versions of MySQL that
it was tried are 5.1 and 6.0a.  


I have searched on the internet and haven't found any answers, and would be
grateful for any help,


I'd recommend using RODBC instead.

The disadvantage is that you'll be at the mercy of whoever wrote the 
ODBC driver, but I believe I've had good experience with the MySQL one.


Duncan Murdoch

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Re: [R] merge table rows (\multirow)

Look at the rgroup and n.rgroup arguments of ?latex
function in Hmisc package or ?print.xtable in xtable package.

On Mon, Jan 12, 2009 at 1:48 PM, Felipe Carrillo
mazatlanmex...@yahoo.com wrote:
 Hi:
 This is what my table should look like:

 Month Week   EstpassageMedFL
 July-27456634
-28256835
-29328736
-30462337
 Aug--31863237
   --32423638
   --33763239
   --34598440
 Sept-35125840
-36423541
-37257342
-38423-43
 I don't want to reshape my dataset, I basically want a latex table showing 
 only one value on the first column instead of repeating the  month over and 
 over. I know this can be done with \multirow if the table is created 
 manually. Since this table is created dynamically on the fly I don't know how 
 to use \multirow like that. Hope is clear. Thanks


 --- On Mon, 1/12/09, Dieter Menne dieter.me...@menne-biomed.de wrote:

 From: Dieter Menne dieter.me...@menne-biomed.de
 Subject: Re: [R] merge table rows (\multirow)
 To: r-h...@stat.math.ethz.ch
 Date: Monday, January 12, 2009, 12:41 AM
 Felipe Carrillo mazatlanmexico at yahoo.com
 writes:

  I need help merging rows.
  I am trying to merge the 'Month' column using
 \multirow. For example for the
 column 'Week' I want July to be
  merged into one row(weeks 27,28,29,30) and so on for
 the following weeks.
  Below, I am creating a PDF using Sweave,
 MikTex,R-2.8.1 and windows XP to show
 an example.


 I don't fully understand how the table should look
 like, mainly
 how the Week display should look like. I suggest that you
 try to
 not do the main formatting in latex/xtable, but rather use
 function reshape or package reshape (below). The following
 should give you a starter, I know it is not fully what you
 want.

 Dieter


 sampDat - Month Week Estpassage MedFL
 July 27   665 34
 July 28   2232   35
 July 29   9241   35
 July 30   28464   35
 Aug  31   41049   35
 Aug 32   82216   35
 Aug 33   230411  35
 Aug 34   358541  35
 Sept 35   747839  35
 Sept 36   459682  36
 Sept 37   609567  36
 Sept 38   979475  36
 Sept 39   837189  36
 DF - read.table(textConnection(sampDat), header = TRUE)

 library(reshape)
 DFm = melt(DF,id=c(Month,Week))
 cast(DFm,variable~Month+Week)

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Re: [R] merge table rows (\multirow)

2009-01-12 Thread Felipe Carrillo

Thanks a lot Dieter, I'll play a little bit with it. Also thanks for the hint 
on how to post a latex reproducible example.

--- On Mon, 1/12/09, Dieter Menne dieter.me...@menne-biomed.de wrote:

 From: Dieter Menne dieter.me...@menne-biomed.de
 Subject: Re: [R] merge table rows (\multirow)
 To: r-help@r-project.org
 Date: Monday, January 12, 2009, 8:32 AM
 Felipe Carrillo wrote:

  I am trying to merge the 'Month' column using
 \multirow. For example for
  the column 'Week' I want July to be merged
 into one row(weeks 27,28,29,30)
  and so on for the following weeks.
  Below, I am creating a PDF using Sweave,
 MikTex,R-2.8.1 and windows XP to
  show an example.

 .. Example removed and partially regenerated below

 Another try; had a bad hour this morning, too many typos. 
 This comes closer to what you want, but not exactly. 
 You might try to manually call format.df which is
 implicitly 
 called in latex, and massage that.
 Again, I do not know if this is easier with xtable.

 If you want to post latex samples, it is not really
 necessary to include it in Sweave. The example
 below works fine for testing.

 Dieter

 library(Hmisc)
 sampDat - Month Week Estpassage MedFL
 July 27  665   34
 July 28  223235
 July 29  924135
 July 30  28464   35
 Aug  31  41049   35
 Aug 32   82216   35
 Aug 33   230411  35
 Aug 34   358541  35
 Sept 35  747839  35
 Sept 36  459682  36
 Sept 37  609567  36
 Sept 38  979475  36
 Sept 39  837189  36
 DF - read.table(textConnection(sampDat), header = TRUE)
 row.names(DF)=DF$Week
 latex(DF[,-2],label=tab:hola,longtable=FALSE,caption='Sample
 table.', rowname=,rgroup=unique(DF$Month),n.rgroup=table(DF$Month))

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 reproducible code.

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[R] Useful books for learning the R software and the S programming language

2009-01-12 Thread Robert Wilk

any useful books for learning the R statistical software?
are they pricey?

and if the books recommended focus on S, how compatible will they be for
someone learning R?

thank you in advance for your help.


P.S.
specialized survey statistical procedures? Is R good at that?

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[R] GarchOxFit Interface

2009-01-12 Thread mohamed elbehi


please send me the 
GarchOxFit Interface
 
 
 
 
thanks
_
[[elided Hotmail spam]]

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[R] Problems with RMySQL and MySQL server version 5.1

2009-01-12 Thread Jeffrey Horner

For those R user's who don't subscribe to R-sig-db and are having 
troubles with the latest RMySQL binary on CRAN, please read the email 
thread at the end of this message.


RMySQL 0.7-2 does work with MySQL 5.1, however the CRAN binary is linked 
against the 5.0 version.


Best,

Jeff

 Original Message 
Subject: Re: [R-sig-DB] Problems with RMySQL and MySQL server version 5.1
Date: Wed, 07 Jan 2009 16:36:48 +
From: Prof Brian Ripley rip...@stats.ox.ac.uk
To: Jeffrey Horner jeff.hor...@vanderbilt.edu
CC: r-sig...@stat.math.ethz.ch
References: 4964cd3d.9000...@vanderbilt.edu

There is a long, long history of precisely this (version mismatches
in MySQL binaries cause crashes).  We had it at 4.x, so I am not
surprised to see it at 5.x now 'x' is no longer unique.

When I distributed binaries, the README on my site pointed this out:
I suggest you ask Uwe to put it in his README, and if possible do a
version check at run time.

Jeffrey Horner wrote:

An FYI to those users having problems with windows RMySQL CRAN binaries.

 Original Message 
Subject: Problems with RMySQL and MySQL server version 5.1
Date: Wed, 7 Jan 2009 09:12:56 +0100
From: Erik Jørgensen erik.jorgen...@agrsci.dk
To: jeff.hor...@vanderbilt.edu

Hi Jeffrey

I have some problems with the use of RMySQL since upgrading to RMySQL
version 0.7-2 and MySQL server version 5.1. The RGUI crashes, when I
e.g. call the function dbGetQuery(conn, sqlstring) (but not e.g.
dbListTables(conn)).

It seems that it is due to incompatibility with MySQL version 5.1.
When I replace the libMySQL.dll file from version 5.1 with the 5.0
version of the file, the problem disappears.

Best regards

Erik Jørgensen
Faculty of Agricultural Sciences,
Aarhus University,
Denmark

PS. R - version 2.8.1 - Windows XP Professional Service Pack3. I
specify the path to the DLL-file manually  via the environment
variable  MYSQL_HOME.




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

--
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Re: [R] RMySQL crashes R

2009-01-12 Thread auburneconomics




Duncan Murdoch-2 wrote:
 
 On 1/12/2009 1:02 PM, auburneconomics wrote:
 This is similar to another current post about RMySQL crashing R, except
 on
 the other post it crashed on connection to the database.  I can
 successfully
 connect, and even can do dbListTables().  But, if I send a query or
 anything
 to actually see the data, R crashes to desktop.  
 
 I have tried this on two computers, one Vista, one XP.  Both crashed.  I
 am
 running 2.8.1, and am using the correct version of DBI and RMySQL.  They
 are
 both loaded with the library() function beforehand.  Versions of MySQL
 that
 it was tried are 5.1 and 6.0a.  
 
 I have searched on the internet and haven't found any answers, and would
 be
 grateful for any help,
 
 I'd recommend using RODBC instead.
 
 The disadvantage is that you'll be at the mercy of whoever wrote the 
 ODBC driver, but I believe I've had good experience with the MySQL one.
 
 Duncan Murdoch
 
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 http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.
 
 

I was hoping to be able to alter a MySQL database within R, which would in
turn immediately update a web page written in PHP.  I am already well versed
in the MySQL/PHP stuff so it would be really neat if I could get this to
work... My hunch is that there is something in a setting somewhere that is
causing this.  I bet its something really stupid (as it always seems to be).

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[R] re tail case-pack ordering problem - can R help?

2009-01-12 Thread gbh22

I'm a programmer, not a mathmatician. I heard about R, and I'm wondering if
anyone can tell me if there is an existing R function that can help with a
problem we're currently trying to find an algorithm for. If R is not the
answer, but you can recommend a known algorithm, that would help a lot!

I'm on a project in a retail corporation, working on a program to assist
retail buyers in creating their orders. The program will be used internally,
and not sold to other companies.

In retail terms, the problem is to determine the best assortment of case
packs to order for a given store, and given a set of desired quantities for
each size of a product. A case pack is a set of sizes that the supplier
assembles. The retailer must order in whole case packs.

For example,

For a given shoe style, in the color brown, a supplier has the following
case packs available.

Sizes -XS S M L XL XXL XXXL
CasePack profiles:
cp1 1 3 5 4 2 1 0
cp2 0 2 2 2 0 0 0
cp3 0 2 3 3 1 0 0
cp4 2 4 5 5 4 1 0

And, for a given store, the retailer has determined the following set the
desired quantites of sizes to purchase. (This determination is based on
sales history, and is not at all based on what the case packs for this item
look like.)

Sizes -XS S M L XL XXL XXXL
4 10 14 12 7 30

The question to be answered is, What set of quantities of the above case
packs will give me the closest distribution to the desired set of sizes?

The output would be a single recommendation for case-pack order quantities,
in a form like the following.

Case packQuantity
cp1 0
cp2 3
cp3 2
cp4 1

There will be some other factors that may operate outside of the above core
problem:

There may be limits on the quantity available on some of the case packs. For
instance, the supplier only has ten available of cp2. It seems this factor
will need to be considered in the core algorithm.
We realize we'll need to apply tolerance levels to the calculation. For
example, for this shoe style, we will accept the solution if it comes to
within 10% (plus or minus) of our desired quantities on each size.
We may also apply a minimum. For example, the solution must have at least 1
of each size having a nonzero quantity.

Thanks for any help!

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Re: [R] Useful books for learning the R software and the S programming language

2009-01-12 Thread Steve_Friedman


Look here to start.

http://www.r-project.org/doc/bib/R-books.html


Steve Friedman Ph. D.
Spatial Statistical Analyst
Everglades and Dry Tortugas National Park
950 N Krome Ave (3rd Floor)
Homestead, Florida 33034

steve_fried...@nps.gov
Office (305) 224 - 4282
Fax (305) 224 - 4147


   
 Robert Wilk 
 dphage...@gmail. 
 com   To 
 Sent by:  r-help@r-project.org
 r-help-boun...@r-  cc 
 project.org   
   Subject 
   [R] Useful books for learning the R 
 01/12/2009 03:36  software and the S programming  
 PM ESTlanguage
   
   
   
   
   
   




any useful books for learning the R statistical software?
are they pricey?

and if the books recommended focus on S, how compatible will they be for
someone learning R?

thank you in advance for your help.


P.S.
specialized survey statistical procedures? Is R good at that?

 [[alternative HTML version deleted]]

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Re: [R] re tail case-pack ordering problem - can R help?

2009-01-12 Thread Stephan Kolassa


Hi gbh22,

your problem looks a lot like a Cutting Stock Problem,
http://en.wikipedia.org/wiki/Cutting_stock_problem

Here, a_ij is your case pack profiles (how many of size j are in case 
pack i), x_i counts how many units of case pack i you want, and q_j 
counts how many units of size j you want. The Cutting Stock Problem then 
minimizes \sum c_ix_i subject to \sum a_ijx_j=q_j for all j, where c_i 
is some penalty parameter for case pack i. Perhaps you could model c_i 
as the price of case pack i, then you would minimize the total cost 
(while disregarding any leftover shoes). I'm sure variations have been 
looked at in the literature. I'd suggest you look at the literature 
cited in the Wikipedia article.


In fact, this problem has been on my radar for a while now, though not 
with high priority. I'm in software for retail, not in retail itself: 
http://www.saf-ag.com. I'd love to discuss any solution approaches with 
you, if you are interested.


Good luck with your case packs!
Stephan Kolassa


gbh22 schrieb:

I'm a programmer, not a mathmatician. I heard about R, and I'm wondering if
anyone can tell me if there is an existing R function that can help with a
problem we're currently trying to find an algorithm for. If R is not the
answer, but you can recommend a known algorithm, that would help a lot!


I'm on a project in a retail corporation, working on a program to assist
retail buyers in creating their orders. The program will be used internally,
and not sold to other companies.


In retail terms, the problem is to determine the best assortment of case
packs to order for a given store, and given a set of desired quantities for
each size of a product. A case pack is a set of sizes that the supplier
assembles. The retailer must order in whole case packs.


For example,

For a given shoe style, in the color brown, a supplier has the following
case packs available.


Sizes -XS   S   M   L  XL  XXL XXXL
CasePack profiles:
cp1  1   3   5   4   2   1   0
cp2  0   2   2   2   0   0   0
cp3  0   2   3   3   1   0   0
cp4  2   4   5   5   4   1   0


And, for a given store, the retailer has determined the following set the
desired quantites of sizes to purchase. (This determination is based on
sales history, and is not at all based on what the case packs for this item
look like.)


Sizes -XS   S   M   L  XL  XXL  XXXL
 4  10  14  12   7   30


The question to be answered is, What set of quantities of the above case
packs will give me the closest distribution to the desired set of sizes?


The output would be a single recommendation for case-pack order quantities,
in a form like the following.


Case packQuantity
cp1 0
cp2 3
cp3 2
cp4 1


There will be some other factors that may operate outside of the above core
problem: 



There may be limits on the quantity available on some of the case packs. For
instance, the supplier only has ten available of cp2. It seems this factor
will need to be considered in the core algorithm.
We realize we'll need to apply tolerance levels to the calculation. For
example, for this shoe style, we will accept the solution if it comes to
within 10% (plus or minus) of our desired quantities on each size.
We may also apply a minimum. For example, the solution must have at least 1
of each size having a nonzero quantity.


Thanks for any help!




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Re: [R] Useful books for learning the R software and the S programming language

2009-01-12 Thread Peter Dalgaard


Robert Wilk wrote:

any useful books for learning the R statistical software?
are they pricey?


Many. Useful depends on the reader, though, so look around. Here's a 
starting point


http://www.r-project.org/doc/bib/R-books.html

(modesty should forbid me to point at item 18 on the list and the fact 
that Amazon US has it currently 19% discounted)


In general R books are cheaper than statistical monographs, but more 
expensive than the large market computer science books.



and if the books recommended focus on S, how compatible will they be for
someone learning R?


Such books are strongly outnumbered by now. One important book from that 
group is Venables+Ripley's Modern Applied Statistics with S explicitly 
addresses R issues.




thank you in advance for your help.


P.S.
specialized survey statistical procedures? Is R good at that?


Not R in itself, but the survey package for it is rumoured to be state 
of the art, and its author has a book on it in its final stages.



--
   O__   Peter Dalgaard Øster Farimagsgade 5, Entr.B
  c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
 (*) \(*) -- University of Copenhagen   Denmark  Ph:  (+45) 35327918
~~ - (p.dalga...@biostat.ku.dk)  FAX: (+45) 35327907

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Re: [R] help on nested mixed effects ANOVA

2009-01-12 Thread Ben Bolker

Mukta Chakraborty chakraborty.mukta at gmail.com writes:

 
 Hello,
 I am trying to run a mixed effects nested ANOVA but none of my codes
 are giving me any meaningful results and I am not sure what I am doing
 wrong. I am a new user on R and would appreciate some help.
 The experimental design is that I have some frogs that have been
 exposed to three acoustic Treatments and I am measuring neural
 activity (egr), in 12 brain regions. Some frogs also called different
 calls than others so I am testing for effect of Call as a fixed
 effect.
 I have Treatment (Trt), Brain Region (Region), and Call (Call) as
 fixed effects and I am modeling Subject (Subj) as a random effect with
 brain region nested within subject. I am using lme4 package and I am
 not sure if that is the correct one to use.
 I also have some missing values as NA as part of the dataset. I have
 also set Subj, Region, and Call as factors using as.factor.
 I have run the model which I am sure is wrong:
 model1-anova(lme(egr ~ Trt* Region, random=~1|Subj/Region,na.action=na.omit))
 
 The output I am getting is:
numDF   denDFF-value   p-value
  Min.   : 1.00   Min.   : 32   Min.   :  2.891   Min.   :0.000e+00
  1st Qu.: 1.75   1st Qu.:197   1st Qu.: 12.120   1st Qu.:0.000e+00
  Median : 6.50   Median :252   Median : 20.275   Median :1.146e-05
  Mean   : 9.00   Mean   :197   Mean   : 87.353   Mean   :1.383e-05
  3rd Qu.:13.75   3rd Qu.:252   3rd Qu.: 95.508   3rd Qu.:2.530e-05
  Max.   :22.00   Max.   :252   Max.   :305.972   Max.   :3.240e-05
 
 Any help on what I am doing wrong on the coding would be very helpful.
 Thanks a lot.
 Mukta


 1.  it would probably be a good idea to ask further questions on this
topic on r-sig-mixed-models instead.

 2. you probably want to look just at the lme() result, and not immediately
wrap it an anova() frame.  I'm a little surprised by your results -- but
since you haven't given a reproducible example I don't know exactly
what's wrong. When I run the lme example in the nlme package and
then do

 anova(fm1)

I get

numDF denDF   F-value p-value
(Intercept) 180 3096.4889  .0001
age 180   85.8464  .0001

which doesn't look like your answer.

3. I don't think you want to use Region as both a random factor and
a fixed factor.  I would have guessed either

lme(egr ~ Trt*Call, random=~1|Subj/Region,na.action=na.omit)) 

OR

lme(egr ~ Trt*Call*Region, random=~1|Subj,na.action=na.omit)) 

I can also imagine that you might be interested in a crossed
design in this case, in which case you should probably use lme4:

library(lme4)
lmer(egr~Trt*Call+(1|Subj)+(1|Region) ...)

[I'm not sure how na.action works for lmer]

 good luck
   Ben Bolker

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[R] roll weekly data to monthly level

2009-01-12 Thread liujb


Dear R users:

I have a data set that looks something like this:
IDtime   y
12/01/20084
12/09/200812
19/01/20088
21/06/20083
23/01/20084
23/09/20089
26/03/20084
31/02/20083
31/10/20088
32/02/20087
32/10/20083

I'd like to sum up the weekly data to the monthly level, so that it looks
something like this:

IDtime   y
12/200816
19/20088
21/20083
23/200813
26/20084
31/200811
32/200810


What is the best way to do it?

Time must be character. How do I truncate a character so that I can remove
the date and only keep the month and year?

Thank you very much in advance.
Julia
-- 
View this message in context: 
http://www.nabble.com/roll-weekly-data-to-monthly-level-tp21424700p21424700.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Useful books for learning the R software and the S programming language

2009-01-12 Thread Carlos Guerra


Robert,

I have Peter's book and I think it can be a very good place to start  
from... dispite the discount... :)


If you like spatial analysis you can try to look for Roger Bivand et  
al. Applied Spatial Data Analysis with R, If you are into something  
else try the Use R collection from Springer... you may find  
something not that pricey that you can use.


Best regards
Carlos

Em 2009/01/12, às 22:07, Peter Dalgaard escreveu:


Robert Wilk wrote:

any useful books for learning the R statistical software?
are they pricey?


Many. Useful depends on the reader, though, so look around. Here's  
a starting point


http://www.r-project.org/doc/bib/R-books.html

(modesty should forbid me to point at item 18 on the list and the  
fact that Amazon US has it currently 19% discounted)


In general R books are cheaper than statistical monographs, but more  
expensive than the large market computer science books.


and if the books recommended focus on S, how compatible will they  
be for

someone learning R?


Such books are strongly outnumbered by now. One important book from  
that group is Venables+Ripley's Modern Applied Statistics with S  
explicitly addresses R issues.



thank you in advance for your help.
P.S.
specialized survey statistical procedures? Is R good at that?


Not R in itself, but the survey package for it is rumoured to be  
state of the art, and its author has a book on it in its final stages.



--
  O__   Peter Dalgaard Øster Farimagsgade 5, Entr.B
 c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen   Denmark  Ph:  (+45)  
35327918
~~ - (p.dalga...@biostat.ku.dk)  FAX: (+45)  
35327907


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Re: [R] Useful books for learning the R software and the S programming language



On Jan 12, 2009, at 3:36 PM, Robert Wilk wrote:


any useful books for learning the R statistical software?
are they pricey?


Compared to medical texts, they are dirt cheap.


and if the books recommended focus on S, how compatible will they be  
for

someone learning R?


Many are available for R these days. It used to be that MASS editions  
1 through 4 by Venables and Ripley were the canonical starting points,  
but in recent years Dalgaard and other have contributed efforts at  
intro and intermediate texts. Cutting and pasting from the R-books page:

http://www.r-project.org/doc/bib/R-books.html

[18]Peter Dalgaard. Introductory Statistics with R.
[49]John Verzani. Using R for Introductory Statistics.
[48]Michael J. Crawley. Statistics: An Introduction using R.
[62] 	William N. Venables and Brian D. Ripley. Modern Applied  
Statistics with S. Fourth Edition. (mentions only S in the title but I  
believe that where differences exist, they are pointed out in this  
edition.)

[63]John Fox. An R and S-Plus Companion to Applied Regression.
[65] 	Frank E. Harrell. Regression Modeling Strategies, with  
Applications to Linear Models, Survival Analysis and Logistic  
Regression. (I don't think Harrell intended this as a tutorial but I  
can't resist a plug. Alzola and Harrell have published a useful book  
length guide:

http://biostat.mc.vanderbilt.edu/twiki/pub/Main/RS/sintro.pdf

And Kuhnert and Venables have also written a book length pdf:
http://cran.r-project.org/doc/contrib/Kuhnert+Venables-R_Course_Notes.zip



thank you in advance for your help.


P.S.
specialized survey statistical procedures? Is R good at that?


See whether Lumley's survey package functions are sufficient.  
Knowledgeable people have opined that they are more complete than what  
is available in SAS.


http://faculty.washington.edu/tlumley/survey/

--
David Winsemius, MD
Heritage Labs





[[alternative HTML version deleted]]

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Re: [R] lm: how are polynomial functions interpreted?

2009-01-12 Thread Carl Witthoft


Well. *_* ,

I think it should have been clear that this was not a question for which 
any code exists.  In fact, I gave two very specific examples of function 
calls.  The entire point of my question was not what's up with my 
(putative) code and data  but rather to try to understand the 
overarching philosophy of the way lm() treats the function it's given.


I do understand the sneaky ways to make it do a linear fit with or 
without forcing the origin.  And, sure, I could have run a data set thru 
a bunch of different quadratic-like functions to try to see what happens.


Let me pick a more complicated example.  The general case of a sin fit 
might be Y = a + b*sin(d*x+phi)  .(where, to be pedantic, x is the only 
data input. All others are coefficients to be found)


If I try  y-lm(yin~I(sin(x))), what is the actual fit function?  And so on.

That's why I was hoping for a more general explanation of what lm() does.



Charles C. Berry wrote:

On Mon, 12 Jan 2009, c...@witthoft.com wrote:

[nothing deleted]

matplot(1:100, lm(rnorm(100)~poly(1:100,4),x=T)$x ) # for example



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


Ahem!


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

..^

Charles C. Berry(858) 534-2098
Dept of Family/Preventive 
Medicine

E mailto:cbe...@tajo.ucsd.eduUC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901





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Re: [R] crash on multiple queries to postgresql db

2009-01-12 Thread Dylan Beaudette

On Friday 09 January 2009, Joe Conway wrote:
 Dylan Beaudette wrote:
  Subsequent calls to:
 
  conn - dbConnect(PgSQL(), host=localhost, dbname=xxx, user=xxx)
  query - dbSendQuery(conn, query_text)
  res - dbGetResult(query)
 
  are resulting in this:
 
  *** glibc detected *** /usr/local/lib/R/bin/exec/R: realloc(): invalid
 
  other attached packages:
  [1] RdbiPgSQL_1.8.0 Rdbi_1.8.0  lattice_0.17-20
 
  Any ideas?

 Well, first off, since you are apparently using RdbiPgSQL from
 bioconductor, you should probably try asking there.

 But in any case, you might try the latest release from bioconductor:
 http://www.bioconductor.org/packages/release/bioc/src/contrib/RdbiPgSQL_1.1
6.0.tar.gz

 or else try the recently released PostgreSQL DBI package on CRAN:
 http://cran.stat.ucla.edu/src/contrib/RPostgreSQL_0.1-3.tar.gz

 HTH,

 Joe

Thanks Joe. I have updated all of my R packages, along with my Bioconductor 
packages... and I still get the odd behavior:

*** glibc detected *** /usr/local/lib/R/bin/exec/R: realloc(): invalid old 
size: 0xb7dad1e0 ***
=== Backtrace: =
/lib/i686/cmov/libc.so.6[0xb7cc3624]
/lib/i686/cmov/libc.so.6[0xb7cc71a1]
/lib/i686/cmov/libc.so.6(realloc+0x106)[0xb7cc7b46]
/usr/local/pgsql/lib/libpq.so.5[0xb7802166]
/usr/local/pgsql/lib/libpq.so.5[0xb780239c]
/usr/local/pgsql/lib/libpq.so.5[0xb77fbd31]
/usr/local/pgsql/lib/libpq.so.5(PQfinish+0x15)[0xb77fbe85]
/usr/local/lib/R/site-library/RdbiPgSQL/libs/RdbiPgSQL.so(PgSQLcloseConnection+0x38)
[0xb783bb98]

Again, here is my session info:

R version 2.8.0 (2008-10-20) 
i686-pc-linux-gnu 

locale:
LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_US.UTF-8;LC_PAPER=en_US.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C

attached base packages:
[1] grid  stats graphics  grDevices utils datasets  methods  
[8] base 

other attached packages:
[1] RdbiPgSQL_1.16.0 Rdbi_1.16.0  lattice_0.17-20


Here is the code that appears to be causing the problems:

# load data from a PostGreSQL database, and return a dataframe
# this will get called whenever the parent function is called
# crashes after 3-5 subsequent calls
load_sensor_data - function(query_text)
{

# conn becomes an object which contains the DB connection:
conn - dbConnect(PgSQL(), host=localhost, dbname=SJER, user=dylan)

## pre-filter out possible bad data:
query - dbSendQuery(conn, query_text)

# fetch data according to query:
res - dbGetResult(query) 

# convert characters to factors
res - transform(res, probe_id = factor(probe_id))
res - transform(res, pedon_id = factor(pedon_id))

# generate a new date-time object in the dataframe
res$date - as.POSIXct(strptime(res$date_time, format=%a %b %e %H:%M:%S %Y))

# cleanup
dbDisconnect(conn)

# doesn't completely fix problem crashing
gc()

# return the resulting dataframe
return(res)
}


Thanks in advance,

Dylan


-- 
Dylan Beaudette
Soil Resource Laboratory
http://casoilresource.lawr.ucdavis.edu/
University of California at Davis
530.754.7341

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Re: [R] lm: how are polynomial functions interpreted?

2009-01-12 Thread roger koenker


[E] Y = a + b*sin(d*x+phi)

isn't a linear model and therefore can't be estimated with lm()  --  
you will need

some heavier artillery.  Linear as in lm() means linear in parameters.

(As it happens, I'm adapting Gordon Smyth's pronyfreq S code for the  
above

problem this afternoon, and have been wondering why someone else hasn't
already done this?  Any clues?


url:www.econ.uiuc.edu/~rogerRoger Koenker
emailrkoen...@uiuc.eduDepartment of Economics
vox: 217-333-4558University of Illinois
fax:   217-244-6678Champaign, IL 61820



On Jan 12, 2009, at 4:57 PM, Carl Witthoft wrote:


Well. *_* ,

I think it should have been clear that this was not a question for  
which any code exists.  In fact, I gave two very specific examples  
of function calls.  The entire point of my question was not what's  
up with my (putative) code and data  but rather to try to  
understand the overarching philosophy of the way lm() treats the  
function it's given.


I do understand the sneaky ways to make it do a linear fit with or  
without forcing the origin.  And, sure, I could have run a data set  
thru a bunch of different quadratic-like functions to try to see  
what happens.


Let me pick a more complicated example.  The general case of a sin  
fit might be Y = a + b*sin(d*x+phi)  .(where, to be pedantic, x is  
the only data input. All others are coefficients to be found)


If I try  y-lm(yin~I(sin(x))), what is the actual fit function?   
And so on.


That's why I was hoping for a more general explanation of what lm()  
does.




Charles C. Berry wrote:

On Mon, 12 Jan 2009, c...@witthoft.com wrote:
[nothing deleted]
matplot(1:100, lm(rnorm(100)~poly(1:100,4),x=T)$x ) # for example


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

Ahem!

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

..^
Charles C. Berry(858) 534-2098
   Dept of Family/ 
Preventive Medicine

E mailto:cbe...@tajo.ucsd.eduUC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego  
92093-0901




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


__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] roll weekly data to monthly level [SEC=UNCLASSIFIED]

2009-01-12 Thread Augusto.Sanabria

Julia,

I had a similar query a while ago which I solved using
a suggestion from Gabor, have a look at:

http://finzi.psych.upenn.edu/R/Rhelp02a/archive/69597.html

Hope it helps,

Augusto



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






-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of liujb
Sent: Tuesday, 13 January 2009 8:58
To: r-help@r-project.org
Subject: [R] roll weekly data to monthly level



Dear R users:

I have a data set that looks something like this:
IDtime   y
12/01/20084
12/09/200812
19/01/20088
21/06/20083
23/01/20084
23/09/20089
26/03/20084
31/02/20083
31/10/20088
32/02/20087
32/10/20083

I'd like to sum up the weekly data to the monthly level, so that it looks 
something like this:

IDtime   y
12/200816
19/20088
21/20083
23/200813
26/20084
31/200811
32/200810


What is the best way to do it?

Time must be character. How do I truncate a character so that I can remove the 
date and only keep the month and year?

Thank you very much in advance.
Julia
--
View this message in context: 
http://www.nabble.com/roll-weekly-data-to-monthly-level-tp21424700p21424700.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] roll weekly data to monthly level

I had trouble getting my output to look like yours until I realized  
that you did not want to sum up the weekly data to the monthly level  
but rather to sum up to the monthly *and* ID level.


 dftag-aggregate(dft$y, list(ID=dft$ID, Month=as.yearmon(dft$time,  
%m/%d/%Y)), FUN=sum)

# in Month order rather than ID, Month order
 dftag[order(dftag$ID, dftag$Month), ]
  IDMonth  x
3  1 Feb 2008 16
7  1 Sep 2008  8
1  2 Jan 2008  3
5  2 Mar 2008 13
6  2 Jun 2008  4
2  3 Jan 2008 11
4  3 Feb 2008 10

--
David Winsemius

On Jan 12, 2009, at 4:57 PM, liujb wrote:



Dear R users:

I have a data set that looks something like this:
IDtime   y
12/01/20084
12/09/200812
19/01/20088
21/06/20083
23/01/20084
23/09/20089
26/03/20084
31/02/20083
31/10/20088
32/02/20087
32/10/20083

I'd like to sum up the weekly data to the monthly level, so that it  
looks

something like this:

IDtime   y
12/200816
19/20088
21/20083
23/200813
26/20084
31/200811
32/200810


What is the best way to do it?

Time must be character. How do I truncate a character so that I can  
remove

the date and only keep the month and year?

Thank you very much in advance.
Julia
--
View this message in context: 
http://www.nabble.com/roll-weekly-data-to-monthly-level-tp21424700p21424700.html
Sent from the R help mailing list archive at Nabble.com.

__
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Re: [R] lm: how are polynomial functions interpreted?

2009-01-12 Thread Bill.Venables

I find the simplest way to interpret a *linear* model formula, as used by lm() 
and aov() is to take the left hand side as specifying the response variable (or 
variables) and to take the right hand side as specifying the *columns of the 
model matrix* in a coded way.  Notice that the parameters are implicit and do 
not occur anywhere in the formula.

To take your example, yin ~ I(sin(x)), [which you could simply write as yin ~ 
sin(x)] would specify yin as the response and the model matrix had two columns 
namely 1 for the intercept term and sin(x)

X = [1 sin(x)]

So the model you would be fitting, in a more conventional notation, would be 

Y = a + b*sin(x) + error.  

lm() and aov() accommodate only *linear* parameters.  You can recoginse a 
linear parameter by the fact that when you differentiate the right hand side of 
the model formula with respect to it, the result does not depend on that 
parameter.

To take your other model

Y = a + b*sin(d*x + phi) + error

(you left out the error, BTW), clearly a and b are linear parameters but d and 
phi are not, so you cannot fit this model directly with lm() or aov().  If you 
knew d and phi, of course, you could fit it since the remaining parameters are 
all linear and you would specify it using  y ~ sin(d*x+phi)  where d and phi 
would need to have values at the time of fitting.

The simplest way to fit this kind of model is to use nls().  You can even 
exploit the fact that a and b are linear parameters by using the plinear 
algorithm, but I'll leave you to sort that one out.  You can also re-write the 
model so that you have just one non-linear parameter, but again, you can sort 
that out.

___

I think the reason why people were perhaps looking a little askance at your 
this kind of question on R help is that there are plenty of books around where 
this sort of issue is really done to death.  The introduction to R from the 
help menu of R is one place where you might start, but there are better ones 
and now plenty of them.

Bill Venables
http://www.cmis.csiro.au/bill.venables/ 


-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of Carl Witthoft
Sent: Tuesday, 13 January 2009 8:58 AM
To: r-help@r-project.org
Subject: Re: [R] lm: how are polynomial functions interpreted?

Well. *_* ,

I think it should have been clear that this was not a question for which 
any code exists.  In fact, I gave two very specific examples of function 
calls.  The entire point of my question was not what's up with my 
(putative) code and data  but rather to try to understand the 
overarching philosophy of the way lm() treats the function it's given.

I do understand the sneaky ways to make it do a linear fit with or 
without forcing the origin.  And, sure, I could have run a data set thru 
a bunch of different quadratic-like functions to try to see what happens.

Let me pick a more complicated example.  The general case of a sin fit 
might be Y = a + b*sin(d*x+phi)  .(where, to be pedantic, x is the only 
data input. All others are coefficients to be found)

If I try  y-lm(yin~I(sin(x))), what is the actual fit function?  And so on.

That's why I was hoping for a more general explanation of what lm() does.



Charles C. Berry wrote:
 On Mon, 12 Jan 2009, c...@witthoft.com wrote:
 
 [nothing deleted]
 
 matplot(1:100, lm(rnorm(100)~poly(1:100,4),x=T)$x ) # for example
 

 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide 
 http://www.R-project.org/posting-guide.html
 
 Ahem!
 
 and provide commented, minimal, self-contained, reproducible code.
 ..^
 
 Charles C. Berry(858) 534-2098
 Dept of Family/Preventive 
 Medicine
 E mailto:cbe...@tajo.ucsd.eduUC San Diego
 http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901
 
 


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Re: [R] roll weekly data to monthly level

Try this. First we read in the data.  In reality you would use
the commented out read line.   Then use chron to convert
the times to chron dates and as.yearmon from zoo to convert
them to year/months.  Finally aggregate and sort.

 Lines - IDtime   y
+ 12/01/20084
+ 12/09/200812
+ 19/01/20088
+ 21/06/20083
+ 23/01/20084
+ 23/09/20089
+ 26/03/20084
+ 31/02/20083
+ 31/10/20088
+ 32/02/20087
+ 32/10/20083

 library(zoo) # as.yearmon
 library(chron) # chron

 # DF - read.table(myfile.dat, header = TRUE, as.is = TRUE)
 DF - read.table(textConnection(Lines), header = TRUE, as.is = TRUE)
 DF$yearmon - as.yearmon(chron(DF$time))
 res - aggregate(DF[y], DF[c(ID, yearmon)], sum)
 res[order(res[, 1], res[, 2]), ]
  ID  yearmon  y
3  1 Feb 2008 16
7  1 Sep 2008  8
1  2 Jan 2008  3
5  2 Mar 2008 13
6  2 Jun 2008  4
2  3 Jan 2008 11
4  3 Feb 2008 10



On Mon, Jan 12, 2009 at 4:57 PM, liujb liujul...@yahoo.com wrote:

 Dear R users:

 I have a data set that looks something like this:
 IDtime   y
 12/01/20084
 12/09/200812
 19/01/20088
 21/06/20083
 23/01/20084
 23/09/20089
 26/03/20084
 31/02/20083
 31/10/20088
 32/02/20087
 32/10/20083

 I'd like to sum up the weekly data to the monthly level, so that it looks
 something like this:

 IDtime   y
 12/200816
 19/20088
 21/20083
 23/200813
 26/20084
 31/200811
 32/200810


 What is the best way to do it?

 Time must be character. How do I truncate a character so that I can remove
 the date and only keep the month and year?

 Thank you very much in advance.
 Julia
 --
 View this message in context: 
 http://www.nabble.com/roll-weekly-data-to-monthly-level-tp21424700p21424700.html
 Sent from the R help mailing list archive at Nabble.com.

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


__
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[R] pmax and sort?

2009-01-12 Thread rkevinburton

I am having a hard time understanding the documentation and I was wondering if 
there would be someone to help clear the cobwebs. 

The documentation for pmax states:

pmax and pmin take one or more vectors (or matrices) as arguments and return a 
single vector giving the ‘parallel’ maxima (or minima) of the vectors. The 
first element of the result is the maximum (minimum) of the first elements of 
all the arguments, the second element of the result is the maximum (minimum) of 
the second elements of all the arguments and so on. Shorter inputs are recycled 
if necessary. attributes (such as names or dim) are transferred from the first 
argument (if applicable). 

This descirption seems to me like what sort would return. Would someone please 
help me understand the differenece between pmax and sort?

Thank you.

Kevin

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[R] pmax and sort?

2009-01-12 Thread rkevinburton

I am having a hard time understanding the documentation and I was wondering if 
there would be someone to help clear the cobwebs. 

The documentation for pmax states:

pmax and pmin take one or more vectors (or matrices) as arguments and return a 
single vector giving the ‘parallel’ maxima (or minima) of the vectors. The 
first element of the result is the maximum (minimum) of the first elements of 
all the arguments, the second element of the result is the maximum (minimum) of 
the second elements of all the arguments and so on. Shorter inputs are recycled 
if necessary. attributes (such as names or dim) are transferred from the first 
argument (if applicable). 

This descirption seems to me like what sort would return. Would someone please 
help me understand the differenece between pmax and sort?

Thank you.

Kevin

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[R] Drawing Polygons with xyplot in lattice

2009-01-12 Thread Josip Dasovic

Hello:

I've come to a dead-end in my search for a solution to a graphing problem that 
I am encountering. I have used xyplot (from the lattice package) successfully 
to plot 36 time-series plots (lines) of under-5 mortality for a set of 
countries in Sub-Sarahan Africa. 

What I would now like to do is to add vertical bands (rectangles) that 
correspond to time-periods during which each of the 36 countries was at war, 
which would be different for each of the 36 panels. The effect I'd like to 
produce is similar that found here: 
http://thinkorthwim.com/2007/09/12/recession/ (but with 36 panels, each of 
which has a different period(s) shaded).

I've done this successfully one plot at a time with the simple plot commmand 
(using polygon after having first drawn the line plot), but am having 
difficulty replicating it with xyplot.

Here is some sample code (with 4 countries and 10 years), which is as far as 
I've been able to get:


library(lattice)
df-data.frame(c(u5cmr=round(runif(40)*100)),country=c(rep(Angola,10), 
rep(Benin,10), rep(Botswana,10), rep(Burkina Faso,10)), 
year=rep(1991:2000,4))
names(df)-c(u5cmr, country, year)
xyplot(u5cmr ~ year | country , ylab = U5 cmr (per thousand), data = df, type 
= l, lty = 1, col =red, xlab=Year, as.table=T, lwd=2)


R version 2.7.2 Patched (2008-09-20 r47259)
on Mac OS 10.5

Thanks,
Josip

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Drawing Polygons with xyplot in lattice

There is an example in the example section of the ?xyplot.zoo help page
in the zoo package.

On Mon, Jan 12, 2009 at 8:16 PM, Josip Dasovic j...@sfu.ca wrote:
 Hello:

 I've come to a dead-end in my search for a solution to a graphing problem 
 that I am encountering. I have used xyplot (from the lattice package) 
 successfully to plot 36 time-series plots (lines) of under-5 mortality for a 
 set of countries in Sub-Sarahan Africa.

 What I would now like to do is to add vertical bands (rectangles) that 
 correspond to time-periods during which each of the 36 countries was at war, 
 which would be different for each of the 36 panels. The effect I'd like to 
 produce is similar that found here: 
 http://thinkorthwim.com/2007/09/12/recession/ (but with 36 panels, each of 
 which has a different period(s) shaded).

 I've done this successfully one plot at a time with the simple plot commmand 
 (using polygon after having first drawn the line plot), but am having 
 difficulty replicating it with xyplot.

 Here is some sample code (with 4 countries and 10 years), which is as far as 
 I've been able to get:


 library(lattice)
 df-data.frame(c(u5cmr=round(runif(40)*100)),country=c(rep(Angola,10), 
 rep(Benin,10), rep(Botswana,10), rep(Burkina Faso,10)), 
 year=rep(1991:2000,4))
 names(df)-c(u5cmr, country, year)
 xyplot(u5cmr ~ year | country , ylab = U5 cmr (per thousand), data = df, 
 type = l, lty = 1, col =red, xlab=Year, as.table=T, lwd=2)


 R version 2.7.2 Patched (2008-09-20 r47259)
 on Mac OS 10.5

 Thanks,
 Josip

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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Can't Destroy Dim Names

Dear all,

I have the following matrix:

 str(mat)
Named chr [1:32268] yQAAA jQAAQ UQAAg FQAAw 1QABA ...
 - attr(*, names)= chr [1:32268] CA CC
CG CT ...


I want to destroy the attribute yielding only this:

 str(mat)
Named chr [1:32268] yQAAA jQAAQ UQAAg FQAAw 1QABA ...


But why this command fail to do it?

 dimnames(mat) - NULL

Btw, the object size for keeping matrix with dim names is greater
than without, right?



- Gundala Viswanath
Jakarta - Indonesia

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] pmax and sort?

2009-01-12 Thread Greg Snow

Here are a couple of quick examples that may help:

 1:10
 [1]  1  2  3  4  5  6  7  8  9 10
 10:1
 [1] 10  9  8  7  6  5  4  3  2  1
 pmax( 10:1, 1:10 )
 [1] 10  9  8  7  6  6  7  8  9 10
 pmin( 1:10, 5 )
 [1] 1 2 3 4 5 5 5 5 5 5

In the first example with pmax, there are 2 vectors being compared (10-1, 
1-10), so first 10 is compared to 1 and since 10 is larger it is the first 
element of the returned vector, then 9 is compared to 2 and 9 becomes the 
second element, this continues till the 10th comparison is 1 vs. 10 and 10 is 
the final element.  The second (pmin) example shows the recycling, each of the 
numbers 1-10 is compared to the number 5, in each case the smaller of the pair 
is returned (1-4 for the first 4 elements, then 5 for the rest).

Hope this helps,

-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
 project.org] On Behalf Of rkevinbur...@charter.net
 Sent: Monday, January 12, 2009 5:55 PM
 To: r-help@r-project.org
 Subject: [R] pmax and sort?
 
 I am having a hard time understanding the documentation and I was
 wondering if there would be someone to help clear the cobwebs.
 
 The documentation for pmax states:
 
 pmax and pmin take one or more vectors (or matrices) as arguments and
 return a single vector giving the ‘parallel’ maxima (or minima) of the
 vectors. The first element of the result is the maximum (minimum) of
 the first elements of all the arguments, the second element of the
 result is the maximum (minimum) of the second elements of all the
 arguments and so on. Shorter inputs are recycled if necessary.
 attributes (such as names or dim) are transferred from the first
 argument (if applicable).
 
 This descirption seems to me like what sort would return. Would someone
 please help me understand the differenece between pmax and sort?
 
 Thank you.
 
 Kevin
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-
 guide.html
 and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Can't Destroy Dim Names

Is this what you want:

 x - c(a=1, b=2, c=3)
 str(x)
 Named num [1:3] 1 2 3
 - attr(*, names)= chr [1:3] a b c
 names(x) - NULL
 x
[1] 1 2 3


On Mon, Jan 12, 2009 at 8:32 PM, Gundala Viswanath gunda...@gmail.com wrote:
 Dear all,

 I have the following matrix:

 str(mat)
 Named chr [1:32268] yQAAA jQAAQ UQAAg FQAAw 1QABA ...
  - attr(*, names)= chr [1:32268] CA CC
 CG CT ...


 I want to destroy the attribute yielding only this:

 str(mat)
 Named chr [1:32268] yQAAA jQAAQ UQAAg FQAAw 1QABA ...


 But why this command fail to do it?

 dimnames(mat) - NULL

 Btw, the object size for keeping matrix with dim names is greater
 than without, right?



 - Gundala Viswanath
 Jakarta - Indonesia

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




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

What is the problem that you are trying to solve?

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Can't Destroy Dim Names

2009-01-12 Thread Greg Snow

Your mat variable is not a matrix, but a vector (a named vector), therefore it 
does not have dimensions or dimnames.

Try names(mat) - NULL

Hope this helps,

-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
 project.org] On Behalf Of Gundala Viswanath
 Sent: Monday, January 12, 2009 6:32 PM
 To: r-h...@stat.math.ethz.ch
 Subject: [R] Can't Destroy Dim Names
 
 Dear all,
 
 I have the following matrix:
 
  str(mat)
 Named chr [1:32268] yQAAA jQAAQ UQAAg FQAAw 1QABA ...
  - attr(*, names)= chr [1:32268] CA CC
 CG CT ...
 
 
 I want to destroy the attribute yielding only this:
 
  str(mat)
 Named chr [1:32268] yQAAA jQAAQ UQAAg FQAAw 1QABA ...
 
 
 But why this command fail to do it?
 
  dimnames(mat) - NULL
 
 Btw, the object size for keeping matrix with dim names is greater
 than without, right?
 
 
 
 - Gundala Viswanath
 Jakarta - Indonesia
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-
 guide.html
 and provide commented, minimal, self-contained, reproducible code.

__
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Re: [R] lm: how are polynomial functions interpreted?

2009-01-12 Thread Carl Witthoft


damn.
My apologies to everyone -- I sent one message and it got destroyed somehow.

Here's the part that was missing, which led to rather a lot of confusion 
on all parts.


The two recent responses to a question about lm suggested

1)  lm(y~poly(x,2))

2) lm(y~I(x^2))

So my question was *supposed* to be related to how lm operated 
differently (if at all) on these two different 'versions' of a quadratic 
fit.  My gut reaction was that y~I(x^2) would not be the same as y~f(x) 
where f(x) is a+bx+cx^2 .


So what I was trying to find out was just how lm() deals with various 
definitions of the orthogonal polynomials its presented with.  Another 
way, maybe, to ask, is: how does one specify to fit exactly to


  a + bx +cx^2  vs

  bx + cx^2   vs
  cx^2
?

Thanks and apologies again to all the people who quite properly 
misunderstood what I was harping on due to the munging of my first post.


Carl


David Winsemius wrote:


On Jan 12, 2009, at 5:57 PM, Carl Witthoft wrote:


Well. *_* ,

I think it should have been clear that this was not a question for 
which any code exists.  In fact, I gave two very specific examples of 
function calls.


Huh? I think most of the readers of the list saw a basically empty 
message body. That was the point of Berry's [nothing deleted]. If you 
are under the belief that this is a continuation of an earlier question, 
then it may not be threading up in the manner you hoped for.




__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Fast way to finding index in Vector

Dear all,

Suppose I have the following vector as repository:

 repo - c(AAA, AAT, AAC, AAG, ATA,ATT)

Given another query vector

 qr - c(AAC, ATT)

is there a way I can find the query index in repository in a fast way.

Giving:

[1] 3 6

Typically the size of  repo is around ~12million element, and
query around ~1 million element.


- Gundala Viswanath
Jakarta - Indonesia

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Re: [R] Help needed for Loading tm package

2009-01-12 Thread Kum-Hoe Hwang

I appreciate alll helpers for R.

I have struggled with Rweka problem related with tm text mining package in R.

My problem was solved when I replaced RWeka_0.3-15.zip with
RWeka_0.3-13.zip under the OS of Win XP.
Now my tm package works fine in R. I don't know why it works.

Thanks all who gave me good help

Kum Hwang Ph.D.

On Mon, Jan 12, 2009 at 5:19 PM, Prof Brian Ripley
rip...@stats.ox.ac.uk wrote:
 How do you account for the fact that the package passed its checks on the
 build machine, e.g.

 http://cran.r-project.org/bin/windows/contrib/2.8/check/RWeka-check.log

 and no one else is reporting this?  (And it does work for me.)

 Blaming your tools is easy, but doing so on a public forum is disrespectful
 and against the rw-FAQ

 http://cran.r-project.org/bin/windows/base/rw-FAQ.html#Can-I-install-packages-into-libraries-in-this-version_003f

 (note the word 'unsupported').

 Uwe Ligges does a great job helping Windows' users by building binary
 packages, and deserves your thanks, not vague and unsubstantiated blame.

 On Mon, 12 Jan 2009, Kum-Hoe Hwang wrote:

 Thank Prof. Brian Ripley for your comments.

 Based on Prof Brian Ripley's comments, I checked Java environments in
 my PC. But I have not solved a tm package problem in Win R software.
 I am not sure but my current conclusion is that the Win-based R binary
 software has definitely a problem with Rweka package or subpackage.

 Should I wait for another upgraded R binary version under the window
 XP, tm' package or Rweka ets?
 Or I'd better migrate to non-window OS such as Linux, etc?

 You definitely shou;d read and follow the FAQ, as this could well be a Java
 incompatibility with the binary package.

 Kum Hwang, Ph.D.

 On Sat, Jan 10, 2009 at 8:06 PM, Prof Brian Ripley
 rip...@stats.ox.ac.uk wrote:

 On Sat, 10 Jan 2009, Kum-Hoe Hwang wrote:

 Howdy Gurus again

 Thanks to  Tony.Breyal, I was able to writing the following script for
 analyzing a text document.
 But I got an error with tm' package. I don't why I got the error from
 the
 R
 script below. I think I followed proccess of R tm manual.

 Please do read the messages you got.  I see

 Error in .jinit(system.file(jar, c(weka.jar, RWeka.jar), package =
 pkgname,  :
  Cannot create Java virtual machine (-1)

 so the problem is with your Java installation and RWeka, not 'tm'.

 First make sure you have a working installation of RWeka -- I suspect you
 do
 not even have Java installed, but it could be a version or path issue
 (but
 very unlikely to be an R issue).


 I use R v2.8.1. and tm_0.3-3.zip under Win XP.

 Thanks in advance,

 Kum Hwang

 # setting directory
 my.path -'C:\\_work\\Daddys\\myProjects\\2009

 defaultProject\\R\\textfile\\'

 # text miner pakacge
 library(tm)

 Loading required package: Snowball
 Loading required package: RWeka
 -
 Error in .jinit(system.file(jar, c(weka.jar, RWeka.jar), package =
 pkgname,  :
  Cannot create Java virtual machine (-1)
 Error : .onLoad failed in 'loadNamespace' for 'RWeka'
 Error: package 'RWeka' could not be loaded

 my.corpurs -Corpus(DirSource(my.path), readerControl =

 list(reader=readPlain))
 Error: could not find function Corpus

 my.tdm - TermDocMatrix(my.corpus)

 Error: could not find function TermDocMatrix

 my.tdm[1,]

 Error: object my.tdm not found


 --
 Kum-Hoe Hwang, Ph.D.

 Phone : 82-31-250-3516
 Email : phdhw...@gmail.com

   [[alternative HTML version deleted]]

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


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




 --
 Kum-Hoe Hwang, Ph.D.

 Phone : 82-31-250-3516
 Email : phdhw...@gmail.com


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




-- 
Kum-Hoe Hwang, Ph.D.

Phone : 82-31-250-3516
Email : phdhw...@gmail.com

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

Re: [R] Fast way to finding index in Vector

Hi Jorge and all,

How can I modified your code when

query size can be bigger than repository,
meaning that it can contain repeats.

e.g. qr - c(AAC, ATT, ATT,AAC, ATT, ATT, AAT, ATT, ATT,  )


Sorry, I should have mentioned this earlier.


- Gundala Viswanath
Jakarta - Indonesia



On Tue, Jan 13, 2009 at 11:11 AM, Jorge Ivan Velez
jorgeivanve...@gmail.com wrote:

 Perhaps
 which(repo%in%qr)
 ?
 HTH,

 Jorge


 On Mon, Jan 12, 2009 at 9:07 PM, Gundala Viswanath gunda...@gmail.com
 wrote:

 Dear all,

 Suppose I have the following vector as repository:

  repo - c(AAA, AAT, AAC, AAG, ATA,ATT)

 Given another query vector

  qr - c(AAC, ATT)

 is there a way I can find the query index in repository in a fast way.

 Giving:

 [1] 3 6

 Typically the size of  repo is around ~12million element, and
 query around ~1 million element.


 - Gundala Viswanath
 Jakarta - Indonesia

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



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

[R] polychoric correlation: issue with coefficient sign

2009-01-12 Thread Dorothee

Hello,

I am running polychoric correlations on a dataset composed of 12 ordinal and
binary variables (N =384), using the polycor package.
One of the association (between 2 dichotomous variables) is very high using
the 2-step estimate (0.933 when polychoric run only between the two
variables; but 0.801 when polychoric run on the 12 variables). The same
correlation run with ML estimate returns a singularity message.

First, I would like to know why the estimations between only the two
dichotomous variables and with all the variables at once (with the 2-step
estimate) returns slightly different results.

Secondly, when i checked back the distribution of these two dichotomous
variables they appear about symmetrically opposed. Therefore, one should
indeed expect a strong association between them, but a negative one, isn't
it? Why does the polychoric correlation returns a positive coefficient? What
does it mean for the rest of the coefficients, should i trust them?

I have to say I'm new to R and not very strong in statistics, I hope I
haven't posted a stupid question...

cheers,
Dorothee
--
View this message in context:
http://www.nabble.com/polychoric-correlation%3A-issue-with-coefficient-sign-tp21425977p21425977.html
Sent from the R help mailing list archive at Nabble.com.

Re: [R] lm: how are polynomial functions interpreted?

Try this to see the X matrix that its fitting y to:

x - 1:10
model.matrix(~ poly(x, 2))
model.matrix(~ I(x^2))



On Mon, Jan 12, 2009 at 9:01 PM, Carl Witthoft c...@witthoft.com wrote:
 damn.
 My apologies to everyone -- I sent one message and it got destroyed somehow.

 Here's the part that was missing, which led to rather a lot of confusion on
 all parts.

 The two recent responses to a question about lm suggested

 1)  lm(y~poly(x,2))

 2) lm(y~I(x^2))

 So my question was *supposed* to be related to how lm operated differently
 (if at all) on these two different 'versions' of a quadratic fit.  My gut
 reaction was that y~I(x^2) would not be the same as y~f(x) where f(x) is
 a+bx+cx^2 .

 So what I was trying to find out was just how lm() deals with various
 definitions of the orthogonal polynomials its presented with.  Another way,
 maybe, to ask, is: how does one specify to fit exactly to

  a + bx +cx^2  vs

  bx + cx^2   vs
  cx^2
 ?

 Thanks and apologies again to all the people who quite properly
 misunderstood what I was harping on due to the munging of my first post.

 Carl


 David Winsemius wrote:

 On Jan 12, 2009, at 5:57 PM, Carl Witthoft wrote:

 Well. *_* ,

 I think it should have been clear that this was not a question for which
 any code exists.  In fact, I gave two very specific examples of function
 calls.

 Huh? I think most of the readers of the list saw a basically empty message
 body. That was the point of Berry's [nothing deleted]. If you are under
 the belief that this is a continuation of an earlier question, then it may
 not be threading up in the manner you hoped for.


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


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

[R] Returning Non-Unique Index with Which (alternatives?)

Dear all,


I tried to find index in repo given  a query with this:

 repo - c(AAA, AAT, AAC, AAG, ATA, ATT)
 qr - c(AAC, ATT, ATT)
 which(repo%in%qr)
[1] 3 6


Note that the query contain repeating elements, yet
the output of which only returns unique.

How can I make it returning

[1] 3 6 6

instead?


- Gundala Viswanath
Jakarta - Indonesia

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

Re: [R] Fast way to finding index in Vector

Is this what you want:

 repo - c(AAA, AAT, AAC, AAG, ATA,ATT)
 qr - c(AAC, ATT, ATT,AAC, ATT, ATT, AAT, ATT, ATT)
 match(qr, repo)
[1] 3 6 6 3 6 6 2 6 6




On Mon, Jan 12, 2009 at 9:22 PM, Gundala Viswanath gunda...@gmail.com wrote:
 Hi Jorge and all,

 How can I modified your code when

 query size can be bigger than repository,
 meaning that it can contain repeats.

 e.g. qr - c(AAC, ATT, ATT,AAC, ATT, ATT, AAT, ATT, ATT,  )


 Sorry, I should have mentioned this earlier.


 - Gundala Viswanath
 Jakarta - Indonesia



 On Tue, Jan 13, 2009 at 11:11 AM, Jorge Ivan Velez
 jorgeivanve...@gmail.com wrote:

 Perhaps
 which(repo%in%qr)
 ?
 HTH,

 Jorge


 On Mon, Jan 12, 2009 at 9:07 PM, Gundala Viswanath gunda...@gmail.com
 wrote:

 Dear all,

 Suppose I have the following vector as repository:

  repo - c(AAA, AAT, AAC, AAG, ATA,ATT)

 Given another query vector

  qr - c(AAC, ATT)

 is there a way I can find the query index in repository in a fast way.

 Giving:

 [1] 3 6

 Typically the size of  repo is around ~12million element, and
 query around ~1 million element.


 - Gundala Viswanath
 Jakarta - Indonesia

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



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




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

What is the problem that you are trying to solve?

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

Re: [R] Fast way to finding index in Vector

Yes Jim, exactly.

BTW, I found from ?match

 Matching for lists is potentially very slow and best avoided
 except in simple cases.

Since I am doing this for million of tags. Is there a faster alternatives?


- Gundala Viswanath
Jakarta - Indonesia



On Tue, Jan 13, 2009 at 12:14 PM, jim holtman jholt...@gmail.com wrote:
 Is this what you want:

 repo - c(AAA, AAT, AAC, AAG, ATA,ATT)
 qr - c(AAC, ATT, ATT,AAC, ATT, ATT, AAT, ATT, ATT)
 match(qr, repo)
 [1] 3 6 6 3 6 6 2 6 6




 On Mon, Jan 12, 2009 at 9:22 PM, Gundala Viswanath gunda...@gmail.com wrote:
 Hi Jorge and all,

 How can I modified your code when

 query size can be bigger than repository,
 meaning that it can contain repeats.

 e.g. qr - c(AAC, ATT, ATT,AAC, ATT, ATT, AAT, ATT, ATT,  )


 Sorry, I should have mentioned this earlier.


 - Gundala Viswanath
 Jakarta - Indonesia



 On Tue, Jan 13, 2009 at 11:11 AM, Jorge Ivan Velez
 jorgeivanve...@gmail.com wrote:

 Perhaps
 which(repo%in%qr)
 ?
 HTH,

 Jorge


 On Mon, Jan 12, 2009 at 9:07 PM, Gundala Viswanath gunda...@gmail.com
 wrote:

 Dear all,

 Suppose I have the following vector as repository:

  repo - c(AAA, AAT, AAC, AAG, ATA,ATT)

 Given another query vector

  qr - c(AAC, ATT)

 is there a way I can find the query index in repository in a fast way.

 Giving:

 [1] 3 6

 Typically the size of  repo is around ~12million element, and
 query around ~1 million element.


 - Gundala Viswanath
 Jakarta - Indonesia

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



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




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

 What is the problem that you are trying to solve?


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

Re: [R] Fast way to finding index in Vector