Re: [R] Start Model for POLYCLASS

2006-07-11 Thread Xiaogang Su
Dear Prof. Kooperberg,

Thanks a lot for your clarification! That is very helpful. I will check
out the polynomial
spline functions in S-Plus that you mentioned. 

Best regards,
-XG Su



Xiaogang Su,  Assistant Professor
Department of Statistics and Actuarial Science
University of Central Florida
Orlando, FL 32816
(407) 823-2940 [O]
[EMAIL PROTECTED]
http://pegasus.cc.ucf.edu/~xsu/

 [EMAIL PROTECTED] 07/10/06 4:01 PM 
That's correct. The option is not there.
You can force it in a starting model - but it can still be deleted,

At some stage Insightful implemented a commercial set of polynomial
spline functions in S-Plus that does have that option. They got it
as far as a beta-library [quoting from my own homepage
http://bear.fhcrc.org/~clk]:

 Doug Clarkson at Insightful Corp. has implemented commercial versions
of hare,
heft, logspline and a variety of related methods, with lots of
additional
features and options. See
http://www.insightful.com/downloads/libraries/default.asp. Look for
S+BEST. 

There are remaining bugs in this code Doug Clarkson is no longer
at Insightful, and the code seems to be pretty much abbandoned.

Charles

Quoting Xiaogang Su [EMAIL PROTECTED]:

 Dear Dr. Graves, 
 
 Thanks for your information and kindness! I really appreciate it.
 According to my findings so far, POLYCLASS doesnot seem to allow for
 variables being forced into the model. Even the commercialized MARS
 doesnot have this function either. However, one ad hoc approach that
 they suggested is to fit linear (or logistic) model and form rediduals
 first and then run MARS with the residuals.  
 
 Best regards,
 -XG Su 
 
 
 
 Xiaogang Su,  Assistant Professor
 Department of Statistics and Actuarial Science
 University of Central Florida
 Orlando, FL 32816
 (407) 823-2940 [O]
 [EMAIL PROTECTED]
 http://pegasus.cc.ucf.edu/~xsu/
 
 
  Spencer Graves [EMAIL PROTECTED] 7/7/2006 5:49:05 AM 
 I have not seen a reply, so I will offer a few suggestions,
 even 
 though I've never used the 'polspline' package.  I scanned several of 
 the help pages and looked in ~\library\polspline' where R is
installed
 
 on my hard drive and found no further documentation, and I found
 nothing 
 new from RSiteSearch(polyclass).  Googling for 'polyclass' led me to
 
 http://bear.fhcrc.org/~clk/soft.html;, the home page for Charles 
 Kooperberg, the author and maintainer.  If it were my problem, I might
 
 try writing him directly at the email address given in 
 help(package=polyclass);  I'm including him as a cc on this reply.
 
 I spent time looking at this, because 'polyclass', 'polymars'
 and 
 'polspline' seem potentially related related to one of my secondary 
 interests.  Unfortunately, I couldn't find sufficient documentation to
 
 allow me to proceed with the time I felt I could afford to invest in 
 this right now.
 
 If it were my problem, before I wrote another email about
 this, I'd 
 first list the function 'polyclass', make a local copy, then work 
 through an example line by line using 'debug(polyclass)'.  This
'debug'
 
 facility is remarkably powerful and easy to use.  I've solved many 
 problems like this using 'debug' in this way.
 
 If that failed to provide the necessary enlightenment, I'd
 submit 
 another post including a self-contained example based on a
modification
 
 of the 'iris' example featured in the 'polyclass' help page.  Your 
 example is NOT self contained, so I would NOT use that.  Using the 
 'iris' example would make it much easier to explain clearly what you 
 want.  It also makes it much easier for someone like me to experiment 
 with alternatives and describe what I did in terms of a tested
 example.
 
 Hope this helps.
 p.s.  I suggest you also review the posting guide! 
 www.R-project.org/posting-guide.html.
 
 Xiaogang Su wrote:
  Dear all, 
  
  I have a question on how to set up the starting model in POLYCLASS
 and
  make sure the terms in the starting model retained in the final
  POLYCLASS model. 
  
  In the function POLYMARS, this can be done using the STARTMODEL
 option.
  See below for example, I started with model 
  y= b0 + b1*X1 + b2*X2 + b3*X4 + b4*X5 + b5*X2*X5 + e
  
  m00 - matrix(c(
   1,  NA, 0, NA, 1, 
   2,  NA, 0, NA, 1, 
   4,  NA, 0, NA, 1, 
   5,  NA, 0, NA, 1,
   2,  NA, 5, NA, 1),nrow = 5, ncol=5, byrow=TRUE);
  
  m2 - polymars(response=PID2$y, predictors=PID2[,1:7], 
  startmodel=m00) 
  summary(m2)  
  
  But I could not figure out how this works for POLYCLASS. There is an
  option FIT in POLYCLASS, which needs to be a POLYCLASS object
though.
 
  
  Any suggestion or information is greatly appreciated. 
  
  Sincerely,
  Xiaogang Su
  
  
  
  Xiaogang Su,  Assistant Professor
  Department of Statistics and Actuarial Science
  University of Central 

[R] Tobit variance covariance matrix

2006-07-11 Thread Leandro Magnusson
Hi,

How can I recover the variance-covariance matrix of the tobit model from 
the variance-covariance of the survreg?
I first used to the survreg function and then I selected the variance 
matrix. However, the last parameter is log(scale) and not the variance 
of the standard deviation of the censored distribution as in the Tobit 
model.
tobit- survreg(Surv(y, y  0, type ='left')~ 0+ z + vh, dist = 'gaussian');
Om   - tobit$var;

Thanks

Leandro

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Re: [R] move axis label text

2006-07-11 Thread Prof Brian Ripley
On Mon, 10 Jul 2006, ivo welch wrote:

 dear R wizards:  sorry, I am stumped.  what is the parameter to just
 move the location of the x-label and y-lable (not the labels on the
 ticks of the x-axis and y-axis)?  probably obvious, but not to me
 right now... regards,  /iaw

They are all done by par(mgp):  ?title does say so.

 'mgp' The margin line (in 'mex' units) for the axis title, axis
  labels and axis line. The default is 'c(3, 1, 0)'.


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

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Re: [R] problem with install gstat package

2006-07-11 Thread Prof Brian Ripley
It is  -l (minus ell) not -1 (minus one)

Note that without actual cut-and-paste of the output, we cannot be sure if 
this is the problem: this is why the posting guide asks for such details.

On Mon, 10 Jul 2006, Jing Gao wrote:

 Hi there,
 I tried to install gstat package to R in Linux. I follow the
 instruction to use command  R CMD INSTALL -1 lib pkgs. But R
 shows syntax error in R CMD.
 Can someone help me with the installation of gstat package?
 Thank you!
 
 Jing

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

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

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[R] Other models of GARCH

2006-07-11 Thread Sumanta Basak




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Hi All,
Can you please tell me whether R can handle other type GARCH models like, 
FIGARCH, E-GARCH, JGR-GARCH etc? Any particular help for these kind of models?

Thanks,
Sumanta Basak.
 

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Hi All,brCan you please tell me whether R can handle other type GARCH 
models like, FIGARCH, E-GARCH, JGR-GARCH etc? Any particular help for these 
kind of models?brbrThanks,brSumanta Basak.br p



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[R] Plot Date

2006-07-11 Thread Sumanta Basak
Hi R-Users,

First of all i apologize if this is too simple for you. I want to plot an index 
of which i have daily data for 10 years. I want to plot specific dates in X 
axis, like 20/05/91 20/12/92 20/07/94   etc. and the index value in 
Y axis. Please suggest me.

Thanks,
Sumanta.


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[R] DIfferent lengths

2006-07-11 Thread Celso Barros
   Dear List members,

   I am having a problem importing SPSS files. For some reason I do not
understand, the variables in the data set come into R with different
lengths, which seems to make regression analysis impossible.

   The strangest thing is this: it SOMETIMES work. I ran the same
commands (always pasting the same thing from RWinEdt) yesterday, and,
although it took a long time for R to do it (and thought it sometimes
crashed in the process), it worked. Still, today, the same commands (pasted
from the same RWinEdt file, using the very same SPSS files) produced this
error message about different lengths.

Does anyone know what the problem could be?

Thanks in advance,

  Celso Barros

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[R] Assistance with dll's to use with dyn.load

2006-07-11 Thread christian.ritter
After having browsed the documentation for a while without discovering what I 
am looking for, maybe one of you would know ...

What I want to do:
I have two fortran files MC.f and ESCA.f. In MC.f there is a call to a routine 
called lpost. This routine (lpost) is defined (among other things) in ESCA.f. 
Under linux, I can do the following:
R CMD SHLIB MC.f
and
R CMD SHLIB ESCA.f
followed by
dyn.load(ESCA.so,local=FALSE)
dyn.load(MC.so)
(see point LINUX for some more)



Under windows, trying to Rcmd SHLIB MC.f will either:
also compile ESCA.f and then run dlltools with ESCA.o yielding one 
MC.dll (if ESCA.f is in the same directory as MC.f)
or fail since the symbol lpost is missing.
In the first case, dyn.load(MC.dll) will then work.



What I would like to have under linux:
MC and ESCA turned into separate shared libraries (.so or .dll) 
be able to 
dyn.load(ESCA.so, local=FALSE); dyn.load(MC.so)
... work work work ...
and then
dyn.unload(ESCA.so); dyn.load(SomethingElse.so) which also contains 
the symbol lpost
(so, I would like to unload the symbol lpost and replace it by 
something else)
in the (good?) old times using S on unix running on a DECstation stuff 
like this was possible, but I never could get it working with R and linux.


In windows, I would like the same operation:
dyn.load(ESCA.dll); dyn.load(MC.dll)
work work work
and then 
dyn.unload(ESCA.dll); dyn.load(SomethingElse.dll) which also 
contains the symbol lpost
is this possible? I see that the parameters local and now are ignored in 
dyn.load for windows but I don't grasp the full meaning.


Summary questions:
1. what are the proper parameters for (Windows) Rcmd.exe SHLIB MC.f such that 
it does not complain about a non existing name lpost?
2. how can I unload the container of lpost such that I can replace it while R 
continues running (under Windows and Linux)?

Sorry for lengthy explanation.



Christian Ritter
Senior Statistical Consultant
Analytical and Statistical Services
Shell Coordination Centre S.A.
Monnet Centre International Laboratory, Avenue Jean Monnet 1, B-1348 
Louvain-La-Neuve, Belgium

Tel: +32 10 477  349 Fax: +32 10 477 219
Email: [EMAIL PROTECTED]
Internet: http://www.analyticalsolutions4u.eu


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[R] Second Partial Derivatives

2006-07-11 Thread Robert Mcfadden
Hello,

Does R have any build-in function which allow me to count second partial
derivatives numerically?

RobertM

 

 


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[R] Table of P values for Fisher's exact test

2006-07-11 Thread jenny tan
Hi,

I have a table of observed counts for various genetic markers. Instead of 
doing Fisher's exact test for each marker one at a time and recording the P 
value manually, is there a script to go through the whole list and generate 
the P value column automatically?

An example of my data:


Counts_CHB and Counts_AA are the observed counts for one allele.
2N_CHB and 2N_AA are the total number of alleles.

geneLocal_Pos   2N_CHB  2N_AA   Counts_CHB  Counts_AA Exact_P
B2M 247590  46  0   5
B2M 353290  44  0   1
FCN2220388  46  0   1
FCN2353690  46  0   9
FCN2402784  46  0   9
FCN2403690  46  0   9
FCN2431890  46  0   9
FCN2439290  46  0   9
FCN2957590  46  0   1

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[R] weights in glmrob

2006-07-11 Thread Celso Barros
Dear List members,


 MY APOLOGIES IF YOU ALREADY RECEIVED THIS. A COLLEAGUE TOLD ME HE
COULD NOT OPEN MY PREVIOUS MESSAGE.


I am runnning a logistic regression that includes sample weights
(expressed by a variable called WEIGHT) using glmob (from robustbase). My
model looks like this:

   B12-glmrob(SERE~COMP+COM,family=binomial,weights=WEIGHT)

   When I include the weights, my standard error columns, as well as the
p-values, show only NAs. If I exclude weights=WEIGHTS, everything works. I
would like to use the weights. Does anyone know what could be the problem?

   Thanks in advance,

   Celso

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Re: [R] Table of P values for Fisher's exact test

2006-07-11 Thread Jacques VESLOT
apply(yourdata[, c(X2N_CHB,X2N_AA,Counts_CHB,Counts_AA)], 1, 
function(x) 
fisher.test(cbind(x[3:4], x[1:2]-x[3:4]))$p.value)
---
Jacques VESLOT

CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex

Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31

http://www-good.ibl.fr
---


jenny tan a écrit :
 Hi,
 
 I have a table of observed counts for various genetic markers. Instead of 
 doing Fisher's exact test for each marker one at a time and recording the P 
 value manually, is there a script to go through the whole list and generate 
 the P value column automatically?
 
 An example of my data:
 
 
 Counts_CHB and Counts_AA are the observed counts for one allele.
 2N_CHB and 2N_AA are the total number of alleles.
 
 gene  Local_Pos   2N_CHB  2N_AA   Counts_CHB  Counts_AA Exact_P
 B2M   247590  46  0   5
 B2M   353290  44  0   1
 FCN2  220388  46  0   1
 FCN2  353690  46  0   9
 FCN2  402784  46  0   9
 FCN2  403690  46  0   9
 FCN2  431890  46  0   9
 FCN2  439290  46  0   9
 FCN2  957590  46  0   1
 
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[R] Coxph function

2006-07-11 Thread Sharon.Mazurel

Dear Sir/Madam,

My name is Sharon Mazurel and I'm busy with my final thesis about
estimating the probability of the success or withdrawal of a merger or
acquisition.

For my model I also use duration analysis and found after doing some
research that the coxph function in R is very useful.

My question is:

In the Surv object you have two arguments, time and event. I have
two events, namely withdrawn and success. 

Does coxph function calculate the coefficients correctly when you put no
event argument into the Surv object?

Thus:

Coxph(Surv(duration)~covariates,data=data) duration=duration of the deal



Best regards,


Sharon Mazurel

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[R] Coxph

2006-07-11 Thread Sharon.Mazurel


Dear all,



My question is:

In the Surv object you have two arguments, time and event. I have
two events, namely withdrawn and success. 
I use no event or status argument in Surv because all my objects die
in my data set.

Does coxph function calculate the coefficients correctly when you put no
event argument into the Surv object?

Thus:

Coxph(Surv(duration)~covariates,data=data) duration=duration of the deal

Duration: is the time till one subject fails or succeed in my research.

Can somebody help me?


Best regards,


Sharon Mazurel


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Re: [R] Second Partial Derivatives

2006-07-11 Thread Rau, Roland
Hi, 

 [mailto:[EMAIL PROTECTED] On Behalf Of Robert Mcfadden
 
 Does R have any build-in function which allow me to count 
 second partial
 derivatives numerically?

library(nlme)
?fdHess

I hope this is the direction you wanted to take.

Best,
Roland



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Re: [R] Coxph

2006-07-11 Thread Heinz Tuechler
From the help page of Surv:
Although unusual, the event indicator can be omitted, in which case all
subjects are assumed to have an event.

That means, you can use coxph that way, _but_ it depends on your model.
Do you really want to model the time on study regardless of the kind of event?

Greetings,
Heinz Tüchler

At 12:22 11.07.2006 +0200, [EMAIL PROTECTED] wrote:


Dear all,



My question is:

In the Surv object you have two arguments, time and event. I have
two events, namely withdrawn and success. 
I use no event or status argument in Surv because all my objects die
in my data set.

Does coxph function calculate the coefficients correctly when you put no
event argument into the Surv object?

Thus:

Coxph(Surv(duration)~covariates,data=data) duration=duration of the deal

Duration: is the time till one subject fails or succeed in my research.

Can somebody help me?


Best regards,


Sharon Mazurel


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Re: [R] Plot Date

2006-07-11 Thread Gabor Grothendieck
Try this.  We first generate some test data, then we plot the
data without the x axis.  We generate a subset of the times
consisting of a sequence of every 7 months and create an
axis with that.  Then we add smaller ticks every month.


library(zoo)

# test data
x - seq(0, 10*365)
z - zoo(x, as.Date(1990-01-01) + x)

plot(z, xaxt = n)

tt - seq(as.Date(1990-05-20), time(z)[length(z)], by = 7 months)
axis(1, tt, format(tt, %d/%m/%y), cex.axis = 0.7, tcl = -0.6)

tt - seq(as.Date(1990-01-20), time(z)[length(z)], by = months)
axis(1, tt, FALSE, tcl = -0.3)


On 7/11/06, Sumanta Basak [EMAIL PROTECTED] wrote:
 Hi R-Users,

 First of all i apologize if this is too simple for you. I want to plot an 
 index of which i have daily data for 10 years. I want to plot specific dates 
 in X axis, like 20/05/91 20/12/92 20/07/94   etc. and the index 
 value in Y axis. Please suggest me.

 Thanks,
 Sumanta.


 -
  Find out what India is talking about on  Yahoo! Answers India.

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[R] Proportional Hazard Function and Competing risks

2006-07-11 Thread Sharon.Mazurel


How can I model coxph() in combination with competing risks

i.e. I have two events and for event the object will leave the data set.
So :

Coxph(Surv(time,event)~) the event is for all my objects 1.

How can I model this?

Sharon
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[R] new object

2006-07-11 Thread Mauricio Cardeal
Hi !

Please, how can I extract n.event and n.risk as a new object from 
example below ?  Thanks in advance.
Mauricio

require(survival)
fit - survfit(Surv(time, status) ~ x, data=aml)

summary(fit)

Call: survfit(formula = Surv(time, status) ~ x, data = aml)

x=Maintained
 time n.risk n.event survival std.err lower 95% CI upper 95% CI
9 11   10.909  0.0867   0.75411.000
   13 10   10.818  0.1163   0.61921.000
   18  8   10.716  0.1397   0.48841.000
   23  7   10.614  0.1526   0.37690.999
   31  5   10.491  0.1642   0.25490.946
   34  4   10.368  0.1627   0.15490.875
   48  2   10.184  0.1535   0.03590.944

x=Nonmaintained
 time n.risk n.event survival std.err lower 95% CI upper 95% CI
5 12   2   0.8333  0.1076   0.64701.000
8 10   2   0.6667  0.1361   0.44680.995
   12  8   1   0.5833  0.1423   0.36160.941
   23  6   1   0.4861  0.1481   0.26750.883
   27  5   1   0.3889  0.1470   0.18540.816
   30  4   1   0.2917  0.1387   0.11480.741
   33  3   1   0.1944  0.1219   0.05690.664
   43  2   1   0.0972  0.0919   0.01530.620
   45  1   1   0.  NA   NA   NA

[[alternative HTML version deleted]]

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Re: [R] new object

2006-07-11 Thread ronggui

sum.out-summary(fit)
names(sum.out)

[1] surv time n.risk   n.event  conf.int std.err
[7] lowerupperstrata   call

sum.out$n.risk

[1] 11 10  8  7  5  4  2 12 10  8  6  5  4  3  2  1

sum.out$n.event

[1] 1 1 1 1 1 1 1 2 2 1 1 1 1 1 1 1

cbind(sum.out$n.risk,sum.out$n.event,sum.out$strata)

 [,1] [,2] [,3]
[1,]   1111
[2,]   1011
[3,]811
[4,]711
[5,]511
[6,]411
[7,]211
[8,]   1222
[9,]   1022
[10,]812
[11,]612
[12,]512
[13,]412
[14,]312
[15,]212
[16,]112


2006/7/11, Mauricio Cardeal [EMAIL PROTECTED]:

Hi !

Please, how can I extract n.event and n.risk as a new object from
example below ?  Thanks in advance.
Mauricio

require(survival)
fit - survfit(Surv(time, status) ~ x, data=aml)

summary(fit)

Call: survfit(formula = Surv(time, status) ~ x, data = aml)

x=Maintained
 time n.risk n.event survival std.err lower 95% CI upper 95% CI
9 11   10.909  0.0867   0.75411.000
   13 10   10.818  0.1163   0.61921.000
   18  8   10.716  0.1397   0.48841.000
   23  7   10.614  0.1526   0.37690.999
   31  5   10.491  0.1642   0.25490.946
   34  4   10.368  0.1627   0.15490.875
   48  2   10.184  0.1535   0.03590.944

x=Nonmaintained
 time n.risk n.event survival std.err lower 95% CI upper 95% CI
5 12   2   0.8333  0.1076   0.64701.000
8 10   2   0.6667  0.1361   0.44680.995
   12  8   1   0.5833  0.1423   0.36160.941
   23  6   1   0.4861  0.1481   0.26750.883
   27  5   1   0.3889  0.1470   0.18540.816
   30  4   1   0.2917  0.1387   0.11480.741
   33  3   1   0.1944  0.1219   0.05690.664
   43  2   1   0.0972  0.0919   0.01530.620
   45  1   1   0.  NA   NA   NA

[[alternative HTML version deleted]]

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--
黄荣贵
Department of Sociology
Fudan University

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[R] Misunderstanding with lines (or elsewhere)

2006-07-11 Thread vincent
Misunderstanding with lines(...) :
http://7d4.com/r/

I would like the y coordinate of the horizontal line to be 1/4,
and also the line to begin at x=0 and end at y=1.

I'm obviously missing something
... so if anybody could help.

Many thanks.


test = function()
{
bmp('test.bmp', width=100, height=100)
m = matrix(0, 2, 2)
par(new=T, fig = c(0,1,0,1), mai=c(0,0,0,0), mar=c(0,0,0,0), bty='n')
image(m)
lines(c(0,1), c(1/4,1/4)) # horizontal line
a = dev.off()
}


The result :
http://7d4.com/r/

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Re: [R] Misunderstanding with lines (or elsewhere)

2006-07-11 Thread Prof Brian Ripley
Think about the coordinate system you are using: if you don't set the 
margins to zero you will see what it is.

I suggest you supply x and y to image() to set the coordinate system to 
what you want it to be.

On Tue, 11 Jul 2006, [EMAIL PROTECTED] wrote:

 Misunderstanding with lines(...) :
 http://7d4.com/r/
 
 I would like the y coordinate of the horizontal line to be 1/4,
 and also the line to begin at x=0 and end at y=1.

And that is what you get.

 I'm obviously missing something
 ... so if anybody could help.
 
 Many thanks.
 
 
 test = function()
 {
 bmp('test.bmp', width=100, height=100)
 m = matrix(0, 2, 2)
 par(new=T, fig = c(0,1,0,1), mai=c(0,0,0,0), mar=c(0,0,0,0), bty='n')
 image(m)
 lines(c(0,1), c(1/4,1/4)) # horizontal line
 a = dev.off()
 }
 
 
 The result :
 http://7d4.com/r/
 
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-- 
Brian D. Ripley,  [EMAIL PROTECTED]
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

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Re: [R] new object

2006-07-11 Thread Jacques VESLOT
  names(summary(fit))
  [1] surv time n.risk   n.event  conf.int std.err
  [7] lowerupperstrata   call
  summary(fit)$n.risk
  [1] 11 10  8  7  5  4  2 12 10  8  6  5  4  3  2  1

---
Jacques VESLOT

CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex

Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31

http://www-good.ibl.fr
---


Mauricio Cardeal a écrit :
 Hi !
 
 Please, how can I extract n.event and n.risk as a new object from 
 example below ?  Thanks in advance.
 Mauricio
 
 require(survival)
 fit - survfit(Surv(time, status) ~ x, data=aml)
 
 summary(fit)
 
 Call: survfit(formula = Surv(time, status) ~ x, data = aml)
 
 x=Maintained
  time n.risk n.event survival std.err lower 95% CI upper 95% CI
 9 11   10.909  0.0867   0.75411.000
13 10   10.818  0.1163   0.61921.000
18  8   10.716  0.1397   0.48841.000
23  7   10.614  0.1526   0.37690.999
31  5   10.491  0.1642   0.25490.946
34  4   10.368  0.1627   0.15490.875
48  2   10.184  0.1535   0.03590.944
 
 x=Nonmaintained
  time n.risk n.event survival std.err lower 95% CI upper 95% CI
 5 12   2   0.8333  0.1076   0.64701.000
 8 10   2   0.6667  0.1361   0.44680.995
12  8   1   0.5833  0.1423   0.36160.941
23  6   1   0.4861  0.1481   0.26750.883
27  5   1   0.3889  0.1470   0.18540.816
30  4   1   0.2917  0.1387   0.11480.741
33  3   1   0.1944  0.1219   0.05690.664
43  2   1   0.0972  0.0919   0.01530.620
45  1   1   0.  NA   NA   NA
 
   [[alternative HTML version deleted]]
 
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Re: [R] Other models of GARCH

2006-07-11 Thread Dirk Eddelbuettel
On Tue, Jul 11, 2006 at 07:41:58AM +0100, Sumanta Basak wrote: 
 Can you please tell me whether R can handle other type GARCH models
 like, FIGARCH, E-GARCH, JGR-GARCH etc? Any particular help for these 
 kind of models?

Yes R (no need to quote it) can: if you provide a (log-)likelihood function,
R can use its optimisers to find those parameter that maximise this
likelihood. 

In case you really wanted to ask if these are pre-programmed for you, then
the answer is no.

Hth, Dirk

-- 
Hell, there are no rules here - we're trying to accomplish something. 
  -- Thomas A. Edison

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Re: [R] --no-save and --save toggle from inside R? + BATCH stderr

2006-07-11 Thread ivo welch
thank you.  exactly what I needed.

can I add it as a suggestion for the next R version to add
an option to quit that prints only one line to stderr (or stdout, but
the real one, not the batch output one) that gives a 1-line job
summary---no error (or nothing), 5 warnings, 5 warnings, fatal
error: blah.  except in rare circumstances, this is likely to be the
user-preferred behavior.

(taking a hand where I have gotten a finger---can I redefine a quit
function that can invoke the original quit() with an option?)

regards,

/iaw

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[R] the problem of install Bugs package in R under linux environment

2006-07-11 Thread Zhijun Liang
Dear All:
  Do you know how to install Bugs package in R under linux environment ?
  I try it ,but it failed to work,it seems that  Bugs package just available in 
windows.
  by the way ,when will the linux version of Bugs package come out ?
  Thank you in advance.
best regards
zhijun

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[R] non positive-definite G matrix in mixed models: bootstrap?

2006-07-11 Thread Bruno L. Giordano
Dear list,
In a mixed model I selected I find a non positive definite random effects 
variance-covariance matrix G, where some parameters are estimated close to 
zero, and related confidence intervals are incredibly large.

Since simplification of the random portion is not an option, for both 
interest in the parameters and significant increase in the model fit, I 
would like to collect unbiased random effects estimates.

I used bootstrap to this purpose, creating a linear model for each cluster 
and bootstraping the variance of the coefficients. Is this procedure 
reasonable? Would it be reasonable in this case to keep the marginal portion 
of the mixed model?
Note that in presence of positive-definite G matrix this bootstrap approach 
and the mixed effect model give highly similar estimates and that in the non 
positive-definite model the bootstrap and mixed model marginal-model 
estimates are highly similar as well.

Thank you
Bruno

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Re: [R] Misunderstanding with lines (or elsewhere)

2006-07-11 Thread vincent
Prof Brian Ripley a écrit :

 Think about the coordinate system you are using: if you don't set the 
 margins to zero you will see what it is.
 I suggest you supply x and y to image() to set the coordinate system to 
 what you want it to be.

Thank you for your answer.
If I have well understood, the right way to proceed
must be :

test = function()
{
m = matrix(0, 2, 2)
par(new=T, fig = c(0,1,0,1), mai=c(0,0,0,0), mar=c(0,0,0,0), bty='n')
x0 = c(0.25 , 0.75)
y0 = x0
image(x=x0, y=y0, z=m)
lines(c(0,1), c(1/4,1/4)) # horizontal line
}

which seems indeed to work.
Thanks.

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[R] Starting Rserve in Java using Runtime Failed

2006-07-11 Thread Shing Hing Man
I have tried using the following piece of Java code to

start Rserve on my Linux PC without success.


try {
String command = R CMD Rserve ;
Process p = Runtime.getRuntime().exec(command);
returnCode = p.waitFor();
if (returnCode != 0) {
errorMessage = Unexpected return code =  +
returnCode;
}

} catch (Exception e) { 
errorMessage = Exception in running : + e;
}


I did not get any non-zero return code or error
message. 
It works when I run 'R CMD Rserve' at the command
line.


My version of R is 2.2.1 and  I am using Rserve
v0.4-3.

Thanks in advance for any assistance!

Shing 

Home page :
  http://uk.geocities.com/matmsh/index.html

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[R] Two Phase Sampling

2006-07-11 Thread Mark Hempelmann
Dear WizaRds,

I tried to construct a two-phase sampling design in Survey just the way 
I hoped understood in Vienna - I was wrong. I think I am too stupid to 
create the correct subset for phase 2. Phase1: Sample 1000 parts with 80 
defective. Phase2: Sample 100 parts out of these 1000 with  15 
defective. 0:ok, 1:defunct. The table below gives the conditional 
sampling values.

Please help me:

library(survey)
ss1 - data.frame(id=1:1000, ph1.x=rep(c(1,0),c(10,990)),
subset=rep(c(1,0),c(100,900)), ph2.y=rep(c(1,0,NA),c(15,85,900)),
n1=rep(1000,1000), n2=rep(100,1000) )
table(ss1$ph1.y, ss1$ph2.x)

 Phase1.x
 Phase2.y  0  1
0 85  0
1  5 10

p2 - twophase(id=list(~id,~id), strata=list(NULL,NULL),
data=ss1, subset=~subset, fpc=list(~n1,~n2))
svymean (~ph2.y, design=p2s)

   mean SE
 ph2.y 0.15  0

However, taking into consideration the 2nd sample, the estimator should be:

ph1.x.bar (phase1)=80/1000=0.08 and ph2.y.bar (phase2)=15/100=0.15 
defect boards, that means y.est=1.5*0.08=0.12 defect boards, since the 
RATIO ESTIMATOR equals 15/10=1.5 defect parts for the ratio of defect 
ph2/defect ph1.

What again did I do wrong? I am positive that the estimator is 12 
defective parts per 100 average, so how do I correctly construct the 
twophase design?

ps: I hope this is not sthg. undergraduates master eloquently...

Thank you so much for your help. I invite you to all the BBQ and beer 
there is in Europe!

Yours always
mark

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[R] help

2006-07-11 Thread Carlos Mauricio Cardeal Mendes
Please, is there any R-list about survival package ?

Thanks
Mauricio

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Re: [R] --no-save and --save toggle from inside R? + BATCH stderr

2006-07-11 Thread ivo welch
the following diff to the shell script invoking R prints an error
message if R terminates with an error code:

112c112,119
   exec sh ${R_HOME}/bin/Rcmd [EMAIL PROTECTED] ;;
---
   sh ${R_HOME}/bin/Rcmd [EMAIL PROTECTED]
   rc=$?
   if [ $rc -ne 0 ]; then
  shift ;
  echo [EMAIL PROTECTED]: Error Return Code: $rc
   fi
   exit $?
   ;;

regards,

/iaw

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Re: [R] --no-save and --save toggle from inside R? + BATCH stderr

2006-07-11 Thread Duncan Murdoch
ivo welch wrote:
 the following diff to the shell script invoking R prints an error
 message if R terminates with an error code:

 112c112,119
exec sh ${R_HOME}/bin/Rcmd [EMAIL PROTECTED] ;;
 ---
   
   sh ${R_HOME}/bin/Rcmd [EMAIL PROTECTED]
   rc=$?
   if [ $rc -ne 0 ]; then
  shift ;
  echo [EMAIL PROTECTED]: Error Return Code: $rc
   fi
   exit $?
   ;;
 
This looks to me like a good suggestion, but I generally don't make 
modifications to platform-specific things on platforms I don't use, so 
I'd suggest posting this to R-devel or on the bug list as a wishlist 
item.  Otherwise it might get lost.


Duncan Murdoch

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Re: [R] non positive-definite G matrix in mixed models: bootstrap?

2006-07-11 Thread Doran, Harold
There is a paper by Rogosa and Saner which shows some equivalences in
what you are doing under certain conditions. They show similarities
between bootstrapping with linear models and how the estimates might be
similar to those obtained from a mixed model.

Rogosa, D. R., and Saner, H. M. (1995). Longitudinal data analysis
examples with random coefficient models. 
Journal of Educational and Behavioral Statistics, 20, 149-170. 

Harold


 -Original Message-
 From: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED] On Behalf Of Bruno 
 L. Giordano
 Sent: Tuesday, July 11, 2006 9:31 AM
 To: r-help@stat.math.ethz.ch
 Subject: [R] non positive-definite G matrix in mixed models: 
 bootstrap?
 
 Dear list,
 In a mixed model I selected I find a non positive definite 
 random effects variance-covariance matrix G, where some 
 parameters are estimated close to zero, and related 
 confidence intervals are incredibly large.
 
 Since simplification of the random portion is not an option, 
 for both interest in the parameters and significant increase 
 in the model fit, I would like to collect unbiased random 
 effects estimates.
 
 I used bootstrap to this purpose, creating a linear model for 
 each cluster and bootstraping the variance of the 
 coefficients. Is this procedure reasonable? Would it be 
 reasonable in this case to keep the marginal portion of the 
 mixed model?
 Note that in presence of positive-definite G matrix this 
 bootstrap approach and the mixed effect model give highly 
 similar estimates and that in the non positive-definite model 
 the bootstrap and mixed model marginal-model estimates are 
 highly similar as well.
 
 Thank you
 Bruno
 
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Re: [R] Proportional Hazard Function and Competing risks

2006-07-11 Thread Heinz Tuechler
Maybe you find that thread helpful:

http://tolstoy.newcastle.edu.au/R/help/00b/1426.html

Heinz

At 12:59 11.07.2006 +0200, [EMAIL PROTECTED] wrote:


How can I model coxph() in combination with competing risks

i.e. I have two events and for event the object will leave the data set.
So :

Coxph(Surv(time,event)~) the event is for all my objects 1.

How can I model this?

Sharon
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Re: [R] Second Partial Derivatives

2006-07-11 Thread Ravi Varadhan
If you'd like to have accurate second derivatives, then check out the
numDeriv package, and in particular, the function hessian.  The
derivatives are based on Richardson extrapolation, and can be evaluated to a
very high degree of accuracy.

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:  [EMAIL PROTECTED]
Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html 
--
 -Original Message-
 From: [EMAIL PROTECTED] [mailto:r-help-
 [EMAIL PROTECTED] On Behalf Of Rau, Roland
 Sent: Tuesday, July 11, 2006 6:44 AM
 To: Robert Mcfadden; r-help@stat.math.ethz.ch
 Subject: Re: [R] Second Partial Derivatives
 
 Hi,
 
  [mailto:[EMAIL PROTECTED] On Behalf Of Robert Mcfadden
 
  Does R have any build-in function which allow me to count
  second partial
  derivatives numerically?
 
 library(nlme)
 ?fdHess
 
 I hope this is the direction you wanted to take.
 
 Best,
 Roland
 
 
 
 --
 This mail has been sent through the MPI for Demographic Rese...{{dropped}}
 
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[R] help on Rserve

2006-07-11 Thread zana adeb
Dear All,
   
  I'm calling R functions from Java by using Rserve, There is an exemple about 
dislay graphics in Java, this exemple is very difficult for me, I don't 
understand it...
  Could someone give me a simple example 
  What I want is putting the graphe in a file to display it after that in a 
browser..
   
  Thank you in advance
   
  XENA


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Re: [R] Misunderstanding with lines (or elsewhere)

2006-07-11 Thread Gavin Simpson
On Tue, 2006-07-11 at 15:39 +0200, [EMAIL PROTECTED] wrote:
 Prof Brian Ripley a écrit :
 
  Think about the coordinate system you are using: if you don't set the 
  margins to zero you will see what it is.
  I suggest you supply x and y to image() to set the coordinate system to 
  what you want it to be.
 
 Thank you for your answer.
 If I have well understood, the right way to proceed
 must be :
 
 test = function()
 {
 m = matrix(0, 2, 2)
 par(new=T, fig = c(0,1,0,1), mai=c(0,0,0,0), mar=c(0,0,0,0), bty='n')
 x0 = c(0.25 , 0.75)
 y0 = x0
 image(x=x0, y=y0, z=m)
 lines(c(0,1), c(1/4,1/4)) # horizontal line
 }
 
 which seems indeed to work.
 Thanks.

You could swap the line:

lines(c(0,1), c(1/4, 1/4))

with:

abline(h = 0.25)

Does the same thing but is more readable IMHO.

Also, I get a warning with your code if no plot is open:

Warning message:
calling par(new=) with no plot

To stop this, drop the new = T (sic) in the call to par. Also, don't use
T when you mean TRUE.

So, in summary this seems to do the same thing as your function:

test = function()
{
m = matrix(0, 2, 2)
par(fig = c(0,1,0,1), mai=c(0,0,0,0), mar=c(0,0,0,0), bty='n')
x0 = c(0.25 , 0.75)
y0 = x0
image(x=x0, y=y0, z=m)
abline(h = 0.25)
}

HTH,

G
-- 
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
 Gavin Simpson [t] +44 (0)20 7679 0522
 ECRC  ENSIS, UCL Geography,  [f] +44 (0)20 7679 0565
 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
 Gower Street, London  [w] http://www.ucl.ac.uk/~ucfagls/cv/
 London, UK. WC1E 6BT. [w] http://www.ucl.ac.uk/~ucfagls/
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%

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Re: [R] --no-save and --save toggle from inside R? + BATCH stderr

2006-07-11 Thread Prof Brian Ripley
This sort of thing is contrary to the Unix spirit.  It gives the user no 
choice but to be told that there is a non-zero error code, whereas it is 
really easy for the user to wrap the command in his/her own script that 
reports the error code in whatever form is desired.

The Unix idea is to think about combining things from your toolkit, not
complicate the tools with endless options (and this would have to be made 
optional as I for one don't want it).

Assuming you are concerned with user errors and not those deep in R, you 
can use R's own error-handling system to set the error status anyway, and 
report as wanted.


On Tue, 11 Jul 2006, Duncan Murdoch wrote:

 ivo welch wrote:
  the following diff to the shell script invoking R prints an error
  message if R terminates with an error code:
 
  112c112,119
 exec sh ${R_HOME}/bin/Rcmd [EMAIL PROTECTED] ;;
  ---

sh ${R_HOME}/bin/Rcmd [EMAIL PROTECTED]
rc=$?
if [ $rc -ne 0 ]; then
   shift ;
   echo [EMAIL PROTECTED]: Error Return Code: $rc
fi
exit $?
;;
  
 This looks to me like a good suggestion, but I generally don't make 
 modifications to platform-specific things on platforms I don't use, so 
 I'd suggest posting this to R-devel or on the bug list as a wishlist 
 item.  Otherwise it might get lost.
 
 
 Duncan Murdoch
 
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-- 
Brian D. Ripley,  [EMAIL PROTECTED]
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

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[R] Multiple tests on 2 way-ANOVA

2006-07-11 Thread Grathwohl, Dominik, LAUSANNE, NRC-BAS
Dear r-helpers,

I have a question about multiple testing.
Here an example that puzzles me:
All matrixes and contrast vectors are presented in treatment contrasts.

1. example:
library(multcomp)
n-60; sigma-20
# n = sample size per group
# sigma standard deviation of the residuals

cov1 - matrix(c(3/4,-1/2,-1/2,-1/2,1,0,-1/2,0,1), nrow = 3, ncol=3, 
byrow=TRUE, 
dimnames = list(c(A, B, C), c(C.1, C.2, C.3)))
# cov1 = variance covariance matrix of the beta coefficients of a 
# 2x2 factorial design (see Piantadosi 2005, p. 509)

cm1 - matrix(c(0, 1, 0, 0, 0, 1), nrow = 2, ncol=3, byrow=TRUE, 
dimnames = list(c(A, B), c(C.1, C.2, C.3)))
# cm1 = contrast matrix for main effects

v1 - csimint(estpar=c(100, 6, 5), df=4*n-3, covm=cov1*sigma^2/n, cmatrix=cm1, 
conf.level=0.95)
summary(v1)

The adjusted p-values are almost the Bonferroni p-values.
If I understood right: You need not to adjust for multiple testing 
on main effects in a 2x2 factorial design 
assuming the absence of interaction. 
I do not think that there is a bug, 
I want to understand, why multcomp does adjust for multiple tests 
having all information about the design of the trial (variance covariance 
matrix)?
Or do I have to introduce somehow more information?

2. example:
And I have second question: How do I proper correct for multiple testing 
if I want to estimate in the presence of interaction the two average main 
effects.
Can some one point me to some literature where I can learn these things?
Here the example, 2x2 factorial with interaction, estimation of average main 
effects:

cov2 - matrix(
c(1,-1,-1, 1,
 -1, 2, 1,-2,
 -1, 1, 2,-2,
  1,-2,-2, 4)
, nrow=4, ncol=4, byrow=TRUE)
cm2 - matrix(c(0, 1, 0, 1/2, 0, 0, 1, 1/2), nrow = 2, ncol=4, byrow=TRUE, 
dimnames = list(c(A, B), c(C.1, C.2, C.3, C.4)))
v2 - csimint(estpar=c(100, 6, 5, 2), df=4*n-4, covm=cov2*sigma^2/n, 
cmatrix=cm2, conf.level=0.95)
summary(v2)

I do not believe that this is the most efficient way for doing this, 
since I made already bad experience with the first example.

My R.version:

platform i386-pc-mingw32
arch i386   
os   mingw32
system   i386, mingw32  
status  
major2  
minor2.1
year 2005   
month12 
day  20 
svn rev  36812  
language R

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Re: [R] Proportional Hazard Function and Competing risks

2006-07-11 Thread Thomas Lumley
On Tue, 11 Jul 2006, [EMAIL PROTECTED] wrote:
 How can I model coxph() in combination with competing risks

 i.e. I have two events and for event the object will leave the data set.
 So :

 Coxph(Surv(time,event)~) the event is for all my objects 1.

 How can I model this?

There is nothing built in. Some proposals for competing risks analysis 
involve Cox models and generalized linear models and so can be implemented 
in R.  It depends on what you want to do -- there isn't a standard 
solution because a lot of what people want to do with competing risks is 
provably impossible.

-thomas

Thomas Lumley   Assoc. Professor, Biostatistics
[EMAIL PROTECTED]   University of Washington, Seattle

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[R] Error With TIFF of Plots

2006-07-11 Thread justin rapp
All,


I am using R on Mac OS X Tiger.  I have assembled some graphs through quartz
and cut and pasted them into MSWord files on my Mac.  I then emailed the
Word document containing the graphs to a Windows computer.  Upon opening the
document, I received an error message telling me that the graphics could not
be compressed.

Is there a way that I can have r create the images in JPEG or another format
so that I can cut and paste them into Word and view them on both Mac and
Windows platforms.

THank you.

jdr

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[R] storing the estimates from lmer

2006-07-11 Thread prabhu bhaga
Dear all,

I'm trying to store/extract  the mean standard error of the fixed effects
parameter and the variance of the random effects parameter from lmer
procedure from mlmre4 package developed by bates n pinheiro. while storing
fixed effects parameter is straight forward, the same is not true for
storing the variance parameter of the random effects. kindly help me

~prabhu

[[alternative HTML version deleted]]

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Re: [R] Proportional Hazard Function and Competing risks

2006-07-11 Thread Ravi Varadhan
Hi,

If you are interested in building regression models for the sub-distribution
functions (or cumulative incidence function), then you may want to look at
the cmprsk package by Gray.  It is based on the article: Fine and Gray
(JASA 1999).  There is also the approach based on multi-state models (see
the work of the Danish group led by Andersen) for modeling state
probabilities.

If you are interested only in the cause-specific hazard models, then you
don't need anything other than the standard Cox relative risk model or
something like that, where events from other causes are simply treated as
censoring.

If, however, you are interested in the net hazard or net probabilities, then
you enter the dangerous and highly controversial realm of the classical
competing risk problem.  You have to overcome a lot of conceptual,
philosophical, and statistical (e.g. identifiability) issues, before
attempting to model it using latent failure time models.  For a cogent
critique of this approach, see Prentice et al. (Biometrics 1978).  Crowder's
(2001) book presents a somewhat more pragmatic view of the classical
competing risks problem.

Hope this helps,
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:  [EMAIL PROTECTED]
Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html 
--

 -Original Message-
 From: [EMAIL PROTECTED] [mailto:r-help-
 [EMAIL PROTECTED] On Behalf Of Thomas Lumley
 Sent: Tuesday, July 11, 2006 10:56 AM
 To: [EMAIL PROTECTED]
 Cc: r-help@stat.math.ethz.ch
 Subject: Re: [R] Proportional Hazard Function and Competing risks
 
 On Tue, 11 Jul 2006, [EMAIL PROTECTED] wrote:
  How can I model coxph() in combination with competing risks
 
  i.e. I have two events and for event the object will leave the data set.
  So :
 
  Coxph(Surv(time,event)~) the event is for all my objects 1.
 
  How can I model this?
 
 There is nothing built in. Some proposals for competing risks analysis
 involve Cox models and generalized linear models and so can be implemented
 in R.  It depends on what you want to do -- there isn't a standard
 solution because a lot of what people want to do with competing risks is
 provably impossible.
 
   -thomas
 
 Thomas Lumley Assoc. Professor, Biostatistics
 [EMAIL PROTECTED] University of Washington, Seattle
 
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 guide.html

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Re: [R] storing the estimates from lmer

2006-07-11 Thread Doran, Harold
You need the VarCorr function. I think you mean that lmer is in the
Matrix package.

 -Original Message-
 From: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED] On Behalf Of prabhu bhaga
 Sent: Tuesday, July 11, 2006 11:19 AM
 To: r-help@stat.math.ethz.ch
 Subject: [R] storing the estimates from lmer
 
 Dear all,
 
 I'm trying to store/extract  the mean standard error of the 
 fixed effects parameter and the variance of the random 
 effects parameter from lmer
 procedure from mlmre4 package developed by bates n pinheiro. 
 while storing fixed effects parameter is straight forward, 
 the same is not true for storing the variance parameter of 
 the random effects. kindly help me
 
 ~prabhu
 
   [[alternative HTML version deleted]]
 
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[R] Question on partial effect

2006-07-11 Thread Guo Wei-Wei
Dear all,

I don't know what's my question is called. I have a performance
variable A, such as sales. And I have another variable B, let's say
establish time of firm. I want to create the third variable that is
sales without the effect of establish time. Maybe it can be called
partial effect problem. I'm not sure.

Does anyone have any suggestion? Thank you in advance!

All the best,
Wei-Wei

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[R] tcltk: help pop-up

2006-07-11 Thread javier garcia-pintado
Hi all;

Please, does anyone have a piece of R-tcltk code that includes a help
pop-up for any item and could send it as an example?

thanks and best regards,

Javier

-- 
Javier García-Pintado
Institute of Earth Sciences Jaume Almera (CSIC)
Lluis Sole Sabaris s/n, 08028 Barcelona
Phone: +34 934095410
Fax:   +34 934110012
e-mail:[EMAIL PROTECTED] 

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Re: [R] DIfferent lengths

2006-07-11 Thread Jing Gao

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[R] problem of fixed-formated output using sprintf

2006-07-11 Thread YIHSU CHEN
Dear R users:

I'm trying to generate a output file with fixed format using function sprintf 
in R. However, the execution time in R is very long even the toy data (smaller 
size df) seems to work fine. The syntax that I used is as follows:

df.fmt - sprintf(%2s%2s%2.4f, df$v1, df$v2, df$v3)
write.table(df.fmt, output.name,...)  

The actual dataset is a df with the dimention of 67944 by 34. I'm wondering 
whether there is an elegant way of doing it. I would like output in a txt file. 

Many thanks.


Yihsu Chen
The Johns Hopkins University

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[R] Date Format

2006-07-11 Thread pierre clauss
Hi everybody,
I need your precious help for, I think, a simple request, but I do not manage 
to solve this.
 
When I use a table function with dates in the rows, the rows are coerced to 
number after the table function.
 
So I need to transform the row names into date format. But I do not manage.
 
Therefore, for an example, I manage to write this :
 
datetest-06/01/2001
datetest-as.Date(datetest,%d/%m/%Y)
datetest-as.numeric(datetest)
 
to get 11328.
 
But I do not obtain the inverse tranformation :
 
datetest-as.Date(datetest,%d/%m/%Y)
 
How do we get this please ?
 
Thanks a lot for your solution.
Pierre.
 
 
 
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Re: [R] Question on partial effect

2006-07-11 Thread Guo Wei-Wei
Thank you, Gavin. I think that might be what I need. But I'm a little
bit wandering what's the scale of resid(mod). Is it
scale(dist)/scale(speed), for example kilometer / (kilometer per
hour)? or something else?

Thank you very much!
Wei-Wei


2006/7/12, Gavin Simpson [EMAIL PROTECTED]:
 On Tue, 2006-07-11 at 23:51 +0800, Guo Wei-Wei wrote:
  Dear all,
 
  I don't know what's my question is called. I have a performance
  variable A, such as sales. And I have another variable B, let's say
  establish time of firm. I want to create the third variable that is
  sales without the effect of establish time. Maybe it can be called
  partial effect problem. I'm not sure.
 
  Does anyone have any suggestion? Thank you in advance!
 
  All the best,
  Wei-Wei

 Do you mean?

 ## dummy data
 A - rnorm(100)
 B - rnorm(100)
 C - resid(lm(A ~ B))

 C now contains the residual variation in A after fitting B.

 e.g. with some real data
 ?cars
 data(cars) # not sure this is needed now, I forget
 mod - lm(dist ~ speed, data  = cars)
 summary(mod)
 partial - resid(mod)

 ## check
 mod2 - lm(dist ~ partial, data = cars)
 summary(mod2)
 ## from the two R^2 form mod1 and mod2 - partial contains dist minus
 ## the effects of speed
  0.6511 + 0.3489
 [1] 1

 HTH

 G
 --
 %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
  Gavin Simpson [t] +44 (0)20 7679 0522
  ECRC  ENSIS, UCL Geography,  [f] +44 (0)20 7679 0565
  Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
  Gower Street, London  [w] http://www.ucl.ac.uk/~ucfagls/cv/
  London, UK. WC1E 6BT. [w] http://www.ucl.ac.uk/~ucfagls/
 %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%



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Re: [R] Date Format

2006-07-11 Thread Gabor Grothendieck
Try this:

library(zoo)
as.Date(11328)

See the Help Desk article in R News 4/1 for more on dates.


On 7/11/06, pierre clauss [EMAIL PROTECTED] wrote:
 Hi everybody,
 I need your precious help for, I think, a simple request, but I do not manage 
 to solve this.

 When I use a table function with dates in the rows, the rows are coerced to 
 number after the table function.

 So I need to transform the row names into date format. But I do not manage.

 Therefore, for an example, I manage to write this :

 datetest-06/01/2001
 datetest-as.Date(datetest,%d/%m/%Y)
 datetest-as.numeric(datetest)

 to get 11328.

 But I do not obtain the inverse tranformation :

 datetest-as.Date(datetest,%d/%m/%Y)

 How do we get this please ?

 Thanks a lot for your solution.
 Pierre.



[[alternative HTML version deleted]]

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Re: [R] storing the estimates from lmer

2006-07-11 Thread Douglas Bates
On 7/11/06, Doran, Harold [EMAIL PROTECTED] wrote:
 You need the VarCorr function. I think you mean that lmer is in the
 Matrix package.

Currently lmer is in the Matrix package.  The plan is to move it back
to the lme4 package when interpackage linking has been added to R -
perhaps as early as the release of R-2.4.0.  Because the lme4 package
automatically loads the Matrix package it is safest to act as if lmer
actually resided in the lme4 package and use

library(lme4)
fm1 - lmer(...



  -Original Message-
  From: [EMAIL PROTECTED]
  [mailto:[EMAIL PROTECTED] On Behalf Of prabhu bhaga
  Sent: Tuesday, July 11, 2006 11:19 AM
  To: r-help@stat.math.ethz.ch
  Subject: [R] storing the estimates from lmer
 
  Dear all,
 
  I'm trying to store/extract  the mean standard error of the
  fixed effects parameter and the variance of the random
  effects parameter from lmer
  procedure from mlmre4 package developed by bates n pinheiro.
  while storing fixed effects parameter is straight forward,
  the same is not true for storing the variance parameter of
  the random effects. kindly help me
 
  ~prabhu
 
[[alternative HTML version deleted]]
 
  __
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  https://stat.ethz.ch/mailman/listinfo/r-help
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  http://www.R-project.org/posting-guide.html
 

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Re: [R] problem of fixed-formated output using sprintf

2006-07-11 Thread jim holtman
Are you sure it is the sprintf vs. write.table?  I constructed a dataframe
of your size and it took less than 1 second to do the sprintf:

 system.time(df.fmt - sprintf(%2s%2s%4.2f, df$V1, df$V2, df$V3))
[1] 0.70 0.00 0.73   NA   NA
 str(df)
`data.frame':   67944 obs. of  34 variables:
 $ V1 : chr  0.2655 0.3721 0.5729 0.9082 ...
 $ V2 : chr  0.2655 0.3721 0.5729 0.9082 ...
 $ V3 : num  0.0642 0.5316 0.5466 0.2798 0.4931 ...
 $ V4 : num  0.8430 0.8279 0.1552 0.5153 0.0725 ...
 $ V5 : num  0.922 0.780 0.811 0.847 0.223 ...


If you want to write this string out to a file, then use 'cat'.  If you add
'\n' to the sprintf, you will get something like this:

 cat(df.fmt[1:5])
0.26550.26550.06
 0.37210.37210.53
 0.57290.57290.55
 0.90820.90820.28
 0.20170.20170.49



On 7/11/06, YIHSU CHEN [EMAIL PROTECTED] wrote:

 Dear R users:

 I'm trying to generate a output file with fixed format using function
 sprintf in R. However, the execution time in R is very long even the toy
 data (smaller size df) seems to work fine. The syntax that I used is as
 follows:

 df.fmt - sprintf(%2s%2s%2.4f, df$v1, df$v2, df$v3)
 write.table(df.fmt, output.name,...)

 The actual dataset is a df with the dimention of 67944 by 34. I'm
 wondering whether there is an elegant way of doing it. I would like output
 in a txt file.

 Many thanks.


 Yihsu Chen
 The Johns Hopkins University

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

What is the problem you are trying to solve?

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Re: [R] Date Format

2006-07-11 Thread Roger D. Peng
Try

structure(11328, class = Date)

or just

class(datetest) - Date

-roger

pierre clauss wrote:
 Hi everybody,
 I need your precious help for, I think, a simple request, but I do not manage 
 to solve this.
  
 When I use a table function with dates in the rows, the rows are coerced to 
 number after the table function.
  
 So I need to transform the row names into date format. But I do not manage.
  
 Therefore, for an example, I manage to write this :
  
 datetest-06/01/2001
 datetest-as.Date(datetest,%d/%m/%Y)
 datetest-as.numeric(datetest)
  
 to get 11328.
  
 But I do not obtain the inverse tranformation :
  
 datetest-as.Date(datetest,%d/%m/%Y)
  
 How do we get this please ?
  
 Thanks a lot for your solution.
 Pierre.
  
  
  
   [[alternative HTML version deleted]]
 
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Re: [R] Question on partial effect

2006-07-11 Thread Gavin Simpson
On Wed, 2006-07-12 at 00:51 +0800, Guo Wei-Wei wrote:
 Thank you, Gavin. I think that might be what I need. But I'm a little
 bit wandering what's the scale of resid(mod). Is it
 scale(dist)/scale(speed), for example kilometer / (kilometer per
 hour)? or something else?
 
 Thank you very much!
 Wei-Wei

The scale of dist - they are just the differences between observed dist
and fitted dist (based on speed). 

mod - lm(dist ~ speed, data = cars)
resid(mod)

 1  2  3  4  5
  3.849460  11.849460  -5.947766  12.052234   2.119825
 6  7  8  9 10
 -7.812584  -3.744993   4.255007  12.255007  -8.677401


# visualise the residuals
plot(resid(mod) ~ dist, data = cars)
abline(h = 0, col = grey)
## length of blue line represents the residual
lines(cars$dist, resid(mod), type = h, col = blue)

So you see that for the 1st residual it is 3.849 ft (the distances are
measured in feet, see ?cars)

Does this help?

G

 
 
 2006/7/12, Gavin Simpson [EMAIL PROTECTED]:
  On Tue, 2006-07-11 at 23:51 +0800, Guo Wei-Wei wrote:
   Dear all,
  
   I don't know what's my question is called. I have a performance
   variable A, such as sales. And I have another variable B, let's say
   establish time of firm. I want to create the third variable that is
   sales without the effect of establish time. Maybe it can be called
   partial effect problem. I'm not sure.
  
   Does anyone have any suggestion? Thank you in advance!
  
   All the best,
   Wei-Wei
 
  Do you mean?
 
  ## dummy data
  A - rnorm(100)
  B - rnorm(100)
  C - resid(lm(A ~ B))
 
  C now contains the residual variation in A after fitting B.
 
  e.g. with some real data
  ?cars
  data(cars) # not sure this is needed now, I forget
  mod - lm(dist ~ speed, data  = cars)
  summary(mod)
  partial - resid(mod)
 
  ## check
  mod2 - lm(dist ~ partial, data = cars)
  summary(mod2)
  ## from the two R^2 form mod1 and mod2 - partial contains dist minus
  ## the effects of speed
   0.6511 + 0.3489
  [1] 1
 
  HTH
 
  G
  --
  %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
   Gavin Simpson [t] +44 (0)20 7679 0522
   ECRC  ENSIS, UCL Geography,  [f] +44 (0)20 7679 0565
   Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
   Gower Street, London  [w] http://www.ucl.ac.uk/~ucfagls/cv/
   London, UK. WC1E 6BT. [w] http://www.ucl.ac.uk/~ucfagls/
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 Gavin Simpson [t] +44 (0)20 7679 0522
 ECRC  ENSIS, UCL Geography,  [f] +44 (0)20 7679 0565
 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
 Gower Street, London  [w] http://www.ucl.ac.uk/~ucfagls/cv/
 London, UK. WC1E 6BT. [w] http://www.ucl.ac.uk/~ucfagls/
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[R] Query about getting averages across a certain parameter in a table

2006-07-11 Thread lalitha viswanath
Hi
I have a table that goes 
data

cluster_ac  clockrate age class
7337 0.9   0.001  alpha_proteins
7888 0.1   0.78   beta proteins

etc

The class column can have 7-8 different unique values
While the clockrate and age columns are floats varying
from 0 to 1.

I wish to get the average clockrate across each of the
classes for this data.

I would appreciate your help regarding the aboe.

Thanks
Lalitha

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[R] detach

2006-07-11 Thread Gabor Grothendieck
We try the following:

search()
as.Date(1)
zoo:::as.Date.numeric

under three circumstances:

1. on a fresh session
2. after issuing library(zoo) noting that as.Date.numeric is provided by zoo
3. after detaching zoo

as.Date(1) fails on #1 but succeeds in #2 and #3.  Should it not fail in #3
since zoo was detached?  Is this how its supposed to work?

(Note that entering zoo:::as.Date.numeric at the console
succeeds in displaying the function in all three cases.)

Here is the console session:

 R.version.string # XP
[1] Version 2.3.1 Patched (2006-06-04 r38279)

 search()
[1] .GlobalEnvpackage:methods   package:stats
[4] package:graphics  package:grDevices package:utils
[7] package:datasets  Autoloads package:base
 as.Date(1)
Error in as.Date.default(1) : do not know how to convert '1' to class Date
 zoo:::as.Date.numeric
function (x, ...)
structure(floor(x + 0.001), class = Date)
environment: namespace:zoo

 library(zoo)
 search()
 [1] .GlobalEnvpackage:zoo   package:methods
 [4] package:stats package:graphics  package:grDevices
 [7] package:utils package:datasets  Autoloads
[10] package:base
 as.Date(1)
[1] 1970-01-02
 zoo:::as.Date.numeric
function (x, ...)
structure(floor(x + 0.001), class = Date)
environment: namespace:zoo

 detach()
 search()
[1] .GlobalEnvpackage:methods   package:stats
[4] package:graphics  package:grDevices package:utils
[7] package:datasets  Autoloads package:base
 as.Date(1)
[1] 1970-01-02
 zoo:::as.Date.numeric
function (x, ...)
structure(floor(x + 0.001), class = Date)
environment: namespace:zoo


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[R] R newbie: logical subsets

2006-07-11 Thread Joshua Tokle
Hello!  I'm a newcomer to R hoping to replace some convoluted database 
code with an R script.  Unfortunately, I haven't been able to figure out 
how to implement the following logic.

Essentially, we have a database of transactions that are coded with a 
geographic locale and a type.  These are being loaded into a data.frame 
with named variables city, type, and price.  E.g., trans$city and all 
that.

We want to calculate mean prices by city and type, AFTER excluding 
outliers.  That is, we want to calculate the mean price in 3 steps:

1. calculate a mean and standard deviation by city and type over all 
transactions
2. create a subset of the original data frame, excluding transactions that 
differ from the relevant mean by more than 2 standard deviations
3. calculate a final mean by city and type based on this subset.

I'm stuck on step 2.  I would like to do something like the following:

fs - list(factor(trans$city), factor(trans$type))
means - tapply(trans$price, fs, mean)
stdevs - tapply(trans$price, fs, sd)

filter - abs(trans$price - means[trans$city, trans$type]) 
 2*stdevs[trans$city, trans$type]

sub - subset(trans, filter)

The above code doesn't work.  What's the correct way to do this?

Thanks,
Josh

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Re: [R] R newbie: logical subsets

2006-07-11 Thread Gabor Grothendieck
Try this, using the built in anscombe data set:

anscombe[!rowSums(abs(scale(anscombe))  2),]



On 7/11/06, Joshua Tokle [EMAIL PROTECTED] wrote:
 Hello!  I'm a newcomer to R hoping to replace some convoluted database
 code with an R script.  Unfortunately, I haven't been able to figure out
 how to implement the following logic.

 Essentially, we have a database of transactions that are coded with a
 geographic locale and a type.  These are being loaded into a data.frame
 with named variables city, type, and price.  E.g., trans$city and all
 that.

 We want to calculate mean prices by city and type, AFTER excluding
 outliers.  That is, we want to calculate the mean price in 3 steps:

 1. calculate a mean and standard deviation by city and type over all
 transactions
 2. create a subset of the original data frame, excluding transactions that
 differ from the relevant mean by more than 2 standard deviations
 3. calculate a final mean by city and type based on this subset.

 I'm stuck on step 2.  I would like to do something like the following:

 fs - list(factor(trans$city), factor(trans$type))
 means - tapply(trans$price, fs, mean)
 stdevs - tapply(trans$price, fs, sd)

 filter - abs(trans$price - means[trans$city, trans$type]) 
 2*stdevs[trans$city, trans$type]

 sub - subset(trans, filter)

 The above code doesn't work.  What's the correct way to do this?

 Thanks,
 Josh

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[R] script problem to obtain pairs of overlap values

2006-07-11 Thread Rogério Rosa da Silva
Dear,

I wrote a code to estimate the overlap between two kernel distributions.
The script must estimates the overlap among each columns of data frame.
With S sampled species (columns) in my data frame, I want obtain
S(S-1)/2 pairs of overlap values between species.
However, the code is not well write at all (only an overlap value is
produced) and I can't find the solution.

To illustrate the calculations, I use the data frame tdon and the
value of the bandwidth h, which was estimated in other part of script.

tdon - data.frame (sp.1=c (5 ,9 ,NA ,5, 11) , sp.2=c (4, 2, 4, NA, 11,
),sp.3=c(5, 4, 2, 6, 13), sp.4=c(3 , 11, NA, 5, 3), sp.5=c(2 ,5 ,2, 9, 9))

 h

[1] 1.047 2.973 0.887 1.520 2.955



Here is the code:

for (i in 1:(nbcol-1)) # nbcol-ncol(tdon)
{tdon1-tdon[,i]
tdon11- subset(tdon1,tdon1!=NA)
fctk1-function(x)
{density (tdon11, bw=h[i], kernel=gaussian)$y}

for (j in (i+1):nbcol)
{tdon2-tdon[,j]
tdon21- subset(tdon2,tdon2!=NA)
fctk2-function(x)
{density (tdon21, bw=h[j], kernel=gaussian)$y}

diffctk-function(x)
{abs(fctk1(x)-fctk2(x))}

intctk- approxfun (diffctk(x), rule=2)
int- integrate(diffctk,-Inf,Inf)$value
 overlap- 1 - 0.5* int
}
}

   
The use of approxfun to integrate the difference in the estimated
density values (my diffctk function) was suggested by Thomas Lumley,
but I'm not sure that I have found the solution or if this solution is correct 
for my problem.

I need that the overlap produce a vector with the length equal to 10, with
all pairs of overlap values.

Any help or advice on improvement for this code will be appreciated.

With kind regards,
   
Rogério

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[R] (no subject)

2006-07-11 Thread Taka Matzmoto
Dear R-users

prob - c(0.5,0.4,0.3,0.1,0.0)
cal - prob * log(prob,base=2)
cal
[1] -0.500 -0.5287712 -0.5210897 -0.3321928NaN

Is there any way to change NaN to zero ?

Thank you

Taka

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Re: [R] (no subject)

2006-07-11 Thread Ben Fairbank
Well, it's a little inelegant, but since the amount of information is
the same in events that are certain to occur and certain not to occur,
you could add the line

prob[prob==0] - 1

after you set up the prob vector, which would take care of the problem.


Ben Fairbank



-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Taka Matzmoto
Sent: Tuesday, July 11, 2006 2:26 PM
To: r-help@stat.math.ethz.ch
Subject: [R] (no subject)

Dear R-users

prob - c(0.5,0.4,0.3,0.1,0.0)
cal - prob * log(prob,base=2)
cal
[1] -0.500 -0.5287712 -0.5210897 -0.3321928NaN

Is there any way to change NaN to zero ?

Thank you

Taka

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[R] August R/Splus course @ 5 locations *** R/Splus Fundamentals and Programming Techniques

2006-07-11 Thread elvis

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[R] 0* log(0) should be zero but NaN

2006-07-11 Thread Taka Matzmoto
Dear R-users

prob - c(0.5,0.4,0.3,0.1,0.0)
cal - prob * log(prob,base=2)
cal
[1] -0.500 -0.5287712 -0.5210897 -0.3321928NaN

Is there any way to change NaN to zero ?

I did come up with this by applying Ripley's relpy to my previous question

cal -prob*log(pmax(prob,0.0001),base=2)

Any suggestion ?

Thank you

Taka

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[R] generating clustered data

2006-07-11 Thread Brian Perron
Hello R folks,

Does anybody have code to share for generating (via simulation) 
clustered data?  The type of data I am looking to simulate would allow 
fitting of a multilevel model with random intercepts.   I looked at the 
mvtnorm package but am not quite sure how to create clusters.  (Can this 
be done by simply changing the seed?)  If somebody could point me where 
to look for the relevant code or perhaps send some sample code, that 
would be great.

Thanks,
Brian

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[R] use of NULL environment is deprecated?

2006-07-11 Thread Patrick Connolly
] version
   _ 
platform   x86_64-unknown-linux-gnu  
arch   x86_64
os linux-gnu 
system x86_64, linux-gnu 
status   
major  2 
minor  3.1   
year   2006  
month  06
day01
svn rev38247 
language   R 
version.string Version 2.3.1 (2006-06-01)


I see in the NEWS file the line:
o   Use of NULL as an environment is deprecated and gives a warning.

Which duly happens.  I get warnings like this:
 Warning message:
use of NULL environment is deprecated

My problem is that I don't know what is being referred to.  A little
birdie tells me that in later versions of R, those warnings will
become errors so I need to work out where they're coming from before I
can use later versions.

My question is: How does one work out which is being referred to by
such a message?  The traceback() function is useful when failure
occurs.  Is there an analagous way of looking into warnings?

TIA

-- 
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.   
   ___Patrick Connolly   
 {~._.~} Great minds discuss ideas
 _( Y )_Middle minds discuss events 
(:_~*~_:)Small minds discuss people  
 (_)-(_)   . Anon
  
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.

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Re: [R] use of NULL environment is deprecated?

2006-07-11 Thread Duncan Murdoch
On 7/11/2006 6:27 PM, Patrick Connolly wrote:
 ] version
_ 
 platform   x86_64-unknown-linux-gnu  
 arch   x86_64
 os linux-gnu 
 system x86_64, linux-gnu 
 status   
 major  2 
 minor  3.1   
 year   2006  
 month  06
 day01
 svn rev38247 
 language   R 
 version.string Version 2.3.1 (2006-06-01)
 
 
 I see in the NEWS file the line:
 o Use of NULL as an environment is deprecated and gives a warning.
 
 Which duly happens.  I get warnings like this:
 Warning message:
 use of NULL environment is deprecated
 
 My problem is that I don't know what is being referred to.  A little
 birdie tells me that in later versions of R, those warnings will
 become errors so I need to work out where they're coming from before I
 can use later versions.
 
 My question is: How does one work out which is being referred to by
 such a message?  The traceback() function is useful when failure
 occurs.  Is there an analagous way of looking into warnings?

options(warn=2) will convert warnings into errors, so traceback will work.

A common situation where I've seen that error is with binary saves from 
earlier versions of R being loaded into current versions.  For example, 
if you installed a package before, but didn't re-install it with 2.3.1, 
or if you are reloading a workspace saved in an earlier version.

Duncan Murdoch

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Re: [R] use of NULL environment is deprecated?

2006-07-11 Thread Sundar Dorai-Raj

Patrick Connolly wrote:
 ] version
_ 
 platform   x86_64-unknown-linux-gnu  
 arch   x86_64
 os linux-gnu 
 system x86_64, linux-gnu 
 status   
 major  2 
 minor  3.1   
 year   2006  
 month  06
 day01
 svn rev38247 
 language   R 
 version.string Version 2.3.1 (2006-06-01)
 
 
 I see in the NEWS file the line:
 o Use of NULL as an environment is deprecated and gives a warning.
 
 Which duly happens.  I get warnings like this:
 
Warning message:
 
 use of NULL environment is deprecated
 
 My problem is that I don't know what is being referred to.  A little
 birdie tells me that in later versions of R, those warnings will
 become errors so I need to work out where they're coming from before I
 can use later versions.
 
 My question is: How does one work out which is being referred to by
 such a message?  The traceback() function is useful when failure
 occurs.  Is there an analagous way of looking into warnings?
 
 TIA
 

Hi, Patrick,

This will happen when you load a package that was created for an earlier 
version of R. If this is your own package, recreate the package using 
R-2.3.1. If it's a package on CRAN, then update.packages() should also 
do the trick.

HTH,

--sundar

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Re: [R] 0* log(0) should be zero but NaN

2006-07-11 Thread jim holtman
 prob - c(0.5,0.4,0.3,0.1,0.0)
 cal - prob * log(prob,base=2)
 cal
[1] -0.500 -0.5287712 -0.5210897 -0.3321928NaN
 cal[is.nan(cal)] - 0
 cal
[1] -0.500 -0.5287712 -0.5210897 -0.3321928  0.000



On 7/11/06, Taka Matzmoto [EMAIL PROTECTED] wrote:

 Dear R-users

 prob - c(0.5,0.4,0.3,0.1,0.0)
 cal - prob * log(prob,base=2)
 cal
 [1] -0.500 -0.5287712 -0.5210897 -0.3321928NaN

 Is there any way to change NaN to zero ?

 I did come up with this by applying Ripley's relpy to my previous question

 cal -prob*log(pmax(prob,0.0001),base=2)

 Any suggestion ?

 Thank you

 Taka

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

What is the problem you are trying to solve?

[[alternative HTML version deleted]]

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Re: [R] use of NULL environment is deprecated?

2006-07-11 Thread Patrick Connolly
On Tue, 11-Jul-2006 at 06:41PM -0400, Duncan Murdoch wrote:

[]

| 
| options(warn=2) will convert warnings into errors, so traceback will work.
| 
| A common situation where I've seen that error is with binary saves from 
| earlier versions of R being loaded into current versions.  For example, 
| if you installed a package before, but didn't re-install it with 2.3.1, 
| or if you are reloading a workspace saved in an earlier version.

Thank you for the explanation.  In this case, it's the latter reason:
the workspace comes from a different computer as well as a different
version of R.

Will saving the objects anew overcome the problem?

(Thanks Sundar also.)

-- 
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.   
   ___Patrick Connolly   
 {~._.~} Great minds discuss ideas
 _( Y )_Middle minds discuss events 
(:_~*~_:)Small minds discuss people  
 (_)-(_)   . Anon
  
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.

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Re: [R] Query about getting averages across a certain parameter in a table

2006-07-11 Thread jim holtman
 x - 
+ cluster_ac  clockrate age class
+ 7337 0.19   0.001  alpha_proteins
+ 7888 0.21   0.78   beta_proteins
+ 7337 0.39   0.001  alpha_proteins
+ 7888 0.41   0.78   beta_proteins
+ 7337 0.59   0.001  alpha_proteins
+ 7888 0.61   0.78   beta_proteins
+ 7337 0.97   0.001  alpha_proteins
+ 7888 0.81   0.78   beta_proteins
+ 7337 0.99   0.001  alpha_proteins
+ 7888 0.1   0.78   beta_proteins
+ 
 df - read.table(textConnection(x), header=TRUE)
 tapply(df$clockrate, df$class, mean)
alpha_proteins  beta_proteins
 0.626  0.428



On 7/11/06, lalitha viswanath [EMAIL PROTECTED] wrote:

 Hi
 I have a table that goes
 data

 cluster_ac  clockrate age class
 7337 0.9   0.001  alpha_proteins
 7888 0.1   0.78   beta proteins

 etc

 The class column can have 7-8 different unique values
 While the clockrate and age columns are floats varying
 from 0 to 1.

 I wish to get the average clockrate across each of the
 classes for this data.

 I would appreciate your help regarding the aboe.

 Thanks
 Lalitha

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

What is the problem you are trying to solve?

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Re: [R] generating clustered data

2006-07-11 Thread Hamilton, Cody

Brian,

What about creating the covariance matrix with the help of the kronecker
product?  For instance, suppose your intercepts are ~ N(0,var1) and your
residual errors are ~ N(0,var2).  Suppose further that you want 10
clusters of 5 observations each.  I believe you can create the overall
covariance matrix with kronecker(diag(10),matrix(var1,5,5)) +
var2*diag(50).  This can then be fed as the variance to the mvtnorm
function.  Hope this helps.

Regards,
   -Cody

-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Brian Perron
Sent: Tuesday, July 11, 2006 15:59 PM
To: r-help@stat.math.ethz.ch
Subject: [R] generating clustered data

Hello R folks,

Does anybody have code to share for generating (via simulation)
clustered data?  The type of data I am looking to simulate would allow
fitting of a multilevel model with random intercepts.   I looked at the
mvtnorm package but am not quite sure how to create clusters.  (Can this

be done by simply changing the seed?)  If somebody could point me where
to look for the relevant code or perhaps send some sample code, that
would be great.

Thanks,
Brian

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Re: [R] 0* log(0) should be zero but NaN [Broadcast]

2006-07-11 Thread Liaw, Andy
Try:

R (cal - prob * log(ifelse(prob == 0, 1, prob), base=2))
[1] -0.500 -0.5287712 -0.5210897 -0.3321928  0.000

Andy 

From: Taka Matzmoto
 
 Dear R-users
 
 prob - c(0.5,0.4,0.3,0.1,0.0)
 cal - prob * log(prob,base=2)
 cal
 [1] -0.500 -0.5287712 -0.5210897 -0.3321928NaN
 
 Is there any way to change NaN to zero ?
 
 I did come up with this by applying Ripley's relpy to my 
 previous question
 
 cal -prob*log(pmax(prob,0.0001),base=2)
 
 Any suggestion ?
 
 Thank you
 
 Taka
 
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Re: [R] use of NULL environment is deprecated?

2006-07-11 Thread Duncan Murdoch
On 7/11/2006 6:59 PM, Patrick Connolly wrote:
 On Tue, 11-Jul-2006 at 06:41PM -0400, Duncan Murdoch wrote:
 
 []
 
 | 
 | options(warn=2) will convert warnings into errors, so traceback will work.
 | 
 | A common situation where I've seen that error is with binary saves from 
 | earlier versions of R being loaded into current versions.  For example, 
 | if you installed a package before, but didn't re-install it with 2.3.1, 
 | or if you are reloading a workspace saved in an earlier version.
 
 Thank you for the explanation.  In this case, it's the latter reason:
 the workspace comes from a different computer as well as a different
 version of R.
 
 Will saving the objects anew overcome the problem?

Yes, it should, though there's a slight risk that someone might have 
meant NULL to mean an empty environment instead of the base environment. 
  This would be a bug with or without the change.

Duncan Murdoch

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Re: [R] Question on partial effect

2006-07-11 Thread Guo Wei-Wei
Than you, Gavin. You helped me out a lot of problems.

Thank you very much!
Wei-Wei




2006/7/12, Gavin Simpson [EMAIL PROTECTED]:
 On Wed, 2006-07-12 at 00:51 +0800, Guo Wei-Wei wrote:
  Thank you, Gavin. I think that might be what I need. But I'm a little
  bit wandering what's the scale of resid(mod). Is it
  scale(dist)/scale(speed), for example kilometer / (kilometer per
  hour)? or something else?
 
  Thank you very much!
  Wei-Wei

 The scale of dist - they are just the differences between observed dist
 and fitted dist (based on speed).

 mod - lm(dist ~ speed, data = cars)
 resid(mod)

  1  2  3  4  5
   3.849460  11.849460  -5.947766  12.052234   2.119825
  6  7  8  9 10
  -7.812584  -3.744993   4.255007  12.255007  -8.677401
 

 # visualise the residuals
 plot(resid(mod) ~ dist, data = cars)
 abline(h = 0, col = grey)
 ## length of blue line represents the residual
 lines(cars$dist, resid(mod), type = h, col = blue)

 So you see that for the 1st residual it is 3.849 ft (the distances are
 measured in feet, see ?cars)

 Does this help?

 G

 
 
  2006/7/12, Gavin Simpson [EMAIL PROTECTED]:
   On Tue, 2006-07-11 at 23:51 +0800, Guo Wei-Wei wrote:
Dear all,
   
I don't know what's my question is called. I have a performance
variable A, such as sales. And I have another variable B, let's say
establish time of firm. I want to create the third variable that is
sales without the effect of establish time. Maybe it can be called
partial effect problem. I'm not sure.
   
Does anyone have any suggestion? Thank you in advance!
   
All the best,
Wei-Wei
  
   Do you mean?
  
   ## dummy data
   A - rnorm(100)
   B - rnorm(100)
   C - resid(lm(A ~ B))
  
   C now contains the residual variation in A after fitting B.
  
   e.g. with some real data
   ?cars
   data(cars) # not sure this is needed now, I forget
   mod - lm(dist ~ speed, data  = cars)
   summary(mod)
   partial - resid(mod)
  
   ## check
   mod2 - lm(dist ~ partial, data = cars)
   summary(mod2)
   ## from the two R^2 form mod1 and mod2 - partial contains dist minus
   ## the effects of speed
0.6511 + 0.3489
   [1] 1
  
   HTH
  
   G
   --
   %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
Gavin Simpson [t] +44 (0)20 7679 0522
ECRC  ENSIS, UCL Geography,  [f] +44 (0)20 7679 0565
Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
Gower Street, London  [w] http://www.ucl.ac.uk/~ucfagls/cv/
London, UK. WC1E 6BT. [w] http://www.ucl.ac.uk/~ucfagls/
   %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
  
  
 
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  Gavin Simpson [t] +44 (0)20 7679 0522
  ECRC  ENSIS, UCL Geography,  [f] +44 (0)20 7679 0565
  Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
  Gower Street, London  [w] http://www.ucl.ac.uk/~ucfagls/cv/
  London, UK. WC1E 6BT. [w] http://www.ucl.ac.uk/~ucfagls/
 %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%



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Re: [R] 0* log(0) should be zero but NaN

2006-07-11 Thread Gabor Grothendieck
Try this:

p - 0:10/10
p * log2(p + !p)

On 7/11/06, Taka Matzmoto [EMAIL PROTECTED] wrote:
 Dear R-users

 prob - c(0.5,0.4,0.3,0.1,0.0)
 cal - prob * log(prob,base=2)
 cal
 [1] -0.500 -0.5287712 -0.5210897 -0.3321928NaN

 Is there any way to change NaN to zero ?

 I did come up with this by applying Ripley's relpy to my previous question

 cal -prob*log(pmax(prob,0.0001),base=2)

 Any suggestion ?

 Thank you

 Taka

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[R] PDF version of Chinese translations of the manual An Introduction to R

2006-07-11 Thread Guohui Ding
Dear All,

   I distributed the HTML style one year ago (
http://www.biosino.org/pages/newhtm/r/schtml and
http://www.biosino.org/pages/newhtm/r/tchtml). Now I polished it, and
rewrote it with Latex. You can download it from the URL:
http://www.biosino.org/R/R-doc/  (
http://www.biosino.org/R/R-doc/files/R-intro_cn.pdf).

   Wish it will be useful for Chinese R users.

   Any suggestion, comment and correction are welcome!
-- 
ADDRESS: Bioinformatics Center, Shanghai Institutes for Biological Sciences,
Chinese Academy of Sciences
320 Yueyang Road, Shanghai 200031, P.R.China
TELEPHONE: 86-21-54920086

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[R] help in vectorization

2006-07-11 Thread Dimitri Szerman
Hi,

I have two data frames. One is like

 dtf = data.frame(y=c(rep(2002,4), rep(2003,5)),
+  m=c(9:12, 1:5),
+  def=c(.74,.75,.76,.78,.80,.82,.85,.85,.87))

and the other

dtf2 = data.frame(y=rep( c(2002,2003),20),
  m=c(trunc(runif(20,1,5)),trunc(runif(20,9,12))),
  inc=rnorm(40,mean=300,sd=150) )

What I want is to divide dtf2$inc by dtf1$def by the levels of y and m
(inc is income and def is the deflator). Any help in vectorizing this is
welcome.

Thank you,
Dimitri

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Re: [R] help in vectorization

2006-07-11 Thread Gabor Grothendieck
Try this:

with(merge(dtf, dtf2), data.frame(y, m, inc = inc/def))

On 7/11/06, Dimitri Szerman [EMAIL PROTECTED] wrote:
 Hi,

 I have two data frames. One is like

  dtf = data.frame(y=c(rep(2002,4), rep(2003,5)),
 +  m=c(9:12, 1:5),
 +  def=c(.74,.75,.76,.78,.80,.82,.85,.85,.87))

 and the other

 dtf2 = data.frame(y=rep( c(2002,2003),20),
  m=c(trunc(runif(20,1,5)),trunc(runif(20,9,12))),
  inc=rnorm(40,mean=300,sd=150) )

 What I want is to divide dtf2$inc by dtf1$def by the levels of y and m
 (inc is income and def is the deflator). Any help in vectorizing this is
 welcome.

 Thank you,
 Dimitri

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[R] PDF version of Chinese translations of the manual An Introduction to R

2006-07-11 Thread Guohui Ding
Dear All,

   I distributed the HTML style one year ago (
http://www.biosino.org/pages/newhtm/r/schtml and
http://www.biosino.org/pages/newhtm/r/tchtml ). Now I polished it, and
rewrote it with Latex. You can download it from the URL:
http://www.biosino.org/R/R-doc/  (
http://www.biosino.org/R/R-doc/files/R-intro_cn.pdf).

   Wish it will be useful for Chinese R users.

   Any suggestion, comment and correction are welcome!

-- 
ADDRESS: Bioinformatics Center, Shanghai Institutes for Biological Sciences,
Chinese Academy of Sciences
320 Yueyang Road, Shanghai 200031, P.R.China
TELEPHONE: 86-21-54920086

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Re: [R] Date Format

2006-07-11 Thread Don MacQueen
If I understand your question correctly, then my suggestion is to try
   table( format(datetest), other.variable)
instead of
   table(datetest, other.variable)

-Don

At 4:23 PM + 7/11/06, pierre clauss wrote:
Hi everybody,
I need your precious help for, I think, a simple request, but I do 
not manage to solve this.

When I use a table function with dates in the rows, the rows are 
coerced to number after the table function.

So I need to transform the row names into date format. But I do not manage.

Therefore, for an example, I manage to write this :

datetest-06/01/2001
datetest-as.Date(datetest,%d/%m/%Y)
datetest-as.numeric(datetest)

to get 11328.

But I do not obtain the inverse tranformation :

datetest-as.Date(datetest,%d/%m/%Y)

How do we get this please ?

Thanks a lot for your solution.
Pierre.



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-- 
-
Don MacQueen
Lawrence Livermore National Laboratory
Livermore, CA, USA

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[R] Rgnome in Ubuntu

2006-07-11 Thread Steve Lane
Hi
I am trying to set up Rgnome in Ubuntu (Dapper) and when I use the 
instructions:

/path/to/gnomeGUI/configure R_HOME=/path/to/R/installation
 make
 make install

I get an error message when I try to 'make' as follows:

/usr/local/lib64/R/etc/Makeconf:5: /make/vars.mk: No such file or directory
make: *** No rule to make target `/make/vars.mk'.  Stop.

Can anyone help me?
Steve

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