date:20110819

You might try using outer to create a matrix that will help out:

 Time - c(1000, 1050, 1100, 1500, 2500, 5000, 6500, 6600, 7000)
 Time
[1] 1000 1050 1100 1500 2500 5000 6500 6600 7000
 ?outer
starting httpd help server ... done
 x - outer(Time, Time, FUN = function(a, b){d - b-a; (d=0)  (d = 1000)})
 x
   [,1]  [,2]  [,3]  [,4]  [,5]  [,6]  [,7]  [,8]  [,9]
 [1,]  TRUE  TRUE  TRUE  TRUE FALSE FALSE FALSE FALSE FALSE
 [2,] FALSE  TRUE  TRUE  TRUE FALSE FALSE FALSE FALSE FALSE
 [3,] FALSE FALSE  TRUE  TRUE FALSE FALSE FALSE FALSE FALSE
 [4,] FALSE FALSE FALSE  TRUE  TRUE FALSE FALSE FALSE FALSE
 [5,] FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE
 [6,] FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE
 [7,] FALSE FALSE FALSE FALSE FALSE FALSE  TRUE  TRUE  TRUE
 [8,] FALSE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE  TRUE
 [9,] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE


This says, reading down the columns, that event 4 occurs after 1, 2 
3 within the window; event 9 occurs after 7  8 within the window;
etc.

On Thu, Aug 18, 2011 at 1:29 PM, jabroesch james.broe...@gmail.com wrote:
 Hello all,
 I have a question which I have been struggling with for several weeks
 now, that I think might be easy for more proficient coders than
 myself. I have a large behavioral dataset, with behaviors and the
 times (milliseconds) that they occurred. Each subject has a separate
 file, and a sample subject file can be generated using the following
 syntax:

 Time - c(1000, 1050, 1100, 1500, 2500, 5000, 6500, 6600, 7000)
 Behavior - c(g, a, s, 5,  z, g, z, g, a)
 mydata - data.frame(Time,Behavior)

 My basic goal is to be able to extract some details about what
 behaviors follow another specific behavior within a time window
 (say1000 milliseconds). I figured out how to determine if one specific
 behavior follows another specific behavior within that window with the
 following syntax.

 TimeG=mydata$Time[mydata$Behavior == g]
 TimeA=mydata$Time[mydata$Behavior == a]
 out=rep(NA, length(TimeG))

 for (i in 1:length(TimeG)){tmp = TimeA-TimeG[i]
 out[i]=(sum(0  tmp  tmp =1000 )0 ) }

 number_of_behaviors-length(TimeG)
 number_of_affectmirroring-sum(out)

 This generates 2 values: the number of times that the target behavior
 g occurred, and the number of times that it was followed by the
 behavior a within 1000 milliseconds.

 Question:
 What I can't seem to figure out is a to generate a count of the number
 of times that multiple different types of behaviors immediately follow
 a specific behavior within 1000 milliseconds. So say the behavior of
 interest is “g” as it is in the example above. I want to determine
 1)what was the next behavior (from a specified list of possible
 behaviors bellow) that followed it within 1000 milliseconds.

 Ideally the output would 1 row with be 13 columns. The first column
 would be the number of times that the target behavior, g in this
 example occurs. The next 12 columns would be the number of times that
 one of the specific behaviors was the next behavior that followed
 within 1000 milliseconds.  So one column for each of these behaviors :
 a s d z x c v q w e r t.

 The two complicating factors are: 1)there might be multiple behaviors
 that followed within 1000 milliseconds, and I only want to count the
 first one; and 2)there are additional behaviors that I would like to
 ignore (like the 5 in the example above).

 Any help or suggestions are appreciated.

 Thank you,
 James Broesch


 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Coding-question-for-behavioral-data-analysis-tp3753151p3753151.html
 Sent from the R help mailing list archive at Nabble.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.





-- 
Jim Holtman
Data Munger Guru

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] Convert week value to date

2011-08-19 Thread Heikki Kaskelma

Folkes, Michael:

 I now realize I could write code to evaluate which of the first 7 days
 in the year is a Monday and then I'd know the start of week 1 in each
 year, and multiply from there.

But note that

 library(surveillance) # ISO week
 isoWeekYear(as.Date(2010-01-01))$ISOWeek
[1] 53

so that 2010-01-01 is actually on the week 53 of the year 2009.
January 4th is always on week 1 of the same year

 isoWeekYear(as.Date(2010-01-04))$ISOWeek
[1] 1

and every now and then there are 53 weeks in a year, not 52.


Heikki Kaskelma
Munkkiniemi

__
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] postscript( does not save the plot

2011-08-19 Thread Alaios

Dear Marc,
I would like to thank you for your answer.
Unfortunately still

setEPS() 
postscript(file=exponcoverapprox.eps)
boxplot(test[30,1:500],exponper[90,1:500],test[150,1:500],test[210,1:500],test[270,1:500],test[330,1:500],test[390,1:500],names=c(1,3,5,8,10,13,15),outline=FALSE,ylim=c(0.01,50),log=y,
 xlab = Xlabel,ylab=Ylabel,boxwex=0.5,pars=list(whisklwd=0,staplelwd=0) )
dev.off()

still does not work, though the 



setEPS() 
postscript(file=exponcoverapprox.eps)
boxplot(test[30,1:500],exponper[90,1:500],test[150,1:500],test[210,1:500],test[270,1:500],test[330,1:500],test[390,1:500],names=c(1,3,5,8,10,13,15),outline=FALSE,ylim=c(0.01,50),log=y,
 xlab = 
Xlabel,ylab=Ylabel,boxwex=0.5)
dev.off()

works fine! It seems that the problem is with the pars=list(.

Just to make it more clear. The dev.off() returns 1 and the file is created. 
The problem is that this file can not be open with any program, while all the 
other .eps files I have and were created by R, with the above methodology work 
really nice.

B.R
Alex




From: Marc Schwartz marc_schwa...@me.com

Cc: R-help@r-project.org R-help@r-project.org
Sent: Wednesday, August 17, 2011 5:48 PM
Subject: Re: [R] postscript( does not save the plot

Not sure what output you get in the first case. You don't need:

  ps.options=setEPS()

just:

  setEPS()


Using:

set.seed(1)
test - matrix(runif(500*500), 500)

setEPS()

postscript(file = exponcoverapprox.eps)

boxplot(test[30, 1:500], test[90, 1:500], test[150, 1:500], test[210, 1:500],
        test[270, 1:500], test[330, 1:500], test[390, 1:500],
        names = c(1, 3, 5, 8, 10, 13, 1), outline = FALSE,
        ylim=c(0.01, 50), log = y, xlab = xvalue, ylab = yvalue,
        boxwex=0.5, pars = list(whisklwd = 0, staplelwd = 0))

dev.off()


I get the attached output which seems to be OK.

Marc



On Aug 17, 2011, at 10:02 AM, Alaios wrote:

 The problem is a bit weird.
 
 This does not work:
 
 ps.options=setEPS()
 postscript(file=exponcoverapprox.eps)
 boxplot(test[30,1:500],test[90,1:500],test[150,1:500],test[210,1:500],test[270,1:500],test[330,1:500],test[390,1:500],names=c(1,3,5,8,10,13,1),outline=FALSE,ylim=c(0.01,50),log=y,
  xlab = xvalue,ylab=yvalue,boxwex=0.5, pars 
 =list(whisklwd=0,staplelwd=0))
 dev.off()
 
 
 
 
 This works
 postscript(file=exponcoverapprox.eps)
 boxplot(test[30,1:500],test[90,1:500],test[150,1:500],test[210,1:500],test[270,1:500],test[330,1:500],test[390,1:500],names=c(1,3,5,8,10,13,1),outline=FALSE,ylim=c(0.01,50),log=y,
  xlab = xvalue,ylab=yvalue,boxwex=0.5)
 dev.off()
 
 
 
 To not bother you with the details, the only difference is the pars 
 =list(whisklwd=0,staplelwd=0) at the end of the boxplot , which I use to 
 remove the whiskers fromt he blot.
 
 B.R
 
 
 
 
 From: Marc Schwartz marc_schwa...@me.com
 
 Cc: R-help@r-project.org R-help@r-project.org
 Sent: Tuesday, August 16, 2011 7:38 PM
 Subject: Re: [R] postscript( does not save the plot
 
 On Aug 16, 2011, at 12:32 PM, Alaios wrote:
 
 Dear all,
 I am using the following code to write the plot to an eps format
 
 postscript(file=test.eps,horizontal=FALSE)
 
 boxplot(test[30,1:500],test[90,1:500],test[150,1:500],test[210,1:500],test[270,1:500],test[330,1:500],test[390,1:500],names=c(1,3,5,8,10,13,1),outline=FALSE,ylim=c(0.01,50),log=y,
  xlab = xvalue,ylab=yvalue,boxwex=0.5, pars 
 =list(whisklwd=0,staplelwd=0))
 
 dev.off()
 
 
 This creates a 6kb eps file, that can not be opened by any program. I tired 
 with photoshop gimp, acrobat reader. This is the normal process I follow to 
 save my plots.
 
 dev.off always returns 1.
 
 and the boxplot function succesfullu does the plot in the screen.
 
 What might be the problem?
 
 I would like to thank you in advance for your help
 
 B.R
 Alex
 
 
 
 You did not create an EPS file. See ?postscript and pay attention to the 
 fourth paragraph under Details:
 
 The postscript produced for a single R plot is EPS (Encapsulated PostScript) 
 compatible, and can be included into other documents, e.g., into LaTeX, using 
 \includegraphics{filename}. For use in this way you will probably want to 
 use setEPS() to set the defaults as horizontal = FALSE, onefile = FALSE, 
 paper = special. Note that the bounding box is for the device region: if 
 you find the white space around the plot region excessive, reduce the margins 
 of the figure region viapar(mar=).
 
 
 HTH,
 
 Marc Schwartz
[[alternative HTML version deleted]]

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

Re: [R] Dates - week and year not day.

Tack on a day of the week (6 as the last day) for a point of reference
for the conversion:

 dates - paste('6.', 0:53, '.2011', sep = '')
 dates
 [1] 6.0.2011  6.1.2011  6.2.2011  6.3.2011  6.4.2011
6.5.2011  6.6.2011
 [8] 6.7.2011  6.8.2011  6.9.2011  6.10.2011 6.11.2011
6.12.2011 6.13.2011
[15] 6.14.2011 6.15.2011 6.16.2011 6.17.2011 6.18.2011
6.19.2011 6.20.2011
[22] 6.21.2011 6.22.2011 6.23.2011 6.24.2011 6.25.2011
6.26.2011 6.27.2011
[29] 6.28.2011 6.29.2011 6.30.2011 6.31.2011 6.32.2011
6.33.2011 6.34.2011
[36] 6.35.2011 6.36.2011 6.37.2011 6.38.2011 6.39.2011
6.40.2011 6.41.2011
[43] 6.42.2011 6.43.2011 6.44.2011 6.45.2011 6.46.2011
6.47.2011 6.48.2011
[50] 6.49.2011 6.50.2011 6.51.2011 6.52.2011 6.53.2011
 as.POSIXct(dates, format = %w.%W.%Y)
 [1] 2011-01-01 EST 2011-01-08 EST 2011-01-15 EST 2011-01-22
EST 2011-01-29 EST
 [6] 2011-02-05 EST 2011-02-12 EST 2011-02-19 EST 2011-02-26
EST 2011-03-05 EST
[11] 2011-03-12 EST 2011-03-19 EDT 2011-03-26 EDT 2011-04-02
EDT 2011-04-09 EDT
[16] 2011-04-16 EDT 2011-04-23 EDT 2011-04-30 EDT 2011-05-07
EDT 2011-05-14 EDT
[21] 2011-05-21 EDT 2011-05-28 EDT 2011-06-04 EDT 2011-06-11
EDT 2011-06-18 EDT
[26] 2011-06-25 EDT 2011-07-02 EDT 2011-07-09 EDT 2011-07-16
EDT 2011-07-23 EDT
[31] 2011-07-30 EDT 2011-08-06 EDT 2011-08-13 EDT 2011-08-20
EDT 2011-08-27 EDT
[36] 2011-09-03 EDT 2011-09-10 EDT 2011-09-17 EDT 2011-09-24
EDT 2011-10-01 EDT
[41] 2011-10-08 EDT 2011-10-15 EDT 2011-10-22 EDT 2011-10-29
EDT 2011-11-05 EDT
[46] 2011-11-12 EST 2011-11-19 EST 2011-11-26 EST 2011-12-03
EST 2011-12-10 EST
[51] 2011-12-17 EST 2011-12-24 EST 2011-12-31 EST NA


On Tue, Aug 16, 2011 at 4:01 AM, holdnatalie osp...@bangor.ac.uk wrote:
 Hi,

 I would be very grateful for some advice. I have read the help pages for
 Date, strptime, etc. All examples seem to use some version of day month year
 as date format. However I have Weekly composite data so ONLY want to input
 the dates as Week.Year (eg 35.2011). strptime seems to show this is possible
 using %W for week (UK convention) and %Y for year. My data is in a df called
 chlorophyll and has a date column. I tried to use the following (after I
 converted to characters using as.character);

 chlorophyll$date - strptime(chlorophyll$date, %W.%Y)

 It recognised the year but replaced the week part with todays date (16TH
 August).

 Any advice?
 Thanks
 Natalie

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Dates-week-and-year-not-day-tp3746591p3746591.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.




-- 
Jim Holtman
Data Munger Guru

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] postscript( does not save the plot



On Aug 19, 2011, at 8:06 AM, Alaios wrote:


Dear Marc,
I would like to thank you for your answer.
Unfortunately still

setEPS()
postscript(file=exponcoverapprox.eps)
boxplot
(test
[30,1
: 
500 
],exponper 
[90,1 
: 
500 
],test 
[150,1 
: 
500 
],test 
[210,1 
: 
500 
],test 
[270,1 
: 
500 
],test 
[330,1 
: 
500 
],test 
[390,1 
: 
500 
],names 
= 
c 
(1 
,3,5,8,10,13,15),outline=FALSE,ylim=c(0.01,50),log=y,  
xlab =  
Xlabel,ylab=Ylabel,boxwex=0.5,pars=list(whisklwd=0,staplelwd=0) )

dev.off()

still does not work, though the


The code you posted does not parse. There is a lissing closing quote  
in the ylab argument.



--
David.





setEPS()
postscript(file=exponcoverapprox.eps)
boxplot 
(test 
[30,1 
: 
500 
],exponper 
[90,1 
: 
500 
],test 
[150,1 
: 
500 
],test 
[210,1 
: 
500 
],test 
[270,1 
: 
500 
],test 
[330,1 
: 
500 
],test 
[390,1 
: 
500 
],names 
= 
c 
(1 
,3,5,8,10,13,15),outline=FALSE,ylim=c(0.01,50),log=y,

xlab =
Xlabel,ylab=Ylabel,boxwex=0.5)
dev.off()

works fine! It seems that the problem is with the pars=list(.

Just to make it more clear. The dev.off() returns 1 and the file is  
created. The problem is that this file can not be open with any  
program, while all the other .eps files I have and were created by  
R, with the above methodology work really nice.


B.R
Alex




From: Marc Schwartz marc_schwa...@me.com

Cc: R-help@r-project.org R-help@r-project.org
Sent: Wednesday, August 17, 2011 5:48 PM
Subject: Re: [R] postscript( does not save the plot

Not sure what output you get in the first case. You don't need:

  ps.options=setEPS()

just:

  setEPS()


Using:

set.seed(1)
test - matrix(runif(500*500), 500)

setEPS()

postscript(file = exponcoverapprox.eps)

boxplot(test[30, 1:500], test[90, 1:500], test[150, 1:500],  
test[210, 1:500],

test[270, 1:500], test[330, 1:500], test[390, 1:500],
names = c(1, 3, 5, 8, 10, 13, 1), outline =  
FALSE,

ylim=c(0.01, 50), log = y, xlab = xvalue, ylab = yvalue,
boxwex=0.5, pars = list(whisklwd = 0, staplelwd = 0))

dev.off()


I get the attached output which seems to be OK.

Marc



On Aug 17, 2011, at 10:02 AM, Alaios wrote:


The problem is a bit weird.

This does not work:

ps.options=setEPS()
postscript(file=exponcoverapprox.eps)
boxplot 
(test 
[30,1 
: 
500 
],test 
[90,1 
: 
500 
],test 
[150,1 
: 
500 
],test 
[210,1 
: 
500 
],test 
[270,1 
: 
500 
],test 
[330,1 
: 
500 
],test 
[390,1 
: 
500 
],names 
= 
c 
(1 
,3,5,8,10,13,1),outline=FALSE,ylim=c(0.01,50),log=y,  
xlab = xvalue,ylab=yvalue,boxwex=0.5, pars

=list(whisklwd=0,staplelwd=0))
dev.off()




This works
postscript(file=exponcoverapprox.eps)
boxplot 
(test 
[30,1 
: 
500 
],test 
[90,1 
: 
500 
],test 
[150,1 
: 
500 
],test 
[210,1 
: 
500 
],test 
[270,1 
: 
500 
],test 
[330,1 
: 
500 
],test 
[390,1 
: 
500 
],names 
= 
c 
(1 
,3,5,8,10,13,1),outline=FALSE,ylim=c(0.01,50),log=y,  
xlab = xvalue,ylab=yvalue,boxwex=0.5)

dev.off()



To not bother you with the details, the only difference is the pars
=list(whisklwd=0,staplelwd=0) at the end of the boxplot , which I  
use to remove the whiskers fromt he blot.


B.R




From: Marc Schwartz marc_schwa...@me.com

Cc: R-help@r-project.org R-help@r-project.org
Sent: Tuesday, August 16, 2011 7:38 PM
Subject: Re: [R] postscript( does not save the plot

On Aug 16, 2011, at 12:32 PM, Alaios wrote:


Dear all,
I am using the following code to write the plot to an eps format

postscript(file=test.eps,horizontal=FALSE)

boxplot 
(test 
[30,1 
: 
500 
],test 
[90,1 
: 
500 
],test 
[150,1 
: 
500 
],test 
[210,1 
: 
500 
],test 
[270,1 
: 
500 
],test 
[330,1 
: 
500 
],test 
[390,1 
: 
500 
],names 
= 
c 
(1 
,3 
,5,8,10,13,1),outline=FALSE,ylim=c(0.01,50),log=y,  
xlab = xvalue,ylab=yvalue,boxwex=0.5, pars  
=list(whisklwd=0,staplelwd=0))


dev.off()


This creates a 6kb eps file, that can not be opened by any  
program. I tired with photoshop gimp, acrobat reader. This is the  
normal process I follow to save my plots.


dev.off always returns 1.

and the boxplot function succesfullu does the plot in the screen.

What might be the problem?

I would like to thank you in advance for your help

B.R
Alex




You did not create an EPS file. See ?postscript and pay attention  
to the fourth paragraph under Details:


The postscript produced for a single R plot is EPS (Encapsulated  
PostScript) compatible, and can be included into other documents,  
e.g., into LaTeX, using \includegraphics{filename}. For use in  
this way you will probably want to use setEPS() to set the defaults  
as horizontal = FALSE, onefile = FALSE, paper = special. Note  
that the bounding box is for the device region: if you find the  
white space around the plot region excessive, reduce the margins of  
the figure region viapar(mar=).



HTH,

Marc Schwartz

[[alternative HTML version deleted]]

__

Re: [R] postscript( does not save the plot

Sometimes when I have a script that does not close out a graphics
device correctly (using PDF), I sometimes have problems opening up the
file.  I use the following command to make sure all graphics devices
are closed before generating plots after a script has not terminated
correctly:

graphics.off()

Try this command before generating your plots.

On Fri, Aug 19, 2011 at 8:06 AM, Alaios ala...@yahoo.com wrote:
 Dear Marc,
 I would like to thank you for your answer.
 Unfortunately still

 setEPS()
 postscript(file=exponcoverapprox.eps)
 boxplot(test[30,1:500],exponper[90,1:500],test[150,1:500],test[210,1:500],test[270,1:500],test[330,1:500],test[390,1:500],names=c(1,3,5,8,10,13,15),outline=FALSE,ylim=c(0.01,50),log=y,
  xlab = Xlabel,ylab=Ylabel,boxwex=0.5,pars=list(whisklwd=0,staplelwd=0) )
 dev.off()

 still does not work, though the



 setEPS()
 postscript(file=exponcoverapprox.eps)
 boxplot(test[30,1:500],exponper[90,1:500],test[150,1:500],test[210,1:500],test[270,1:500],test[330,1:500],test[390,1:500],names=c(1,3,5,8,10,13,15),outline=FALSE,ylim=c(0.01,50),log=y,
  xlab =
 Xlabel,ylab=Ylabel,boxwex=0.5)
 dev.off()

 works fine! It seems that the problem is with the pars=list(.

 Just to make it more clear. The dev.off() returns 1 and the file is created. 
 The problem is that this file can not be open with any program, while all the 
 other .eps files I have and were created by R, with the above methodology 
 work really nice.

 B.R
 Alex



 
 From: Marc Schwartz marc_schwa...@me.com

 Cc: R-help@r-project.org R-help@r-project.org
 Sent: Wednesday, August 17, 2011 5:48 PM
 Subject: Re: [R] postscript( does not save the plot

 Not sure what output you get in the first case. You don't need:

   ps.options=setEPS()

 just:

   setEPS()


 Using:

 set.seed(1)
 test - matrix(runif(500*500), 500)

 setEPS()

 postscript(file = exponcoverapprox.eps)

 boxplot(test[30, 1:500], test[90, 1:500], test[150, 1:500], test[210, 1:500],
         test[270, 1:500], test[330, 1:500], test[390, 1:500],
         names = c(1, 3, 5, 8, 10, 13, 1), outline = FALSE,
         ylim=c(0.01, 50), log = y, xlab = xvalue, ylab = yvalue,
         boxwex=0.5, pars = list(whisklwd = 0, staplelwd = 0))

 dev.off()


 I get the attached output which seems to be OK.

 Marc



 On Aug 17, 2011, at 10:02 AM, Alaios wrote:

 The problem is a bit weird.

 This does not work:

 ps.options=setEPS()
 postscript(file=exponcoverapprox.eps)
 boxplot(test[30,1:500],test[90,1:500],test[150,1:500],test[210,1:500],test[270,1:500],test[330,1:500],test[390,1:500],names=c(1,3,5,8,10,13,1),outline=FALSE,ylim=c(0.01,50),log=y,
  xlab = xvalue,ylab=yvalue,boxwex=0.5, pars
 =list(whisklwd=0,staplelwd=0))
 dev.off()




 This works
 postscript(file=exponcoverapprox.eps)
 boxplot(test[30,1:500],test[90,1:500],test[150,1:500],test[210,1:500],test[270,1:500],test[330,1:500],test[390,1:500],names=c(1,3,5,8,10,13,1),outline=FALSE,ylim=c(0.01,50),log=y,
  xlab = xvalue,ylab=yvalue,boxwex=0.5)
 dev.off()



 To not bother you with the details, the only difference is the pars
 =list(whisklwd=0,staplelwd=0) at the end of the boxplot , which I use to 
 remove the whiskers fromt he blot.

 B.R



 
 From: Marc Schwartz marc_schwa...@me.com

 Cc: R-help@r-project.org R-help@r-project.org
 Sent: Tuesday, August 16, 2011 7:38 PM
 Subject: Re: [R] postscript( does not save the plot

 On Aug 16, 2011, at 12:32 PM, Alaios wrote:

 Dear all,
 I am using the following code to write the plot to an eps format

 postscript(file=test.eps,horizontal=FALSE)

 boxplot(test[30,1:500],test[90,1:500],test[150,1:500],test[210,1:500],test[270,1:500],test[330,1:500],test[390,1:500],names=c(1,3,5,8,10,13,1),outline=FALSE,ylim=c(0.01,50),log=y,
  xlab = xvalue,ylab=yvalue,boxwex=0.5, pars 
 =list(whisklwd=0,staplelwd=0))

 dev.off()


 This creates a 6kb eps file, that can not be opened by any program. I tired 
 with photoshop gimp, acrobat reader. This is the normal process I follow to 
 save my plots.

 dev.off always returns 1.

 and the boxplot function succesfullu does the plot in the screen.

 What might be the problem?

 I would like to thank you in advance for your help

 B.R
 Alex



 You did not create an EPS file. See ?postscript and pay attention to the 
 fourth paragraph under Details:

 The postscript produced for a single R plot is EPS (Encapsulated PostScript) 
 compatible, and can be included into other documents, e.g., into LaTeX, 
 using \includegraphics{filename}. For use in this way you will probably 
 want to use setEPS() to set the defaults as horizontal = FALSE, onefile = 
 FALSE, paper = special. Note that the bounding box is for the device 
 region: if you find the white space around the plot region excessive, reduce 
 the margins of the figure region viapar(mar=).


 HTH,

 Marc Schwartz
        [[alternative HTML version deleted]]


 __

Re: [R] postscript( does not save the plot

2011-08-19 Thread Alaios

Dear al,
I would like to thank you for your replies.

I have tried with graphics.off() but did not help too.

I am also sorry that my example was not reproducible

So this one



setEPS() 
postscript(file=mytest.eps)
boxplot(test[30,1:500],test[90,1:500],test[150,1:500],test[210,1:500],test[270,1:500],test[330,1:500],test[390,1:500],names=c(1,3,5,8,10,13,15),outline=FALSE,ylim=c(0.01,50),log=y,
 xlab = Number of Sensors/km^2,ylab=Percentage of Coverage Error Estimations 
(log scale),boxwex=0.5)
dev.off()


always saves to a eps  file that works

while this one



setEPS() 
postscript(file=mytest.eps)
boxplot(test[30,1:500],test[90,1:500],test[150,1:500],test[210,1:500],test[270,1:500],test[330,1:500],test[390,1:500],names=c(1,3,5,8,10,13,15),outline=FALSE,ylim=c(0.01,50),log=y,
 xlab = Number of Sensors/km^2,ylab=Percentage of Coverage Error 
Estimations (log scale),boxwex=0.5,pars=list(whisklwd=0,staplelwd=0))
dev.off()
 

will create an eps file that is not saved. I think is the last part of the 
boxplot pars=list(whisklwd=0,staplelwd=0) that creates that.

What should I do to debug this?

B.R
Alex





# Trashes
test-exponper

setEPS() 
postscript(file=mytest.eps)
boxplot(test[30,1:500],test[90,1:500],test[150,1:500],test[210,1:500],test[270,1:500],test[330,1:500],test[390,1:500],names=c(1,3,5,8,10,13,15),outline=FALSE,ylim=c(0.01,50),log=y,
 xlab = Number of Sensors/km^2,ylab=Percentage of Coverage Error Estimations 
(log scale),boxwex=0.5)
# legend(topright,c(Exponential))
dev.off()





From: jim holtman jholt...@gmail.com

Cc: Marc Schwartz marc_schwa...@me.com; R-help@r-project.org 
R-help@r-project.org
Sent: Friday, August 19, 2011 2:27 PM
Subject: Re: [R] postscript( does not save the plot

Sometimes when I have a script that does not close out a graphics
device correctly (using PDF), I sometimes have problems opening up the
file.  I use the following command to make sure all graphics devices
are closed before generating plots after a script has not terminated
correctly:

graphics.off()

Try this command before generating your plots.


 Dear Marc,
 I would like to thank you for your answer.
 Unfortunately still

 setEPS()
 postscript(file=exponcoverapprox.eps)
 boxplot(test[30,1:500],exponper[90,1:500],test[150,1:500],test[210,1:500],test[270,1:500],test[330,1:500],test[390,1:500],names=c(1,3,5,8,10,13,15),outline=FALSE,ylim=c(0.01,50),log=y,
  xlab = Xlabel,ylab=Ylabel,boxwex=0.5,pars=list(whisklwd=0,staplelwd=0) )
 dev.off()

 still does not work, though the



 setEPS()
 postscript(file=exponcoverapprox.eps)
 boxplot(test[30,1:500],exponper[90,1:500],test[150,1:500],test[210,1:500],test[270,1:500],test[330,1:500],test[390,1:500],names=c(1,3,5,8,10,13,15),outline=FALSE,ylim=c(0.01,50),log=y,
  xlab =
 Xlabel,ylab=Ylabel,boxwex=0.5)
 dev.off()

 works fine! It seems that the problem is with the pars=list(.

 Just to make it more clear. The dev.off() returns 1 and the file is created. 
 The problem is that this file can not be open with any program, while all the 
 other .eps files I have and were created by R, with the above methodology 
 work really nice.

 B.R
 Alex



 
 From: Marc Schwartz marc_schwa...@me.com

 Cc: R-help@r-project.org R-help@r-project.org
 Sent: Wednesday, August 17, 2011 5:48 PM
 Subject: Re: [R] postscript( does not save the plot

 Not sure what output you get in the first case. You don't need:

   ps.options=setEPS()

 just:

   setEPS()


 Using:

 set.seed(1)
 test - matrix(runif(500*500), 500)

 setEPS()

 postscript(file = exponcoverapprox.eps)

 boxplot(test[30, 1:500], test[90, 1:500], test[150, 1:500], test[210, 1:500],
         test[270, 1:500], test[330, 1:500], test[390, 1:500],
         names = c(1, 3, 5, 8, 10, 13, 1), outline = FALSE,
         ylim=c(0.01, 50), log = y, xlab = xvalue, ylab = yvalue,
         boxwex=0.5, pars = list(whisklwd = 0, staplelwd = 0))

 dev.off()


 I get the attached output which seems to be OK.

 Marc



 On Aug 17, 2011, at 10:02 AM, Alaios wrote:

 The problem is a bit weird.

 This does not work:

 ps.options=setEPS()
 postscript(file=exponcoverapprox.eps)
 boxplot(test[30,1:500],test[90,1:500],test[150,1:500],test[210,1:500],test[270,1:500],test[330,1:500],test[390,1:500],names=c(1,3,5,8,10,13,1),outline=FALSE,ylim=c(0.01,50),log=y,
  xlab = xvalue,ylab=yvalue,boxwex=0.5, pars
 =list(whisklwd=0,staplelwd=0))
 dev.off()




 This works
 postscript(file=exponcoverapprox.eps)
 boxplot(test[30,1:500],test[90,1:500],test[150,1:500],test[210,1:500],test[270,1:500],test[330,1:500],test[390,1:500],names=c(1,3,5,8,10,13,1),outline=FALSE,ylim=c(0.01,50),log=y,
  xlab = xvalue,ylab=yvalue,boxwex=0.5)
 dev.off()



 To not bother you with the details, the only difference is the pars
 =list(whisklwd=0,staplelwd=0) at the end of the boxplot , which I use to 
 remove the whiskers fromt he blot.

 B.R



 
 From: Marc Schwartz

Re: [R] postscript( does not save the plot



On Aug 19, 2011, at 9:08 AM, Alaios wrote:


Dear al,
I would like to thank you for your replies.

I have tried with graphics.off() but did not help too.

I am also sorry that my example was not reproducible


It has never been reproducible because you have ignored the request 3  
days ago to supply the 'test' object. You have also ignores the  
request that you post sessionInfo()


--

David Winsemius, MD
West Hartford, CT

__
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] Concatenate two strings in one in a string matrix

Hello

Many thanks.

* is not a typo.   The output is a description of a nonlinear system so
terms such as y(k-1)*y(k-2) are allowed.   I wonder whether  could be
ignored so that the outputs such as y_{01}(k-003)* would not show up.

Cheers

Ed


On Thu, Aug 18, 2011 at 3:40 PM, David Winsemius dwinsem...@comcast.netwrote:


 On Aug 18, 2011, at 2:35 PM, Eduardo Mendes wrote:

  Dear R-Users

 I have the following matrix

  out$desc  [,1][,2]

 [1,]   
 [2,] y_{01}(k-001) 
 [3,] y_{01}(k-002) 
 [4,] y_{01}(k-003) 
 [5,] u_{01}(k-001) 
 [6,] u_{01}(k-002) 
 [7,] u_{01}(k-003) 
 [8,] y_{01}(k-001) y_{01}(k-001)
 [9,] y_{01}(k-001) y_{01}(k-002)
 [10,] y_{01}(k-001) y_{01}(k-003)
 [11,] y_{01}(k-001) u_{01}(k-001)


 and need to concatenate each line to a single string.  Something like

 [2,] y_{01}(k-001)   - [2,] y_{01}(k-001)

 [11,] y_{01}(k-001) u_{01}(k-001) - [11,]
 y_{01}(k-001)*u_{01}(k-001)

 Is there a way to do it without going through every column?


 apply(out$desc, 1, paste, collapse=)

 It is ambiguous what you want for a delimiter. In one case you used , and
 another you used *. I used .


 --

 David Winsemius, MD
 West Hartford, CT



[[alternative HTML version deleted]]

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

Re: [R] More efficient option to append()?

2011-08-19 Thread Paul Hiemstra

 On 08/17/2011 10:53 PM, Alex Ruiz Euler wrote:
 Dear R community,

 I have a 2 million by 2 matrix that looks like this:

 x-sample(1:15,200, replace=T)
 y-sample(1:10*1000, 200, replace=T)
   x y
 [1,] 10  4000
 [2,]  3  1000
 [3,]  3  4000
 [4,]  8  6000
 [5,]  2  9000
 [6,]  3  8000
 [7,]  2 1
 (...)


 The first column is a population expansion factor for the number in the
 second column (household income). I want to expand the second column
 with the first so that I end up with a vector beginning with 10
 observations of 4000, then 3 observations of 1000 and so on. In my mind
 the natural approach would be to create a NULL vector and append the
 expansions:

 myvar-NULL
 myvar-append(myvar, replicate(x[1],y[1]), 1)

 for (i in 2:length(x)) {
 myvar-append(myvar,replicate(x[i],y[i]),sum(x[1:i])+1)
 }

 to end with a vector of sum(x), which in my real database corresponds
 to 22 million observations.

 This works fine --if I only run it for the first, say, 1000
 observations. If I try to perform this on all 2 million observations
 it takes long, way too long for this to be useful (I left it running
 11 hours yesterday to no avail).


 I know R performs well with operations on relatively large vectors. Why
 is this so inefficient? And what would be the smart way to do this?

Hi Alex,

The other reply already gave you the R way of doing this while avoiding
the for loop. However, there is a more general reason why your for loop
is terribly inefficient. A small set of examples:

largeVector = runif(10e4)
outputVector = NULL
system.time(for(i in 1:length(largeVector)) {
outputVector = append(outputVector, largeVector[i] + 1)
})
#   user  system elapsed
 # 6.591   0.168   6.786

The problem in this code is that outputVector keeps on growing and
growing. The operating system needs to allocate more and more space as
the object grows. This process is really slow. Several (much) faster
alternatives exist:

# Pre-allocating the outputVector
outputVector = rep(0,length(largeVector))
system.time(for(i in 1:length(largeVector)) {
outputVector[i] = largeVector[i] + 1
})
#   user  system elapsed
# 0.178   0.000   0.178
# speed up of 37 times, this will only increase for large
# lengths of largeVector

# Using apply functions
system.time(outputVector - sapply(largeVector, function(x) return(x + 1)))
#   user  system elapsed
#  0.124   0.000   0.125
# Even a bit faster

# Using vectorisation
system.time(outputVector - largeVector + 1)
#   user  system elapsed
#  0.000   0.000   0.001
# Practically instant, 6780 times faster than the first example

It is not always clear which method is most suitable and which performs
best. At least they all perform much, much better than the naive option
of letting outputVector grow.

cheers,
Paul

 Thanks in advance.
 Alex

 __
 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.


-- 
Paul Hiemstra, Ph.D.
Global Climate Division
Royal Netherlands Meteorological Institute (KNMI)
Wilhelminalaan 10 | 3732 GK | De Bilt | Kamer B 3.39
P.O. Box 201 | 3730 AE | De Bilt
tel: +31 30 2206 494

http://intamap.geo.uu.nl/~paul
http://nl.linkedin.com/pub/paul-hiemstra/20/30b/770

__
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] More efficient option to append()?

2011-08-19 Thread Paul Hiemstra

 On 08/18/2011 07:46 AM, Timothy Bates wrote:
 This takes a few seconds to do 1 million lines, and remains explicit/for loop 
 form

 numberofSalaryBands = 100 # 200
 x= sample(1:15,numberofSalaryBands, replace=T)
 y= sample((1:10)*1000, numberofSalaryBands, replace=T)
 df   = data.frame(x,y)
 finalN   = sum(df$x)
 myVar= rep(NA, finalN)
 outIndex = 1
 i= 1
 for (i in 1:numberofSalaryBands) {
   kount = df$x[i]
   myVar[outIndex:(outIndex+kount-1)] = rep(df$y[i], kount) # Make x[i] 
 copies of value y[i]

For posterity, the problem in the code of the OP was that myVar was
continuously growing. This required the operating system to continuously
create more space for myVar, which is a very slow process. In this
example you preallocate the space needed for myVar by creating an object
of the appropriate length before the for loop.

So, in my opinion, for loops and append should be avoided like the plague!

my 2cts :)

Paul

   outIndex = outIndex+kount
 }
 head(myVar)
 plyr::count(myVar)


 On Aug 18, 2011, at 12:17 AM, Alex Ruiz Euler wrote:


 Dear R community,

 I have a 2 million by 2 matrix that looks like this:

 x-sample(1:15,200, replace=T)
 y-sample(1:10*1000, 200, replace=T)
  x y
 [1,] 10  4000
 [2,]  3  1000
 [3,]  3  4000
 [4,]  8  6000
 [5,]  2  9000
 [6,]  3  8000
 [7,]  2 1
 (...)


 The first column is a population expansion factor for the number in the
 second column (household income). I want to expand the second column
 with the first so that I end up with a vector beginning with 10
 observations of 4000, then 3 observations of 1000 and so on. In my mind
 the natural approach would be to create a NULL vector and append the
 expansions:

 myvar-NULL
 myvar-append(myvar, replicate(x[1],y[1]), 1)

 for (i in 2:length(x)) {
 myvar-append(myvar,replicate(x[i],y[i]),sum(x[1:i])+1)
 }

 to end with a vector of sum(x), which in my real database corresponds
 to 22 million observations.

 This works fine --if I only run it for the first, say, 1000
 observations. If I try to perform this on all 2 million observations
 it takes long, way too long for this to be useful (I left it running
 11 hours yesterday to no avail).


 I know R performs well with operations on relatively large vectors. Why
 is this so inefficient? And what would be the smart way to do this?

 Thanks in advance.
 Alex

 __
 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.


-- 
Paul Hiemstra, Ph.D.
Global Climate Division
Royal Netherlands Meteorological Institute (KNMI)
Wilhelminalaan 10 | 3732 GK | De Bilt | Kamer B 3.39
P.O. Box 201 | 3730 AE | De Bilt
tel: +31 30 2206 494

http://intamap.geo.uu.nl/~paul
http://nl.linkedin.com/pub/paul-hiemstra/20/30b/770

__
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] More efficient option to append()?

2011-08-19 Thread Paul Hiemstra

 As I already stated in my reply to your earlier post:

resending the answer for the archives of the mailing list...

Hi Alex,

The other reply already gave you the R way of doing this while avoiding
the for loop. However, there is a more general reason why your for loop
is terribly inefficient. A small set of examples:

largeVector = runif(10e4)
outputVector = NULL
system.time(for(i in 1:length(largeVector)) {
outputVector = append(outputVector, largeVector[i] + 1)
})
#   user  system elapsed
 # 6.591   0.168   6.786

The problem in this code is that outputVector keeps on growing and
growing. The operating system needs to allocate more and more space as
the object grows. This process is really slow. Several (much) faster
alternatives exist:

# Pre-allocating the outputVector
outputVector = rep(0,length(largeVector))
system.time(for(i in 1:length(largeVector)) {
outputVector[i] = largeVector[i] + 1
})
#   user  system elapsed
# 0.178   0.000   0.178
# speed up of 37 times, this will only increase for large
# lengths of largeVector

# Using apply functions
system.time(outputVector - sapply(largeVector, function(x) return(x + 1)))
#   user  system elapsed
#  0.124   0.000   0.125
# Even a bit faster

# Using vectorisation
system.time(outputVector - largeVector + 1)
#   user  system elapsed
#  0.000   0.000   0.001
# Practically instant, 6780 times faster than the first example

It is not always clear which method is most suitable and which performs
best. At least they all perform much, much better than the naive option
of letting outputVector grow.

cheers,
Paul



On 08/17/2011 11:17 PM, Alex Ruiz Euler wrote:

 Dear R community,

 I have a 2 million by 2 matrix that looks like this:

 x-sample(1:15,200, replace=T)
 y-sample(1:10*1000, 200, replace=T)
   x y
 [1,] 10  4000
 [2,]  3  1000
 [3,]  3  4000
 [4,]  8  6000
 [5,]  2  9000
 [6,]  3  8000
 [7,]  2 1
 (...)


 The first column is a population expansion factor for the number in the
 second column (household income). I want to expand the second column
 with the first so that I end up with a vector beginning with 10
 observations of 4000, then 3 observations of 1000 and so on. In my mind
 the natural approach would be to create a NULL vector and append the
 expansions:

 myvar-NULL
 myvar-append(myvar, replicate(x[1],y[1]), 1)

 for (i in 2:length(x)) {
 myvar-append(myvar,replicate(x[i],y[i]),sum(x[1:i])+1)
 }

 to end with a vector of sum(x), which in my real database corresponds
 to 22 million observations.

 This works fine --if I only run it for the first, say, 1000
 observations. If I try to perform this on all 2 million observations
 it takes long, way too long for this to be useful (I left it running
 11 hours yesterday to no avail).


 I know R performs well with operations on relatively large vectors. Why
 is this so inefficient? And what would be the smart way to do this?

 Thanks in advance.
 Alex

 __
 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.


-- 
Paul Hiemstra, Ph.D.
Global Climate Division
Royal Netherlands Meteorological Institute (KNMI)
Wilhelminalaan 10 | 3732 GK | De Bilt | Kamer B 3.39
P.O. Box 201 | 3730 AE | De Bilt
tel: +31 30 2206 494

http://intamap.geo.uu.nl/~paul
http://nl.linkedin.com/pub/paul-hiemstra/20/30b/770

__
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] strange convention for time zone names

2011-08-19 Thread Denis Chabot

Hi,

My time zone in Montreal is Standard time zone:UTC/GMT -5 hours  (see 
http://www.timeanddate.com/worldclock/city.html?n=165).

Yet, in R (POSIXct objects) I must specify the opposite, i.e. UTC+5:

dateMontreal = as.POSIXct(2011-01-15 05:00:00, tz=EST)
dateMontreal2 = as.POSIXct(2011-01-15 05:00:00, tz=UTC+5)
wrongdateMontreal = as.POSIXct(2011-01-15 05:00:00, tz=UTC-5)

dateLondon = as.POSIXct(2011-01-15 10:00:00, tz=UTC0)
difftime(dateMontreal, dateLondon)
Time difference of 0 secs

difftime(dateMontreal2, dateLondon)
Time difference of 0 secs

difftime(wrongdateMontreal, dateLondon)
Time difference of -10 hours

Is there a reason for this counter-intuitive convention?

Denis
R version 2.13.1 (2011-07-08)
Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit)

locale:
[1] fr_CA.UTF-8/fr_CA.UTF-8/C/C/fr_CA.UTF-8/fr_CA.UTF-8

attached base packages:
[1] stats graphics  grDevices utils datasets  methods   base 
__
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] Concatenate two strings in one in a string matrix



On Aug 19, 2011, at 9:45 AM, Eduardo Mendes wrote:


Hello

Many thanks.

* is not a typo.   The output is a description of a nonlinear system  
so terms such as y(k-1)*y(k-2) are allowed.   I wonder whether   
could be ignored so that the outputs such as y_{01}(k-003)* would  
not show up.



Paste with *  then remove trailing *'s:

ifelse( grepl(*$, result), sub(*,, result), result)


Cheers

Ed


On Thu, Aug 18, 2011 at 3:40 PM, David Winsemius dwinsem...@comcast.net 
 wrote:


On Aug 18, 2011, at 2:35 PM, Eduardo Mendes wrote:

Dear R-Users

I have the following matrix

out$desc  [,1][,2]
[1,]   
[2,] y_{01}(k-001) 
[3,] y_{01}(k-002) 
[4,] y_{01}(k-003) 
[5,] u_{01}(k-001) 
[6,] u_{01}(k-002) 
[7,] u_{01}(k-003) 
[8,] y_{01}(k-001) y_{01}(k-001)
[9,] y_{01}(k-001) y_{01}(k-002)
[10,] y_{01}(k-001) y_{01}(k-003)
[11,] y_{01}(k-001) u_{01}(k-001)


and need to concatenate each line to a single string.  Something like

[2,] y_{01}(k-001)   - [2,] y_{01}(k-001)

[11,] y_{01}(k-001) u_{01}(k-001) - [11,] y_{01}(k-001)*u_{01} 
(k-001)


Is there a way to do it without going through every column?

apply(out$desc, 1, paste, collapse=)

It is ambiguous what you want for a delimiter. In one case you used  
, and another you used *. I used .



--

David Winsemius, MD
West Hartford, CT




David Winsemius, MD
West Hartford, CT

__
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] Concatenate two strings in one in a string matrix



On Aug 19, 2011, at 10:21 AM, David Winsemius wrote:



On Aug 19, 2011, at 9:45 AM, Eduardo Mendes wrote:


Hello

Many thanks.

* is not a typo.   The output is a description of a nonlinear  
system so terms such as y(k-1)*y(k-2) are allowed.   I wonder  
whether  could be ignored so that the outputs such as y_{01} 
(k-003)* would not show up.



Paste with *  then remove trailing *'s:

ifelse( grepl(*$, result), sub(*,, result), result)


Rather (but still untested)

ifelse( grepl(\\*$, result), sub(\\*,, result), result)

--
david.



Cheers

Ed


On Thu, Aug 18, 2011 at 3:40 PM, David Winsemius dwinsem...@comcast.net 
 wrote:


On Aug 18, 2011, at 2:35 PM, Eduardo Mendes wrote:

Dear R-Users

I have the following matrix

out$desc  [,1][,2]
[1,]   
[2,] y_{01}(k-001) 
[3,] y_{01}(k-002) 
[4,] y_{01}(k-003) 
[5,] u_{01}(k-001) 
[6,] u_{01}(k-002) 
[7,] u_{01}(k-003) 
[8,] y_{01}(k-001) y_{01}(k-001)
[9,] y_{01}(k-001) y_{01}(k-002)
[10,] y_{01}(k-001) y_{01}(k-003)
[11,] y_{01}(k-001) u_{01}(k-001)


and need to concatenate each line to a single string.  Something like

[2,] y_{01}(k-001)   - [2,] y_{01}(k-001)

[11,] y_{01}(k-001) u_{01}(k-001) - [11,] y_{01}(k-001)*u_{01} 
(k-001)


Is there a way to do it without going through every column?

apply(out$desc, 1, paste, collapse=)

It is ambiguous what you want for a delimiter. In one case you used  
, and another you used *. I used .



--

David Winsemius, MD
West Hartford, CT




David Winsemius, MD
West Hartford, CT

__
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.


David Winsemius, MD
West Hartford, CT

__
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 with format()

2011-08-19 Thread Michael Karol

R Users:

  Can anyone please help me with the following:
  I'm unclear as to how to get format to do what I want.
  I've tried the following and get unexpected results.

Input.
val-321.6

format(val, digits=1)
format(val, digits=2) 
format(val, digits=3)
format(val, digits=4)
format(val, digits=5)

Output

[1] 322
[1] 322
[1] 322
[1] 321.6
[1] 321.6

Whereas I would expect to get...

[1] 300
[1] 320
[1] 322
[1] 321.6
[1] 321.60

since digits is defined as the number of significant digits.  The number 
321.6 shown to 1 significant digit should be 300., not 322 which is 3 
significant digits!  Likewise for the other cases.

Can anyone explain what format() is doing? 


Regards, 
Michael

__
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] Help with format()



On Aug 19, 2011, at 10:23 AM, Michael Karol wrote:


R Users:

 Can anyone please help me with the following:
 I'm unclear as to how to get format to do what I want.
 I've tried the following and get unexpected results.

Input.
val-321.6

format(val, digits=1)
format(val, digits=2)
format(val, digits=3)
format(val, digits=4)
format(val, digits=5)

Output

[1] 322
[1] 322
[1] 322
[1] 321.6
[1] 321.6

Whereas I would expect to get...

[1] 300
[1] 320
[1] 322
[1] 321.6
[1] 321.60

since digits is defined as the number of significant digits.   
The number 321.6 shown to 1 significant digit should be 300., not  
322 which is 3 significant digits!  Likewise for the other cases.


Can anyone explain what format() is doing?


As you are observing the significance test only starts to kick in to  
the right of the decimal, but it does not add or extend  
significance to numbers that don't have the specified extent. (I agree  
the help page is not clear on these points.) If you want to force a  
particular size then you would want sprintf or formatC (as the help  
page for format does link to.)





Regards,
Michael

__
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.


David Winsemius, MD
West Hartford, CT

__
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] Leading zeros

2011-08-19 Thread Vasco Cadavez

Hello,
I have a dataset with an Id columns like: 
4/3003
55/333
66/22
 I want to put leading zeros to get:
0004/3003
00055/333
66/22

How can I solve this?
Thanks

Vasco Cadavez 


[[alternative HTML version deleted]]

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

Re: [R] Leading zeros



On Aug 19, 2011, at 11:12 AM, Vasco Cadavez wrote:


Hello,
I have a dataset with an Id columns like:
4/3003
55/333
66/22
I want to put leading zeros to get:
0004/3003
00055/333
66/22

How can I solve this?


?sprintf
?formatC

--

David Winsemius, MD
West Hartford, CT

__
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] Leading zeros



On Aug 19, 2011, at 11:17 AM, David Winsemius wrote:



On Aug 19, 2011, at 11:12 AM, Vasco Cadavez wrote:


Hello,
I have a dataset with an Id columns like:
4/3003
55/333
66/22
I want to put leading zeros to get:
0004/3003
00055/333
66/22

How can I solve this?


?sprintf
?formatC

I may have been too quick. Padding with leading zeros using sprintf is  
described for numeric but not for character types. There are several  
character padding funcitons when you search:


http://search.r-project.org/cgi-bin/namazu.cgi?query=pad+character++max=100result=normalsort=scoreidxname=functionsidxname=Rhelp08idxname=Rhelp10idxname=Rhelp02





--

David Winsemius, MD
West Hartford, CT

__
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.


David Winsemius, MD
West Hartford, CT

__
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] Error in read.dcf(file = tmpf) : Line starting 'head ...' is malformed!

2011-08-19 Thread Benoit Bruneau

Dear R-Users,

I'm trying to setup a personal repository for a few packages I'm working on.
I am on R-Forge but I still need to have various versions of my package that
R-Forge does not build (for R 2.8.1 for example).

So I followed the instructions in this document:
Hhttp://
cran.r-project.org/doc/manuals/R-admin.html#Setting-up-a-package-repository

and used this function as recommended:
*write_PACKAGES()*

Now, when I try to install or update my package, I get:

*install.packages(bmisc, repos=http://www.benoitr.comze.com/R;)**
 Error in read.dcf(file = tmpf) : Line starting 'head ...' is malformed!*

Can someone help me with this ?


Benoit
---
R: 2.13.1

[[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.

[R] AFT model time-dependent with weibull distribution

2011-08-19 Thread javier palacios

Dear R-community,

I have tried to estimate an accelerated failure time(AFT) and proportional
hazard (PH) parametric survival model with time-independent  and
time-dependent covariates. For that purpose, I have used the eha package. 


Please, consider this example: 

weibullph - phreg(Surv(sta,time,S) ~ TDC1 + TIC1, dist=weibull,
data.frame=Data)  

weibullaft-aftreg(Surv(sta,time,S) ~ TDC1 + TIC1, dist=weibull,
data.frame=Data)

## aftreg gives error when I add ID argument...

Error in aftreg.fit(X, Y, dist, strats, offset, init, shape, id, control,  : 
  Overlapping intervals for id  2
 
From help(aftreg): id If there are more than one spell per individual, it is
essential to keep spells together by the id argument. This allows for
time-varying covariates.


data table: 

Data 
   S sta time  TDC1 total_time TIC1  ID
A 1   0   1  48.50   1  1   1
B 0   0   1  65.96   2   1   2 
B 1   1   2  65.08   2  12
C 0   0   1  0.002   4   3 
C 1   1   2  0.002   4   3 
D 0   0   1 72.742   5   4
D 1   1   2 72.522   5   4
E 0   0   1 61.84 2  35
E 0   1   2 60.562   35 
F 0   0   1 35.044   26
F 0   1   2 36.974   26 
F 0   2   3 37.924   26
F 1   3   4 39.014   26

time - time to event 
sta - starting time 
TDC - time dependent covariates
TIC - time independent covariate 
total_time - total time at risk 
ID - ID

1- What happens if the ID is not included? 
2- How can I solve this error?
3- Why the phreg function does not need an ID?


Thanks,

Javier


--
View this message in context: 
http://r.789695.n4.nabble.com/AFT-model-time-dependent-with-weibull-distribution-tp3755079p3755079.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.

[R] ATSP to TSP reformulation

2011-08-19 Thread Ian Deters

Greetings,

I am having trouble getting the function reformulate_ATSP_as_TSP to work for
me.  I have provided a simple example of some of the code I've been using. 
In particular, I'm not sure why I'm getting the error 

Error in dimnames(tsp) - list(lab, lab) : 
  length of 'dimnames' [1] not equal to array extent

since I created the object ATSP with a valid square matrix.  Is there
something simple I'm missing?  The code is below.  Thank you for your time.

x = array(0, dim=c(4,4))
 fix(x)
 x
 col1 col2 col3 col4
[1,]0123
[2,]10   115
[3,]2406
[4,]3560
 library(TSP)
 example = ATSP(x)
 example2 = reformulate_ATSP_as_TSP(example, infeasible = 1000, cheap =
 .0001)
Error in dimnames(tsp) - list(lab, lab) : 
  length of 'dimnames' [1] not equal to array extent

--
View this message in context: 
http://r.789695.n4.nabble.com/ATSP-to-TSP-reformulation-tp3755143p3755143.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.

[R] How to generate piecewise cubic spline with many knots?

2011-08-19 Thread xuyongdeng

Hi all,


I have a series of intra-day data. The variables exhibit a typical daily
pattern over the day. I need to diurnallly adjust the data. It takes the
follow form


1, regress y on a piecewise cubic spline of x with knots (a1,a2,a3,a4...). x
is the time of a day.
2, divide original series by the spline forecast.

I know there are some functions to do the spline, like bs, splint, I do not
know which one is exactly I need.

Does anyway give me some suggestion?

Thanks.

--
View this message in context: 
http://r.789695.n4.nabble.com/How-to-generate-piecewise-cubic-spline-with-many-knots-tp3755419p3755419.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.

[R] Multiple Traveling Salesperson Problem

2011-08-19 Thread Ian Deters

While R has library TSP to help solve traveling salesperson problems, does
anyone know if it has any libraries to help solve multiple traveling
salesperson problems?  For instance, suppose one is planning school bus
routes and one has multiple buses.  Thank you for your time.

--
View this message in context: 
http://r.789695.n4.nabble.com/Multiple-Traveling-Salesperson-Problem-tp3755151p3755151.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.

Re: [R] How to use PC1 of PCA and dim1 of MCA as a predictor in logistic regression model for data reduction

2011-08-19 Thread Mark Difford

On Aug 19, 2011 khosoda wrote:

 I used x10.homals4$objscores[, 1] as a predictor for logistic regression 
 as in the same way as PC1 in PCA. 
 Am I going the right way?

Hi Kohkichi,

Yes, but maybe explore the sets= argument (set Response as the target
variable and the others as the predictor variables). Then use Dim1 scores.
Also think about fitting a rank-1 restricted model, combined with the sets=
option.

See the vignette to the package and look at

@ARTICLE{MIC98,
  author = {Michailides, G. and de Leeuw, J.},
  title = {The {G}ifi system of descriptive multivariate analysis},
  journal = {Statistical Science},
  year = {1998},
  volume = {13},
  pages = {307--336},
  abstract = {}
}

Regards, Mark.

-
Mark Difford (Ph.D.)
Research Associate
Botany Department
Nelson Mandela Metropolitan University
Port Elizabeth, South Africa
--
View this message in context: 
http://r.789695.n4.nabble.com/How-to-use-PC1-of-PCA-and-dim1-of-MCA-as-a-predictor-in-logistic-regression-model-for-data-reduction-tp3750251p3755163.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.

[R] help Dxy and C-index calculation

2011-08-19 Thread Chafika.MAZOUNI

Dear professor,

 

I am currently using Design package and the cph formula for assessing
multivariable analysis.

 

I am tryng to get the C-index for my survival model based on Dxy
coefficient.

 

I am confused since there is a negative value.

 

Do I need to used the absolute Dxy ?

 

  index.orig training test optimism index.corrected  n

Dxy   -0.341357727 -0.344002740 -0.341357727 -0.002645013
-0.338712715 40

R2 0.084694141  0.095440176  0.079077594  0.016362582
0.068331560 40

Slope  1.0  1.0  0.897999711  0.102000289
0.897999711 40

D  0.033368588  0.038458457  0.030983429  0.007475027
0.025893561 40

U -0.002890981 -0.002968495  0.003412045 -0.006380539
0.003489558 40

Q  0.036259569  0.041426951  0.027571385  0.013855566
0.022404003 40

 

Many thanks

 

Dr MAZOUNI

Institut Gustave Roussy

Department of breast surgical oncology

Villejuif, France


[[alternative HTML version deleted]]

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

Re: [R] splitting sample names

2011-08-19 Thread 1Rnwb

Thanks to all of you for the suggestions and corrections.
Sharad

--
View this message in context: 
http://r.789695.n4.nabble.com/splitting-sample-names-tp3753712p3755297.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.

Re: [R] Labelling all variables at once (using Hmisc label)

2011-08-19 Thread Monsieur Do

Indeed, as David pointed out, all the portion that used courier font (all the 
good stuff) was absent from the email posting. 


Thanks for your answers.
[[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.

[R] require(dataset) for example.

2011-08-19 Thread Ken Hutchison

Dear R Users
 Any idea if there exists any one dimensional Cox Process datasets in R?
'Spatstat' is very comprehensive but doesn't seem to have any examples of 1D
(time series) Doubly Stochastic Poisson Process data. (I am aware it can be
simulated)
 Thank you,
   Ken

[[alternative HTML version deleted]]

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

Re: [R] Leading zeros



Copying list one what was sent in reply. Anybody have a better solution?


On Aug 19, 2011, at 11:57 AM, Vasco Cadavez wrote:


Thanks,

A solution can be by substring to remove the /
then numeric will be ok! What you think?

How can I remove the /



with sub or gsub:

 sprintf(%010.0f, as.integer(gsub(/,,  
c(4/3003,55/333,66/22)) ))

[1] 043003 055333 006622

--
David.



Thanks

Vasco Cadavez   

- Menssagem Original -
De:
David Winsemius dwinsem...@comcast.net

Para:
David Winsemius dwinsem...@comcast.net
Cópia:
Vasco Cadavez vcadavez@ipbpt, r-help@r-project.org
Enviado:
Fri, 19 Aug 2011 11:51:08 -0400
Assunto:
Re: [R] Leading zeros



On Aug 19, 2011, at 11:17 AM, David Winsemius wrote:


 On Aug 19, 2011, at 11:12 AM, Vasco Cadavez wrote:

 Hello,
 I have a dataset with an Id columns like:
 4/3003
 55/333
 66/22
 I want to put leading zeros to get:
 0004/3003
 00055/333
 66/22

 How can I solve this?

 ?sprintf
 ?formatC

I may have been too quick. Padding with leading zeros using sprintf  
is

described for numeric but not for character types. There are severa


__
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] Leading zeros

On Fri, Aug 19, 2011 at 9:19 AM, David Winsemius dwinsem...@comcast.net wrote:

 Copying list one what was sent in reply. Anybody have a better solution?


No sure my solution is better, but it avoids the integer conversion
and retains the /.
I wrote a function that padds entries of input character vector with
zeros to final length 'length'.

padZeros = function(x, length)
{
  n = length(x);
  out = rep(, n);
  for (i in 1:length(x))
  {
l = nchar(x[i]);
out[i] = paste( paste(rep(0, length-l), collapse = ), x[i], sep = );
  }
  out
}

 padZeros(c(33/22, 4/50005, 6644/2233), length = 12)
[1] 00033/22 04/50005 0006644/2233

HTH,

Peter

__
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] some questions about stepwise cox regression model and heterogeneity in survival analysis

2011-08-19 Thread asaki1986

Hello, all users here:

Recently i am doing a project of survival analysis. I collect the
characteristics of patients and have got some factors which are related with
the cancer.

When i come to overcome relations between genotypes (Snps) and
survival time via stratified analysis, i come across the problem of how to
use coxph(survival package) to performe multivariate cox regression analysis
with adjustment for some potential confounders. I have not found any
suggestions of the survival package manual.

And when i want to calculate the heterogeneity between genotypes and
other stratified factors, i can not find something command like this (only
survdiff for log-rank test, but no command for heterogeneity test).

At last, stepwise cox regression need to use a significance level of
0.05 for entering and 0.10 for removing the explanatory variables. First how
can i do stepwise cox regression analysis using survival package. And how
can i limit the significance level?

Could anyone here offer me the solution? Thank a lot, it will be
appreciate of your kindly answers!

--
View this message in context:
http://r.789695.n4.nabble.com/some-questions-about-stepwise-cox-regression-model-and-heterogeneity-in-survival-analysis-tp3755591p3755591.html
Sent from the R help mailing list archive at Nabble.com.

[R] sign of the y axis in partialPlot for randomForest regression

2011-08-19 Thread Cédric Mondy

Hi everybody,

I used randomForest to regress invertebrates abundances in least impaired
river reaches from some environmental parameters. Then I used these models
to predict invertebrates abundances in impaired reaches.

Now I would like to model the deviation (observation - prediction) with a
set of chemical parameters to see if the deviations from predictions could
be explained with water chemistry.

I did built the model, and I used partialPlot to  depict the patterns
between individual water chemistry parameters and deviation from prediction.

I know that the range and the values indicated on the y-axis do not
correspond to the 'raw' deviation. However, since my deviations could be
positive (i.e. greater abundance than expected) or negative (i.e. lower
abundance than expected), I would like to know if the signs of the y-axis do
correspond to those of my deviation values: does a negative value in the
y-axis really correspond to a negative value of deviation.


Thanks for your help

Cédric 

--
View this message in context: 
http://r.789695.n4.nabble.com/sign-of-the-y-axis-in-partialPlot-for-randomForest-regression-tp3755583p3755583.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.

[R] adding text to a plot created with strat.plot() from package rioja

2011-08-19 Thread Jason Paul Joines

I have a plot created with strat.plot() from package rioja.  When 
the plot is created with scale.percent=FALSE, each x axes is labeled at 
0 and its maximum.  However, when scale.percent=TRUE, the x axes are not 
labeled.  I need to use scale.percent=TRUE and I need labels for the x axes.


I have been able to add labels to the x axes with mtext but it is 
very tedious to find the correct position.  Is there a better way to do 
this or a better way to find the desired coordinates than trial and error?



Jason
===

__
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] installing packages systemwide

2011-08-19 Thread Mary Kindall

I installed some downloaded packages in R. I always do
$sudo R CMD INSTALL anRpackage.tar.gz


By default it is storing these packages into my directory
/home/mary/R/x86_64-pc-linux-gnu-library/2.13/.

However I want them to be systemwide into /usr/local/lib/R/site-library/
folder.

I tried
$sudo R
R install.packages(anRpackage, dep=TRUE)

I did not succeed into getting them install in req folder.
Any idea?



-- 
-
Mary Kindall
Yorktown Heights, NY
USA

[[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.

[R] Writing non-graphic (text) output to PDF

2011-08-19 Thread Ed Heaton

Hi, friends.

I keep coming to you because I'm so new to R and can't seem to figure out
some simple things.  Sorry.

Consider the following code.  I want to load a table and write out the
structure to a PDF document.  I just can't seem to manage writing
non-graphic output to PDF.  Any help?  I've tried several functions, but
nothing worked.  All I get is the title.

# **
# Load the DEBT table.
  debt - readRDS(T:/R.Data/Debt.rData)
  dim(debt)
# Open the debt.pdf file for graphics output.
  pdf(
file=paste(
  R:/DAS/DMS/FedDebt
 ,DataDiscovery
 ,DistributionAnalysis
 ,Report
 ,Debt.pdf
 ,sep=/
)
  )
# ==
# Write the debt structucture to the output PDF.
  plot.new()
  title(DEBT)
  str(debt)
# ==
  dev.off() # Turn off the PDF device.
# ** End of Program 



Ed

Ed Heaton
Project Manager, Sr. SAS Developer
Data and Analytic Solutions, Inc.
10318 Yearling Drive
Rockville, MD 20850
Office: 301-520-7414
ehea...@dasconsultants.com
www.dasconsultants.com http://www.dasconsultants.com/ 
CMMI ML-2, SBA 8(a)  SDB, WBE (WBENC), MBE (VA  MD)

e...@heaton.name

(Re: http://www.r-project.org/posting-guide.html)

__
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] how to merge distance data based on location

2011-08-19 Thread Matthew Keller

Hi all,

I have two data frames, two columns each, 1000s of rows. Each row
represents a segment of the genome where a deletion has occurred.
First column is start position of the deletion in genomic distance,
second is end position.

So, e.g., first 3 rows of data frame A is:
1003 1023
5932 6120
12348 12689

first 3 rows of data frame B is:
852 5305
1010 1015
8500 9500
1 13000

I want to merge based on distance, such that each row will be
deletions that overlap. So I'd like:
1003 1023852 5305   1010 1015
5932 6120
8500 9500
12348 126891 13000

Does anyone have ideas about how to accomplish this?
Thank you,

Matthew Keller

-- 
Matthew C Keller
Asst. Professor of Psychology
University of Colorado at Boulder
www.matthewckeller.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] 3D surface plot

2011-08-19 Thread Duncan Murdoch


On 11-08-16 9:50 PM, Eric Heupel wrote:

I have what is probably a noob question, but

I am trying to create a 3d plot to illustrate the range of values for the 
following simple function:

A = B*(C/D)

B, C, and D are independent variables whose range are equal (e.g. 1 to 3 
inclusive)

I figure it's not possible to map the surface of A on the 3d space defined by 
B, C and D but I would like to create a surface defined by the lower and upper 
limits of the variables - that is to say a rectangle with corners at (1,1,1), 
(2,3,2), (3,3,3) and (3,2,2) with a color map displayed on it corresponding to 
the values of A and a color key to the side of that.

I have been able to wrap my head part way around persp and wireframe and can 
create a surface for A~B*(C/D) in either, but have not managed to create either 
the


The contour3d function in misc3d might do what you want.  See the examples.

Duncan Murdoch

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

Re: [R] installing packages systemwide

2011-08-19 Thread Yihui Xie

Take a look at:

R CMD INSTALL --help

and you will realize that you need to specify the library path, e.g. R
CMD INSTALL anRpackage --library=/usr/local/...

or take a look at ?install.packages and use the second argument, e.g.
install.packages('anRpackage', lib = '/usr/local/...')


Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA



On Fri, Aug 19, 2011 at 12:10 PM, Mary Kindall mary.kind...@gmail.com wrote:
 I installed some downloaded packages in R. I always do
 $sudo R CMD INSTALL anRpackage.tar.gz


 By default it is storing these packages into my directory
 /home/mary/R/x86_64-pc-linux-gnu-library/2.13/.

 However I want them to be systemwide into /usr/local/lib/R/site-library/
 folder.

 I tried
 $sudo R
 R install.packages(anRpackage, dep=TRUE)

 I did not succeed into getting them install in req folder.
 Any idea?



 --
 -
 Mary Kindall
 Yorktown Heights, NY
 USA

        [[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.


__
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] [R-sig-Debian] installing packages systemwide

2011-08-19 Thread Michael Dewey


At 18:10 19/08/2011, Mary Kindall wrote:


I installed some downloaded packages in R. I always do
$sudo R CMD INSTALL anRpackage.tar.gz


By default it is storing these packages into my directory
/home/mary/R/x86_64-pc-linux-gnu-library/2.13/.

However I want them to be systemwide into /usr/local/lib/R/site-library/
folder.

I tried
$sudo R
R install.packages(anRpackage, dep=TRUE)

I did not succeed into getting them install in req folder.


There is a parameter lib to install.packages. Does that do what you would like?


Any idea?



--
-
Mary Kindall
Yorktown Heights, NY
USA

[[alternative HTML version deleted]]

___
R-SIG-Debian mailing list
r-sig-deb...@r-project.org
https://stat.ethz.ch/mailman/listinfo/r-sig-debian


Michael Dewey
i...@aghmed.fsnet.co.uk
http://www.aghmed.fsnet.co.uk/home.html

__
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] Concatenate two strings in one in a string matrix

Many thanks.

The still untested worked.

Cheers

Ed


On Fri, Aug 19, 2011 at 11:23 AM, David Winsemius dwinsem...@comcast.netwrote:


 On Aug 19, 2011, at 10:21 AM, David Winsemius wrote:


 On Aug 19, 2011, at 9:45 AM, Eduardo Mendes wrote:

  Hello

 Many thanks.

 * is not a typo.   The output is a description of a nonlinear system so
 terms such as y(k-1)*y(k-2) are allowed.   I wonder whether  could be
 ignored so that the outputs such as y_{01}(k-003)* would not show up.

  Paste with *  then remove trailing *'s:

 ifelse( grepl(*$, result), sub(*,, result), result)


 Rather (but still untested)


 ifelse( grepl(\\*$, result), sub(\\*,, result), result)

 --
 david.


  Cheers

 Ed


 On Thu, Aug 18, 2011 at 3:40 PM, David Winsemius dwinsem...@comcast.net
 wrote:

 On Aug 18, 2011, at 2:35 PM, Eduardo Mendes wrote:

 Dear R-Users

 I have the following matrix

 out$desc  [,1][,2]
 [1,]   
 [2,] y_{01}(k-001) 
 [3,] y_{01}(k-002) 
 [4,] y_{01}(k-003) 
 [5,] u_{01}(k-001) 
 [6,] u_{01}(k-002) 
 [7,] u_{01}(k-003) 
 [8,] y_{01}(k-001) y_{01}(k-001)
 [9,] y_{01}(k-001) y_{01}(k-002)
 [10,] y_{01}(k-001) y_{01}(k-003)
 [11,] y_{01}(k-001) u_{01}(k-001)


 and need to concatenate each line to a single string.  Something like

 [2,] y_{01}(k-001)   - [2,] y_{01}(k-001)

 [11,] y_{01}(k-001) u_{01}(k-001) - [11,]
 y_{01}(k-001)*u_{01}(k-001)

 Is there a way to do it without going through every column?

 apply(out$desc, 1, paste, collapse=)

 It is ambiguous what you want for a delimiter. In one case you used ,
 and another you used *. I used .


 --

 David Winsemius, MD
 West Hartford, CT



 David Winsemius, MD
 West Hartford, CT

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


 David Winsemius, MD
 West Hartford, CT



[[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.

[R] gsub for numeric characters in string

2011-08-19 Thread Rebecca Gray

Dear all,

I have what is a bit of a confusing question, so I hope that I can explain
clearly. Thank you for your help in advance.

I would like to do a replacement procedure on several strings, but the way
that I am currently going about it is not working.

I have defined len, which is a series comprising the lengths of different
items, all preceded by a colon.

 len
[1] :328 :154  :135
[4] :147 :30  :50
[7] :252 :45  ;

'lenplustate is a series that is comprised of an attribute of each item
preceding the colon, followed by the length (as defined in len).

 lenplusstate
[1] 1:328  1:154   4:135
[4] NA:147 3:30   2:50
[7] NA:252 NA:45  NA;

tree is a string that gives the specific relationship (via parenthetical
notation) among the different items. Note that the lengths are included in
this tree (following the colon), and the name of each item (1-5) precedes a
colon. However, not every colon is preceded with a name (because there are
internal nodes in the tree structure).

 tree
[1] (*1*:328,((*5*:154,*2*:135):147,(*3*:30,*4*:50):252):45);

I would like to replace the length with the lengthplusstate in the tree,
while removing the names, so that it looks like this:

 theoreticalnewtree
[1] (*1*:328,((*1*:154,*4*:135)NA:147,(*3*:30,*2*:50)NA:252)NA:45);

I am using this code:

  for (j in all)
newtree - gsub(ln[j], lnplusstate[j], tree)

However, I end up with this:

 newtree
[1] (*11*:328,((*51*:154,*24*:135)NA:147,(*33*:30,*42*:50)NA:252)NA:45);


that is, I have not removed the names from the string so that now the state
information is wrong.

If anyone can help with the proper code to get rid of the names in tree
while replacing the length with lengthplustate, I would appreciate it.

I apologize if this was unclear.

All the best,

Rebecca

[[alternative HTML version deleted]]

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

Re: [R] how to merge distance data based on location

2011-08-19 Thread Jean V Adams

r-help-boun...@r-project.org wrote on 08/19/2011 12:15:39 PM:

 [image removed] 
 
 [R] how to merge distance data based on location
 
 Matthew Keller 
 
 to:
 
 r help
 
 08/19/2011 12:18 PM
 
 Sent by:
 
 r-help-boun...@r-project.org
 
 Hi all,
 
 I have two data frames, two columns each, 1000s of rows. Each row
 represents a segment of the genome where a deletion has occurred.
 First column is start position of the deletion in genomic distance,
 second is end position.
 
 So, e.g., first 3 rows of data frame A is:
 1003 1023
 5932 6120
 12348 12689
 
 first 3 rows of data frame B is:
 852 5305
 1010 1015
 8500 9500
 1 13000

The first row of data frame B describes a deletion that fully envelopes 
the deletion described in the second row.  Does this make sense?

 
 I want to merge based on distance, such that each row will be
 deletions that overlap. So I'd like:
 1003 1023852 5305   1010 1015
 5932 6120
 8500 9500
 12348 126891 13000

Would you mind describing what you plan to do with the resulting merged 
data frame?  I ask because there may be some approach (other than data 
frame merging) that might serve your needs better.

What if the second row of data frame B was
1025 1038
it would still overlap with the first row of B, but it wouldn't overlap 
with the first row of A.  How would you want your merged data frame to 
look?

 
 Does anyone have ideas about how to accomplish this?
 Thank you,
 
 Matthew Keller
 
 -- 
 Matthew C Keller
 Asst. Professor of Psychology
 University of Colorado at Boulder
 www.matthewckeller.com
 

Jean


`·.,,  (((º   `·.,,  (((º   `·.,,  (((º

Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East Steinfest Road
Antigo, WI 54409  USA
[[alternative HTML version deleted]]

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

Re: [R] Leading zeros

2011-08-19 Thread Ole Peter Smith

-- Forwarded message --
From: Ole Peter Smith ole@gmail.com
Date: Fri, Aug 19, 2011 at 1:40 PM
Subject: Re: [R] Leading zeros
To: David Winsemius dwinsem...@comcast.net


I'm all new to R, assisting the last days of topics from the sideline.
I am, however, a longterm programmer.

This question makes me ask myself - and now here - is there any string
split-function, ex some thing like:

split('/',2000/3000/4) -- (2000,3000,4)

After this patting with zero is a breeze, sprintf('%010d',...

0le

On Fri, Aug 19, 2011 at 1:19 PM, David Winsemius dwinsem...@comcast.net wrote:

 Copying list one what was sent in reply. Anybody have a better solution?

 On Aug 19, 2011, at 11:57 AM, Vasco Cadavez wrote:

 Thanks,

 A solution can be by substring to remove the /
 then numeric will be ok! What you think?

 How can I remove the /

 with sub or gsub:

 sprintf(%010.0f, as.integer(gsub(/,, c(4/3003,55/333,66/22))
 ))
 [1] 043003 055333 006622

 --
 David.


 Thanks

 Vasco Cadavez

 - Menssagem Original -
 De:
 David Winsemius dwinsem...@comcast.net

 Para:
 David Winsemius dwinsem...@comcast.net
 Cópia:
 Vasco Cadavez vcadavez@ipbpt, r-help@r-project.org
 Enviado:
 Fri, 19 Aug 2011 11:51:08 -0400
 Assunto:
 Re: [R] Leading zeros



 On Aug 19, 2011, at 11:17 AM, David Winsemius wrote:

 
  On Aug 19, 2011, at 11:12 AM, Vasco Cadavez wrote:
 
  Hello,
  I have a dataset with an Id columns like:
  4/3003
  55/333
  66/22
  I want to put leading zeros to get:
  0004/3003
  00055/333
  66/22
 
  How can I solve this?
 
  ?sprintf
  ?formatC
 
 I may have been too quick. Padding with leading zeros using sprintf is
 described for numeric but not for character types. There are severa

 __
 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.




--
                                            /
                                         ( O O )
=oOO==(_)==OOo=
      God does not care about our mathematical difficulties.
                    He integrates empirically - Einstein
                                .oooO   Oooo.
==(     )=(     )=
                                     \  (     )  /
                                      \_)   (_/
===
                        Ole Peter Smith, IME, UFG
    http://www.mat.ufg.br/docentes/olepeter - ole at mat.ufg.br
===
                   Life sure is a Mystery to be Lived
                      Not a Problem to be Solved
===



-- 
                                            /
                                         ( O O )
=oOO==(_)==OOo=
      God does not care about our mathematical difficulties.
                    He integrates empirically - Einstein
                                .oooO   Oooo.
==(     )=(     )=
                                     \  (     )  /
                                      \_)   (_/
===
                        Ole Peter Smith, IME, UFG
    http://www.mat.ufg.br/docentes/olepeter - ole at mat.ufg.br
===
                   Life sure is a Mystery to be Lived
                      Not a Problem to be Solved
===
__
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] chisq.test(): standardized (adjusted) Pearson residuals

2011-08-19 Thread Stephen Davies

I'm using chisq.test() on a matrix of categorical data, and I see that the
residuals attribute of the returned object will give me the Pearson residuals.
That's cool. However, what I'd really like is the standardized (adjusted)
Pearson residuals, which have a N(0,1) distribution. Is there a way to do that
in R (other than by me programming it myself?)

__
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] R and Sweave

2011-08-19 Thread danielepippo

Hi everybody.

 I'm trying to use R with Sweave but I have a problem perhaps with the
directory path of sweave in R.
The windows path is this:
C:\Program Files (x86)\R\R-2.9.2\share\texmf\Sweave

When I run the latex file with R, the program works well, without any
errors, but when I create the pdf file, the Scode doesn't work. I think the
problem is about the path.
Can anybody suggest to me anything to do?

Thanks very much
Cheers

--
View this message in context: 
http://r.789695.n4.nabble.com/R-and-Sweave-tp3755837p3755837.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.

Re: [R] Leading zeros

2011-08-19 Thread R. Michael Weylandt

There is a strsplit() function (syntax is (stringToBeSplit, splitAt) ) that
may be of use. I haven't followed the thread so I don't know how well it
handles the original problem.

Michael Weylandt

On Fri, Aug 19, 2011 at 12:41 PM, Ole Peter Smith ole@gmail.com wrote:

 -- Forwarded message --
 From: Ole Peter Smith ole@gmail.com
 Date: Fri, Aug 19, 2011 at 1:40 PM
 Subject: Re: [R] Leading zeros
 To: David Winsemius dwinsem...@comcast.net


 I'm all new to R, assisting the last days of topics from the sideline.
 I am, however, a longterm programmer.

 This question makes me ask myself - and now here - is there any string
 split-function, ex some thing like:

 split('/',2000/3000/4) -- (2000,3000,4)

 After this patting with zero is a breeze, sprintf('%010d',...

 0le

 On Fri, Aug 19, 2011 at 1:19 PM, David Winsemius dwinsem...@comcast.net
 wrote:
 
  Copying list one what was sent in reply. Anybody have a better solution?
 
  On Aug 19, 2011, at 11:57 AM, Vasco Cadavez wrote:
 
  Thanks,
 
  A solution can be by substring to remove the /
  then numeric will be ok! What you think?
 
  How can I remove the /
 
  with sub or gsub:
 
  sprintf(%010.0f, as.integer(gsub(/,, c(4/3003,55/333,66/22))
  ))
  [1] 043003 055333 006622
 
  --
  David.
 
 
  Thanks
 
  Vasco Cadavez
 
  - Menssagem Original -
  De:
  David Winsemius dwinsem...@comcast.net
 
  Para:
  David Winsemius dwinsem...@comcast.net
  Cópia:
  Vasco Cadavez vcadavez@ipbpt, r-help@r-project.org
  Enviado:
  Fri, 19 Aug 2011 11:51:08 -0400
  Assunto:
  Re: [R] Leading zeros
 
 
 
  On Aug 19, 2011, at 11:17 AM, David Winsemius wrote:
 
  
   On Aug 19, 2011, at 11:12 AM, Vasco Cadavez wrote:
  
   Hello,
   I have a dataset with an Id columns like:
   4/3003
   55/333
   66/22
   I want to put leading zeros to get:
   0004/3003
   00055/333
   66/22
  
   How can I solve this?
  
   ?sprintf
   ?formatC
  
  I may have been too quick. Padding with leading zeros using sprintf is
  described for numeric but not for character types. There are severa
 
  __
  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.
 



 --
 /
  ( O O )
 =oOO==(_)==OOo=
   God does not care about our mathematical difficulties.
 He integrates empirically - Einstein
 .oooO   Oooo.
 ==( )=( )=
  \  ( )  /
   \_)   (_/
 ===
 Ole Peter Smith, IME, UFG
 http://www.mat.ufg.br/docentes/olepeter - ole at mat.ufg.br
 ===
Life sure is a Mystery to be Lived
   Not a Problem to be Solved
 ===



 --
 /
  ( O O )
 =oOO==(_)==OOo=
   God does not care about our mathematical difficulties.
 He integrates empirically - Einstein
 .oooO   Oooo.
 ==( )=( )=
  \  ( )  /
   \_)   (_/
 ===
 Ole Peter Smith, IME, UFG
 http://www.mat.ufg.br/docentes/olepeter - ole at mat.ufg.br
 ===
Life sure is a Mystery to be Lived
   Not a Problem to be Solved
 ===
 __
 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.


[[alternative HTML version deleted]]

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

Re: [R] COXPH TIME-DEPENDENT

2011-08-19 Thread Daniel Malter

B is the specification for time-varying covariates. Otherwise, your model
will think that each row is one independent observation that either had an
event or was censored at time or total_time.

HTH,
Daniel



javier palacios wrote:
 
 Dear R-community, 
 
 which of the following two formats is correct? Are both correct? 
 
 Please, consider this example: 
 
 data table: 
 
 Data
S sta time  TDC1 total_time
 A 1   0   1  48.50   1
 B 0   0   1  65.96   2
 B 1   1   2  65.08   2
 C 0   0   1  0.002
 C 1   1   2  0.002
 D 0   0   1 72.742
 D 1   1   2 72.522
 E 0   0   1 61.84 2
 E 0   1   2 60.562
 F 0   0   1 35.044
 F 0   1   2 36.974
 F 0   2   3 37.924
 F 1   3   4 39.014
 
 time - time to event
 sta - starting time
 TDC - time dependent covariates
 total_time - total time at risk
 
 option A
 
 coxph(Surv(time,S) ~ time_dependent_covariates,
 data=data.frame(Data))
  
 option B
 
 coxph(Surv(sta,time,S) ~ time_dependent_covariates,
 data=data.frame(Data))
 
 option C
 
 coxph(Surv(total_time,S) ~ time_dependent_covariates,
 data=data.frame(Data))
 
 
 How can time at risk be visualized in the coxph output?
 
 Best regards, 
 
 Javier
 

--
View this message in context: 
http://r.789695.n4.nabble.com/COXPH-TIME-DEPENDENT-tp3754837p3755852.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.

Re: [R] chisq.test(): standardized (adjusted) Pearson residuals



On Aug 19, 2011, at 1:28 PM, Stephen Davies wrote:

I'm using chisq.test() on a matrix of categorical data, and I see  
that the
residuals attribute of the returned object will give me the  
Pearson residuals.
That's cool. However, what I'd really like is the standardized  
(adjusted)
Pearson residuals, which have a N(0,1) distribution. Is there a way  
to do that

in R (other than by me programming it myself?)


?scale

--

David Winsemius, MD
West Hartford, CT

__
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] Hmisc::rcorr on a 'data.frame'?

Dear all
?Hmisc::rcorr states that it takes as main argument a numeric
matrix. But is it normal that it fails in such an ugly way on a data
frame? (See below.) If the function didn't attempt any conversion to a
matrix, I would have expected it to state that in the error message
that it didn't accept 'data.frame' objects in its input. Also, I
vaguely remember having used in the past rcorr() on data frames.

Regards
Liviu

 require(Hmisc)
 rcorr(mtcars[ , 1:4])
Error in storage.mode(x) - if (.R.) double else single :
  (list) object cannot be coerced to type 'double'
 rcorr(as.matrix(mtcars[ , 1:4]))
   mpg   cyl  disphp
mpg   1.00 -0.85 -0.85 -0.78
cyl  -0.85  1.00  0.90  0.83
disp -0.85  0.90  1.00  0.79
hp   -0.78  0.83  0.79  1.00

n= 32


P
 mpg cyl disp hp
mpg   0   00
cyl   0   00
disp  0   00
hp0   0   0




-- 
Do you know how to read?
http://www.alienetworks.com/srtest.cfm
http://goodies.xfce.org/projects/applications/xfce4-dict#speed-reader
Do you know how to write?
http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail

__
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] display only the top-right half of a correlation matrix?

Dear all
Is there an easy way to display only one half (top-right or
bottom-left) of a correlation matrix?

 require(Hmisc)
 rcorr(as.matrix(mtcars[ , 1:4]))
   mpg   cyl  disphp
mpg   1.00 -0.85 -0.85 -0.78
cyl  -0.85  1.00  0.90  0.83
disp -0.85  0.90  1.00  0.79
hp   -0.78  0.83  0.79  1.00

n= 32


P
 mpg cyl disp hp
mpg   0   00
cyl   0   00
disp  0   00
hp0   0   0


Since the two sides are identical, there is little value in having
both displayed at the same time. Moreover, it considerably slows down
the inspection of the results.

Thank you
Liviu



-- 
Do you know how to read?
http://www.alienetworks.com/srtest.cfm
http://goodies.xfce.org/projects/applications/xfce4-dict#speed-reader
Do you know how to write?
http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail

__
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] gsub for numeric characters in string

On Fri, Aug 19, 2011 at 11:11 AM, Rebecca Gray atlas...@gmail.com wrote:
 Dear all,

 I have what is a bit of a confusing question, so I hope that I can explain
 clearly. Thank you for your help in advance.

 I would like to do a replacement procedure on several strings, but the way
 that I am currently going about it is not working.

 I have defined len, which is a series comprising the lengths of different
 items, all preceded by a colon.

 len
 [1] :328 :154  :135
 [4] :147 :30  :50
 [7] :252 :45  ;

 'lenplustate is a series that is comprised of an attribute of each item
 preceding the colon, followed by the length (as defined in len).

 lenplusstate
 [1] 1:328  1:154   4:135
 [4] NA:147 3:30   2:50
 [7] NA:252 NA:45  NA;

 tree is a string that gives the specific relationship (via parenthetical
 notation) among the different items. Note that the lengths are included in
 this tree (following the colon), and the name of each item (1-5) precedes a
 colon. However, not every colon is preceded with a name (because there are
 internal nodes in the tree structure).

 tree
 [1] (*1*:328,((*5*:154,*2*:135):147,(*3*:30,*4*:50):252):45);

 I would like to replace the length with the lengthplusstate in the tree,
 while removing the names, so that it looks like this:

 theoreticalnewtree
 [1] (*1*:328,((*1*:154,*4*:135)NA:147,(*3*:30,*2*:50)NA:252)NA:45);


I can help you, but what is the name of each item? I thought it was
the index of the item in the len and lenplusstate variables, but that
apparently is not the case. You have to specify the names as well.

Peter

__
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] display only the top-right half of a correlation matrix?

2011-08-19 Thread Jean V Adams

r-help-boun...@r-project.org wrote on 08/19/2011 01:50:48 PM:

 [image removed] 
 
 [R] display only the top-right half of a correlation matrix?
 
 Liviu Andronic 
 
 to:
 
 r-help@r-project.org Help
 
 08/19/2011 01:55 PM
 
 Sent by:
 
 r-help-boun...@r-project.org
 
 Dear all
 Is there an easy way to display only one half (top-right or
 bottom-left) of a correlation matrix?

See ?lower.tri

 
  require(Hmisc)
  rcorr(as.matrix(mtcars[ , 1:4]))
mpg   cyl  disphp
 mpg   1.00 -0.85 -0.85 -0.78
 cyl  -0.85  1.00  0.90  0.83
 disp -0.85  0.90  1.00  0.79
 hp   -0.78  0.83  0.79  1.00
 
 n= 32
 
 
 P
  mpg cyl disp hp
 mpg   0   00
 cyl   0   00
 disp  0   00
 hp0   0   0
 
 
 Since the two sides are identical, there is little value in having
 both displayed at the same time. Moreover, it considerably slows down
 the inspection of the results.
 
 Thank you
 Liviu
 

Jean

`·.,,  (((º   `·.,,  (((º   `·.,,  (((º

Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East Steinfest Road
Antigo, WI 54409  USA
[[alternative HTML version deleted]]

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

Re: [R] display only the top-right half of a correlation matrix?

On Fri, Aug 19, 2011 at 11:50 AM, Liviu Andronic landronim...@gmail.com wrote:
 Dear all
 Is there an easy way to display only one half (top-right or
 bottom-left) of a correlation matrix?

 require(Hmisc)
 rcorr(as.matrix(mtcars[ , 1:4]))
       mpg   cyl  disp    hp
 mpg   1.00 -0.85 -0.85 -0.78
 cyl  -0.85  1.00  0.90  0.83
 disp -0.85  0.90  1.00  0.79
 hp   -0.78  0.83  0.79  1.00


Use as.dist: here's an example.

 x = matrix(rnorm(5*100), 100, 5)
 as.dist(cor(x))
  1 2 3 4
2 -2.892981e-06
3  2.873711e-02  1.002969e-02
4 -5.803705e-02  4.022733e-02 -6.154211e-02
5  1.137083e-01 -8.065676e-02 -9.279316e-02 -8.201583e-02

__
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] rms:fastbw variable selection differences with AIC .vs. p value methods

2011-08-19 Thread Rob James

I want to employ a parsimonious model to draw nomograms, as the full 
model is too complex to draw nomograms readily (several interactions of 
continuous variables).  However, one interesting variable stays or 
leaves based on whether I choose p value or AIC options to 
fastbw().  My question boils down to this: Is there a theoretical reason 
to prefer one over another?



Consider:

fastbw(model94c, aic=1e10)

 DeletedChi-Sq d.f. P  Residual d.f. P  AIC
 ToD  0.11  3   0.99030.11   3   0.9903   -5.89
 Experience * ToD 2.56  3   0.46462.67   6   0.8487   -9.33
 Experience * Assoc   0.45  2   0.79703.13   8   0.9262  -12.87
 RatePressure 2.99  3   0.39396.11  11   0.8658  -15.89
 DW_height_t  2.92  3   0.40479.03  14   0.8293  -18.97
 TBV * Experience 3.46  3   0.3260   12.49  17   0.7698  -21.51
 Experience * Sex 0.05  1   0.8153   12.54  18   0.8181  -23.46
 Experience   0.18  1   0.67212.72  19   0.8526  -25.28
 Sex  1.19  1   0.2745   13.91  20   0.8348  -26.09
 Assoc6.09  2   0.0475   20.01  22   0.5826  -23.99
 Experience * Pulse  10.53  3   0.0146   30.53  25   0.2049  -19.47
 Sex * ToD   18.24  3   0.0004   48.77  28   0.0088   -7.23
 PulsePressure   21.15  3   0.0001   69.92  31   0.00017.92
 Race19.87  2   0.   89.79  33   0.   23.79
 Pulse   25.31  3   0.  115.09  36   0.   43.09
 Age * Experience   202.80  3   0.  317.89  39   0.  239.89
 TBV282.41  3   0.  600.30  42   0.  516.30
 Location   310.19 14   0.  910.50  56   0.  798.50
 Age809.64  3   0. 1720.13  59   0. 1602.13



The ordering of variables is expected, and is consistent with the 
substantial knowledge I have about the outcome.


The problematic variable is Sex * TOD .  When I use p value as the rule, 
with an SLS of 0.01, the variable is retained, but when I use AIC , 
Sex*TOD is not retained. This reflects the fact that while the Sex*TOD 
interaction is theoretically interesting, the AIC value is negative and 
relatively small in magnitude, even as the p value skirts below 0.01.   
Is this judgement territory or are their statistical considerations that 
should be invoked? Caveats?




Is there a theoretical reason to choose AIC over p value methods, or is 
either acceptable?


__
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] More efficient option to append()?

2011-08-19 Thread Alex Ruiz Euler



Thanks for the code corrections. I see how for loops, append and
naively populating a NULL vector can be so resource consuming. I tried
the codes with 20 million observations in the following machine:

processor   : 7
cpu family  : 6
model name  : Intel(R) Core(TM) i7 CPU   Q 720  @ 1.60GHz
cpu MHz : 933.000
cache size  : 6144 KB


First I tried Timothy's code and left it running for half an hour and I
had to interrupt the command at

Timing stopped at: 1033.516 829.147 1845.648 


Then Dennis' option:
 
 user  system elapsed 
 25.793   0.224  25.784 

And for Paul's option, using a vector of length 20 million I had to
stop at: 

Timing stopped at: 850.577 8.868 851.464 


Not very efficient for relatively large vectors. I have also read
that using {} instead of () to wrap for example {x+1} works
faster, as do working directly with matrices instead of dataframes.

Thanks for your input.

Alex




On Fri, 19 Aug 2011 13:58:09 +
Paul Hiemstra paul.hiems...@knmi.nl wrote:

  As I already stated in my reply to your earlier post:
 
 resending the answer for the archives of the mailing list...
 
 Hi Alex,
 
 The other reply already gave you the R way of doing this while avoiding
 the for loop. However, there is a more general reason why your for loop
 is terribly inefficient. A small set of examples:
 
 largeVector = runif(10e4)
 outputVector = NULL
 system.time(for(i in 1:length(largeVector)) {
 outputVector = append(outputVector, largeVector[i] + 1)
 })
 #   user  system elapsed
  # 6.591   0.168   6.786
 
 The problem in this code is that outputVector keeps on growing and
 growing. The operating system needs to allocate more and more space as
 the object grows. This process is really slow. Several (much) faster
 alternatives exist:
 
 # Pre-allocating the outputVector
 outputVector = rep(0,length(largeVector))
 system.time(for(i in 1:length(largeVector)) {
 outputVector[i] = largeVector[i] + 1
 })
 #   user  system elapsed
 # 0.178   0.000   0.178
 # speed up of 37 times, this will only increase for large
 # lengths of largeVector
 
 # Using apply functions
 system.time(outputVector - sapply(largeVector, function(x) return(x + 1)))
 #   user  system elapsed
 #  0.124   0.000   0.125
 # Even a bit faster
 
 # Using vectorisation
 system.time(outputVector - largeVector + 1)
 #   user  system elapsed
 #  0.000   0.000   0.001
 # Practically instant, 6780 times faster than the first example
 
 It is not always clear which method is most suitable and which performs
 best. At least they all perform much, much better than the naive option
 of letting outputVector grow.
 
 cheers,
 Paul
 
 
 
 On 08/17/2011 11:17 PM, Alex Ruiz Euler wrote:
 
  Dear R community,
 
  I have a 2 million by 2 matrix that looks like this:
 
  x-sample(1:15,200, replace=T)
  y-sample(1:10*1000, 200, replace=T)
x y
  [1,] 10  4000
  [2,]  3  1000
  [3,]  3  4000
  [4,]  8  6000
  [5,]  2  9000
  [6,]  3  8000
  [7,]  2 1
  (...)
 
 
  The first column is a population expansion factor for the number in the
  second column (household income). I want to expand the second column
  with the first so that I end up with a vector beginning with 10
  observations of 4000, then 3 observations of 1000 and so on. In my mind
  the natural approach would be to create a NULL vector and append the
  expansions:
 
  myvar-NULL
  myvar-append(myvar, replicate(x[1],y[1]), 1)
 
  for (i in 2:length(x)) {
  myvar-append(myvar,replicate(x[i],y[i]),sum(x[1:i])+1)
  }
 
  to end with a vector of sum(x), which in my real database corresponds
  to 22 million observations.
 
  This works fine --if I only run it for the first, say, 1000
  observations. If I try to perform this on all 2 million observations
  it takes long, way too long for this to be useful (I left it running
  11 hours yesterday to no avail).
 
 
  I know R performs well with operations on relatively large vectors. Why
  is this so inefficient? And what would be the smart way to do this?
 
  Thanks in advance.
  Alex
 
  __
  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.
 
 
 -- 
 Paul Hiemstra, Ph.D.
 Global Climate Division
 Royal Netherlands Meteorological Institute (KNMI)
 Wilhelminalaan 10 | 3732 GK | De Bilt | Kamer B 3.39
 P.O. Box 201 | 3730 AE | De Bilt
 tel: +31 30 2206 494
 
 http://intamap.geo.uu.nl/~paul
 http://nl.linkedin.com/pub/paul-hiemstra/20/30b/770

__
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] R and Sweave

2011-08-19 Thread Ista Zahn

On Fri, Aug 19, 2011 at 2:23 PM, danielepippo dan...@hotmail.it wrote:
 Hi everybody.

  I'm trying to use R with Sweave but I have a problem perhaps with the
 directory path of sweave in R.
 The windows path is this:
 C:\Program Files (x86)\R\R-2.9.2\share\texmf\Sweave

 When I run the latex file with R, the program works well, without any
 errors, but when I create the pdf file, the Scode doesn't work. I think the
 problem is about the path.
 Can anybody suggest to me anything to do?

Please post the type of LaTeX installation you are using (MikTeX?) and
the version, along with the error you get when you try to compile to
document. It might be as easy as adding C:\Program Files
(x86)\R\R-2.9.2\share\texmf\Sweave to the MikTeX roots directory, but
more information is needed.

Also update to the latest R version if possible. Version 2.9 is two
years old by now, and two years is a long time in R-land.

Best,
Ista


 Thanks very much
 Cheers

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/R-and-Sweave-tp3755837p3755837.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.




-- 
Ista Zahn
Graduate student
University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.org

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

Re: [R] Leading zeros

Here is yet another way of prepending leading zeros on a string:

 x - c('123/1234', '234/12', '21342342134/34', '99')
 n - 10  # upto 10 leading zeros (max length of string)
 leading - paste(rep('0', n), collapse = '')
 # add up to 10 zeros and then truncate to max length, but at least keep 
 original
 substring(paste(leading, x, sep = '')
+, pmin(11, nchar(x) + 1)  # starting location
+, n + nchar(x)  # ending location
+)
[1] 00123/1234 234/12 21342342134/34 99

On Fri, Aug 19, 2011 at 12:41 PM, Ole Peter Smith ole@gmail.com wrote:
 -- Forwarded message --
 From: Ole Peter Smith ole@gmail.com
 Date: Fri, Aug 19, 2011 at 1:40 PM
 Subject: Re: [R] Leading zeros
 To: David Winsemius dwinsem...@comcast.net


 I'm all new to R, assisting the last days of topics from the sideline.
 I am, however, a longterm programmer.

 This question makes me ask myself - and now here - is there any string
 split-function, ex some thing like:

 split('/',2000/3000/4) -- (2000,3000,4)

 After this patting with zero is a breeze, sprintf('%010d',...

 0le

 On Fri, Aug 19, 2011 at 1:19 PM, David Winsemius dwinsem...@comcast.net 
 wrote:

 Copying list one what was sent in reply. Anybody have a better solution?

 On Aug 19, 2011, at 11:57 AM, Vasco Cadavez wrote:

 Thanks,

 A solution can be by substring to remove the /
 then numeric will be ok! What you think?

 How can I remove the /

 with sub or gsub:

 sprintf(%010.0f, as.integer(gsub(/,, c(4/3003,55/333,66/22))
 ))
 [1] 043003 055333 006622

 --
 David.


 Thanks

 Vasco Cadavez

 - Menssagem Original -
 De:
 David Winsemius dwinsem...@comcast.net

 Para:
 David Winsemius dwinsem...@comcast.net
 Cópia:
 Vasco Cadavez vcadavez@ipbpt, r-help@r-project.org
 Enviado:
 Fri, 19 Aug 2011 11:51:08 -0400
 Assunto:
 Re: [R] Leading zeros



 On Aug 19, 2011, at 11:17 AM, David Winsemius wrote:

 
  On Aug 19, 2011, at 11:12 AM, Vasco Cadavez wrote:
 
  Hello,
  I have a dataset with an Id columns like:
  4/3003
  55/333
  66/22
  I want to put leading zeros to get:
  0004/3003
  00055/333
  66/22
 
  How can I solve this?
 
  ?sprintf
  ?formatC
 
 I may have been too quick. Padding with leading zeros using sprintf is
 described for numeric but not for character types. There are severa

 __
 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.




 --
                                             /
                                          ( O O )
 =oOO==(_)==OOo=
       God does not care about our mathematical difficulties.
                     He integrates empirically - Einstein
                                 .oooO   Oooo.
 ==(     )=(     )=
                                      \  (     )  /
                                       \_)   (_/
 ===
                         Ole Peter Smith, IME, UFG
     http://www.mat.ufg.br/docentes/olepeter - ole at mat.ufg.br
 ===
                    Life sure is a Mystery to be Lived
                       Not a Problem to be Solved
 ===



 --
                                             /
                                          ( O O )
 =oOO==(_)==OOo=
       God does not care about our mathematical difficulties.
                     He integrates empirically - Einstein
                                 .oooO   Oooo.
 ==(     )=(     )=
                                      \  (     )  /
                                       \_)   (_/
 ===
                         Ole Peter Smith, IME, UFG
     http://www.mat.ufg.br/docentes/olepeter - ole at mat.ufg.br
 ===
                    Life sure is a Mystery to be Lived
                       Not a Problem to be Solved
 ===
 __
 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
Data Munger Guru

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] display only the top-right half of a correlation matrix?

On Fri, Aug 19, 2011 at 9:02 PM, Peter Langfelder
peter.langfel...@gmail.com wrote:
 Use as.dist: here's an example.

Seems promising, but for one issue: I would like to keep the diagonal
and thus specify 'diag=T', but then as.dist() replaces the diagonal
values with zero. (See below.) Is there a way to prevent it from doing
that? Either keep the original values, or not display anything in the
diagonal (as for the upper part)?

Regards
Liviu


 (xb - rcorr(as.matrix(mtcars[ , 1:4])))
   mpg   cyl  disphp
mpg   1.00 -0.85 -0.85 -0.78
cyl  -0.85  1.00  0.90  0.83
disp -0.85  0.90  1.00  0.79
hp   -0.78  0.83  0.79  1.00

n= 32


P
 mpg cyl disp hp
mpg   0   00
cyl   0   00
disp  0   00
hp0   0   0
 round(as.dist(xb$r, T), 2)
   mpg   cyl  disphp
mpg   0.00
cyl  -0.85  0.00
disp -0.85  0.90  0.00
hp   -0.78  0.83  0.79  0.00


__
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] display only the top-right half of a correlation matrix?

On Fri, Aug 19, 2011 at 12:32 PM, Liviu Andronic landronim...@gmail.com wrote:
 On Fri, Aug 19, 2011 at 9:02 PM, Peter Langfelder
 peter.langfel...@gmail.com wrote:
 Use as.dist: here's an example.

 Seems promising, but for one issue: I would like to keep the diagonal
 and thus specify 'diag=T', but then as.dist() replaces the diagonal
 values with zero. (See below.) Is there a way to prevent it from doing
 that? Either keep the original values, or not display anything in the
 diagonal (as for the upper part)?

if as.dist doesn't work, use brute force:

x = matrix(rnorm(5*100), 100, 5)
mat = signif(cor(x), 2);
mat[lower.tri(mat)] = 

data.frame(mat)

Peter

__
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] Windows 7 issues with installing packages and setting library paths




On 19.08.2011 13:32, Christoph Scherber wrote:

Dear all,

I am forced to work in an environment without administrator rights.



When using R2.13.1 on Windows 7 (64-Bit), I found that I can´t install or 
update any packages due to missing writing permissions.

I managed to get full access to a directory on my C:\ drive now - but how do I 
specify that all libraries shall be installed into this directory?

In Rcmd_environ I have the following entries:
## from R.sh
R_SHARE_DIR=C:\\Program Files\\R\\R-2.13.1\share
R_INCLUDE_DIR=C:\\Program Files\\R\\R-2.13.1\share\include
R_DOC_DIR=C:\\Program Files\\R\\R-2.13.1\share\doc
R_ARCH=
R_LIBS_USER=C:\\Program Files\\R\\R-2.13.1\\library
R_LIBS=C:\\Program Files\\R\\R-2.13.1\\library


In Rprofile.site I have the following entries:

.Library.site=C:\\Program Files\\R\\R-2.13.1\\library
.Library=C:\\Program Files\\R\\R-2.13.1\\library
.libPaths=C:\\Program Files\\R\\R-2.13.1\\library



Forget about anything your wrote above and delete what you entered.
A clean installation should provide you with a personal library for 
packages anyway. If that is not the case, you can also set an 
environment variable R_LIBS_USER whjere you can point to any directory 
(=library) where you'd like to install packages.


See ?Startup and ?.libPaths and of course the R Installation and 
Administration manual for more help.


Best,
Uwe Ligges






What else do I need to change?

When I start up R, I get the following error message:

Error: cannot change value of locked binding for '.Library'

When calling .libPaths, I still get the wrong path:
winfs-uni.top.gwdg.de/cscherb1$/R/R-2.13.1/library


R has been installed at
C:\\Program Files\\R

but for some reason it still uses
winfs-uni.top.gwdg.de/cscherb1$/R

as the default directory for libraries (where I don´t have write permissions 
for some unknown reasons)

What can I do to change the default library installation location?

Any help would be greatly appreciated!

Many thanks and best wishes
Christoph








__
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.

Re: [R] More efficient option to append()?




On 19.08.2011 15:50, Paul Hiemstra wrote:

  On 08/17/2011 10:53 PM, Alex Ruiz Euler wrote:

Dear R community,

I have a 2 million by 2 matrix that looks like this:

x-sample(1:15,200, replace=T)
y-sample(1:10*1000, 200, replace=T)
   x y
[1,] 10  4000
[2,]  3  1000
[3,]  3  4000
[4,]  8  6000
[5,]  2  9000
[6,]  3  8000
[7,]  2 1
(...)


The first column is a population expansion factor for the number in the
second column (household income). I want to expand the second column
with the first so that I end up with a vector beginning with 10
observations of 4000, then 3 observations of 1000 and so on. In my mind
the natural approach would be to create a NULL vector and append the
expansions:

myvar-NULL
myvar-append(myvar, replicate(x[1],y[1]), 1)

for (i in 2:length(x)) {
myvar-append(myvar,replicate(x[i],y[i]),sum(x[1:i])+1)
}

to end with a vector of sum(x), which in my real database corresponds
to 22 million observations.

This works fine --if I only run it for the first, say, 1000
observations. If I try to perform this on all 2 million observations
it takes long, way too long for this to be useful (I left it running
11 hours yesterday to no avail).


I know R performs well with operations on relatively large vectors. Why
is this so inefficient? And what would be the smart way to do this?


Hi Alex,

The other reply already gave you the R way of doing this while avoiding
the for loop. However, there is a more general reason why your for loop
is terribly inefficient. A small set of examples:

largeVector = runif(10e4)
outputVector = NULL
system.time(for(i in 1:length(largeVector)) {



Please do teach people to use seq_along(largeVector) rather than 
1:length(largeVector) (the latter is not save in case of length 0 objects).


Uwe Ligges



 outputVector = append(outputVector, largeVector[i] + 1)
})
#   user  system elapsed
  # 6.591   0.168   6.786

The problem in this code is that outputVector keeps on growing and
growing. The operating system needs to allocate more and more space as
the object grows. This process is really slow. Several (much) faster
alternatives exist:

# Pre-allocating the outputVector
outputVector = rep(0,length(largeVector))
system.time(for(i in 1:length(largeVector)) {
 outputVector[i] = largeVector[i] + 1
})
#   user  system elapsed
# 0.178   0.000   0.178
# speed up of 37 times, this will only increase for large
# lengths of largeVector

# Using apply functions
system.time(outputVector- sapply(largeVector, function(x) return(x + 1)))
#   user  system elapsed
#  0.124   0.000   0.125
# Even a bit faster

# Using vectorisation
system.time(outputVector- largeVector + 1)
#   user  system elapsed
#  0.000   0.000   0.001
# Practically instant, 6780 times faster than the first example

It is not always clear which method is most suitable and which performs
best. At least they all perform much, much better than the naive option
of letting outputVector grow.

cheers,
Paul


Thanks in advance.
Alex

__
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.

Re: [R] R and Sweave




On 19.08.2011 21:15, Ista Zahn wrote:

On Fri, Aug 19, 2011 at 2:23 PM, danielepippodan...@hotmail.it  wrote:

Hi everybody.

  I'm trying to use R with Sweave but I have a problem perhaps with the
directory path of sweave in R.
The windows path is this:
C:\Program Files (x86)\R\R-2.9.2\share\texmf\Sweave

When I run the latex file with R, the program works well, without any
errors, but when I create the pdf file, the Scode doesn't work. I think the
problem is about the path.
Can anybody suggest to me anything to do?


Please post the type of LaTeX installation you are using (MikTeX?) and
the version, along with the error you get when you try to compile to
document. It might be as easy as adding C:\Program Files
(x86)\R\R-2.9.2\share\texmf\Sweave to the MikTeX roots directory, but
more information is needed.

Also update to the latest R version if possible. Version 2.9 is two
years old by now, and two years is a long time in R-land.


The actual problem is that LaTeX has still problems with paths 
containing spaces.


Uwe Ligges




Best,
Ista



Thanks very much
Cheers

--
View this message in context: 
http://r.789695.n4.nabble.com/R-and-Sweave-tp3755837p3755837.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.







__
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: how to generate counts from generalized poisson distribution

2011-08-19 Thread Chee Chen

Dear All,
Is there a simulator that can generate observations from a generalized poisson 
distribution?
Thanks and regards,
Chee
[[alternative HTML version deleted]]

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

Re: [R] display only the top-right half of a correlation matrix?

On Fri, Aug 19, 2011 at 9:38 PM, Peter Langfelder
peter.langfel...@gmail.com wrote:
 if as.dist doesn't work, use brute force:

 x = matrix(rnorm(5*100), 100, 5)
 mat = signif(cor(x), 2);
 mat[lower.tri(mat)] = 

 data.frame(mat)

Yes, brute force works. This isn't quite how I wanted to do this, but
the following seems to work for me.

Thanks all
Liviu

require(Hmisc)
print.rcorr -
function (x, upper=FALSE, ...)
{
r - format(round(x$r, 2))
if(!is.null(upper)) r[if(!upper) upper.tri(r) else lower.tri(r)] - ''
print(data.frame(r))
n - x$n
if (all(n == n[1, 1]))
cat(\nn=, n[1, 1], \n\n)
else {
cat(\nn\n)
print(n)
}
cat(\nP\n)
P - x$P
P - ifelse(P  0.0001, 0, P)
p - format(round(P, 4))
p[is.na(P)] - 
if(!is.null(upper)) p[if(!upper) upper.tri(p) else lower.tri(p)] - ''
print(p, quote = FALSE)
invisible()
}

 (xb - rcorr(as.matrix(mtcars[ , 1:4])))
   mpg   cyl  disphp
mpg   1.00
cyl  -0.85  1.00
disp -0.85  0.90  1.00
hp   -0.78  0.83  0.79  1.00

n= 32


P
 mpg cyl disp hp
mpg
cyl   0
disp  0   0
hp0   0   0

__
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] R and Sweave

2011-08-19 Thread Ista Zahn

On Fri, Aug 19, 2011 at 4:25 PM, Uwe Ligges
lig...@statistik.tu-dortmund.de wrote:


 On 19.08.2011 21:15, Ista Zahn wrote:

 On Fri, Aug 19, 2011 at 2:23 PM, danielepippodan...@hotmail.it  wrote:

 Hi everybody.

  I'm trying to use R with Sweave but I have a problem perhaps with the
 directory path of sweave in R.
 The windows path is this:
 C:\Program Files (x86)\R\R-2.9.2\share\texmf\Sweave

 When I run the latex file with R, the program works well, without any
 errors, but when I create the pdf file, the Scode doesn't work. I think
 the
 problem is about the path.
 Can anybody suggest to me anything to do?

 Please post the type of LaTeX installation you are using (MikTeX?) and
 the version, along with the error you get when you try to compile to
 document. It might be as easy as adding C:\Program Files
 (x86)\R\R-2.9.2\share\texmf\Sweave to the MikTeX roots directory, but
 more information is needed.

 Also update to the latest R version if possible. Version 2.9 is two
 years old by now, and two years is a long time in R-land.

 The actual problem is that LaTeX has still problems with paths containing
 spaces.

That may be the problem, but I don't think it's a certainty; Sweave
works on my system (with MikTeX 2.9) even though the R texmf directory
path contains spaces. All I had to do was add C:\Program
Files\R\R-2.13.1\share\texmf to the list of MikTeX root directories.

Best,
Ista

 Uwe Ligges



 Best,
 Ista


 Thanks very much
 Cheers

 --
 View this message in context:
 http://r.789695.n4.nabble.com/R-and-Sweave-tp3755837p3755837.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.








-- 
Ista Zahn
Graduate student
University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.org

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

[R] Build a package - check error

Dear R-users

I am slowly migrating my mex files (MATLAB - Fortran and C) to R.  To get my
own functions available on R section I have decided to learn how to build  a
R package. I choose a simple example with a few Fortran and R functions
(wrapper).

The fortran sources are located at src and the R functions at R (as
recommended).  The building process went ok but R CMD check did not. The
error mgs was

Error in dyn.load(fortran.so) :
  unable to load shared object
'/home/eduardo/R_packages/test.Rcheck/fortran.so':

Although I can see that R cannot find the compiled fortran code I do not
know what to do.   I believe it is something to do with the following lines
in  the R-wrapper file

if (!is.loaded('calnpr'))
   dyn.load(fortran.so)

How to add the path so that once the package is installed the compiled
fortran code can be found?

Many thanks

Cheers

Ed

[[alternative HTML version deleted]]

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

Re: [R] Error in read.dcf(file = tmpf) : Line starting 'head ...' is malformed!




On 19.08.2011 16:45, Benoit Bruneau wrote:

Dear R-Users,

I'm trying to setup a personal repository for a few packages I'm working on.
I am on R-Forge but I still need to have various versions of my package that
R-Forge does not build (for R 2.8.1 for example).

So I followed the instructions in this document:
Hhttp://
cran.r-project.org/doc/manuals/R-admin.html#Setting-up-a-package-repository

and used this function as recommended:
*write_PACKAGES()*

Now, when I try to install or update my package, I get:

*install.packages(bmisc, repos=http://www.benoitr.comze.com/R;)**

Error in read.dcf(file = tmpf) : Line starting 'head  ...' is malformed!*


Can someone help me with this ?



Yes:
R-2.8.1 under Windows will look into
 http://www.benoitr.comze.com/R/bin/windows/contrib/2.8/...
for packages - and that does not exist. You just provide a binary 
directory For R-2.13.x


You have to build a binary with R-2.8.x for the /2.8/ repository.
It is cumbersome to support so many binary versions since you have to 
build a binary of your package for each of the R versions you want to 
support.


Best,
Uwe Ligges






Benoit
---
R: 2.13.1

[[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.


__
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] Build a package - check error




On 19.08.2011 22:53, Eduardo Mendes wrote:

Dear R-users

I am slowly migrating my mex files (MATLAB - Fortran and C) to R.  To get my
own functions available on R section I have decided to learn how to build  a
R package. I choose a simple example with a few Fortran and R functions
(wrapper).

The fortran sources are located at src and the R functions at R (as
recommended).  The building process went ok but R CMD check did not. The
error mgs was

Error in dyn.load(fortran.so) :
   unable to load shared object
'/home/eduardo/R_packages/test.Rcheck/fortran.so':

Although I can see that R cannot find the compiled fortran code I do not
know what to do.   I believe it is something to do with the following lines
in  the R-wrapper file

if (!is.loaded('calnpr'))
dyn.load(fortran.so)



1. If the package is called calnpr, the shared library is also called 
that way.

2. you have to provide the path to the shared library.

See ?.First.lib for how to do it in a package without NAMESPACE (and 
note that NAMESPACES are forced for the next R release).


Best,
Uwe Ligges



How to add the path so that once the package is installed the compiled
fortran code can be found?

Many thanks

Cheers

Ed

[[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.


__
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] AFT model time-dependent with weibull distribution

2011-08-19 Thread Göran Broström

On Fri, Aug 19, 2011 at 2:55 PM, javier palacios xpfen...@gmail.com wrote:
 Dear R-community,

 I have tried to estimate an accelerated failure time(AFT) and proportional
 hazard (PH) parametric survival model with time-independent  and
 time-dependent covariates. For that purpose, I have used the eha package.


 Please, consider this example:

 weibullph - phreg(Surv(sta,time,S) ~ TDC1 + TIC1, dist=weibull,
 data.frame=Data)

 weibullaft-aftreg(Surv(sta,time,S) ~ TDC1 + TIC1, dist=weibull,
 data.frame=Data)

 ## aftreg gives error when I add ID argument...

 Error in aftreg.fit(X, Y, dist, strats, offset, init, shape, id, control,  :
  Overlapping intervals for id  2

Does not happen to me.


 From help(aftreg): id If there are more than one spell per individual, it is
 essential to keep spells together by the id argument. This allows for
 time-varying covariates.


 data table:

 Data
   S sta time  TDC1 total_time TIC1          ID
 A 1   0   1  48.50       1              1           1
 B 0   0   1  65.96       2               1           2
 B 1   1   2  65.08       2              1            2
 C 0   0   1  0.00        2               4           3
 C 1   1   2  0.00        2               4           3
 D 0   0   1 72.74        2               5           4
 D 1   1   2 72.52        2               5           4
 E 0   0   1 61.84         2              3            5
 E 0   1   2 60.56        2               3            5
 F 0   0   1 35.04        4               2            6
 F 0   1   2 36.97        4               2            6
 F 0   2   3 37.92        4               2            6
 F 1   3   4 39.01        4               2            6

Duplicated rownames are no allowed.


 time - time to event
 sta - starting time
 TDC - time dependent covariates
 TIC - time independent covariate
 total_time - total time at risk
 ID - ID

 1- What happens if the ID is not included?
Read the documentation
 2- How can I solve this error?
Read the documentation
 3- Why the phreg function does not need an ID?
Read the documentation (and a good text book on survival analysis)

And please read the posting guide; the elementary information is missing.


 Thanks,

 Javier


 --
 View this message in context: 
 http://r.789695.n4.nabble.com/AFT-model-time-dependent-with-weibull-distribution-tp3755079p3755079.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.




-- 
Göran Broström

__
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] Plot label symbols and superscript

2011-08-19 Thread Marlin Keith Cox

I was unable to find an answer to my problem.  I would like to label
the y axis of a plot with a rate and would like to use a dot (•)
rather than a multiplication sign (x).


ylab = quote(Speed~(cmxsec^2))


Thanks in advance.

keith


-- 
M. Keith Cox, Ph.D.
Alaska NOAA Fisheries, National Marine Fisheries Service
Auke Bay Laboratories
17109 Pt. Lena Loop Rd.
Juneau, AK 99801
keith@noaa.gov
marlink...@gmail.com
U.S. (907) 789-6603

__
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] Build a package - check error

Hi

I have modified the path to

dyn.load(paste(Sys.getenv(R_LIBS_USER),/fortran/src/fortran.so,sep=))


and the package could installed, loaded and the lines with dyn.load worked.
It does not look like a pretty solution but works on my linux (I am not sure
if it works on my mac or windows).

I am not sure if this is what you meant but as I have no clue what
.First.lib does or NAMESPACES means this is the best I come up with.

Please correct me if I am wrong.

Many thanks

Ed



On Fri, Aug 19, 2011 at 6:03 PM, Uwe Ligges lig...@statistik.tu-dortmund.de
 wrote:



 On 19.08.2011 22:53, Eduardo Mendes wrote:

 Dear R-users

 I am slowly migrating my mex files (MATLAB - Fortran and C) to R.  To get
 my
 own functions available on R section I have decided to learn how to build
  a
 R package. I choose a simple example with a few Fortran and R functions
 (wrapper).

 The fortran sources are located at src and the R functions at R (as
 recommended).  The building process went ok but R CMD check did not. The
 error mgs was

 Error in dyn.load(fortran.so) :
   unable to load shared object
 '/home/eduardo/R_packages/**test.Rcheck/fortran.so':

 Although I can see that R cannot find the compiled fortran code I do not
 know what to do.   I believe it is something to do with the following
 lines
 in  the R-wrapper file

 if (!is.loaded('calnpr'))
dyn.load(fortran.so)



 1. If the package is called calnpr, the shared library is also called that
 way.
 2. you have to provide the path to the shared library.

 See ?.First.lib for how to do it in a package without NAMESPACE (and note
 that NAMESPACES are forced for the next R release).

 Best,
 Uwe Ligges


  How to add the path so that once the package is installed the compiled
 fortran code can be found?

 Many thanks

 Cheers

 Ed

[[alternative HTML version deleted]]

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



[[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.

[R] How to add horizontal lines above bar graph to display p-values?

2011-08-19 Thread Sébastien Vigneau

Hi,

I would like to draw horizontal lines above a bar graph, in order to display
the p-values of a Fisher test. Here is an
examplehttp://thejns.org/action/showPopup?citid=citart1id=f3-1060501doi=10.3171%2Fped.2007.106.6.501of
the type of display I would like to have. Is there a way to draw the
horizontal lines and write their associated p-values in R?

Thanks for you help!

Sebastien Vigneau

[[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.

[R] Calculating p-value for 1-tailed test in a linear model

2011-08-19 Thread Andrew Campomizzi

Hello,

I'm having trouble figuring out how to calculate a p-value for a 1-tailed
test of beta_1 in a linear model fit using command lm.  My model has only 1
continuous, predictor variable.  I want to test the null hypothesis beta_1
is = 0.  I can calculate the p-value for a 2-tailed test using the code
2*pt(-abs(t-value), df=degrees.freedom), where t-value and degrees.freedom
are values provided in the summary of the lm.  The resulting p-value is the
same as provided by the summary of the lm for beta_1.  I'm unsure how to
change my calculation of the p-value for a 1-tailed test.

Thanks for your assistance,

Andy


[[alternative HTML version deleted]]

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

Re: [R] COXPH TIME-DEPENDENT

2011-08-19 Thread javier palacios

Thanks, 
Javier

--
View this message in context: 
http://r.789695.n4.nabble.com/COXPH-TIME-DEPENDENT-tp3754837p3756128.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.

Re: [R] help: how to generate counts from generalized poisson distribution

2011-08-19 Thread Michael Lerch

On Fri, Aug 19, 2011 at 04:43:35PM -0400, Chee Chen wrote:
 Dear All,
 Is there a simulator that can generate observations from a generalized 
 poisson distribution?
 Thanks and regards,
 Chee

You can use rzigp() in the ZIGP package. rzigp(n,mu,phi,omega=0) for
generalized poisson.

Michael

__
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] Auto key legend does not match plot

2011-08-19 Thread markm0705

Dear R-help members.  I am an 'R-learner' (about 6 hours so far) using the
lattice library to create a ranked dotplot and am colour coding the dots by
a variable called Commodity.  However when i use autokey to make a legend
the size (cex) and symbol (pch) do not match what is on the dotplot.  
Code is below and image attached 

library(lattice)
Cal_dat - read.table(Calibration2.dat,header = TRUE,sep = \t,)
dotplot(reorder(Label.yr, Resc_Gt)~ Resc_Gt,groups=Commodity,
data=Cal_dat,cex=1.5, pch=19,aspect=xy,
auto.key=list(space=right,title=Commodity)) 

Any assistance appreciated
http://r.789695.n4.nabble.com/file/n3756245/Ranked_boxplot_by_commodity.png 

--
View this message in context: 
http://r.789695.n4.nabble.com/Auto-key-legend-does-not-match-plot-tp3756245p3756245.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.

[R] Lattice help: Dotplot

2011-08-19 Thread sw1

With Dotplot, I'm trying to make a figure that will ultimately have the same
x-axis (which will be my response variable and the error bars), but the
y-axis will consist of a different label for every point. Here's my code:

Dotplot(fTaxonGrouped ~ Cbind(normSlope,normLwr,normUpr)|fGroup,
groups=fEpoch, 
pch=c(17,15,19), col=c(3:1), 
scales = list(
y = list(relation = sliced,axs = r, alternating = 
0,tick.number=10),
x = list(tick.number = 6)
))

It ALMOST does exactly what I. There are six small graphs, and each
datapoint has a unique value on the y-axis (integers), but once I try to
change the y-axis tick labels to the list containing the names of each
datapoint (characters), it lists only the first 10 names for all 6 graphs --
that is, it repeats the names, so they are no longer unique and are also
mislabeled. 

If I change the relation to same, the correct datapoints are plotted for
their appropriate group and hence appropriate plot, and every point is
labeled correctly, but because only about 10 data points apply to each
group, I have 6 plots with about 10 points each but all 40 names listed in
every one.

Essentially, I need the graph the results when relation is set to same,
but with blank labels removed (like when relation is set to sliced) and
thus scaled correctly, but the labels cannot be changed into integers, which
is what happens when relation is set to slice.

Thanks

--
View this message in context: 
http://r.789695.n4.nabble.com/Lattice-help-Dotplot-tp3756134p3756134.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.

Re: [R] Writing non-graphic (text) output to PDF

2011-08-19 Thread baptiste auguie

Hi,

Try this,

library(gridExtra)
example(grid.table)

or addtable2plot() in plotrix, or textplot() in gplots, or Hmisc using
latex, or Sweave, ...

HTH,

baptiste

PS: please read the posting guide


On 20 August 2011 05:14, Ed Heaton heat...@comcast.net wrote:
 Hi, friends.

 I keep coming to you because I'm so new to R and can't seem to figure out
 some simple things.  Sorry.

 Consider the following code.  I want to load a table and write out the
 structure to a PDF document.  I just can't seem to manage writing
 non-graphic output to PDF.  Any help?  I've tried several functions, but
 nothing worked.  All I get is the title.

 # **
 # Load the DEBT table.
  debt - readRDS(T:/R.Data/Debt.rData)
  dim(debt)
 # Open the debt.pdf file for graphics output.
  pdf(
    file=paste(
      R:/DAS/DMS/FedDebt
     ,DataDiscovery
     ,DistributionAnalysis
     ,Report
     ,Debt.pdf
     ,sep=/
    )
  )
 # ==
 # Write the debt structucture to the output PDF.
  plot.new()
  title(DEBT)
  str(debt)
 # ==
  dev.off() # Turn off the PDF device.
 # ** End of Program 



 Ed

 Ed Heaton
 Project Manager, Sr. SAS Developer
 Data and Analytic Solutions, Inc.
 10318 Yearling Drive
 Rockville, MD 20850
 Office: 301-520-7414
 ehea...@dasconsultants.com
 www.dasconsultants.com http://www.dasconsultants.com/
 CMMI ML-2, SBA 8(a)  SDB, WBE (WBENC), MBE (VA  MD)

 e...@heaton.name

 (Re: http://www.r-project.org/posting-guide.html)

 __
 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.

Re: [R] Auto key legend does not match plot

2011-08-19 Thread Weidong Gu

It is most likely due to your ordering of y values. You need to write
key manually to reflect the change. Without providing reproduciable
data, you may not get specific help.

Weidong Gu

On Fri, Aug 19, 2011 at 6:23 PM, markm0705 markm0...@gmail.com wrote:
 Dear R-help members.  I am an 'R-learner' (about 6 hours so far) using the
 lattice library to create a ranked dotplot and am colour coding the dots by
 a variable called Commodity.  However when i use autokey to make a legend
 the size (cex) and symbol (pch) do not match what is on the dotplot.
 Code is below and image attached

 library(lattice)
 Cal_dat - read.table(Calibration2.dat,header = TRUE,sep = \t,)
 dotplot(reorder(Label.yr, Resc_Gt)~ Resc_Gt,groups=Commodity,
        data=Cal_dat,cex=1.5, pch=19,aspect=xy,
        auto.key=list(space=right,title=Commodity))

 Any assistance appreciated
 http://r.789695.n4.nabble.com/file/n3756245/Ranked_boxplot_by_commodity.png

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Auto-key-legend-does-not-match-plot-tp3756245p3756245.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.


__
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] Plot label symbols and superscript



On Aug 19, 2011, at 6:36 PM, Marlin Keith Cox wrote:


I was unable to find an answer to my problem.  I would like to label
the y axis of a plot with a rate and would like to use a dot (•)
rather than a multiplication sign (x).


ylab = quote(Speed~(cmxsec^2))



?plotmath  # seemed like the logical place to look
# yep, there it is.

plot(1,1, ylab = quote(Speed~(cm%.%sec^2)) )

--

David Winsemius, MD
West Hartford, CT

__
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] Leading zeros

2011-08-19 Thread Jeff Newmiller

x - rep(00,2)
y - c(23/45,67/8)
substr(x,1+nchar(x)-nchar(y), nchar(x)) - y
x

---
Jeff Newmiller The . . Go Live...
DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
--- 
Sent from my phone. Please excuse my brevity.

David Winsemius dwinsem...@comcast.net wrote:


Copying list one what was sent in reply. Anybody have a better solution?

 On Aug 19, 2011, at 11:57 AM, Vasco Cadavez wrote:

 Thanks,

 A solution can be by substring to remove the /
 then numeric will be ok! What you think?

 How can I remove the /

with sub or gsub:

 sprintf(%010.0f, as.integer(gsub(/,, 
c(4/3003,55/333,66/22)) ))
[1] 043003 055333 006622

-- 
David.


 Thanks

 Vasco Cadavez

 - Menssagem Original -
 De:
 David Winsemius dwinsem...@comcast.net

 Para:
 David Winsemius dwinsem...@comcast.net
 CÃ³pia:
 Vasco Cadavez vcadavez@ipbpt, r-help@r-project.org
 Enviado:
 Fri, 19 Aug 2011 11:51:08 -0400
 Assunto:
 Re: [R] Leading zeros



 On Aug 19, 2011, at 11:17 AM, David Winsemius wrote:

 
  On Aug 19, 2011, at 11:12 AM, Vasco Cadavez wrote:
 
  Hello,
  I have a dataset with an Id columns like:
  4/3003
  55/333
  66/22
  I want to put leading zeros to get:
  0004/3003
  00055/333
  66/22
 
  How can I solve this?
 
  ?sprintf
  ?formatC
 
 I may have been too quick. Padding with leading zeros using sprintf 
 is
 described for numeric but not for character types. There are severa

_

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.


[[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.

[R] a Question regarding glm for linear regression

2011-08-19 Thread Andra Isan

Hello All,
I have a question about glm in R. I would like to fit a model with glm 
function, I have a vector y (size n) which is my response variable and I have 
matrix X which is by size (n*f) where f is the number of features or columns. I 
have about 80 features, and when I fit a model using the following formula, 
glmfit = glm(y ~ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 + 
x13 + x14 + x15 + x16 + x17 + x18 + x19 + x20 + x21 + x22 + x23 + x24 + x25 + 
x26 + x27 + x28 + x29 + x30 + x31 + x32 + x33 + x34 + x35 + x36 + x37 + x38 + 
x39 + x40 + x41 + x42 + x43 + x44 + x45 + x46 + x47 + x48 + x49 + x50 + x51 + 
x52 + x53 + x54 + x55 + x56 + x57 + x58 + x59 + x60 + x61 + x62 + x63 + x64 + 
x65 + x66 + x67 + x68 + x69 + x70 + x71 + x72 + x73 + x74 + x75 + x76 + x77 + 
x78 + x79 + x80)
it gives me an error Error in eval(expr, envir, enclos) : object 'x3' not 
found which I dont know why I am given those errors. The other thing is that 
when I use the glm.fit, I can get coefficients without any errors. So, I am 
not sure what is going on and if glm.fit is the same as glm, can I use glm.fit 
instead of glm?

Thanks a lot, 
Andra


__
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] a Question regarding glm for linear regression

2011-08-19 Thread Steven Kennedy

convert your matrix to a data frame:
df - as.data.frame(mymatrix)

then you can simplify your formula and specify where the data is coming from:
glm.fit - glm(y~., data=df)

the . in the formula means all columns in your dataframe (except
y, if it is in df)




On Sat, Aug 20, 2011 at 10:43 AM, Andra Isan andra_i...@yahoo.com wrote:
 Hello All,
 I have a question about glm in R. I would like to fit a model with glm 
 function, I have a vector y (size n) which is my response variable and I have 
 matrix X which is by size (n*f) where f is the number of features or columns. 
 I have about 80 features, and when I fit a model using the following formula,
 glmfit = glm(y ~ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 
 + x13 + x14 + x15 + x16 + x17 + x18 + x19 + x20 + x21 + x22 + x23 + x24 + x25 
 + x26 + x27 + x28 + x29 + x30 + x31 + x32 + x33 + x34 + x35 + x36 + x37 + x38 
 + x39 + x40 + x41 + x42 + x43 + x44 + x45 + x46 + x47 + x48 + x49 + x50 + x51 
 + x52 + x53 + x54 + x55 + x56 + x57 + x58 + x59 + x60 + x61 + x62 + x63 + x64 
 + x65 + x66 + x67 + x68 + x69 + x70 + x71 + x72 + x73 + x74 + x75 + x76 + x77 
 + x78 + x79 + x80)
 it gives me an error Error in eval(expr, envir, enclos) : object 'x3' not 
 found which I dont know why I am given those errors. The other thing is that 
 when I use the glm.fit, I can get coefficients without any errors. So, I am 
 not sure what is going on and if glm.fit is the same as glm, can I use 
 glm.fit instead of glm?

 Thanks a lot,
 Andra


 __
 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.

Re: [R] Lattice help: Dotplot



On Aug 19, 2011, at 5:07 PM, sw1 wrote:


With Dotplot,


Are you sure that you are using lattice? Maybe you out to look more  
closely at:


?Dotplot



I'm trying to make a figure that will ultimately have the same
x-axis (which will be my response variable and the error bars), but  
the
y-axis will consist of a different label for every point. Here's my  
code:


Dotplot(fTaxonGrouped ~ Cbind(normSlope,normLwr,normUpr)|fGroup,
groups=fEpoch,
pch=c(17,15,19), col=c(3:1),
scales = list(
		y = list(relation = sliced,axs = r, alternating =  
0,tick.number=10),

x = list(tick.number = 6)
))

It ALMOST does exactly what I. There are six small graphs, and each
datapoint has a unique value on the y-axis (integers), but once I  
try to

change the y-axis tick labels to the list containing the names of each
datapoint (characters), it lists only the first 10 names for all 6  
graphs --
that is, it repeats the names, so they are no longer unique and are  
also

mislabeled.

If I change the relation to same, the correct datapoints are  
plotted for

their appropriate group and hence appropriate plot, and every point is
labeled correctly, but because only about 10 data points apply to each
group, I have 6 plots with about 10 points each but all 40 names  
listed in

every one.

Essentially, I need the graph the results when relation is set to  
same,
but with blank labels removed (like when relation is set to  
sliced) and
thus scaled correctly, but the labels cannot be changed into  
integers, which

is what happens when relation is set to slice.



--

David Winsemius, MD
West Hartford, CT

__
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] Labelling all variables at once (using Hmisc label)

2011-08-19 Thread Frank Harrell

Sorry about the nabble problem.  At any rate, do require(Hmisc) then ?label
to see how to associate a vector of labels with all the variables in a data
frame at once.
Frank

do999 wrote:
 
 Indeed, as David pointed out, all the portion that used courier font (all
 the good stuff) was absent from the email posting. 
 
 
 Thanks for your answers.
   [[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.
 


-
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
View this message in context: 
http://r.789695.n4.nabble.com/Labelling-all-variables-at-once-using-Hmisc-label-tp3745660p3756459.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.

Re: [R] Hmisc::rcorr on a 'data.frame'?

2011-08-19 Thread Frank Harrell

I don't see anything wrong with using as.matrix.  The documentation doesn't
say it will support a data frame.
Frank

Liviu Andronic wrote:
 
 Dear all
 ?Hmisc::rcorr states that it takes as main argument a numeric
 matrix. But is it normal that it fails in such an ugly way on a data
 frame? (See below.) If the function didn't attempt any conversion to a
 matrix, I would have expected it to state that in the error message
 that it didn't accept 'data.frame' objects in its input. Also, I
 vaguely remember having used in the past rcorr() on data frames.
 
 Regards
 Liviu
 
 require(Hmisc)
 rcorr(mtcars[ , 1:4])
 Error in storage.mode(x) - if (.R.) double else single :
   (list) object cannot be coerced to type 'double'
 rcorr(as.matrix(mtcars[ , 1:4]))
mpg   cyl  disphp
 mpg   1.00 -0.85 -0.85 -0.78
 cyl  -0.85  1.00  0.90  0.83
 disp -0.85  0.90  1.00  0.79
 hp   -0.78  0.83  0.79  1.00
 
 n= 32
 
 
 P
  mpg cyl disp hp
 mpg   0   00
 cyl   0   00
 disp  0   00
 hp0   0   0
 
 
 
 
 -- 
 Do you know how to read?
 http://www.alienetworks.com/srtest.cfm
 http://goodies.xfce.org/projects/applications/xfce4-dict#speed-reader
 Do you know how to write?
 http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail
 
 __
 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.
 


-
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
View this message in context: 
http://r.789695.n4.nabble.com/Hmisc-rcorr-on-a-data-frame-tp3755868p3756462.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.

Re: [R] help Dxy and C-index calculation

2011-08-19 Thread Frank Harrell

Replace Design with rms (for general reasons not related to your question;  
Design is about to be obsolete).


Negate Dxy.  The linear predictor for the Cox model is relative log  
hazard.  Higher hazard means shorter survival time.  For other survival  
models the model is stated in terms of survival time.


Frank


On Fri, 19 Aug 2011 08:30:50 -0500, chafika.mazo...@igr.fr  
chafika.mazo...@igr.fr wrote:



Dear professor,

I am currently using Design package and the cph formula for assessing  
multivariable analysis.


I am tryng to get the C-index for my survival model based on Dxy  
coefficient.


I am confused since there is a negative value.

Do I need to used the absolute Dxy ?

  index.orig training test optimism index.corrected  n
Dxy   -0.341357727 -0.344002740 -0.341357727 -0.002645013 
-0.338712715 40
R2 0.084694141  0.095440176  0.079077594  0.016362582  
0.068331560 40
Slope  1.0  1.0  0.897999711  0.102000289  
0.897999711 40
D  0.033368588  0.038458457  0.030983429  0.007475027  
0.025893561 40
U -0.002890981 -0.002968495  0.003412045 -0.006380539  
0.003489558 40
Q  0.036259569  0.041426951  0.027571385  0.013855566  
0.022404003 40


Many thanks

Dr MAZOUNI
Institut Gustave Roussy
Department of breast surgical oncology
Villejuif, France



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

__
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] a Question regarding glm for linear regression