Re: [R] Loop question?
Here's a toy example which you can apply the logic of: dfr - expand.grid(1:3,1:2) results - apply(dfr, 1, sum) dfr[results==4,] On 25 January 2013 22:19, Andras Farkas motyoc...@yahoo.com wrote: Dear All I have the following data (somewhat simplyfied): TINF -1 a -c(500,750,1000,1250,1500,1750,2000) b -c(8,12,18,24,36,48,60,72,96) following function: infcprodessa -function (D, tin, tau, ts) (D * (1 - exp(-0.048 * tin))/(tin * (0.048*79) * (1 - exp(-0.048 * tau * exp(-0.048 * (ts - tin)) z -sapply(1:1, function(n) infcprodessa(1000,TINF,12,12-TINF)) is there a way to select the combination of respective a and b values that would result in a calculated z that is between 15 and 20? In this case the a would be 1000 and the b would be 12 (other combinations are also possible), but how could I automatically find them? perhaps a loop? Apreciate the help, Sincerely, Andras [[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] Loop question?
Sorry Jeff, probably the new version of Yahoo mail doing the html, I switched back to the older one hope that takes care of the problem. Let me clarify the code below: TINF -1 a -c(500,750,1000,1250,1500,1750,2000) b -c(8,12,18,24,36,48,60,72,96) infcprodessa -function (D, tin, tau, ts) (D * (1 - exp(-0.048 * tin))/(tin * (0.048*79) * (1 - exp(-0.048 * tau * exp(-0.048 * (ts - tin)) z -sapply(1:1, function(n) infcprodessa(a,TINF,b,b-TINF)) so I am looking to find the combination of a = 1000 and b = 12, which iz a value for z that would fall between 15 and 20. Sometimes though there may be more than one combinations that will meet criteria, in that case I would like to select the combination based on the smallest a value that has the respective b value to meet the criteria, sorry for the confusion, thanks, Andras --- On Sat, 1/26/13, Jeff Newmiller jdnew...@dcn.davis.ca.us wrote: From: Jeff Newmiller jdnew...@dcn.davis.ca.us Subject: Re: [R] Loop question? To: Andras Farkas motyoc...@yahoo.com, r-help@r-project.org r-help@r-project.org Date: Saturday, January 26, 2013, 2:09 AM Please read the Posting Guide no html email reproducible example please In general, you can use expand.grid to generate all combinations of inputs, compute results as a vector just as long as the expand.grid data frame has rows, and identify which results meet your criteria by a logical test, and use that test to identify which input combinations worked. Provide a working starting point and someone might give you working code as an answer. (where do a and b come into your problem?) --- 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. Andras Farkas motyoc...@yahoo.com wrote: Dear All � I have the following data (somewhat simplyfied): � TINF -1 a -c(500,750,1000,1250,1500,1750,2000) b -c(8,12,18,24,36,48,60,72,96) � following function: � infcprodessa -function (D, tin, tau, ts) � (D * (1 - exp(-0.048 * tin))/(tin * (0.048*79) * (1 - exp(-0.048 * tau * exp(-0.048 * (ts - tin)) z -sapply(1:1, function(n) infcprodessa(1000,TINF,12,12-TINF)) � is there a way to select the combination of respective a and b values that would result in a calculated z that is between 15 and 20? In this case the a would be 1000 and the b would be 12 (other combinations are also possible), but how could I automatically find them? perhaps a loop? � Apreciate the help, � Sincerely, � Andras [[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] Loop question?
On 26-01-2013, at 12:31, Andras Farkas motyoc...@yahoo.com wrote: Sorry Jeff, probably the new version of Yahoo mail doing the html, I switched back to the older one hope that takes care of the problem. Let me clarify the code below: TINF -1 a -c(500,750,1000,1250,1500,1750,2000) b -c(8,12,18,24,36,48,60,72,96) infcprodessa -function (D, tin, tau, ts) (D * (1 - exp(-0.048 * tin))/(tin * (0.048*79) * (1 - exp(-0.048 * tau * exp(-0.048 * (ts - tin)) z -sapply(1:1, function(n) infcprodessa(a,TINF,b,b-TINF)) so I am looking to find the combination of a = 1000 and b = 12, which iz a value for z that would fall between 15 and 20. Sometimes though there may be more than one combinations that will meet criteria, in that case I would like to select the combination based on the smallest a value that has the respective b value to meet the criteria, ab - expand.grid(a=a,b=b) z - unlist(sapply(seq_len(nrow(ab)), function(k) infcprodessa(ab[k,1],TINF,ab[k,2],ab[k,2]-TINF))) z.target - which(z=15 z=20) ab[z.target,] z[z.target] which.min(ab[z.target,][,a]) Berend __ 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] Loop question?
The unlist-sapply-seq_len bit is unnecessarily convoluted, since the infcprodessa function can accept vector inputs. z - infcprodessa( ab$a, TINF, ab$b, ab$b-TINF ) possibles - ab[ z = 15 z = 20, ] possibles[ which.min( possibles$a ), ] --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Berend Hasselman b...@xs4all.nl wrote: On 26-01-2013, at 12:31, Andras Farkas motyoc...@yahoo.com wrote: Sorry Jeff, probably the new version of Yahoo mail doing the html, I switched back to the older one hope that takes care of the problem. Let me clarify the code below: TINF -1 a -c(500,750,1000,1250,1500,1750,2000) b -c(8,12,18,24,36,48,60,72,96) infcprodessa -function (D, tin, tau, ts) (D * (1 - exp(-0.048 * tin))/(tin * (0.048*79) * (1 - exp(-0.048 * tau * exp(-0.048 * (ts - tin)) z -sapply(1:1, function(n) infcprodessa(a,TINF,b,b-TINF)) so I am looking to find the combination of a = 1000 and b = 12, which iz a value for z that would fall between 15 and 20. Sometimes though there may be more than one combinations that will meet criteria, in that case I would like to select the combination based on the smallest a value that has the respective b value to meet the criteria, ab - expand.grid(a=a,b=b) z - unlist(sapply(seq_len(nrow(ab)), function(k) infcprodessa(ab[k,1],TINF,ab[k,2],ab[k,2]-TINF))) z.target - which(z=15 z=20) ab[z.target,] z[z.target] which.min(ab[z.target,][,a]) Berend __ 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] Loop question?
Dear All I have the following data (somewhat simplyfied): TINF -1 a -c(500,750,1000,1250,1500,1750,2000) b -c(8,12,18,24,36,48,60,72,96) following function: infcprodessa -function (D, tin, tau, ts) (D * (1 - exp(-0.048 * tin))/(tin * (0.048*79) * (1 - exp(-0.048 * tau * exp(-0.048 * (ts - tin)) z -sapply(1:1, function(n) infcprodessa(1000,TINF,12,12-TINF)) is there a way to select the combination of respective a and b values that would result in a calculated z that is between 15 and 20? In this case the a would be 1000 and the b would be 12 (other combinations are also possible), but how could I automatically find them? perhaps a loop? Apreciate the help, Sincerely, Andras [[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] Loop question?
Please read the Posting Guide no html email reproducible example please In general, you can use expand.grid to generate all combinations of inputs, compute results as a vector just as long as the expand.grid data frame has rows, and identify which results meet your criteria by a logical test, and use that test to identify which input combinations worked. Provide a working starting point and someone might give you working code as an answer. (where do a and b come into your problem?) --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Andras Farkas motyoc...@yahoo.com wrote: Dear All � I have the following data (somewhat simplyfied): � TINF -1 a -c(500,750,1000,1250,1500,1750,2000) b -c(8,12,18,24,36,48,60,72,96) � following function: � infcprodessa -function (D, tin, tau, ts) � (D * (1 - exp(-0.048 * tin))/(tin * (0.048*79) * (1 - exp(-0.048 * tau * exp(-0.048 * (ts - tin)) z -sapply(1:1, function(n) infcprodessa(1000,TINF,12,12-TINF)) � is there a way to select the combination of respective a and b values that would result in a calculated z that is between 15 and 20? In this case the a would be 1000 and the b would be 12 (other combinations are also possible), but how could I automatically find them? perhaps a loop? � Apreciate the help, � Sincerely, � Andras [[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] Loop question
Dear Sebastian, The following will create the names paste(sb,1:5,sep=) paste(sw,1:5,sep=) paste(Lw,1:5,sep=) paste(Lb,1:5,sep=) Then you can easily combine and/or order them in R. Hope this helps. Ozgur - Ozgur ASAR Research Assistant Middle East Technical University Department of Statistics 06531, Ankara Turkey Ph: 90-312-2105309 http://www.stat.metu.edu.tr/people/assistants/ozgur/ -- View this message in context: http://r.789695.n4.nabble.com/Loop-question-tp4631896p4631900.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] Loop question
Note that in R = 2.15 you can also use paste0 for this operation more efficiently. Michael On Thu, May 31, 2012 at 1:58 AM, Özgür Asar oa...@metu.edu.tr wrote: Dear Sebastian, The following will create the names paste(sb,1:5,sep=) paste(sw,1:5,sep=) paste(Lw,1:5,sep=) paste(Lb,1:5,sep=) Then you can easily combine and/or order them in R. Hope this helps. Ozgur - Ozgur ASAR Research Assistant Middle East Technical University Department of Statistics 06531, Ankara Turkey Ph: 90-312-2105309 http://www.stat.metu.edu.tr/people/assistants/ozgur/ -- View this message in context: http://r.789695.n4.nabble.com/Loop-question-tp4631896p4631900.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] Loop question
Hello,
I have a dataframe (Lx) with 5 Lb, and 5 Lw variables. I want to
create several variables according to the loop presented below (data
attached).
Lx - read.csv(Lx.csv, header=T, sep=,)
for (i in 1:20) {
Lx$sb1[i] - Lx$Lb1[i+1]/Lx$Lb1[i]
Lx$sb2[i] - Lx$Lb2[i+1]/Lx$Lb2[i]
Lx$sb3[i] - Lx$Lb3[i+1]/Lx$Lb3[i]
Lx$sb4[i] - Lx$Lb4[i+1]/Lx$Lb4[i]
Lx$sb5[i] - Lx$Lb5[i+1]/Lx$Lb5[i]
Lx$sw1[i] - Lx$Lw1[i+1]/Lx$Lw1[i]
Lx$sw2[i] - Lx$Lw2[i+1]/Lx$Lw2[i]
Lx$sw3[i] - Lx$Lw3[i+1]/Lx$Lw3[i]
Lx$sw4[i] - Lx$Lw4[i+1]/Lx$Lw4[i]
Lx$sw5[i] - Lx$Lw5[i+1]/Lx$Lw5[i]
}
How I could also create the variable names (letters b and w, and
numbers from 1 to 5 in s and L variables) using a loop in R?
In Stata I can use this command:
foreach r in b w {;
foreach s of numlist 0(5)40 {;
foreach ed of numlist 1/5 {;
local f = `s'+5;
scalar define rL`r'`ed'_`f'over`s' = L`r'`ed'_`f' / L`r'`ed'_`s';
};
};
};
but I do not how to do it in R.
Thank you in advance!
--
Sebastián Daza
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[R] loop question
Dear all,
I am trying to implement the following in a loop:
g - cbind(c(1, 2, 3), c(1, 2, 3))
h - cbind(c(1, 2, 3), c(1, 2, 3))
c - cbind(g[,1]*h[,1], g[,2]*h[,2])
g-rowSums(c)
My attempt looks like this but does not produce the desired results as
above.
for (i in 1:2) {g- rowSums(cbind(g[,i]*h[,i]))}
How do I do this correctly?
Many thanks,
Thomas
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[R] loop question
Dear all,
I am trying to set up a list with 1:c objects each meant to capture the
coefficients for one coefficient and 100 replications. I receive the
following error message:
Error in betaboot[[p]] : subscript out of bounds.
My code is below. Where is my mistake?
Many thanks,
Thomas
_
betaboot-list(NULL)
for (i in 1:c) {
betaboot[[i]]-cbind()
}
num - 100 # this is the number of bootstraps
for (i in 1:num) {
[BOOTSTRAP]
coef.temp - coef(model.temp, data=newdata)
for (p in 1:c){
betaboot[[p]] - cbind(betaboot[[p]], coef.temp[,p])
}
}
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Re: [R] loop question
Dear Thomas,
On Tue, Apr 5, 2011 at 8:33 AM, Thomas thomas.tri...@cantab.net wrote:
Dear all,
I am trying to set up a list with 1:c objects each meant to capture the
coefficients for one coefficient and 100 replications. I receive the
following error message:
Error in betaboot[[p]] : subscript out of bounds.
My code is below. Where is my mistake?
Many thanks,
Thomas
_
betaboot-list(NULL)
if you know the number of bootstraps (which you seem to later on), a
preferred way to instatiate the list would be:
betaboot - vector(mode = list, length = yourlength)
for (i in 1:c) {
because c() is such an important function, I would strongly
encourage you not to use it also as a variable.
betaboot[[i]]-cbind()
Don't use this to build an empty list.
}
num - 100 # this is the number of bootstraps
for (i in 1:num) {
[BOOTSTRAP]
coef.temp - coef(model.temp, data=newdata)
for (p in 1:c){
betaboot[[p]] - cbind(betaboot[[p]], coef.temp[,p])
This should work assuming betaboot is instatiated properly. That
said, it looks like you have a nested for loop and then just keep
cbind()ing each element of betaboot bigger and bigger. You may get a
performance increase if you also instantiate each matrix/dataframe
inside betaboot. Then the call would become something like:
betaboot[[i]][,p] - coef.temp[,p]
that is, you can use a chained series of extraction operators to get
to the appropriate column in the matrix/dataframe inside the
appropriate list element. Then rather than constantly using cbind(),
you just place coef.temp[,p] where you want it. The only requirement
is that you know the sizes of the matrices/dataframes going in so you
can create empty ones from the get go.
Cheers,
Josh
}
}
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--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
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Re: [R] Loop question
Brendan,
Matrix is atomic. Once you define t1 in matrix, t1[1]=0 rather than the
whole column. I would just convert t1 to a data frame, which is a special
list, by adding t1- data.frame(t1). Now t1[1] represents the whole column.
Then you can use your loop to add more columns.
Jun
On Fri, Apr 17, 2009 at 9:12 PM, Brendan Morse morse.bren...@gmail.comwrote:
Hi everyone, I am trying to accomplish a small task that is giving me quite
a headache. I would like to automatically generate a series of matrices and
give them successive names. Here is what I thought at first:
t1-matrix(0, nrow=250, ncol=1)
for(i in 1:10){
t1[i]-rnorm(250)
}
What I intended was that the loop would create 10 different matrices with a
single column of 250 values randomly selected from a normal distribution,
and that they would be labeled t11, t12, t13, t14 etc.
Can anyone steer me in the right direction with this one?
Thanks!
Brendan
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--
Jun Shen PhD
PK/PD Scientist
BioPharma Services
Millipore Corporation
15 Research Park Dr.
St Charles, MO 63304
Direct: 636-720-1589
[[alternative HTML version deleted]]
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Re: [R] Loop question
On Fri, Apr 17, 2009 at 10:12 PM, Brendan Morse morse.bren...@gmail.com wrote:
...I would like to automatically generate a series of matrices and
give them successive names. Here is what I thought at first:
t1-matrix(0, nrow=250, ncol=1)
for(i in 1:10){
t1[i]-rnorm(250)
}
What I intended was that the loop would create 10 different matrices with a
single column of 250 values randomly selected from a normal distribution,
and that they would be labeled t11, t12, t13, t14 etc.
Very close. But since you've started out with a *matrix* t1, your
assignments to t1[i] will assign to parts of the matrix. To correct
this, all you need to do is initialize t1 as a *list of matrices* or
(even better) as an *empty list*, like this:
t1 - list()
and then assign to *elements* of the list (using [[ ]] notation), not
to *sublists* of the list (which is what [ ] notation means in R),
like this:
for(i in 1:10){
t1[[i]] - rnorm(250)
}
Is that what you had in mind?
-s
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[R] Loop question
Hi everyone, I am trying to accomplish a small task that is giving me
quite a headache. I would like to automatically generate a series of
matrices and give them successive names. Here is what I thought at
first:
t1-matrix(0, nrow=250, ncol=1)
for(i in 1:10){
t1[i]-rnorm(250)
}
What I intended was that the loop would create 10 different matrices
with a single column of 250 values randomly selected from a normal
distribution, and that they would be labeled t11, t12, t13, t14 etc.
Can anyone steer me in the right direction with this one?
Thanks!
Brendan
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Re: [R] Loop question
Dear Brendan,
One way could be either
bigt - sapply(1:10,function(x) rnorm(250))
colnames(bigt) - paste('t',1:10,sep=)
bigt
or
bigt2 - NULL
for(i in 1:10) bigt2 - cbind( bigt2, rnorm(250) )
colnames(bigt2) - paste('t',1:10,sep=)
bigt2
or
bigt3 - matrix(rnorm(250*10),ncol=10)
colnames(bigt3) - paste('t',1:10,sep=)
bigt3
HTH,
Jorge
On Fri, Apr 17, 2009 at 10:12 PM, Brendan Morse morse.bren...@gmail.comwrote:
Hi everyone, I am trying to accomplish a small task that is giving me quite
a headache. I would like to automatically generate a series of matrices and
give them successive names. Here is what I thought at first:
t1-matrix(0, nrow=250, ncol=1)
for(i in 1:10){
t1[i]-rnorm(250)
}
What I intended was that the loop would create 10 different matrices with a
single column of 250 values randomly selected from a normal distribution,
and that they would be labeled t11, t12, t13, t14 etc.
Can anyone steer me in the right direction with this one?
Thanks!
Brendan
__
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http://www.R-project.org/posting-guide.html
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[[alternative HTML version deleted]]
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Re: [R] Loop question
Brendan Morse wrote:
Hi everyone, I am trying to accomplish a small task that is giving me
quite a headache. I would like to automatically generate a series of
matrices and give them successive names. Here is what I thought at
first:
t1-matrix(0, nrow=250, ncol=1)
for(i in 1:10){
t1[i]-rnorm(250)
}
What I intended was that the loop would create 10 different matrices
with a single column of 250 values randomly selected from a normal
distribution, and that they would be labeled t11, t12, t13, t14 etc.
I think you're basically looking for
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-turn-a-string-into-a-variable_003f
or
http://wiki.r-project.org/rwiki/doku.php?id=tips:data-misc:create_var_names
but see the comments in both places that indicate why it may be easier to do
this
as a list.
for(i in 1:10){
assign(paste(t1,i,sep=),matrix(rnorm(250)))
}
--
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
