Re: [R] Permutations with replacement (final final final)
Hi Daniel,
Turns out, your code, however simple, is quite elegant for my needs
(sometimes I overanalyze :O). Here is my last code to do what I need to
do:
### begin R code
# generate a matrix of ten thousand rows of 1-8
z - t(matrix(rep(1:8,1),8,1))
library(Matrix)
# use the R sample function in a loop to sample each line with replacement
zcomb=Matrix()
for (i in 1:dim(z)[1]) {
z1 - t(matrix(sample(z,8,replace=TRUE)))
zcomb = rbind(zcomb,z1)
}
zcomb
### end R code #
Regards,
Jesse A. Canchola
Daniel Nordlund [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
08/18/2006 05:16 PM
To
'Jesse Albert Canchola' [EMAIL PROTECTED], 'r-help'
r-help@stat.math.ethz.ch
cc
Subject
Re: [R] Permutations with replacement
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]
On Behalf Of Jesse Albert Canchola
Sent: Friday, August 18, 2006 1:02 PM
To: r-help
Subject: [R] Permutations with replacement
Is there a simple function or process that will create permutations with
replacement?
I know that using the combinat package
## begin R code ##
library(combinat)
m - t(array(unlist(permn(3)), dim = c(3, 6)))
# we can get the permutations, for example 3!=6
# gives us
m
[,1] [,2] [,3]
[1,]123
[2,]132
[3,]312
[4,]321
[5,]231
[6,]213
## end R code ##
I'd like to include the with replacement possibilities such as
1,1,3
1,1,2
2,3,3
Isn't what you want just sampling with replacement?
x - c(1,2,3)
sample(x,3,replace=TRUE)
Hope this is helpful,
Dan
Dan Nordlund
Bothell, WA USA
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Re: [R] Permutations with replacement
Thanks, David. That worked fabulously!
Here is the R code for the hypercube test example:
## begin R code
library(combinat)
x - rep(3,3) # for partitions of 3 units into the three classes {1,2,3}
hcube(x, scale=1, transl=0)
### end R code
For the larger one I want (i.e., 8^8), I will take a random sample of
10,000 from the 16,777,216 possibilities.
Regards,
Jesse Canchola
[EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
08/18/2006 01:33 PM
To
Jesse Albert Canchola [EMAIL PROTECTED], r-help
r-help@stat.math.ethz.ch
cc
Subject
Re: [R] Permutations with replacement
If you also want 1,1,1 and so on, the number of these is n^n,
(n choices for each of n slots.)
In that case, you could use hcube from combinat.
David L. Reiner
Rho Trading Securities, LLC
Chicago IL 60605
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Jesse Albert
Canchola
Sent: Friday, August 18, 2006 3:26 PM
To: r-help
Subject: [R] Permutations with replacement
Is there a simple function or process that will create a matrix of
permutations with replacement?
I know that using the combinat package
## begin R code ##
library(combinat)
m - t(array(unlist(permn(3)), dim = c(3, 6)))
# we can get the permutations, for example 3!=6
# gives us
m
[,1] [,2] [,3]
[1,]123
[2,]132
[3,]312
[4,]321
[5,]231
[6,]213
## end R code ##
I'd like to include the with replacement possibilities such as
1,1,3
1,1,2
2,3,3
and so on. This will eventually be done on 8!=40,320 rather than the
development version using 3! as above.
If no function exists (I've Googled on CRAN with no palpable luck), then
perhaps this is more of a bootstrap type problem.
Thanks for your help in advance,
Jesse Canchola
___
The information contained in this e-mail is for the exclusive use of the
intended recipient(s) and may be confidential, proprietary, and/or
legally privileged. Inadvertent disclosure of this message does not
constitute a waiver of any privilege. If you receive this message in
error, please do not directly or indirectly use, print, copy, forward,
or disclose any part of this message. Please also delete this e-mail
and all copies and notify the sender. Thank you.
For alternate languages please go to http://bayerdisclaimer.bayerweb.com
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Permutations with replacement
Is there a simple function or process that will create permutations with replacement? I know that using the combinat package ## begin R code ## library(combinat) m - t(array(unlist(permn(3)), dim = c(3, 6))) # we can get the permutations, for example 3!=6 # gives us m [,1] [,2] [,3] [1,]123 [2,]132 [3,]312 [4,]321 [5,]231 [6,]213 ## end R code ## I'd like to include the with replacement possibilities such as 1,1,3 1,1,2 2,3,3 and so on. This will eventually be done on 8!=40,320 rather than the development version using 3! as above. If no function exists (I've Googled on CRAN with no palpable luck), then perhaps this is more of a bootstrap type problem. Thanks for your help in advance, Jesse Canchola ___ The information contained in this e-mail is for the exclusive use of the intended recipient(s) and may be confidential, proprietary, and/or legally privileged. Inadvertent disclosure of this message does not constitute a waiver of any privilege. If you receive this message in error, please do not directly or indirectly use, print, copy, forward, or disclose any part of this message. Please also delete this e-mail and all copies and notify the sender. Thank you. For alternate languages please go to http://bayerdisclaimer.bayerweb.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Permutations with replacement
Is there a simple function or process that will create a matrix of permutations with replacement? I know that using the combinat package ## begin R code ## library(combinat) m - t(array(unlist(permn(3)), dim = c(3, 6))) # we can get the permutations, for example 3!=6 # gives us m [,1] [,2] [,3] [1,]123 [2,]132 [3,]312 [4,]321 [5,]231 [6,]213 ## end R code ## I'd like to include the with replacement possibilities such as 1,1,3 1,1,2 2,3,3 and so on. This will eventually be done on 8!=40,320 rather than the development version using 3! as above. If no function exists (I've Googled on CRAN with no palpable luck), then perhaps this is more of a bootstrap type problem. Thanks for your help in advance, Jesse Canchola ___ The information contained in this e-mail is for the exclusive use of the intended recipient(s) and may be confidential, proprietary, and/or legally privileged. Inadvertent disclosure of this message does not constitute a waiver of any privilege. If you receive this message in error, please do not directly or indirectly use, print, copy, forward, or disclose any part of this message. Please also delete this e-mail and all copies and notify the sender. Thank you. For alternate languages please go to http://bayerdisclaimer.bayerweb.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Permutations with replacement
If you also want 1,1,1 and so on, the number of these is n^n, (n choices for each of n slots.) In that case, you could use hcube from combinat. David L. Reiner Rho Trading Securities, LLC Chicago IL 60605 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Jesse Albert Canchola Sent: Friday, August 18, 2006 3:26 PM To: r-help Subject: [R] Permutations with replacement Is there a simple function or process that will create a matrix of permutations with replacement? I know that using the combinat package ## begin R code ## library(combinat) m - t(array(unlist(permn(3)), dim = c(3, 6))) # we can get the permutations, for example 3!=6 # gives us m [,1] [,2] [,3] [1,]123 [2,]132 [3,]312 [4,]321 [5,]231 [6,]213 ## end R code ## I'd like to include the with replacement possibilities such as 1,1,3 1,1,2 2,3,3 and so on. This will eventually be done on 8!=40,320 rather than the development version using 3! as above. If no function exists (I've Googled on CRAN with no palpable luck), then perhaps this is more of a bootstrap type problem. Thanks for your help in advance, Jesse Canchola ___ The information contained in this e-mail is for the exclusive use of the intended recipient(s) and may be confidential, proprietary, and/or legally privileged. Inadvertent disclosure of this message does not constitute a waiver of any privilege. If you receive this message in error, please do not directly or indirectly use, print, copy, forward, or disclose any part of this message. Please also delete this e-mail and all copies and notify the sender. Thank you. For alternate languages please go to http://bayerdisclaimer.bayerweb.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list 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] Permutations with replacement
-Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Jesse Albert Canchola Sent: Friday, August 18, 2006 1:02 PM To: r-help Subject: [R] Permutations with replacement Is there a simple function or process that will create permutations with replacement? I know that using the combinat package ## begin R code ## library(combinat) m - t(array(unlist(permn(3)), dim = c(3, 6))) # we can get the permutations, for example 3!=6 # gives us m [,1] [,2] [,3] [1,]123 [2,]132 [3,]312 [4,]321 [5,]231 [6,]213 ## end R code ## I'd like to include the with replacement possibilities such as 1,1,3 1,1,2 2,3,3 Isn't what you want just sampling with replacement? x - c(1,2,3) sample(x,3,replace=TRUE) Hope this is helpful, Dan Dan Nordlund Bothell, WA USA __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
