I have a question regarding the samr package.
For a 2 class unpaired problem, with sample 1 of size N1 and sample 2
of size N2, samr computes at most (N1+N2)! permutations of the two
samples (if the user-supplied parameter nperms allows it). However,
there are only (N1+N2)/(N1!*N2!) DISTINCT
Dear R Users,
How can I randomize a matrix (with contains only 0 and 1) with the
constraint that margin totals (columns and rows sums) are kept constant?
Some have called this swap permutation or something like that.
The principle is to find bloc of
10
01
and change it for
01
10
there can be
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
]
[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
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]
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
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
-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
Dear R users,
This is a second summary of the permutation problem I previously posted.
This summary restates the problem as well as the solution.
First of all thanks to everyone including Erich, Robin, Gabor, Christian,
Ingmar and others for your suggestions.
With the help of an off-list
On Wed, Jul 14, 2004 at 07:42:49PM +0200, Jordi Altirriba Gutiérrez wrote:
Ive 12 elements in blocks of 3 elements and I want only to make
permutations inter-blocks (no intra-blocks) (sorry if the terminology is
not accurate)
I am not a mathematician, but this sounds to me a little bit
hi again
what a stimulating R discussion! This is R-help at its very best!
I think I understand why you don't want pure inter-block permutations.
My solution would be to realize that weeding out forbidden permutations is
quite difficult and time-consuming (also as several people have pointed out
Dear R users,
First of all, I want to thank the algorithms , time and suggestions to
Rolf, Robert, Marc, Gabor, Adaikalavan, Cliff, Robin, Erich and Fernando.
I want to sum up a little bit all the e-mails, the algorithms and the
results of a test for 1000 permutations (in my last e-mail is
On 7/14/04 7:42 PM, Jordi Altirriba Gutiérrez [EMAIL PROTECTED]
wrote:
4 5 6 | 2 1 3 | 7 8 9 | 10 11 12 NO because it's an intra-block
permutation of permutation 3
- -
10 1 7 | 4 8 7 | 5 6 12 | 3 2 9YES---Xth permutation
1 10 7 | 4 8 7 | 5 6 12 | 3 2 9NO because
Jordi:
If I understand you well, the function below may do what you asked for.
It is not clear to me from your posting wether e.g.
1 2 4 3 5 6 7 8 910 11 12
and
1 4 2 3 5 6 7 8 910 11 12
should count as differente permutations, i.e., wether once one pair of
elements
Perhaps what you want might better be described as
ordered partitions?
Is what you want the following:
We study sequences of length 12 and divide them in
4 segments
position 1 2 3, position 4 5 6,
position 7 8 9, position 10 11 12,
Find all permutation sequences of the numbers 1 to 12
with the
Dear R users,
First of all, thanks to Rolf, Brad, Robin, Erich, Fernando and Adaikalavan
for your time and suggestions.
Ive been testing some algorithms (sorry for the delay, Im very slow, and
Im a completely beginner in Rs world).
First, the Robin algorithm.
I think that there is a
As Erich points out, there is some question as to
what the original problem really is but lets
assume its as Erich describes. Then, to get a
random ordered permutation we just get a random
permutation of 12 elements and sort the intra-block
elements like this:
I think the issue here is in the two keywords - permutations or sample.
AFAIK, permutations should return all admissible (by some rule)
combinations. If this is a large number, as some have pointed out, then
one essentially takes a _sample_ of all admissible combinations. Since
you earlier
Ramasamy [EMAIL PROTECTED]
To: Jordi Altirriba Gutiérrez [EMAIL PROTECTED]
CC: [EMAIL PROTECTED], R-help [EMAIL PROTECTED]
Subject: Re: [R] Permutations
Date: Wed, 14 Jul 2004 18:00:49 +0100
I think the issue here is in the two keywords - permutations or sample.
AFAIK, permutations should return all
On Wed, 2004-07-14 at 08:06, Rolf Turner wrote:
In respect of generating random ``restricted'' permutations, it
occurred to me as I was driving home last night If one is going
to invoke some kind of ``try again if it doesn't work procedure''
then one might as well keep it simple:
Gutirrez altirriba at hotmail.com
: CC: rksh at soc.soton.ac.uk, R-help r-help at stat.math.ethz.ch
: Subject: Re: [R] Permutations
: Date: Wed, 14 Jul 2004 18:00:49 +0100
:
: I think the issue here is in the two keywords - permutations or sample.
:
: AFAIK, permutations should return all admissible
Ramasamy ramasamy at cancer.org.uk
: : To: Jordi Altirriba Gutirrez altirriba at hotmail.com
: : CC: rksh at soc.soton.ac.uk, R-help r-help at stat.math.ethz.ch
: : Subject: Re: [R] Permutations
: : Date: Wed, 14 Jul 2004 18:00:49 +0100
: :
: : I think the issue here is in the two keywords
Dear R users,
Im a beginner user of R and Ive a problem with permutations that I dont
know how to solve. Ive 12 elements in blocks of 3 elements and I want only
to make permutations inter-blocks (no intra-blocks) (sorry if the
terminology is not accurate), something similar to:
1 2 3 | 4 5
On Tue, 2004-07-13 at 14:07, Jordi Altirriba Gutirrez wrote:
Dear R users,
Im a beginner user of R and Ive a problem with permutations that I dont
know how to solve. Ive 12 elements in blocks of 3 elements and I want only
to make permutations inter-blocks (no intra-blocks) (sorry if the
On Tue, 2004-07-13 at 14:29, Marc Schwartz wrote:
On Tue, 2004-07-13 at 14:07, Jordi Altirriba Gutirrez wrote:
Dear R users,
Im a beginner user of R and Ive a problem with permutations that I dont
know how to solve. Ive 12 elements in blocks of 3 elements and I want only
to make
Marc Schwartz wrote (in response to a question from Jordi Altirriba):
You can use the permutations() function in the 'gregmisc' package on
CRAN:
# Assuming you installed 'gregmisc' and used library(gregmisc)
# First create 'groups' consisting of the four blocks
groups - c(1 2 3, 4 5 6, 7
PROTECTED]
Sent: Tuesday, July 13, 2004 3:07 PM
To: [EMAIL PROTECTED]
Subject: [R] Permutations
Dear R users,
I'm a beginner user of R and I've a problem with permutations that I don't
know how to solve. I've 12 elements in blocks of 3 elements and I want only
to make permutations inter-blocks
! http://www.R-project.org/posting-guide.html
From [EMAIL PROTECTED] Tue Jul 13 17:19:55 2004
From: Baskin, Robert [EMAIL PROTECTED]
To: =?iso-8859-1?Q?=27Jordi_Altirriba_Guti=E9rrez=27?= [EMAIL PROTECTED],
[EMAIL PROTECTED]
Subject: RE: [R] Permutations
Date: Tue, 13 Jul 2004 16:12:46
On Tue, 2004-07-13 at 15:02, Rolf Turner wrote:
Marc Schwartz wrote (in response to a question from Jordi Altirriba):
snip
This does not solve the problem that was posed. It only permutes the
blocks, and does not allow for swapping between blocks. For instance
it does produce the
Dang!!! Forget it. My random restricted permutation generator
doesn't ``quite'' work. Further testing --- which I should've done
before posting (sigh) --- reveals that it can get into a situation in
which there's nothing left to sample from, and it still needs to fill
out the permutation. I.e.
For what it's worth, here is a mild revision of my restr.perm()
function, which seems NOT to fall over. I.e. it appears to
``reliably'' generate restricted permutations.
Whether these are genuinely ***random*** restricted permutations
(i.e. does each restricted permutation of 1:12 have the same
Dear all
If I consider
d- c(1,2,3,4)
N- 4
n- 2
out1- matrix(0,N^n,n)
z-1
for(i in 1:N)
{
for(j in 1:N)
{
out1[z,1] = d[i]
out1[z,2] = d[j]
z- z+1
}
}
library(gregmisc)
out2- permutations(N,n,d,T,T)
I have that out1==out2. Ok
Now, if I consider
d- c(1,2,3,4)
N- 4
n- 3
out1-
lamack lamack wrote:
Dear all
If I consider
d- c(1,2,3,4)
N- 4
n- 2
out1- matrix(0,N^n,n)
z-1
for(i in 1:N)
{
for(j in 1:N)
{
out1[z,1] = d[i]
out1[z,2] = d[j]
z- z+1
}
}
library(gregmisc)
out2- permutations(N,n,d,T,T)
I have that out1==out2. Ok
Now, if
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