[R] permutation within rows of a matrix

[Peter Ehlers]

I must be missing something. What's wrong with

  t(apply(mat, 1, sample))

?

Peter Ehlers

On 2011-11-16 12:12, Gavin Simpson wrote:
> On Wed, 2011-11-16 at 14:29 -0500, R. Michael Weylandt wrote:
>> Suppose your matrix is called X.
>>
>> ? sample
>> X[sample(nrow(X)),]
>
> That will shuffle the rows at random, not permute within the rows.
>
> Here is an alternative, first using one of my packages (permute -
> shameful promotion ;-) !:
>
> mat<- matrix(sample(0:1, 100, replace = TRUE), ncol = 10)
>
> require(permute)
> perms<- shuffleSet(10, nset = 10)
> ## permute mat
> t(sapply(seq_len(nrow(perms)),
>           function(i, perms, mat) mat[i, perms[i,]],
>           mat = mat, perms = perms))
>
> If you don't want to use permute, then you can do this via standard R
> functions
>
> perms<- t(replicate(nrow(mat), sample(ncol(mat))))
> ## permute mat
> t(sapply(seq_len(nrow(perms)),
>           function(i, perms, mat) mat[i, perms[i,]],
>           mat = mat, perms = perms))
>
> HTH
>
> G
>
>> Michael
>>
>> On Wed, Nov 16, 2011 at 11:45 AM, Juan Antonio Balbuena<balbuena at uv.es>  wrote:
>>> Hello
>>> This is probably a basic question but I am quite new to R.
>>>
>>> I need to permute elements within rows of a binary matrix, such as
>>>
>>>      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
>>>   [1,]    0    0    0    0    1    0    0    0    0     0
>>>   [2,]    0    0    1    1    0    0    0    1    1     0
>>>   [3,]    0    1    0    0    0    0    1    0    0     0
>>>   [4,]    0    0    0    0    0    0    1    1    0     0
>>>   [5,]    0    0    0    1    0    0    0    0    1     0
>>>   [6,]    0    0    1    1    0    0    0    0    0     1
>>>   [7,]    0    0    0    0    0    0    0    0    0     0
>>>   [8,]    1    1    0    1    0    0    0    1    0     1
>>>   [9,]    1    0    0    1    0    1    0    1    0     0
>>> [10,]    0    0    0    0    0    0    0    1    0     1
>>>
>>>
>>> That is, elements within each row are permuted freely and independently from
>>> the other rows.
>>>
>>> I see that is is workable by creating a array for each row, performing
>>> sample and binding the arrays again, but I wonder whether there is a more
>>> efficient way of doing the trick.
>>>
>>> Any help will be much appreciated.
>>>
>>>
>>>
>>> --
>>> View this message in context: http://r.789695.n4.nabble.com/permutation-within-rows-of-a-matrix-tp4076989p4076989.html
>>> Sent from the R help mailing list archive at Nabble.com.
>>>
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>>
>> ______________________________________________
>> R-help at 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.
>>
>