[R] rearrange set of items randomly
(Ted Harding)
ted.harding at wlandres.net
Tue Nov 8 10:30:35 CET 2011
On 08-Nov-11 08:59:38, Ted Harding wrote:
> On 08-Nov-11 07:46:15, flokke wrote:
>> Sorry, but I dont think that I get what you mean (I am a
>> quite new user though)
>> I hae a data file of a sample, so I cannot use random numbers,
>> the only thing I want to do is to randomly reassign the order
>> of the items.
>
> The simplest method of randomly re-ordering is to use
> sample() to re-arrange (1:N) randomly, where N is the
> number of items in the data. Then use the result to
> access the items. Example:
>
> D <- data.frame(X1=c(1.1,2.1,3.1,4.1),
> X2=c(1.2,2.2,3.2,4.2))
> N <- nrow(D)
> ix <- sample((1:N))
> D
> # X1 X2
> # 1 1.1 1.2
> # 2 2.1 2.2
> # 3 3.1 3.2
> # 4 4.1 4.2
> N
> # [1] 4
> ix
> # [1] 3 2 4 1
> D[ix,]
> # X1 X2
> # 3 3.1 3.2
> # 2 2.1 2.2
> # 4 4.1 4.2
> # 1 1.1 1.2
>
> Note that the defaults for sample() are (see ?sample):
>
> For 'sample' the default for 'size' is the number
> of items inferred from the first argument, so that
> 'sample(x)' generates a random permutation of the
> elements of 'x' (or '1:x').
>
> and the default for the 'replace' option is "FALSE",
> so sample((1:N)) samples N from (1:N) without replacement,
> i.e. a random permutation.
>
> Ted.
While I am at it, an alternative to this use of sample()
is to use order() to find the permutation which re-arranges
a set of random numbers into increasing order. This in effect
returns a random permutation of (1:N). Hence, instead of
"ix <- sample(1:N))" in the above, you could use:
ix <- order(runif(N))
Ted.
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Date: 08-Nov-11 Time: 09:30:33
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