Well, you can create the expand.grid data frame programmatically via:
u <- 1:3
len <- length(u)
v <-do.call(expand.grid, split(rep(u,len),rep(seq_len(len),e=len)))
And then you can use unique.array to get the unique rows after the sort:
unique(t(apply(v,1,sort)))
However, I agree with your sentiments. Not only does this seem
inelegant, but it will not scale well.
I would imagine a recursive approach would be more efficient -- as
then only the sets you need would be produced and there'd be no
sorting, etc. -- but I have neither the time nor interest to work it
out.
... and I bet someone already has done this in some R package anyway.
Cheers,
Bert
On Thu, Nov 7, 2013 at 9:04 AM, Ted Harding <ted.hard...@wlandres.net> wrote:
> On 07-Nov-2013 13:38:29 Konstantin Tretiakov wrote:
>> Hello!
>>
>> I need to obtain all possible combinations with replacement when
>> order is not important.
>> E.g. I have a population x{1,2,3}.
>> So I can get (choose(3+3-1,3)=) 10 combinations from this population
>> with 'size=3'.
>> How can I get a list of all that combinations?
>>
>> I have tried 'expand.grid()' and managed to get only samples where
>> order is important.
>> 'combn()' gave me samples without replacement.
>>
>> Best regards,
>> Konstantin Tretyakov.
>
> >From your description I infer that, from {1,2,3}, you want the result:
>
> 1 1 1
> 1 1 2
> 1 1 3
> 1 2 2
> 1 2 3
> 1 3 3
> 2 2 2
> 2 2 3
> 2 3 3
> 3 3 3
>
> The following will do that:
>
> u <- c(1,2,3)
> unique(t(unique(apply(expand.grid(u,u,u),1,sort),margin=1)))
>
> # [,1] [,2] [,3]
> # [1,] 1 1 1
> # [2,] 1 1 2
> # [3,] 1 1 3
> # [4,] 1 2 2
> # [5,] 1 2 3
> # [6,] 1 3 3
> # [7,] 2 2 2
> # [9,] 2 3 3
> #[10,] 3 3 3
>
> There may be a simpler way!
> Ted.
>
> -------------------------------------------------
> E-Mail: (Ted Harding) <ted.hard...@wlandres.net>
> Date: 07-Nov-2013 Time: 17:04:50
> This message was sent by XFMail
>
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--
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
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