all combinations with replacement not ordered
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.Harding at 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.Harding at 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