Re. partitions()

Question

Why is

julia> collect(partitions(1,2))
0-element Array{Any,1}

returned instead of

2-element Array{Any,1}:
 [0,1]
 [1,0]

and do I really have to

x = collect(partitions(n,m));
y = Array(Int64,length(x),length(x[1]));
for i in 1:length(x)
    for j in 1:length(x[1])
        y[i,j] = x[i][j];
    end
end

to convert the result to a two-dimensional array?

Answer 1

Here's another approach to your problem that I think is a little simpler, using the Combinatorics.jl library:

multisets(n, k) = map(A -> [sum(A .== i) for i in 1:n],
                      with_replacement_combinations(1:n, k))

This allocates a bunch of memory, but I think your current approach does too. Maybe it would be useful to make a first-class version and add it to Combinatorics.jl.

Examples:

julia> multisets(2, 1)
2-element Array{Array{Int64,1},1}:
 [1,0]
 [0,1]

julia> multisets(3, 5)
21-element Array{Array{Int64,1},1}:
 [5,0,0]
 [4,1,0]
 [4,0,1]
 [3,2,0]
 [3,1,1]
 [3,0,2]
 [2,3,0]
 [2,2,1]
 [2,1,2]
 [2,0,3]
 ⋮      
 [1,2,2]
 [1,1,3]
 [1,0,4]
 [0,5,0]
 [0,4,1]
 [0,3,2]
 [0,2,3]
 [0,1,4]
 [0,0,5]

The argument order is backwards from yours to match mathematical convention. If you prefer the other way, that can easily be changed.