CODEWITHSUNDEEP
list-comprehensionjulia
Pigna
Pigna

Reputation: 2924

All combinations of all possible lengths in Julia

For instance, given the input n = 3 I want to obtain:

[(1,),(2,),(3,),(1,2),(1,3),(2,3),(1,2,3)]

I tried a python-like syntax:

combs = [comb for x in collect(1:n) for comb in combinations(collect(1:n),x)]

but I got the following error message:

ERROR: LoadError: syntax: expected ]

I've also tried this:

combs = [comb for comb in vcat([combinations(collect(1:n),x) for x in collect(1:n)])]

but I got:

[Base.Combinations{Array{Int64,1}}([1,2,3],1),Base.Combinations{Array{Int64,1}}([1,2,3],2),Base.Combinations{Array{Int64,1}}([1,2,3],3)]

How to get the result I want?

Upvotes: 2

Views: 774

Answers (1)

Dan Getz
Dan Getz

Reputation: 18217

Is

n = 3
vcat([collect(combinations(1:n,i)) for i=1:n]...)

OK?

Output:

7-element Array{Array{Int64,1},1}:
 [1]    
 [2]    
 [3]    
 [1,2]  
 [1,3]  
 [2,3]  
 [1,2,3]

OTHER METHODS:

Another method is [65-findin(bits(i),'1') for i=1:(2^n-1)], which points the way to a very efficient implementation along the lines of:

tmp = BitVector(sizeof(Int)*8)
[begin tmp.chunks[1]=i; find(tmp) end for i=1:(2^n-1)]

although it uses BitVector internals, which might be obscure.

For memory efficiency:

using Iterators
chain(([combinations(1:n,i) for i=1:n])...) |> collect

(could be used directly as iterator in a for loop). But with Iterators one could use:

drop(subsets(1:n),1) |> collect

which is readable.

Upvotes: 6

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