How about this? combos(1, L) -> [[X] || X <...

2010-04-21T03:24:29.353-07:00

How about this?

combos(1, L) -> [[X] || X <-L];
combos(K, L) when K == length(L) -> [L];
combos(K, [H|T]) ->
[[H | Subcombos] || Subcombos <- combos(K-1, T)]
++(combos(K, T)).

combos(L) ->
lists:foldl(
fun(K, Acc) -> Acc++(combos(K, L)) end,
[[]],
lists:seq(1, length(L))
).

(explanation here)

I guess your version is faster, though.

[...] was feeling nostalgic and went back to see h...

2008-10-21T09:32:58.000-07:00

[...] was feeling nostalgic and went back to see how I calculated combinations in Erlang and combinations in Ruby. I wanted to see if there was an efficient way to do it without resorting [...]

Comments on Seth Ladd's Blog: Calculating Combinations the Erlang Way

How about this? combos(1, L) -> [[X] || X <...

[...] was feeling nostalgic and went back to see h...