How do I have multiple generators for list comprehensions in F#?

Question

I am trying to write something like this

[(x,y)|x<- [1,2,3], y <- [’a’,’b’]]
 => [(1,’a’),(1,’b’),(2,’a’),(2,’b’),(3,’a’),(3,’b’)]

possible duplicate: http://stackoverflow.com/questions/935996/calculating-the-cartesian-product-of-a-list-of-numbers-with-f — BrokenGlass, Mar 17 '11 at 03:36

score 5 · Accepted Answer · answered Mar 17 '11 at 03:36

5

[for x in [1;2;3] do
 for y in ['a';'b'] do
 yield x,y]

answered Mar 17 '11 at 03:36

kvb

score 1 · Answer 2 · answered Mar 17 '11 at 05:46

1

just another fun way

[1;2;3] |> List.map ( fun X -> ['a';'b'] |> List.map (fun A -> X,A) )

answered Mar 17 '11 at 05:46

cnd

3

Or rather `[1;2;3] |> List.collect ( fun X -> ['a';'b'] |> List.map (fun A -> X,A) )` – Ed'ka Mar 17 '11 at 08:12
@Ed'ka yes, it's must be better. – cnd Mar 17 '11 at 08:18

score -2 · Answer 3 · answered Mar 17 '11 at 03:33

-2

F# equivalent.

List.zip [1;2;3] ['a';'b';'c']

answered Mar 17 '11 at 03:33

Nyi Nyi

1

Not quite - this pairs up elements in corresponding positions (and therefore requires lists of the same length); the question is looking for the Cartesian product of the two lists. – kvb Mar 17 '11 at 03:37
@kvb - Thanks for pointing out. I need to be more careful about comprehending question next time. – Nyi Nyi Mar 17 '11 at 04:47

3 Answers3