Applicative Functor on Lists

Question

I understand that following will be:

[(+2),(+1)]<*>[1,2,3] == [3,4,5,2,3,4]

I also understand that fmap is implemented as map. But how could I mentally map this computation in my head? The first time I saw this I assumed the following:

[(*2),(+1)]<*>[1,2,3] == [4,5,6]

The second time around I though about: [[3,4,5],[2,3,4]]. But on the other hand <*> returns an f b so I knew it wouldn't be possible. So my question is, what are the mental steps to make sense out of this?

Louis Wasserman · Accepted Answer · 2014-12-17T01:37:36.657

19

fs <*> xs is more-or-less [f x | f <- fs, x <- xs]. Monads have the Applicative instance

fM <*> xM = do 
  f <- fM
  x <- xM
  return (f x)

which corresponds to the list comprehension pretty directly.

edited Dec 17 '14 at 01:37

answered Dec 17 '14 at 00:41

Louis Wasserman

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score 8 · Answer 2 · edited Jan 06 '15 at 03:13

8

To remember this you might find it easier to simply imagine <*> to be × (Cartesian product):

[a, b, c] × [x, y, z] == [a x, a y, a z, b x, b y, ...]

edited Jan 06 '15 at 03:13

dfeuer

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answered Dec 17 '14 at 04:13

Shoe

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And in fact, `(,) <$> xs <*> ys` produces a representation of `xs × ys`. – dfeuer Jan 06 '15 at 03:12

score 1 · Answer 3 · answered Nov 03 '18 at 19:41

1

Interesting. Does an applicative always have to be the cartesian product? Or is the strategy really a matter of choice. For instance could a "zip" of two lists also be valid? ie:

[F1,F2,F3] <*> [a,b,c] == [F1 a, F2 b, F3 c]

answered Nov 03 '18 at 19:41

Sri

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It could. This is implemented in Control.Applicative.ZipList. – Alistair Wall Apr 05 '20 at 10:47

Applicative Functor on Lists

3 Answers3