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haskellarrow-abstraction

Is there something like `map2 :: (i -> a) -> (i -> b) -> [i] -> [(a,b)]`?

I just wrote functions like this up to map4 just because they seem useful:

map2 :: Functor f => (i -> a) -> (i -> b) -> f i -> f (a,b)
map2 f1 f2 = fmap $ \i -> (f1 i, f2 i)

Before I continue to map8 i thought I'd ask if there is something similar in some standard module. Hayoo doesn't seem to know any function that has the signature above.

Note: I already found Control.Arrow.&&& which reduces the above to:

map2 f1 f2 = fmap (f1 &&& f2)

But there doesn't seem to be a similar function for a fanout more than two.

Solution

  • (->) i is an applicative functor, so you can write (&&&) as

    f &&& g = (,) <$> f <*> g

    and you could write map3 as

    map3 f1 f2 f3 = map ((,,) <$> f1 <*> f2 <*> f3)

    except that it isn't shorter than

    map3 f1 f2 f3 = map $ \i -> (f1 i, f2 i, f3 i)

    But thanks to Gabriel's tip, this is shorter:

    map3 f1 f2 f3 = map (liftA3 (,,) f1 f2 f3)