Is there a Haskell lens function for "zipping" same-length tuples?

Question

I would like to be able to combine two tuples of the same length using a function, similar to the zipWith function from base. For example, for the case of length 3 tuples:

zipTupleWith f (a0,a1,a2) (b0,b1,b2) = (f a0 b0, f a1 b1, f a2 b2)

Though I would like a single function that works for any length.

I have made a function zipTupleWith using the lens package:

zipTupleWith :: (Each s1 t1 a a, Each s2 s2 b b) => (a -> b -> a) -> s1 -> s2 -> t1
zipTupleWith f a b = a & partsOf each %~ flip (zipWith f) (b ^.. each)

This can zipWith two tuples so long as the function is of type a -> b -> a. This restriction is because of the partsOf function being used on the argument a.

There are 3 reasons that I'm not happy with my solution:

1. I would like to be able to use a function of type a -> b -> c, allowing things like zipTuple = zipTupleWith (,).
2. The above implementation will not catch errors caused by tuples of different lengths being passed in (zipTupleWith (+) (1,2,3) (100,100) = (101, 102, 3) - I would like this to be a compilation error).
3. It creates an intermediary list (b ^.. each).

So, is there a way of doing this using optics? I saw that the tuple-th package can do this, but I would prefer to avoid adding another dependency just for this, and Template Haskell seems excessive for what I'm doing.

Answer 1

I know you asked for a lens-based approach, but if you only have small tuples, you can implement what you want with a type class without too much trouble. Consider for instance:

class ZipTuple as a where
  type TupOf as x :: *
  zipTuple :: (a -> b -> c) -> as -> TupOf as b -> TupOf as c

instance ZipTuple (a,a) a where
  type TupOf (a,a) b = (b,b)
  zipTuple f (a1,a2) (b1,b2) = (f a1 b1, f a2 b2)

instance ZipTuple (a,a,a) a where
  type TupOf (a,a,a) b = (b,b,b)
  zipTuple f (a1,a2,a3) (b1,b2,b3) = (f a1 b1, f a2 b2, f a3 b3)

There may be a more elegant way to write this, but the pattern is straightforward. It should be easy to add instances for whatever length tuples you want.

---

If you want arbitrarily long tuples but don't want template haskell, there's also the Generics route. Here's a solution that zips based on the shape of the generic representation:

import GHC.Generics

class TupleZipG fa fb a b c | fa -> a, fb -> b where
  type Out fa fb a b c :: (* -> *)
  tupleZipG :: (a -> b -> c) -> fa x -> fb x -> Out fa fb a b c x

instance (TupleZipG l1 l2 a b c, TupleZipG r1 r2 a b c) => TupleZipG (l1 :*: r1) (l2 :*: r2) a b c where
  type Out (l1 :*: r1) (l2 :*: r2) a b c = Out l1 l2 a b c :*: Out r1 r2 a b c
  tupleZipG f (l1 :*: r1) (l2 :*: r2) = tupleZipG f l1 l2 :*: tupleZipG f r1 r2

instance TupleZipG (S1 m (Rec0 a)) (S1 m' (Rec0 b)) a b c where
  type Out (S1 m (Rec0 a)) (S1 m' (Rec0 b)) a b c = S1 m (Rec0 c)
  tupleZipG f (M1 (K1 a)) (M1 (K1 b)) = M1 $ K1 $ f a b

instance TupleZipG fa fb a b c => TupleZipG (D1 m fa) (D1 m' fb) a b c where
  type Out (D1 m fa) (D1 m' fb) a b c = D1 m (Out fa fb a b c)
  tupleZipG f (M1 a) (M1 b) = M1 $ tupleZipG f a b

instance TupleZipG fa fb a b c => TupleZipG (C1 m fa) (C1 m' fb) a b c where
  type Out (C1 m fa) (C1 m' fb) a b c = C1 m (Out fa fb a b c)
  tupleZipG f (M1 a) (M1 b) = M1 $ tupleZipG f a b

tupleZip
  :: (TupleZipG (Rep as) (Rep bs) a b c, Generic cs, Generic as,
      Generic bs, Out (Rep as) (Rep bs) a b c ~ Rep cs) =>
     (a -> b -> c) -> as -> bs -> cs
tupleZip f t1 t2 = to $ tupleZipG f (from t1) (from t2)

Warning: Type inference isn't great with this Generic approach.

Answer 2

It looks like you could do something like this:

zipTupleWith :: (Each s s a a, Each t v b c, Each t s b a)
  => (a -> b -> c) -> s -> t -> v
zipTupleWith f s t = t & unsafePartsOf each %~ zipWith f (s ^.. each)

giving:

> zipTupleWith replicate (1,2,3) ('a','b','c')
("a","bb","ccc")
> zipTupleWith (+) (1,2) (3,4,5)
-- type error

The trick here is the "extra" contraint Each t s b a. The other two contraints are implied by the two uses of each -- basically, Each s s a a is implied by the expression s ^.. each that extracts all the as from s, while Each t v b c is implied by t & unsafePartsOf each %~ ... for some ... :: [b] -> [c]. But adding the otherwise unnecessary constraint Each t s b a enforces equal tuple length at the type level by asserting that IF every b in t was replaced with an a, the result would be an s.

Note that lens isn't doing anything magical here. There's an Each instance for a bunch of different sized tuples, and there's just enough information in the type class to trick it into defining zipTupleWith in a really ugly, roundabout manner.

It would be much more straightforward to just define the type class you need directly, as per @DDub's answer.