Question

I’m really new to Haskell and I’m stuck on trying to map the first item of each pair in a list.

Obviously this works:

map :: (a -> b) -> [a] -> [b]
map f xs = [f x | x <- xs]

But how do I get it to work for

map :: (a -> b) -> [(a, Int)] -> [b]

I just want it to ignore the Int values for now and apply f to a like it does in the first example. I’ve been trying for ages now so thanks for any help.

Answer 1

Well, assuming you don’t want to use the build-in function map, starting from this:

map :: (a -> b) -> [a] -> [b]
map f xs = [f x | x <- xs]

To accept a list of type [(a, Int)] and use just the a, you can pattern match the tuple:

map :: (a -> b) -> [(a, Int)] -> [b]
map f xs = [f x | (x, y) <- xs]

If you want to keep the Int, you can put it back together afterwards:

map :: (a -> b) -> [(a, Int)] -> [(b, Int)]
map f xs = [(f x, y) | (x, y) <- xs]

But all of this is a bit redundant. You can do the same by changing the argument to the original, generic map:

map :: (a -> b) -> [a] -> [b]
map f xs = [f x | x <- xs]

mapFst :: (a -> b) -> [(a, Int)] -> [b]
mapFst f xs = map (f . fst) xs

mapOnFirst :: (a -> b) -> [(a, Int)] -> [(b, Int)]
mapOnFirst f xs = map (\(x,y) -> (f x, y)) xs

For the third version, the standard library’s module Control.Arrow gives you a function called first that can be used to get the same effect:

mapOnFirst :: (a -> b) -> [(a, Int)] -> [(b, Int)]
mapOnFirst f xs = map (first f) xs

Neat, huh?