I have two functions:
fun1 :: Int -> [Int]
fun2 :: [Int] -> [Int]
fun2 accept Int list and apply fun1 to each element of this list with help map. But fun1 return [Int]. So, I have type conflict. How to solve my problem?
3 Answers
You likely want a combination of map and concat to achieve it. Assuming fun1 and fun2 are something like this:
fun1 :: Int -> [Int]
fun1 x = [x,x]
fun2 :: [Int] -> [Int]
fun2 = map (+ 1)
solution :: [Int] -> [Int]
solution xs = concat $ map fun1 (fun2 xs)
Or as suggested by @CarstenKonig, you can use concatMap
solution2 :: [Int] -> [Int]
solution2 xs = concatMap fun1 $ fun2 xs
which can be further simplified to:
solution2 :: [Int] -> [Int]
solution2 = concatMap fun1 . fun2
3 Comments
Random Dev
based on "fun2 accept Int list and apply fun1 to each element of this" I would think that the OP really just wants
fun2 = concatMap fun1 though - but yeah that's another one that type-checks ;)
Sibi
@CarstenKönig I felt this was the natural answer to OP :) But yes, I may be wrong.
Random Dev
that's the problem (not with your answer) - the question is not really good but the OP don't seem to be able to clarify right now
The ability to transform [[a]] to [a] is what make List (among other thing) a Monad.
Therefore you can use the do notation :
fun2 xs = do
x <- xs
fun1 x
which can the be rewritten fun2 xs = xs >>= fun1 or even better
fun2 = (>>= fun1)
Comments
Another way is with list comprehension. Mapping fun1 over the list xs is
fun2' xs = [ fun1 x | x <- xs ]
-- = map fun1 xs
-- = do x <- xs -- for each x in xs:
-- return (fun1 x) -- yield (fun1 x)
which indeed would have a different type than what you wanted.
To flatten it one step further we do
fun2 xs = [ y | x <- xs, y <- fun1 x ]
-- = concatMap fun1 xs
-- = do x <- xs -- for each x in xs:
-- y <- fun1 x -- for each y in (fun1 x):
-- return y -- yield y
fun2 :: [Int] -> [[Int]]concatMapinstead ofmap(as infun2 = concatMap fun1) -concatMap fun1will applyfun1to each element in the input list but will flatten all the resulting lists into one by concatenating them