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Replacing explicit recursion with higher-order functions

I have the following code that is designed to receive a list aand a list b'and returns all pairs [(a, b)] such that

  • Each aand each is bdisplayed only once in each pairing.
  • Each pair (a, b)satisfies a certain condition cond, i.e. cond :: a -> b -> Bool.

For example, the result for lists [1, 2] [x, y, z] should be

[[(1, x), (2, y)]
 [(1, x), (2, z)]
 [(1, y), (2, x)]
 [(1, y), (2, z)]
 [(1, z), (2, x)]
 [(1, z), (2, y)]]

Here is some (somewhat abstracted) code that does the job with explicit recursion, but I would like to replace it with a word or something similar. Any tips?

someFn :: [a] -> [b] -> [ [(a, b)] ]
someFn [] _ = []
someFn (a : as) bs = [ [(a,b)] ++ rest | b <- bs, rest <- someFn as (bs \\ [b]), cond a b]
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2 answers

, , . , , ,

foo :: [a] -> [b] -> (a -> b -> Bool)-> [(a,b)]
foo x y with = filter (uncurry with) [(a,b) | a <- x, b <- y]

[ ]

, ()

bar :: [a] -> [b] -> [[(a,b)]]
bar xs ys = map (zip xs) $ permutations ys


biz :: (a -> b -> Bool) -> [[(a,b)]] -> [[(a,b)]]
biz = map . filter . uncurry
+5

foldr :

delete :: Int -> [a] -> [a]
delete index xs = let (ys, _:zs) = splitAt index xs in ys ++ zs

ifoldr :: (Int -> a -> b -> b) -> b -> [a] -> b
ifoldr f acc xs = foldr (\(a, b) c -> f a b c) acc $ zip [0..] xs

someFn :: (a -> b -> Bool) -> [a] -> [b] -> [[(a,b)]]
someFn _ [] _ = [[]]
someFn cond (a:as) bs = ifoldr (\index b acc -> if cond a b
    then concat [map ((a,b):) . someFn cond as $ delete index bs, acc]
    else acc) [] bs

, someFn _ [] _ = [[]], someFn :: (a -> b -> Bool) -> [a] -> [b] -> [[(a,b)]].

someFn :

someFn (\a b -> True) [1,2] "xyz"

-- [[(1,'x'),(2,'y')],
--  [(1,'x'),(2,'z')],
--  [(1,'y'),(2,'x')],
--  [(1,'y'),(2,'z')],
--  [(1,'z'),(2,'x')],
--  [(1,'z'),(2,'y')]]

someFn (\a b -> case (a,b) of (1,'x') -> False
                              (2,'y') -> False
                              otherwise -> True) [1,2] "xyz"

-- [[(1,'y'),(2,'x')],
--  [(1,'y'),(2,'z')],
--  [(1,'z'),(2,'x')]]

, .

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