Sustitución de listas

3 Comentarios

1. Haskell

   sustituye :: Eq a => [a] -> [a] -> [a] -> [a] sustituye _ _ [] = [] sustituye xs ys (z:zs) | take u (z:zs) == xs = ys ++ (sustituye xs ys (drop u (z:zs))) | otherwise = z : sustituye xs ys zs where u = length xs

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   sustituye :: Eq a => [a] -> [a] -> [a] -> [a] sustituye _ _ [] = [] sustituye xs ys (z:zs)     | take u (z:zs) == xs = ys ++ (sustituye xs ys (drop u (z:zs)))       | otherwise           = z : sustituye xs ys zs     where u = length xs

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2. Recursion simple:

   Haskell

   sustituye xs yz = rec [] xs where rec _ [] [] = yz rec acc _ [] = acc rec _ [] zs = yz ++ rec [] xs zs rec acc (w:ws) (z:zs) | w == z = rec (acc ++ [z]) ws zs | otherwise = acc ++ z: rec [] xs zs

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   sustituye xs yz = rec [] xs    where rec _   [] [] = yz          rec acc _  [] = acc          rec _   [] zs = yz ++ rec [] xs zs          rec acc (w:ws) (z:zs) | w == z    = rec (acc ++ [z]) ws zs                                | otherwise = acc ++ z: rec [] xs zs

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3. Un fold estricto bastante feo:

   Haskell

   sustituye1 :: Eq a => [a] -> [a] -> [a] -> [a] sustituye1 xs ys = snd . foldl' f ((xs,[]),[]) where f ((i:is,acc),zs) x | i /= x = reset $ zs++[x] | i == x && null is = reset $ acc++ys | otherwise = ((is,acc),zs++[x]) reset zs = ((xs,zs),zs)

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   sustituye1 :: Eq a => [a] -> [a] -> [a] -> [a] sustituye1 xs ys = snd . foldl' f ((xs,[]),[])    where f ((i:is,acc),zs) x | i /= x            = reset $ zs++[x]                              | i == x && null is = reset $ acc++ys                              | otherwise         = ((is,acc),zs++[x])          reset zs = ((xs,zs),zs)

   Con Data.List.Split todo es mucho mas facil:

   Haskell

   sustituye2 :: Eq a => [a] -> [a] -> [a] -> [a] sustituye2 xs ys = intercalate ys . splitOn xs

   1 2	sustituye2 :: Eq a => [a] -> [a] -> [a] -> [a] sustituye2 xs ys = intercalate ys . splitOn xs

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