Replace Strings in Haskell

Question

I am looking for a way to replace occurrences of strings in Haskell. I have this working for single pairs. My current function is implemented like so:

replaceList :: (Eq a) => [a] -> ([a],[a]) -> [a]
replaceList z@(x:xs) (a,b)
| take (length a) z == a = b ++ strt z (length a)
| otherwise = x : replaceList xs (a,b)

In this context, strt is just a function that returns a list starting at a certain index. This function works. replaceList "Dragon" ("ragon","odo") will return "I am a dodo". However, I am looking for a way to make this function accept a list of these tupples. For instance:

replaceList "I am a dragon" [("I","You"),("am","are"),("a dragon","awesome")] returning "You are awesome".

The methods I have tried so far have been to map a partially applied replaceList over a list of tupples, but that returns a list of each individual element changed. I also tried using this code:

replaceInfinity :: (Eq a) => [a] -> [([a],[a])] -> [a]
replaceInfinity x [] = x
replaceInfinity x ((a,b):ys) = (flip replaceList (a,b)) . (replaceInfinity x ys)

but that fails to compile. I'm relatively new to Haskell and am using this to rewrite an old language preprocessor. I can't understand the logic of how to implement this type of function.

Could someone tell me the tactic to get to the answer, or even implement the function for me so I could learn how it would be done? If it helps, I have the entire source file I've been using to play around with this here: http://hpaste.org/85363

גלעד ברקן · Accepted Answer · 2013-04-07 18:04:54Z

1

The smallest change to your code that I could think of is:

replaceInfinity :: Eq a => [a] -> [([a], [a])] -> [a]
replaceInfinity x []     = x
replaceInfinity x (y:ys) = replaceInfinity (replaceList x y) ys

OUTPUT:
*Main> replaceInfinity "I am a dragon" [("I","You"),("am","are"),("a dragon","awesome")]
"You are awesome"

On second thought:

replaceInfinity x ((a,b):ys) = (flip replaceInfinity ys) . (replaceList x) $ (a,b)

Or more succinctly:

replaceInfinity x (y:ys) = flip replaceInfinity ys . replaceList x $ y

edited Apr 7, 2013 at 18:04

answered Apr 7, 2013 at 17:35

גלעד ברקן

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2 Comments

Nathaniel Pisarski Over a year ago

Thanks groovy. Coming from imperative languages with somewhat easier data structures and iteration, I just couldn't wrap my head around how to do it, even though I got close. I guess I kind of misunderstood composition, because replaceInfinity ( replaceList x y) ys was what I was trying to do with it.

גלעד ברקן Over a year ago

@Dymatic I got the compositional statement to work this way: replaceInfinity x [] = x; replaceInfinity x (y:ys) = flip replaceInfinity ys . replaceList x $ y

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