Break string into substring in haskell

Question

I am having a problem with my function, I want to do a pattern matching on a string with my function but I have a problem with breaking the string into substrings. I want for a string like "ccaabbccaacc" and a regular expresion like "a*b*c*" to obtain ["cc", "aabbcc", "aacc", ""], a list with the breaked subtring. I have made this function which returns all the parts of the substring

parts :: [a] -> [[[a]]]
parts [ ] = [[ ]]
parts [c] = [[[c]]]
parts (c : cs) = concat [[(c : p) : ps ,[c] : p : ps] | p : ps <- parts cs]

but when I apply my matchs function overt all the results it returns more that i want and i don't know how to filter the results. Can somene help me?

While it is generally a good idea to break such a problem into easy steps like you do here, I think this is an example where it's more effective to write one recursive function that does the job entirely. Have you tried that, yet? — leftaroundabout
– leftaroundabout, Commented Apr 28, 2013 at 13:11

Will Ness · Accepted Answer · 2013-04-28 20:56:34Z

1

I take it that this question is about parsing - that you want to break a string up into maximal chunks matching a given regexp, e.g. "a*b*c*".

That's like iterated lexer application, only with regexp. Assuming the existence of function

reglex :: String -> String -> (String, String)

that takes a regexp string, an input string, and returns a pair of longest matching prefix, and the rest of the input string, we can write

import Control.Arrow

parts reg str = ($ ("",str)) $
   iterate (reglex reg . snd) >>>
   tail >>>
   span (not.null.fst)

and then do something with the result of this.

answered Apr 28, 2013 at 20:56

Will Ness

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2 Comments

Anderson Green Over a year ago

Does Haskell have any built-in function for extracting a substring between two specified indices (from one specific string)?

Will Ness Over a year ago

we can do this with repeated application of splitAt: between i j str = let (a,b)=splitAt i str; (c,d)=splitAt (j-i) b in c, zero-based, non-inclusive of j. "simplified" as btw i j s = fst . splitAt (j-i) . snd . splitAt i $ s.

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