2008/7/22 Dmitri O.Kondratiev :

On the side: The more I use Haskell - the more I like it ! It helps me think about the problem I solve much more clearly then when I use imperative language.

If I want to replace a substring in a string, then I would search my string left to right, looking for any occurrence of the substring. If I find such an occurrence, I would replace it and continue searching from immediately after the replacement. This algorithm can be directly expressed in Haskell. More efficient algorithms do exist. replaceStr :: String -> String -> String -> String replaceStr [] old new = [] replaceStr str old new = loop str where loop [] = [] loop str = let (prefix, rest) = splitAt n str in if old == prefix -- found an occurrence? then new ++ loop rest -- yes: replace it else head str : loop (tail str) -- no: keep looking n = length old

Ronald, Your algoritm is more simple and so it is better, I agree. My algorithm is different and consists of two steps: 1) Split source string into a list of chunks not containing substring to be replaced. 2) Concatenate chunks inserting new substring in between chuncks. With this approach I get a 'bonus' function of spliting string into chunks :) * Tue Jul 22 16:38:53 EDT 2008 **Ronald Guida* oddron at gmail.comwrote: * * If I want to replace a substring in a string, then I would search my string left to right, looking for any occurrence of the substring. If I find such an occurrence, I would replace it and continue searching from immediately after the replacement. This algorithm can be directly expressed in Haskell. More efficient algorithms do exist. replaceStr :: String -> String -> String -> String replaceStr [] old new = [] replaceStr str old new = loop str where loop [] = [] loop str = let (prefix, rest) = splitAt n str in if old == prefix -- found an occurrence? then new ++ loop rest -- yes: replace it else head str : loop (tail str) -- no: keep looking n = length old On Tue, Jul 22, 2008 at 8:21 PM, Dmitri O.Kondratiev wrote:

Roberto thanks! Shame on me, to post code without enough testing :( Yet, thanks to your comments *I think* I have found the bugs you wrote about and now my code works, please see corrected version below. Extra substring at the end was a result of using foldr with initial element of []. I fixed this with foldl and first chunk as its initial element. Incomplete substitution in case of duplicate elements in the pattern was a bug in my 'takeOut' function that I have also fixed. Stiil the problem that I have not yet designed solution for is when a substring to replace extends from the end of a string to the next string. In other words - first part of substring ends the first string and second part of substring starts the second string. My algorithm currently does not account for such a case.

On the side: The more I use Haskell - the more I like it ! It helps me think about the problem I solve much more clearly then when I use imperative language.

Corrected code:

-- replace all occurances of "123" with "58" in a string: test = replStr "abc123def123gh123ikl" "123" "58"

{-- In a string replace all occurances of an 'old' substring with a 'new' substring --} replStr str old new = foldl ((\newSub before after -> before ++ newSub ++ after) new) firstChunk otherChunks where chunks = splitStr str old firstChunk = head chunks otherChunks = tail chunks {-- Split string into a list of chunks. Chunks are substrings located in a string between 'sub' substrings --} splitStr str sub = mkChunkLst str sub [] where -- mkChunkLst 'src string' 'substr-to-extract' 'list of chunks' -- makes list of chunks located between 'substr-to-extract' pieces in src string mkChunkLst [] _ chunkLst = chunkLst mkChunkLst str sub chunkLst = mkChunkLst after sub (chunkLst ++ [chunk]) where (chunk, _, after) = takeOut str sub [] []

{-- Take out substring from a string. String is divided into: "before substr" ++ "match" ++ "after substr" where 'match' is substring to split out --}

takeOut after [] before match = (before, match, after) takeOut [] _ before match = (before, match, []) takeOut (x:xs) (y:ys) before match | x == y = takeOut xs ys before (match ++ [x]) | otherwise = takeOut xs (y:ys) (before ++ [x]) []

On Tue, Jul 22, 2008 at 7:39 PM, Roberto Zunino wrote:

...

Dmitri O.Kondratiev wrote:

...

I wrote my own version, please criticize:

-- replace all occurances of "123" with "58" in a string: test = replStr "abc123def123gh123ikl" "123" "58"

This is a tricky problem: first of all, you fail your own test! ;-)

*Main> test "abc58def58gh58ikl58"

(Note the extra 58 at the end.)

Other common pitfalls:

*Main> replStr "abc1123def" "123" "58" "abc1158def58"

(extra 1 ?)

*Main> replStr "abc12123def" "123" "58" "abc121258def58"

(extra 12 ?)

A useful function from Data.List: stripPrefix

(Of course, there are more efficient string match algorithms)

Regards, Zun.

-- Dmitri O. Kondratiev dokondr@gmail.com http://www.geocities.com/dkondr

-- Dmitri O. Kondratiev dokondr@gmail.com http://www.geocities.com/dkondr