[Haskell-cafe] Newbie: Replacing substring?
Dmitri O.Kondratiev
What standard library function can be used to replace substring in a string
(or sub-list in a list) ?
I wrote my own version, please criticize:
-- 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 = foldr ((\newSub before after -> before ++ newSub ++
after) new) [] chunks
where chunks = splitStr str old
{--
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 ++ match ++ [x]) []
--
Dmitri O. Kondratiev
dok...@gmail.com
http://www.geocities.com/dkondr
Roberto Zunino
> 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.
_______________________________________________
Haskell-Cafe mailing list
Haskel...@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe
Dmitri O.Kondratiev
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 <zun...@di.unipi.it> 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.
>
--
Ronald Guida
> 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
Chaddaï Fouché
> 2008/7/22 Dmitri O.Kondratiev <dok...@gmail.com>:
>> 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.
Your idea but expressed in a more elegant fashion (maybe...) :
replace :: (Eq a) => [a] -> [a] -> [a] -> [a]
replace _ _ [] = []
replace old new xs@(y:ys) =
case stripPrefix old xs of
Nothing -> y : replace old new ys
Just ys' -> new ++ replace old new ys'
--
Jedaď
Dmitri O.Kondratiev
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.com<haskell-cafe%40haskell.org?Subject=%5BHaskell-cafe%5D%20Newbie%3A%20Replacing%20substring%3F&In-Reply-To=53396d9e0807220921o74630a51ub435420f37f21bf9%40mail.gmail.com>wrote:
*
*
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 <dok...@gmail.com>
wrote: