How to split a string in Haskell?

Question

Is there a standard way to split a string in Haskell?

lines and words work great from splitting on a space or newline, but surely there is a standard way to split on a comma?

I couldn't find it on Hoogle.

To be specific, I'm looking for something where split "," "my,comma,separated,list" returns ["my","comma","separated","list"].

Answer 1

Remember that you can look up the definition of Prelude functions!

http://www.haskell.org/onlinereport/standard-prelude.html

Looking there, the definition of words is,

words   :: String -> [String]
words s =  case dropWhile Char.isSpace s of
                      "" -> []
                      s' -> w : words s''
                            where (w, s'') = break Char.isSpace s'

So, change it for a function that takes a predicate:

wordsWhen     :: (Char -> Bool) -> String -> [String]
wordsWhen p s =  case dropWhile p s of
                      "" -> []
                      s' -> w : wordsWhen p s''
                            where (w, s'') = break p s'

Then call it with whatever predicate you want!

main = print $ wordsWhen (==',') "break,this,string,at,commas"

Answer 2

There is a package for this called split.

cabal install split

Use it like this:

ghci> import Data.List.Split
ghci> splitOn "," "my,comma,separated,list"
["my","comma","separated","list"]

It comes with a lot of other functions for splitting on matching delimiters or having several delimiters.

Answer 3

If you use Data.Text, there is splitOn:

http://hackage.haskell.org/packages/archive/text/0.11.2.0/doc/html/Data-Text.html#v:splitOn

This is built in the Haskell Platform.

So for instance:

import qualified Data.Text as T
main = print $ T.splitOn (T.pack " ") (T.pack "this is a test")

or:

{-# LANGUAGE OverloadedStrings #-}

import qualified Data.Text as T
main = print $ T.splitOn " " "this is a test"

Answer 4

Use Data.List.Split, which uses split:

[me@localhost]$ ghci
Prelude> import Data.List.Split
Prelude Data.List.Split> let l = splitOn "," "1,2,3,4"
Prelude Data.List.Split> :t l
l :: [[Char]]
Prelude Data.List.Split> l
["1","2","3","4"]
Prelude Data.List.Split> let { convert :: [String] -> [Integer]; convert = map read }
Prelude Data.List.Split> let l2 = convert l
Prelude Data.List.Split> :t l2
l2 :: [Integer]
Prelude Data.List.Split> l2
[1,2,3,4]

Answer 5

Without importing anything a straight substitution of one character for a space, the target separator for words is a space. Something like:

words [if c == ',' then ' ' else c|c <- "my,comma,separated,list"]

or

words let f ',' = ' '; f c = c in map f "my,comma,separated,list"

You can make this into a function with parameters. You can eliminate the parameter character-to-match my matching many, like in:

 [if elem c ";,.:-+@!$#?" then ' ' else c|c <-"my,comma;separated!list"]

Answer 6

In the module Text.Regex (part of the Haskell Platform), there is a function:

splitRegex :: Regex -> String -> [String]

which splits a string based on a regular expression. The API can be found at Hackage.

Answer 7

Try this one:

import Data.List (unfoldr)

separateBy :: Eq a => a -> [a] -> [[a]]
separateBy chr = unfoldr sep where
  sep [] = Nothing
  sep l  = Just . fmap (drop 1) . break (== chr) $ l

Only works for a single char, but should be easily extendable.

Answer 8

split :: Eq a => a -> [a] -> [[a]]
split d [] = []
split d s = x : split d (drop 1 y) where (x,y) = span (/= d) s

E.g.

split ';' "a;bb;ccc;;d"
> ["a","bb","ccc","","d"]

A single trailing delimiter will be dropped:

split ';' "a;bb;ccc;;d;"
> ["a","bb","ccc","","d"]

Answer 9

I find this simpler to understand:

split :: Char -> String -> [String]
split c xs = case break (==c) xs of 
  (ls, "") -> [ls]
  (ls, x:rs) -> ls : split c rs

Answer 10

I started learning Haskell yesterday, so correct me if I'm wrong but:

split :: Eq a => a -> [a] -> [[a]]
split x y = func x y [[]]
    where
        func x [] z = reverse $ map (reverse) z
        func x (y:ys) (z:zs) = if y==x then 
            func x ys ([]:(z:zs)) 
        else 
            func x ys ((y:z):zs)

gives:

*Main> split ' ' "this is a test"
["this","is","a","test"]

or maybe you wanted

*Main> splitWithStr  " and " "this and is and a and test"
["this","is","a","test"]

which would be:

splitWithStr :: Eq a => [a] -> [a] -> [[a]]
splitWithStr x y = func x y [[]]
    where
        func x [] z = reverse $ map (reverse) z
        func x (y:ys) (z:zs) = if (take (length x) (y:ys)) == x then
            func x (drop (length x) (y:ys)) ([]:(z:zs))
        else
            func x ys ((y:z):zs)

Answer 11

I don’t know how to add a comment onto Steve’s answer, but I would like to recommend the
  GHC libraries documentation,
and in there specifically the
  Sublist functions in Data.List

Which is much better as a reference, than just reading the plain Haskell report.

Generically, a fold with a rule on when to create a new sublist to feed, should solve it too.

Answer 12

Example in the ghci:

>  import qualified Text.Regex as R
>  R.splitRegex (R.mkRegex "x") "2x3x777"
>  ["2","3","777"]

Answer 13

In addition to the efficient and pre-built functions given in answers I'll add my own which are simply part of my repertory of Haskell functions I was writing to learn the language on my own time:

-- Correct but inefficient implementation
wordsBy :: String -> Char -> [String]
wordsBy s c = reverse (go s []) where
    go s' ws = case (dropWhile (\c' -> c' == c) s') of
        "" -> ws
        rem -> go ((dropWhile (\c' -> c' /= c) rem)) ((takeWhile (\c' -> c' /= c) rem) : ws)

-- Breaks up by predicate function to allow for more complex conditions (\c -> c == ',' || c == ';')
wordsByF :: String -> (Char -> Bool) -> [String]
wordsByF s f = reverse (go s []) where
    go s' ws = case ((dropWhile (\c' -> f c')) s') of
        "" -> ws
        rem -> go ((dropWhile (\c' -> (f c') == False)) rem) (((takeWhile (\c' -> (f c') == False)) rem) : ws)

The solutions are at least tail-recursive so they won't incur a stack overflow.

Answer 14

I am far late but would like to add it here for those interested, if you're looking for a simple solution without relying on any bloated packages:

split :: String -> String -> [String]
split _ "" = []
split delim str =
  split' "" str []
  where
    dl = length delim

    split' :: String -> String -> [String] -> [String]
    split' h t f
      | dl > length t = f ++ [h ++ t]
      | delim == take dl t = split' "" (drop dl t) (f ++ [h])
      | otherwise = split' (h ++ take 1 t) (drop 1 t) f

Answer 15

So many answers, but I don't like them all. I don't know Haskell actually, but I wrote much shorter and (as I think) cleaner version for 5 minutes;

splitString :: Char -> [Char] -> [[Char]]
splitString _ [] = []
splitString sep str = 
    let (left, right) = break (==sep) str 
    in left : splitString sep (drop 1 right)