In Haskell, how do you trim whitespace from the beginning and end of a string?

Question

How do you trim whitespace from the start and end of a string?

trim \"  abc \" 

=>

\"abc\"

Edit:

Ok, let me be a little cleare

Answer 1

After this question was asked (circa 2012) Data.List got dropWhileEnd making this a lot easier:

trim = dropWhileEnd isSpace . dropWhile isSpace

Answer 2

From: http://en.wikipedia.org/wiki/Trim_(programming)#Haskell

import Data.Char (isSpace)

trim :: String -> String
trim = f . f
   where f = reverse . dropWhile isSpace

Answer 3

Inefficient but easy to understand and paste in where needed:

strip = lstrip . rstrip
lstrip = dropWhile (`elem` " \t")
rstrip = reverse . lstrip . reverse

Answer 4

I know this is an old post, but I saw no solutions that implemented good old fold.

First strip the leading white-space using dropWhile. Then, using foldl' and a simple closure, you can analyze the rest of the string in one pass, and based on that analysis, pass that informative parameter to take, without needing reverse:

import Data.Char (isSpace)
import Data.List (foldl')

trim :: String -> String
trim s = let
  s'    = dropWhile isSpace s
  trim' = foldl'
            (\(c,w) x -> if isSpace x then (c,w+1)
                         else (c+w+1,0)) (0,0) s'
  in
   take (fst trim') s'

Variable c keeps track of combined white and non white-space that should be absorbed, and variable w keeps track of right side white-space to be stripped.

Test Runs:

print $ trim "      a   b c    "
print $ trim "      ab c    "
print $ trim "    abc    "
print $ trim "abc"
print $ trim "a bc    "

Output:

"a   b c"
"ab c"
"abc"
"abc"
"a bc"

Answer 5

If you have serious text processing needs then use the text package from hackage:

> :set -XOverloadedStrings
> import Data.Text
> strip "  abc   "
"abc"

If you're too stubborn to use text and don't like the inefficiency of the reverse method then perhaps (and I mean MAYBE) something like the below will be more efficient:

import Data.Char

trim xs = dropSpaceTail "" $ dropWhile isSpace xs

dropSpaceTail maybeStuff "" = ""
dropSpaceTail maybeStuff (x:xs)
        | isSpace x = dropSpaceTail (x:maybeStuff) xs
        | null maybeStuff = x : dropSpaceTail "" xs
        | otherwise       = reverse maybeStuff ++ x : dropSpaceTail "" xs


> trim "  hello this \t should trim ok.. .I  think  ..  \t "
"hello this \t should trim ok.. .I  think  .."

I wrote this on the assumption that the length of spaces would be minimal, so your O(n) of ++ and reverse is of little concern. But once again I feel the need to say that if you actually are concerned about the performance then you shouldn't be using String at all - move to Text.

EDIT making my point, a quick Criterion benchmark tells me that (for a particularly long string of words with spaces and ~200 pre and post spaces) my trim takes 1.6 ms, the trim using reverse takes 3.5ms, and Data.Text.strip takes 0.0016 ms...

Answer 6

You can combine Data.Text's strip with it's un/packing functions to avoid having overloaded strings:

import qualified Data.Text as T

strip  = T.unpack . T.strip . T.pack
lstrip = T.unpack . T.stripStart . T.pack
rstrip = T.unpack . T.stripEnd . T.pack

Testing it:

> let s = "  hello  "
> strip s
"hello"
> lstrip s
"hello  "
> rstrip s
"  hello"