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

Along the lines of what other people have suggested, you can avoid having to reverse your string by using:

import Data.Char (isSpace)

dropFromTailWhile _ [] = []
dropFromTailWhile p item
  | p (last items) = dropFromTailWhile p $ init items
  | otherwise      = items

trim :: String -> String
trim = dropFromTailWhile isSpace . dropWhile isSpace

Answer 2

For sure, Data.Text is better for performance. But, as was mentioned, it's just fun to do it with lists. Here is a version that rstrip's the string in single pass (without reverse and ++) and supports infinite lists:

rstrip :: String -> String
rstrip str = let (zs, f) = go str in if f then [] else zs
    where
        go [] = ([], True)
        go (y:ys) =
            if isSpace y then
                let (zs, f) = go ys in (y:zs, f)
            else
                (y:(rstrip ys), False)

p.s. as for infinite lists, that will work:

List.length $ List.take n $ rstrip $ cycle "abc  "

and, for obvious reason, that will not (will run forever):

List.length $ List.take n $ rstrip $ 'a':(cycle " ")

Answer 3

This should be right about O(n), I believe:

import Data.Char (isSpace)

trim :: String -> String
-- Trimming the front is easy. Use a helper for the end.
trim = dropWhile isSpace . trim' []
  where
    trim' :: String -> String -> String
    -- When finding whitespace, put it in the space bin. When finding
    -- non-whitespace, include the binned whitespace and continue with an
    -- empty bin. When at the end, just throw away the bin.
    trim' _ [] = []
    trim' bin (a:as) | isSpace a = trim' (bin ++ [a]) as
                     | otherwise = bin ++ a : trim' [] as

Answer 4

Nowadays the MissingH package ships with a strip function:

import           Data.String.Utils

myString = "    foo bar    "
-- strip :: String -> String
myTrimmedString = strip myString
-- myTrimmedString == "foo bar"

So if the conversion from String to Text and back does not make sense in your situation, you could use the function above.

Answer 5

I don't know anything about the runtime or efficiency but what about this:

-- entirely input is to be trimmed
trim :: String -> String
trim = Prelude.filter (not . isSpace')

-- just the left and the right side of the input is to be trimmed
lrtrim :: String -> String
lrtrim = \xs -> rtrim $ ltrim xs
  where
    ltrim = dropWhile (isSpace')
    rtrim xs
      | Prelude.null xs = []
      | otherwise = if isSpace' $ last xs
                    then rtrim $ init xs
                    else xs 

-- returns True if input equals ' '
isSpace' :: Char -> Bool
isSpace' = \c -> (c == ' ')

A solution without using any other module or library than the Prelude.

Some tests:

>lrtrim ""
>""

>lrtrim "       "
>""

>lrtrim "haskell       "
>"haskell"

>lrtrim "      haskell       "
>"haskell"

>lrtrim "     h  a  s k e   ll       "
>"h  a  s k e   ll"

It could be runtime O(n).

But I actually don't know it because I don't know the runtimes of the functions last and init. ;)

Answer 6

Another (std) solution

import System.Environment
import Data.Text

strip :: String -> IO String
strip = return . unpack . Data.Text.strip . pack

main = getLine >>= Main.strip >>= putStrLn