How do you trim whitespace from the start and end of a string?
trim " abc "
=>
"abc"
Edit:
Ok, let me be a little clearer. I did not understand that string literals were treated so differently from Strings.
I would like to do this:
import qualified Data.Text as T
let s :: String = " abc "
in T.strip s
Is this possible in Haskell? I am using -XOverloadedStrings but that appears only to work for literals.
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...