 Levenshtein Distance in Haskell

Seeing how lately the levenschtein distance is so popular, I thought I’d add a haskell version to the mix.

The code below is pretty self-explanatory, using the typical memoization pattern. I do wonder, however, what would happen if somehow GHC supported lazily-resizable arrays. Perhaps then you could do this on infinite lists and only ask the levenschtein distance up to a certain index.

module Lev whereimport Data.Array

levenshtein :: (Eq a) => [a] -> [a] -> Integer
levenshtein xs ys = arr ! (lenx, leny)
where arr :: (Array (Int, Int) Integer)
arr                   = array ((0,0), (lenx, leny))  [((ix, iy), worker ix iy) |
ix <- [0..lenx], iy <- [0..leny] ]
lenx                  = length xs
leny                  = length ys
arrx                  = listArray (1, lenx) xs
arry                  = listArray (1, leny) ys
worker 0      iy      = fromIntegral iy
worker ix     0       = fromIntegral ix
worker ix     iy      = minimum [1 + arr ! (ix-1, iy), 1 + arr ! (ix, iy-1),
cost ix iy + arr ! (ix-1, iy-1)]
cost ix iy            = if (arrx ! ix) == (arry ! iy) then 0 else 1

1. jorpic says:

this is a little bit shorter and faster:

levenshtein sa sb = last \$ foldl (transform sa) [0..length sa] sb
where
transform str xs@(x:xs’) c = res
where
res = x+1 : zipWith4 compute str xs xs’ res
compute c’ x y z = minimum [y+1, z+1, x + if c' == c then 0 else 1]

2. Christophe Poucet says:

Interesting code Rather cryptic, I must admit.

3. lf says:

if c’ == c then 0 else 1

can also be written

fromEnum (c /= c’)

if you don’t like inline ifs 4. Anonymous says:

Small improvements on jorpic’s version:

levenshtein sa sb = last \$ foldl transform [0..length sa] sb where
transform xs@(x:xs’) c = scanl compute (x+1) (zip3 sa xs xs’) where
compute z (c’, x, y) = minimum [y+1, z+1, x + fromEnum (c' /= c)]

5. gasche says:

What is the complexity of your implementation, given (li ! n) is in O(n) ? I would expect O(n^3) instead of the usual O(n^2).

6. gasche says:

Sorry, I just realized you were using Data.Array, not list, so you have constant-time random access.

7. Grazer says:

It is also possible to have a much faster algorithm for similar sequences. See Lloyd Allison’s haskell implementation of a O(N*d) solution.