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POPython faster than compiled Haskell?
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copied!<p>I have a simple script written in both Python and Haskell. It reads a file with 1,000,000 newline separated integers, parses that file into a list of integers, quick sorts it and then writes it to a different file sorted. This file has the same format as the unsorted one. Simple.</p> <p>Here is Haskell:</p> <pre><code>quicksort :: Ord a => [a] -> [a] quicksort [] = [] quicksort (p:xs) = (quicksort lesser) ++ [p] ++ (quicksort greater) where lesser = filter (< p) xs greater = filter (>= p) xs main = do file <- readFile "data" let un = lines file let f = map (\x -> read x::Int ) un let done = quicksort f writeFile "sorted" (unlines (map show done)) </code></pre> <p>And here is Python:</p> <pre class="lang-python prettyprint-override"><code>def qs(ar): if len(ar) == 0: return ar p = ar[0] return qs([i for i in ar if i < p]) + [p] + qs([i for i in ar if i > p]) def read_file(fn): f = open(fn) data = f.read() f.close() return data def write_file(fn, data): f = open('sorted', 'w') f.write(data) f.close() def main(): data = read_file('data') lines = data.split('\n') lines = [int(l) for l in lines] done = qs(lines) done = [str(l) for l in done] write_file('sorted', "\n".join(done)) if __name__ == '__main__': main() </code></pre> <p>Very simple. Now I compile the Haskell code with</p> <pre><code>$ ghc -O2 --make quick.hs </code></pre> <p>And I time those two with:</p> <pre><code>$ time ./quick $ time python qs.py </code></pre> <p>Results:</p> <p>Haskell:</p> <pre><code>real 0m10.820s user 0m10.656s sys 0m0.154s </code></pre> <p>Python:</p> <pre><code>real 0m9.888s user 0m9.669s sys 0m0.203s </code></pre> <p>How can Python possibly be faster than native code Haskell?</p> <p>Thanks</p> <p><strong>EDIT</strong>:</p> <ul> <li>Python version: 2.7.1</li> <li>GHC version: 7.0.4</li> <li>Mac OSX, 10.7.3</li> <li>2.4GHz Intel Core i5</li> </ul> <p>List generated by</p> <pre class="lang-python prettyprint-override"><code>from random import shuffle a = [str(a) for a in xrange(0, 1000*1000)] shuffle(a) s = "\n".join(a) f = open('data', 'w') f.write(s) f.close() </code></pre> <p>So all numbers are unique.</p>

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