interesting. seems that "if ' ' in source:" is a highly optimized code
as it is even faster than "if str.find(' ') != -1:' when I assume they
end up in the same C loops ?
Ron Adam wrote:
> Guess again... Is this the results below what you were expecting?
>
> Notice the join adds a space to the end if the source string is a single
> word. But I allowed for that by adding one in the same case for the
> index method.
>
> The big win I was talking about was when no spaces are in the string.
> The index can then just return the replacement.
>
> These are relative percentages of time to each other. Smaller is better.
>
> Type 1 = no spaces
> Type 2 = space at 10% of length
> Type 3 = space at 90% of length
>
> Type: Length
>
> Type 1: 10 split/join: 317.38% index: 31.51%
> Type 2: 10 split/join: 212.02% index: 47.17%
> Type 3: 10 split/join: 186.33% index: 53.67%
> Type 1: 100 split/join: 581.75% index: 17.19%
> Type 2: 100 split/join: 306.25% index: 32.65%
> Type 3: 100 split/join: 238.81% index: 41.87%
> Type 1: 1000 split/join: 1909.40% index: 5.24%
> Type 2: 1000 split/join: 892.02% index: 11.21%
> Type 3: 1000 split/join: 515.44% index: 19.40%
> Type 1: 10000 split/join: 3390.22% index: 2.95%
> Type 2: 10000 split/join: 2263.21% index: 4.42%
> Type 3: 10000 split/join: 650.30% index: 15.38%
> Type 1: 100000 split/join: 3342.08% index: 2.99%
> Type 2: 100000 split/join: 1175.51% index: 8.51%
> Type 3: 100000 split/join: 677.77% index: 14.75%
> Type 1: 1000000 split/join: 3159.27% index: 3.17%
> Type 2: 1000000 split/join: 867.39% index: 11.53%
> Type 3: 1000000 split/join: 679.47% index: 14.72%
>
>
>
>
> import time
> def test(func, source):
> t = time.clock()
> n = 6000000/len(source)
> s = ''
> for i in xrange(n):
> s = func(source, "replace")
> tt = time.clock()-t
> return s, tt
>
> def replace_word1(source, newword):
> """Replace the first word of source with newword."""
> return newword + " " + " ".join(source.split(None, 1)[1:])
>
> def replace_word2(source, newword):
> """Replace the first word of source with newword."""
> if ' ' in source:
> return newword + source[source.index(' '):]
> return newword + ' ' # space needed to match join results
>
>
> def makestrings(n):
> s1 = 'abcdefghij' * (n//10)
> i, j = n//10, n-n//10
> s2 = s1[:i] + ' ' + s1[i:] + 'd.' # space near front
> s3 = s1[:j] + ' ' + s1[j:] + 'd.' # space near end
> return [s1,s2,s3]
>
> for n in [10,100,1000,10000,100000,1000000]:
> for sn,s in enumerate(makestrings(n)):
> r1, t1 = test(replace_word1, s)
> r2, t2 = test(replace_word2, s)
> assert r1 == r2
> print "Type %i: %-8i split/join: %.2f%% index: %.2f%%" \
> % (sn+1, n, t1/t2*100.0, t2/t1*100.0)
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