Thanks for the answer. Run it with a big number: f2(9230821908403878236537264867326482638462035732098490238409328947638257462745672354627356427356742563256745362772537532756732673275732,9)
I had this result: 8.1015883211e-06 1.06158743591e-05 512 2009/8/26 Mark Tolonen <metolone+gm...@gmail.com<metolone%2bgm...@gmail.com> > > > "Cevahir Demirkiran" <cevo...@gmail.com> wrote in message > news:3f74e020908251648k7b391a09g78b155507b2f2...@mail.gmail.com... > > Hi, >> >> I would like to do a floor division by a power of 2 in python to make it >> faster than / (modular division in 2.x). >> However, it is slower. >> What is the reason of that? >> I checked them via: >> >> def f2(x,n): >> t1 = clock() >> r = x/pow(2,n) >> t2 = clock() >> print (t2-t1) >> print r >> t2 = clock() >> r = x>>n >> t3 = clock() >> print (t3-t2) >> print r >> >> > It's not slower on my system, but note the inconsistent results also: > > f2(1024,5) >>>> >>> 3.47396033483e-06 > 32 > 2.19077375482e-06 > 32 > >> f2(1024,5) >>>> >>> 4.84135603429e-06 > 32 > 3.08499440393e-06 > 32 > >> f2(1024,5) >>>> >>> 4.6782844052e-06 > 32 > 3.77604384028e-06 > 32 > > Time it with timeit... > > C:\>python -m timeit -n 10000000 -s x=1024 "x>>5" > 10000000 loops, best of 3: 0.113 usec per loop > > C:\>python -m timeit -n 10000000 -s x=1024 "x/pow(2,5)" > 10000000 loops, best of 3: 0.468 usec per loop > > Right-shift is over 4x faster. > > -Mark > > > -- > http://mail.python.org/mailman/listinfo/python-list > -- Cevahir Demirkiran
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