[Python-Dev] PEP 393 decode() oddity (original) (raw)

Paul Moore p.f.moore at gmail.com
Sun Mar 25 21:12:11 CEST 2012

On 25 March 2012 19:51, Serhiy Storchaka <storchaka at gmail.com> wrote:

Anyone can test.

$ ./python -m timeit -s 'enc = "latin1"; import codecs; d = codecs.getdecoder(enc); x = ("\u0020" * 100000).encode(enc)' 'd(x)' 10000 loops, best of 3: 59.4 usec per loop $ ./python -m timeit -s 'enc = "latin1"; import codecs; d = codecs.getdecoder(enc); x = ("\u0080" * 100000).encode(enc)' 'd(x)' 10000 loops, best of 3: 28.4 usec per loop The results are fairly stable (±0.1 µsec) from run to run. It looks funny thing.

Hmm, yes. I see the same results. Odd.

PS D:\Data> py -3.3 -m timeit -s "enc = 'latin1'; import codecs; d = codecs.getdecoder(enc); x = ('\u0020' * 100000).encode(enc)" "d(x)" 10000 loops, best of 3: 37.3 usec per loop PS D:\Data> py -3.3 -m timeit -s "enc = 'latin1'; import codecs; d = codecs.getdecoder(enc); x = ('\u0080' * 100000).encode(enc)" "d(x)" 100000 loops, best of 3: 18 usec per loop PS D:\Data> py -3.3 -m timeit -s "enc = 'latin1'; import codecs; d = codecs.getdecoder(enc); x = ('\u0020' * 100000).encode(enc)" "d(x)" 10000 loops, best of 3: 37.6 usec per loop PS D:\Data> py -3.3 -m timeit -s "enc = 'latin1'; import codecs; d = codecs.getdecoder(enc); x = ('\u0080' * 100000).encode(enc)" "d(x)" 100000 loops, best of 3: 18.3 usec per loop PS D:\Data> py -3.3 -m timeit -s "enc = 'latin1'; import codecs; d = codecs.getdecoder(enc); x = ('\u0020' * 100000).encode(enc)" "d(x)" 10000 loops, best of 3: 37.8 usec per loop PS D:\Data> py -3.3 -m timeit -s "enc = 'latin1'; import codecs; d = codecs.getdecoder(enc); x = ('\u0080' * 100000).encode(enc)" "d(x)" 100000 loops, best of 3: 18.3 usec per loop

Paul.

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