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Date: Sun, 19 Aug 2012 14:05:48 -0400
From: Dave Angel <d@davea.name>
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To: Blind Anagram <noname@nowhere.com>
Subject: Re: How do I display unicode value stored in a string variable using ord()
References: <f801e06f-f7b2-4aca-b352-66856a939746@googlegroups.com><308df2af-abe7-4043-b199-0a39f440e0ab@googlegroups.com> <502f8a2a$0$29978$c3e8da3$5496439d@news.astraweb.com> <XI2dncR-HMTzgazNnZ2dnUVZ8jCdnZ2d@brightview.co.uk>
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On 08/19/2012 01:03 PM, Blind Anagram wrote:
> "Steven D'Aprano"  wrote in message
> news:502f8a2a$0$29978$c3e8da3$5496439d@news.astraweb.com...
>
> On Sat, 18 Aug 2012 01:09:26 -0700, wxjmfauth wrote:
>
> [...]
> If you can consistently replicate a 100% to 1000% slowdown in string
> handling, please report it as a performance bug:
>
> http://bugs.python.org/
>
> Don't forget to report your operating system.
>
> ====================================================
> For interest, I ran your code snippets on my laptop (Intel core-i7
> 1.8GHz) running Windows 7 x64.
>
> Running Python from a Windows command prompt,  I got the following on
> Python 3.2.3 and 3.3 beta 2:
>
> python33\python" -m timeit "('abc' * 1000).replace('c', 'de')"
> 10000 loops, best of 3: 39.3 usec per loop
> python33\python" -m timeit "('ab…' * 1000).replace('…', '……')"
> 10000 loops, best of 3: 51.8 usec per loop
> python33\python" -m timeit "('ab…' * 1000).replace('…', 'x…')"
> 10000 loops, best of 3: 52 usec per loop
> python33\python" -m timeit "('ab…' * 1000).replace('…', 'œ…')"
> 10000 loops, best of 3: 50.3 usec per loop
> python33\python" -m timeit "('ab…' * 1000).replace('…', '€…')"
> 10000 loops, best of 3: 51.6 usec per loop
> python33\python" -m timeit "('XYZ' * 1000).replace('X', 'éç')"
> 10000 loops, best of 3: 38.3 usec per loop
> python33\python" -m timeit "('XYZ' * 1000).replace('Y', 'p?')"
> 10000 loops, best of 3: 50.3 usec per loop
>
> python32\python" -m timeit "('abc' * 1000).replace('c', 'de')"
> 10000 loops, best of 3: 24.5 usec per loop
> python32\python" -m timeit "('ab…' * 1000).replace('…', '……')"
> 10000 loops, best of 3: 24.7 usec per loop
> python32\python" -m timeit "('ab…' * 1000).replace('…', 'x…')"
> 10000 loops, best of 3: 24.8 usec per loop
> python32\python" -m timeit "('ab…' * 1000).replace('…', 'œ…')"
> 10000 loops, best of 3: 24 usec per loop
> python32\python" -m timeit "('ab…' * 1000).replace('…', '€…')"
> 10000 loops, best of 3: 24.1 usec per loop
> python32\python" -m timeit "('XYZ' * 1000).replace('X', 'éç')"
> 10000 loops, best of 3: 24.4 usec per loop
> python32\python" -m timeit "('XYZ' * 1000).replace('Y', 'p?')"
> 10000 loops, best of 3: 24.3 usec per loop
>
> This is an average slowdown by a factor of close to 2.3 on 3.3 when
> compared with 3.2.
>

Using your measurement numbers, I get an average of 1.95, not 2.3



-- 

DaveA