How do I display unicode value stored in a string variable using ord()

[88888 Dihedral]

On Monday, August 20, 2012 1:03:34 AM UTC+8, Blind Anagram wrote:
> "Steven D'Aprano"  wrote in message 
> 
> news:502f8a2a$0$29978$c3e8da3$5496439d at 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.
> 
> 
> 
> I am not posting this to perpetuate this thread but simply to ask whether, 
> 
> as you suggest, I should report this as a possible problem with the beta?

Un, another set of functions for seeding up ASCII string othe pertions 
might be needed. But it is better that Python 3.3 supports unicode strings
to be easy to be used by people in different languages first.

Anyway I think Cython and Pyrex can be used to tackle this problem.