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Groups > comp.lang.python > #32030

Re: get each pair from a string.

From Mark Lawrence <breamoreboy@yahoo.co.uk>
Subject Re: get each pair from a string.
Date 2012-10-24 13:14 +0100
References (2 earlier) <1a8e1d4b-dddc-48c6-aec8-8041bb559a30@u4g2000pbo.googlegroups.com> <177aa300-9863-44f3-94da-7bc11bff2eda@i7g2000pbf.googlegroups.com> <7fdadf96-2ec2-43c7-a5e2-74228c06faca@googlegroups.com> <mailman.2674.1351010504.27098.python-list@python.org> <ae632dbf-de0f-4b9a-91e5-1288f216f13f@googlegroups.com>
Newsgroups comp.lang.python
Message-ID <mailman.2752.1351080768.27098.python-list@python.org> (permalink)

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On 24/10/2012 10:32, wxjmfauth@gmail.com wrote:
> Le mardi 23 octobre 2012 18:41:45 UTC+2, Ian a écrit :
>> On Tue, Oct 23, 2012 at 12:47 AM,  <wxjmfauth@gmail.com> wrote:
>>
>>
>>
>> In any case, have you been following the progress of issue 16061?
>>
>> There is a patch for the str.replace regression that you identified,
>>
>> which results in better performance across the board than 3.2.  So
>>
>> although 3.3 is slower than 3.2 on this one microbenchmark, 3.4 will
>>
>> not be.
>
> Yes.
> A workaround to solve one of the corner cases of something
> which is wrong by design.
>
> jmf
>

Do you get paid money for acting as a complete moron or do you do it for 
free?

-- 
Cheers.

Mark Lawrence.

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Thread

Re: get each pair from a string. rusi <rustompmody@gmail.com> - 2012-10-22 09:19 -0700
  Re: get each pair from a string. rusi <rustompmody@gmail.com> - 2012-10-22 21:59 -0700
    Re: get each pair from a string. wxjmfauth@gmail.com - 2012-10-22 23:47 -0700
      Re: get each pair from a string. Neil Cerutti <neilc@norwich.edu> - 2012-10-23 12:30 +0000
      Re: get each pair from a string. Mark Lawrence <breamoreboy@yahoo.co.uk> - 2012-10-23 13:44 +0100
      Re: get each pair from a string. Ian Kelly <ian.g.kelly@gmail.com> - 2012-10-23 10:41 -0600
        Re: get each pair from a string. wxjmfauth@gmail.com - 2012-10-24 02:32 -0700
          Re: get each pair from a string. Mark Lawrence <breamoreboy@yahoo.co.uk> - 2012-10-24 13:14 +0100
        Re: get each pair from a string. wxjmfauth@gmail.com - 2012-10-24 02:32 -0700

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