Re: String concatenation - which is the fastest way ?

References	(4 earlier) <20110811064030.GB4990@host.pgf.com.pl> <CAPTjJmqiiKcT7Zg-XkbrQ4r-keU9+1Kj=+sEwa4xp==2EG-5aQ@mail.gmail.com> <20110811134613.GE4990@host.pgf.com.pl> <CAPTjJmr7qDeGU+ttEXsdBOASqZ5w1mYgkg18o+RtFp55HHp8Ew@mail.gmail.com> <20110811143923.GG4990@host.pgf.com.pl>
Date	2011-08-12 10:20 +0100
Subject	Re: String concatenation - which is the fastest way ?
From	Chris Angelico <rosuav@gmail.com>
Newsgroups	comp.lang.python
Message-ID	<mailman.2212.1313140844.1164.python-list@python.org> (permalink)

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On Thu, Aug 11, 2011 at 3:39 PM,  <przemolicc@poczta.fm> wrote:
> On Thu, Aug 11, 2011 at 02:48:43PM +0100, Chris Angelico wrote:
>> List of strings. Take it straight from your Oracle interface and work
>> with it directly.
>
> Can I use this list in the following way ?
> subprocess_1 - run on list between 1 and 10000
> subprocess_2 - run on list between 10001 and 20000
> subprocess_3 - run on list between 20001 and 30000
> etc
> ...

Yep! You use list slicing. Working with smaller numbers for an example:

>>> ltrs=['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
>>> ltrs[:10]
['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']
>>> ltrs[10:20]
['k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't']
>>> ltrs[20:]
['u', 'v', 'w', 'x', 'y', 'z']

(I actually created that list as "list(string.ascii_lowercase)" for
what that's worth.)

Slice operations are quite efficient.

Chris Angelico

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Re: String concatenation - which is the fastest way ? Chris Angelico <rosuav@gmail.com> - 2011-08-12 10:20 +0100

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