String concatenation - which is the fastest way ?

Chris Angelico rosuav at
Fri Aug 12 05:20:41 EDT 2011

On Thu, Aug 11, 2011 at 3:39 PM,  <przemolicc at> 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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