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| Date | Fri, 12 Aug 2011 10:20:41 +0100 |
| Subject | Re: String concatenation - which is the fastest way ? |
| From | Chris Angelico <rosuav@gmail.com> |
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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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