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Groups > comp.lang.python > #11246
| Date | 2011-08-12 03:15 +0100 |
|---|---|
| From | MRAB <python@mrabarnett.plus.com> |
| Subject | Re: Processing a large string |
| References | <b16af723-854c-449d-8b45-565d73579e17@br5g2000vbb.googlegroups.com> |
| Newsgroups | comp.lang.python |
| Message-ID | <mailman.2201.1313115355.1164.python-list@python.org> (permalink) |
On 12/08/2011 03:03, goldtech wrote:
> Hi,
>
> Say I have a very big string with a pattern like:
>
> akakksssk3dhdhdhdbddb3dkdkdkddk3dmdmdmd3dkdkdkdk3asnsn.....
>
> I want to split the sting into separate parts on the "3" and process
> each part separately. I might run into memory limitations if I use
> "split" and get a big array(?) I wondered if there's a way I could
> read (stream?) the string from start to finish and read what's
> delimited by the "3" into a variable, process the smaller string
> variable then append/build a new string with the processed data?
>
> Would I loop it and read it char by char till a "3"...? Or?
>
You could write a generator like this:
def split(string, sep):
pos = 0
try:
while True:
next_pos = string.index(sep, pos)
yield string[pos : next_pos]
pos = next_pos + 1
except ValueError:
yield string[pos : ]
string = "akakksssk3dhdhdhdbddb3dkdkdkddk3dmdmdmd3dkdkdkdk3asnsn..."
for part in split(string, "3"):
print(part)
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Processing a large string goldtech <goldtech@worldpost.com> - 2011-08-11 19:03 -0700
Re: Processing a large string MRAB <python@mrabarnett.plus.com> - 2011-08-12 03:15 +0100
Re: Processing a large string Steven D'Aprano <steve+comp.lang.python@pearwood.info> - 2011-08-12 12:30 +1000
Re: Processing a large string Nobody <nobody@nowhere.com> - 2011-08-12 05:11 +0100
Re: Processing a large string Peter Otten <__peter__@web.de> - 2011-08-12 10:39 +0200
Re: Processing a large string goldtech <goldtech@worldpost.com> - 2011-08-12 06:36 -0700
Re: Processing a large string Peter Otten <__peter__@web.de> - 2011-08-12 16:48 +0200
Re: Processing a large string Paul Rudin <paul.nospam@rudin.co.uk> - 2011-08-28 20:18 +0100
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