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Groups > comp.lang.python > #74147
| References | <d580e76b-793e-435d-917b-613ae912a93f@googlegroups.com> |
|---|---|
| Date | 2014-07-07 21:38 -0600 |
| Subject | Re: finditer |
| From | Jason Friedman <jsf80238@gmail.com> |
| Newsgroups | comp.lang.python |
| Message-ID | <mailman.11615.1404790710.18130.python-list@python.org> (permalink) |
On Mon, Jul 7, 2014 at 1:19 AM, gintare <g.statkute@gmail.com> wrote:
> If smbd has time, maybe you could advice how to accomplish this task in faster way.
>
> I have a text = """ word{vb}
> wordtransl {vb}
>
> sent1.
>
> sent1trans.
>
> sent2
>
> sent2trans... """
>
> I need to match once wordtransl, and than many times repeating patterns consisting of sent and senttrans.
You might try itertools.groupby
(https://docs.python.org/3/library/itertools.html#module-itertools).
text = """ word{vb}
wordtransl {vb}
sent1
sent1trans
sent2
sent2trans
"""
import itertools
import re
result_list = list()
lines = text.split("\n")
for line in lines[:]:
if line.startswith("sent"):
break
lines.pop(0)
def is_start(x):
pattern = re.compile(r"sent\d+$")
if re.search(pattern, x):
return True
for key, mygroup in itertools.groupby(lines, is_start):
result_list.append(list(mygroup))
print(result_list)
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finditer gintare <g.statkute@gmail.com> - 2014-07-07 00:19 -0700 Re: finditer Jason Friedman <jsf80238@gmail.com> - 2014-07-07 21:38 -0600
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