Python regular expression match with file extension

Question

I want to use Python regular expression utility to find the files which has this pattern:

000014_L_20111026T194932_1.txt
000014_L_20111026T194937_2.txt
...
000014_L_20111026T194928_12.txt

So the files I want have an underscore '_' followed by a number (1 or more digits) and then followed by '.txt' as the extension. I used the following regular expression but it didn't match the above names:

match = re.match('_(\d+)\.txt$', file)

What should be the correct regex to match the file names?

see: docs.python.org/2/library/re.html#search-vs-match

BlackVegetable
– BlackVegetable

2013-04-18 15:49:32 +00:00
Commented Apr 18, 2013 at 15:49 — BlackVegetable
– BlackVegetable, Commented Apr 18, 2013 at 15:49
Thanks. I get used to Perl style regex :)

tonga
– tonga

2013-04-18 16:01:23 +00:00
Commented Apr 18, 2013 at 16:01 — tonga
– tonga, Commented Apr 18, 2013 at 16:01

Martijn Pieters · Accepted Answer · 2013-04-18 15:46:17Z

14

You need to use .search() instead; .match() anchors to the start of the string. Your pattern is otherwise fine:

>>> re.search('_(\d+)\.txt$', '000014_L_20111026T194928_12.txt')
<_sre.SRE_Match object at 0x10e8b40a8>
>>> re.search('_(\d+)\.txt$', '000014_L_20111026T194928_12.txt').group(1)
'12'

answered Apr 18, 2013 at 15:46

Martijn Pieters

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2 Comments

tonga Over a year ago

Thanks Martijin. This works. So even if I use '$' to indicate from the end of string, .match function doesn't search from the end?

Martijn Pieters Over a year ago

@tonga: No; $ is an anchor, it'll only match at the end of the string; it won't dictate where the search begins. You should see .match() as adding an implicit ^ to your pattern.

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