Python regex findall OR

Question

I am compiling the following pattern:

pattern = re.compile("media.+\.(aac|ts)")

My idea is to obtain .ts and .aac media files contained in a string. The media file names can be media-u9xuxtkay_213.aac or media-u9xuxtkay_213.ts

According to this accepted answer Python regular expressions OR you can use ( | ) as an OR But I dont see how that's an accepted answer since it doenst seem to work to me:

In [23]: s
Out[23]: 'Sent from my iPhone'

In [24]: patt = re.compile("Sent from my (iPhone|iPod)")

In [25]: patt.findall(s)
Out[25]: ['iPhone']

So I call the findall and I get this:

In [37]: media
Out[37]: 'media-u9xuxtkay_213.aac'

In [38]: pattern = re.compile("media.+\.(aac|ts)")

In [39]: pattern.findall(media)
Out[39]: ['aac']

I should get a media-u9xuxtkay_213.aac instead just aac. The same way the accepted answer should return Sent from my iPhone instead just iPhone

No, if you use it with `re.finditer` and grab `.group()`. But there is a more convenient thing - a non-capturing group. — Wiktor Stribiżew, Oct 19 '16 at 14:05
Might be duplicated but the question you linked is impossible to find through the search engine with that title when looking for this issue. — lapinkoira, Oct 19 '16 at 14:06
I found it with "re.findall wierd behavior" - top result. It is becoming a very frequent, repetitive issue. BTW, what search words did you use? — Wiktor Stribiżew, Oct 19 '16 at 14:06
typed the same thing, doesnt even appear in the first page, but "ok" — lapinkoira, Oct 19 '16 at 14:10
I looked for "python regex or" to see how to use it properly and the first question it's a wrong one which is not doing what OP asked and still is accepted http://stackoverflow.com/questions/8609597/python-regular-expressions-or — lapinkoira, Oct 19 '16 at 14:11
The question I linked has 19k views, the one you linked has barely 200, so guess where the search engine are leading to — lapinkoira, Oct 19 '16 at 14:12
Well, the second answer in the thread you linked to provides the regex compatible with re.findall. So, it can also be used to close this question. — Wiktor Stribiżew, Oct 19 '16 at 14:19

score 10 · Accepted Answer · answered Oct 19 '16 at 14:04

10

The parentheses act as a "capture group", denoting what the regex should select from the string. You can use (?: ) to make it a non-capturing group, so it gets the whole string instead.

re.compile(r"media.+\.(?:aac|ts)")

answered Oct 19 '16 at 14:04

ASGM

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Python regex findall OR

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