Getting file extension using pattern matching in python

Question

I am trying to find the extension of a file, given its name as a string. I know I can use the function os.path.splitext but it does not work as expected in case my file extension is .tar.gz or .tar.bz2 as it gives the extensions as gz and bz2 instead of tar.gz and tar.bz2 respectively.
So I decided to find the extension of files myself using pattern matching.

print re.compile(r'^.*[.](?P<ext>tar\.gz|tar\.bz2|\w+)$').match('a.tar.gz')group('ext')
>>> gz            # I want this to come as 'tar.gz'
print re.compile(r'^.*[.](?P<ext>tar\.gz|tar\.bz2|\w+)$').match('a.tar.bz2')group('ext')
>>> bz2           # I want this to come 'tar.bz2'

I am using (?P<ext>...) in my pattern matching as I also want to get the extension.

Please help.

What if name="hi.c.java" in case you want .java alone right ? — Kracekumar, Jun 29 '11 at 18:22
for the time being, yes. But, I should be able to add more in the regex pattern later if I want to. — Pushpak Dagade, Jun 29 '11 at 18:23

score 21 · Answer 1 · answered Jun 29 '11 at 18:21

21

root,ext = os.path.splitext('a.tar.gz')
if ext in ['.gz', '.bz2']:
   ext = os.path.splitext(root)[1] + ext

_{Some people, when confronted with a problem, think "I know, I'll use regular expressions." Now they have two problems.}

answered Jun 29 '11 at 18:21

phihag

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Ok that will work in this case but I want to solve it using python regular expressions. – Pushpak Dagade Jun 29 '11 at 18:22
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@Guanidene: If it's homework, mark the question homework. If it's not homework, don't use a regular expression when the function's already been written, debugged and works. – S.Lott Jun 29 '11 at 18:23
@S.Lott - It is no homework, I want to tackle the problem using regex thats all. If just solving was the aim, I could have done it long before as phihag says. – Pushpak Dagade Jun 29 '11 at 18:25
@phihag - one more reason why I wish to use regex - It is compact. If I go your way, I will unnecessarily need 3 lines (and which also makes my code clumsy), while using regex I can get everything in a single line! – Pushpak Dagade Jun 29 '11 at 18:43
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@Guanidene More compact does not equal more readable and maintainable. Also, why is a complicated regular expression less clumsy than three lines even non-programmers could understand? Anyway, to each his own. – phihag Jun 29 '11 at 18:49

score 5 · Accepted Answer · answered Jun 29 '11 at 18:25

5

>>> print re.compile(r'^.*[.](?P<ext>tar\.gz|tar\.bz2|\w+)$').match('a.tar.gz').group('ext')
gz
>>> print re.compile(r'^.*?[.](?P<ext>tar\.gz|tar\.bz2|\w+)$').match('a.tar.gz').group('ext')
tar.gz
>>>

The ? operator tries to find the minimal match, so instead of .* eating ".tar" as well, .*? finds the minimal match that allows .tar.gz to be matched.

answered Jun 29 '11 at 18:25

Omri Barel

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I really thank you for this! This thing took a lot of my time! – Pushpak Dagade Jun 29 '11 at 18:29

score 3 · Answer 3 · answered Jun 29 '11 at 18:31

3

I have idea which is much easier than breaking your head with regex,sometime it might sound stupid too.
name="filename.tar.gz" extensions=('.tar.gz','.py') [x for x in extensions if name.endswith(x)]

answered Jun 29 '11 at 18:31

Kracekumar

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Suppose there exists a file extension `.gz` (assume) too which I may want to match. So in this case your tuple will be `extensions=('.gz','.tar.gz','.py')` (and name=`filenmae.tar.gz`) and if I execute this - `[x for x in extensions if name.endswith(x)]` it will wrongly match `gz` when I want to match it with `tar.gz`. Dude, what I want is a universal solution, not a data specific solution! – Pushpak Dagade Jun 29 '11 at 18:36
There is a option for that ,so place tar.gz first in the list if it matches return,but this method will not work once you place gz before tar.gz.>>> extensions=('tar.gz','gz','py') `>>> name 'set.tar.gz' >>> def test(): ... for x in extensions: ... if name.endswith(x): ... return x ... return ' ' >>> test() 'tar.gz' >>> ` – Kracekumar Jun 29 '11 at 18:43
@Guanidene:Yea I will not remember agreed,but you can leave a comment,might be regex is a correct solution and it is,but while answering the question I started by saying it might be stupid and there is a option,I am not arguing this is right . – Kracekumar Jun 29 '11 at 18:50

score 3 · Answer 4 · answered Feb 24 '12 at 09:37

Starting from phihags answer:

DOUBLE_EXTENSIONS = ['tar.gz','tar.bz2'] # Add extra extensions where desired.

def guess_extension(filename):
    """
    Guess the extension of given filename.
    """
    root,ext = os.path.splitext(filename)
    if any([filename.endswith(x) for x in DOUBLE_EXTENSIONS]):
        root, first_ext = os.path.splitext(root)
        ext = first_ext + ext
    return root, ext

score 2 · Answer 5 · answered Nov 16 '16 at 05:52

2

this is simple and works on both single and multiple extensions

In [1]: '/folder/folder/folder/filename.tar.gz'.split('/')[-1].split('.')[0]
Out[1]: 'filename'

In [2]: '/folder/folder/folder/filename.tar'.split('/')[-1].split('.')[0]
Out[2]: 'filename'

In [3]: 'filename.tar.gz'.split('/')[-1].split('.')[0]
Out[3]: 'filename'

answered Nov 16 '16 at 05:52

themachinist

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In [4]: '/folder/folder/folder/filename.tar.gz'.split('/')[-1].split('.')[1:] Out[4]: ['tar', 'gz'] Works well for me! – four43 Sep 25 '18 at 14:49
A variant I used of this to get the filename and file extension: `f_name, f_ext = os.path.splitext(os.path.basename(body).split("/")[-1])` – Chioke Aarhus Jul 26 '22 at 00:27

score 1 · Answer 6 · edited May 25 '16 at 02:12

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Continuing from phihags answer to generic remove all double or triple extensions such as CropQDS275.jpg.aux.xml use while '.' in:

tempfilename, file_extension = os.path.splitext(filename)
while '.' in tempfilename:
     tempfilename, tempfile_extension = os.path.splitext(tempfilename)
     file_extension = tempfile_extension + file_extension

edited May 25 '16 at 02:12

Daisy Leigh Brenecki

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answered May 25 '16 at 00:53

John Lang

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Getting file extension using pattern matching in python

6 Answers6

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