Check if string ends with one of the strings from a list

Question

Though not widely known, str.endswith also accepts a tuple. You don't need to loop.

>>> 'test.mp3'.endswith(('.mp3', '.avi'))
True

Just use:

if file_name.endswith(tuple(extensions)):

Take an extension from the file and see if it is in the set of extensions:

>>> import os
>>> extensions = set(['.mp3','.avi'])
>>> file_name = 'test.mp3'
>>> extension = os.path.splitext(file_name)[1]
>>> extension in extensions
True

Using a set because time complexity for lookups in sets is O(1) (docs).

There is two ways: regular expressions and string (str) methods.

String methods are usually faster ( ~2x ).

import re, timeit
p = re.compile('.*(.mp3|.avi)$', re.IGNORECASE)
file_name = 'test.mp3'
print(bool(t.match(file_name))
%timeit bool(t.match(file_name)

792 ns ± 1.83 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

file_name = 'test.mp3'
extensions = ('.mp3','.avi')
print(file_name.lower().endswith(extensions))
%timeit file_name.lower().endswith(extensions)

274 ns ± 4.22 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

another way which can return the list of matching strings is

sample = "alexis has the control"
matched_strings = filter(sample.endswith, ["trol", "ol", "troll"])
print matched_strings
['trol', 'ol']

I just came across this, while looking for something else.

I would recommend to go with the methods in the os package. This is because you can make it more general, compensating for any weird case.

You can do something like:

import os

the_file = 'aaaa/bbbb/ccc.ddd'

extensions_list = ['ddd', 'eee', 'fff']

if os.path.splitext(the_file)[-1] in extensions_list:
    # Do your thing.

I have this:

def has_extension(filename, extension):

    ext = "." + extension
    if filename.endswith(ext):
        return True
    else:
        return False