Check if list items contains substrings from another list

Solution 1:

If you just want a test, join the target list into a string and test each element of bad like so:

>>> my_list = ['abc-123', 'def-456', 'ghi-789', 'abc-456', 'def-111', 'qwe-111']
>>> bad = ['abc', 'def']
>>> [e for e in bad if e in '\n'.join(my_list)]
['abc', 'def']

From your question, you can test each element as a sub string against the each element of the other this way:

>>> [i for e in bad for i in my_list if e in i]
['abc-123', 'abc-456', 'def-456', 'def-111']

It is fast (in comparison to one of the other methods):

>>> def f1():
...    [item for item in my_list if any(x in item for x in bad)]
... 
>>> def f2():
...    [i for e in bad for i in my_list if e in i]
... 
>>> timeit.Timer(f1).timeit()
5.062238931655884
>>> timeit.Timer(f2).timeit()
1.35371994972229

From your comment, here is how you get the elements that do not match:

>>> set(my_list)-{i for e in bad for i in my_list if e in i}
{'ghi-789', 'qwe-111'}

Solution 2:

In [4]: filter(lambda item: any(x in item for x in bad), my_list)
Out[4]: ['abc-123', 'def-456', 'abc-456', 'def-111']

or

In [13]: [item for item in my_list if any(x in item for x in bad)]
Out[13]: ['abc-123', 'def-456', 'abc-456', 'def-111']

Solution 3:

some_list = ['abc-123', 'def-456', 'ghi-789', 'abc-456']
bad = ['abc', 'def']
for s in some_list:
    for item in bad:
       if item in s:
          print 'Found ', s

It's simple, works fine and fast( only if your list is not very big.)