I have a list of strings, from which I want to locate every line that has 'http://' in it, but does not have 'lulz', 'lmfao', '.png', or any other items in a list of strings in it. How would I go about this?
My instincts tell me to use regular expressions, but I have a moral objection to witchcraft.
Here is an option that is fairly extensible if the list of strings to exclude is large:
exclude = ['lulz', 'lmfao', '.png']
filter_func = lambda s: 'http://' in s and not any(x in s for x in exclude)
matching_lines = filter(filter_func, string_list)
List comprehension alternative:
matching_lines = [line for line in string_list if filter_func(line)]
lambda allows you to define the function inline instead of setting up a variable filter_func; but you could just as easily write def filter_func(s): return 'http://' in s and not any(x in s for x in exclude). Remember, functions are objects.
lambda. There is no reason to prefer it to a def here.This is almost equivalent to F.J's solution, but uses generator expressions instead of lambda expressions and the filter function:
haystack = ['http://blah', 'http://lulz', 'blah blah', 'http://lmfao']
exclude = ['lulz', 'lmfao', '.png']
http_strings = (s for s in haystack if s.startswith('http://'))
result_strings = (s for s in http_strings if not any(e in s for e in exclude))
print list(result_strings)
When I run this it prints:
['http://blah']
result_strings = [s for s in haystack if s.startswith('http://') and not any(e in s for e in exclude)]. It needs a line break to fit 80 columns (per most style guides), but I would argue it is slightly easier to follow than the two-generator version. timeit also reports that this is a fair bit faster, and also slightly faster than F.J's filter version (which, IMO, is the hardest to follow of the three).Try this:
for s in strings:
if 'http://' in s and not 'lulz' in s and not 'lmfao' in s and not '.png' in s:
# found it
pass
Other option, if you need your options more flexible:
words = ('lmfao', '.png', 'lulz')
for s in strings:
if 'http://' in s and all(map(lambda x, y: x not in y, words, list(s * len(words))):
# found it
pass
