CODEWITHSUNDEEP
python
directedition
directedition

Reputation: 11705

If string does not contain any of list of strings in python

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.

Upvotes: 11

Views: 18539

Answers (3)

Pablo Santa Cruz
Pablo Santa Cruz

Reputation: 181350

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

Upvotes: 2

srgerg
srgerg

Reputation: 19329

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']

Upvotes: 3

Andrew Clark
Andrew Clark

Reputation: 208545

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)]

Upvotes: 14

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