Check if each element in a list match with a string in Python?

Question

I am writing a simple if else loop to check if a string match with multiple words like this:

if "word1" in data or "word2" in data or "word3" in data:
    ....

I am not sure if we have a more comprehensive way to process this kind of comparison ?

Thank you very much

Steven Rumbalski · Accepted Answer · 2012-07-30 18:29:34Z

10

if any(word in data for word in ('word1', 'word2', 'word3')):
    ...

If you run into performance issues, you may want to convert data to a set before running the comparisons.

answered Jul 30, 2012 at 18:29

Steven Rumbalski

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1 Comment

mrblue Over a year ago

actually, inspectorG4dget's solution worked for another case. In mine, this is the best. Thank you very much.

JBernardo · Accepted Answer · 2012-07-30 18:29:26Z

8

You can do:

if any(x in data for x in ('word1', 'word2', 'word3')):

answered Jul 30, 2012 at 18:29

JBernardo

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Comments

mgilson · Accepted Answer · 2012-07-30 18:33:03Z

7

Why not a set intersect?

if set(["word1", "word2","word3"]) & set(data):
    # do stuff!

edited Jul 30, 2012 at 18:33

mgilson

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answered Jul 30, 2012 at 18:30

inspectorG4dget

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9 Comments

mgilson Over a year ago

Clever (+1), but again, don't use a list, use a tuple. (There's no reason for a list here and a tuple is faster to generate)

mgilson Over a year ago

and actually, set(data).intersection(("word1","word2","word3")) might be slightly faster as there is no need to create 2 sets for this operation.

inspectorG4dget Over a year ago

@mgilson: I seem to be off my game with the formatting today. Thank you for cleaning up my posts (I believe this is the 2nd or 3rd one today that you have fixed the formatting in)

BrenBarn Over a year ago

Note that this won't work if data was a string containing "word1", "word2", etc., as substrings (which it might have been based on the OP's code).

Steven Rumbalski Over a year ago

There is one performance implication: the intersection of two sets does not shortcut in the same way as any. This is probably not an issue here.

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