if allwords in title: match

Question

Using python3, i have a list of words like: ['foot', 'stool', 'carpet']

these lists vary in length from 1-6 or so. i have thousands and thousands of strings to check, and it is required to make sure that all three words are present in a title. where: 'carpet stand upon the stool of foot balls.' is a correct match, as all the words are present here, even though they are out of order.

ive wondered about this for a long time, and the only thing i could think of was some sort of iteration like:

for word in list: if word in title: match!

but this give me results like 'carpet cleaner' which is incorrect. i feel as though there is some sort of shortcut to do this, but i cant seem to figure it out without using excessivelist(), continue, break or other methods/terminology that im not yet familiar with. etc etc.

Would you want `recarpeted footstool` to match? – Mad Physicist Aug 24 '16 at 18:22 — Mad Physicist, Aug 24 '16 at 18:22

alecxe · Accepted Answer · 2016-08-24T18:19:00.510

4

You can use all():

words = ['foot', 'stool', 'carpet']
title = "carpet stand upon the stool of foot balls."

matches = all(word in title for word in words)

Or, inverse the logic with not any() and not in:

matches = not any(word not in title for word in words)

edited Aug 24 '16 at 18:19

answered Aug 24 '16 at 17:58

alecxe

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You might want to split if substrings should not be matched. – Padraic Cunningham Aug 24 '16 at 18:20
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@PadraicCunningham. Also to avoid things like `recarpeted` from matching. – Mad Physicist Aug 24 '16 at 18:21
Good points, thanks! Nltk's *word tokenizer* might be a bit better option to split into words and handle all the punctuation. – alecxe Aug 24 '16 at 18:24
Here's a simpler option that just strips the punctuation using vanilla Python: http://stackoverflow.com/a/17951315/2988730 – Mad Physicist Aug 24 '16 at 18:28

if allwords in title: match

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