CODEWITHSUNDEEP
pythonregex
DrWolfe
DrWolfe

Reputation: 159

How use regular expressions to compare an input with a set?

I would like a regular expression python code to:

I am no good at all with regular expressions.

Upvotes: 0

Views: 176

Answers (3)

John Machin
John Machin

Reputation: 82992

Your self-answer would be better using a set rather than that loop.

Using i for a text variable and n for an index is very counter-intuitive. And keywords_found is a misnomer.

Try this:

>>> import re
>>> keywords = set(('cars', 'jewelry', 'gas'))
>>> pattern = re.compile('[a-z]+', re.IGNORECASE)
>>> txt = 'GAS, CaRs, Jewelrys'
>>> text_words = set(pattern.findall(txt.lower()))
>>> print "keywords:", keywords
keywords: set(['cars', 'gas', 'jewelry'])
>>> print "text_words:", text_words
text_words: set(['cars', 'gas', 'jewelrys'])
>>> print "text words in keywords:", text_words & keywords
text words in keywords: set(['cars', 'gas'])
>>> print "text words NOT in keywords:", text_words - (text_words & keywords)
text words NOT in keywords: set(['jewelrys'])
>>> print "keywords NOT in text words:", keywords - (text_words & keywords)
keywords NOT in text words: set(['jewelry'])

Upvotes: 0

DrWolfe
DrWolfe

Reputation: 159

After much google searching, and with trial an error, I a created a solution that works to separate multiple words from the input of characters.

import re

keywords = ('cars', 'jewelry', 'gas')
pattern = re.compile('[a-z]+', re.IGNORECASE)
txt = 'GAS, CaRs, Jewelrys'

keywords_found = pattern.findall(txt.lower())

n = 0

for i in keywords_found:
    if i in keywords:
        print keywords_found[n]
    n = n + 1

Upvotes: 0

Ignacio Vazquez-Abrams
Ignacio Vazquez-Abrams

Reputation: 799082

Why bother?

>>> 'FOO'.lower() in set(('foo', 'bar', 'baz'))
True
>>> 'Quux'.lower() in set(('foo', 'bar', 'baz'))
False

Upvotes: 6

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