How to detect if string is matched with some of many patterns in regex

Question

i have a problem with my python script, i want to make some regex function. Now i've done make it but how to detect if the string is matched with some patterns ?

I give an example, i have one string and two patterns of regex. Now how do i know if the string is matched with first pattern or second pattern ?

This is my script.

#!/usr/bin/python
import re
words = ["(@sosiora\*+([1-5]*)(\W|)+(@[a-z]*)(\W|))",
         "((@[a-z]*)(\W|)@sosiora\*+([1-5]*)+(\W|))"
        ]
pattern = re.compile('|'.join(words), flags=re.IGNORECASE)
text = "@sosiora*4 @samsungID"

m = pattern.findall(text)

print m;

if m:
    if len(m[0][1]) > 1:
        print 'Rating is not accepted : ' +m[0][1]
        print 'String: ' + text
    else :
        print 'Found %d matches' % len(m)
        print 'String: ' + text
        print 'Rating: ' + m[0][1]
        print 'Target: ' + m[0][3]
        print 'Pattern: ' // this is output to show the pattern
else:
    print 'rate is not found'

So, the output what i want is just Please help me, thank you.

Ali SAID OMAR · Accepted Answer · 2015-08-06 15:39:53Z

2

You can loop over your patterns:

def _get_pos_and_group(s):
    words = ["(@sosiora\*+([1-5]*)(\W|)+(@[a-z]*)(\W|))",
             "((@[a-z]*)(\W|)@sosiora\*+([1-5]*)+(\W|))"
            ]

    for i, w in enumerate(words):
        if re.search(w, s):
            return i, re.search(w, s).group(0)
    return -1, None



print(_get_pos_and_group("@sosiora*4 @samsungID"))

(0, '@sosiora*4 @samsung')

answered Aug 6, 2015 at 15:39

Ali SAID OMAR

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1 Comment

Caleth Over a year ago

This also has the benefit of not combining 2 patterns that each are malformed, but when combined become something else - eg first pattern lacks a ) and second lacks a (

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