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I am trying out the difflib library. I have two lists: L_1 and L_2 containing strings. I want to know, if those sequences are similar (order is not important).

L_1 = ["Bob", "Mary", "Hans"]
L_2 = ["Bob", "Marie", "Háns"]

should be ok. But 

L_1 = ["Nirdosch", "Mary", "Rolf"]
L_2 = ["Bob", "Marie", "Háns"]

should not be ok.

I came up with the idea of iterating over the first list L_1 and to match every element of L_1 by the method

difflib.get_close_matches()

against the second list L_2. If there was a match with a ratio bigger then let's say 0.7 remove it from L_2 and continue. But I doubt it is a good plan. Is there a better one?

Aufwind
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1 Answers1

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I would do something like:

import difflib

L_1 = ["Bob", "Mary", "Hans"]
L_2 = ["Bob", "Marie", "Hans"]

def similiarity(L_1, L_2):
    L_1 = set(intern(w) for w in L_1)
    L_2 = set(intern(w) for w in L_2)

    to_match = L_1.difference( L_2)
    against = L_2.difference(L_1)
    for w in to_match:
        res = difflib.get_close_matches(w, against)
        if len(res):
            against.remove( res[0] )
    return (len(L_2)-len(against)) / (len(L_1))

print similiarity(L_1,L_2)
fabrizioM
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  • Nice solution - however I would in general implement my own comparison based on the Levenshtein distance. – Andreas Jung Mar 30 '11 at 17:06
  • I thought difflib uses Levenshtein under the hood – fabrizioM Mar 30 '11 at 17:13
  • First: Thanks for your answer! I read up on intern(), but I didn't really get what it means. Would you be so kind and give me a hint? – Aufwind Mar 30 '11 at 17:31
  • if you create two strings with the same value, those strings are different objects. With intern the second time you try to create a string that have already been created, it will return the *same* string object. In this way comparing two strings is just a matter of object address, constant, no matter how long is the string – fabrizioM Mar 30 '11 at 17:40