如何使用 difflib 有效地比较两个字符串列表字典？

Question

. Answers to this question are eligible for a +50 reputation bounty. tristan wants to to this question.

I have two large dictionaries of lists. All the elements of the lists are strings. I want to compare all against all and compute their respective similarity - but the naive way I use is obviously very slow and does not scale at all:

import numpy as np
import difflib  

first_dict = {"first1" : ["aa", "bb","cc", "dd"], "first2" : ["ff", "gg"]}

second_dict = {"second1" : ["cc", "dd", "jj", "aa", "bb"], "second2" : ["ff", "gg"], "second3" : ["hh", "ii"]}  

mat = np.empty((len(second_dict), len(first_dict)))

for i, second in enumerate(second_dict.keys()):
    for j, first in enumerate(first_dict.keys()):
        sm = difflib.SequenceMatcher(None, sorted(first_dict[first]), sorted(second_dict[second]))
        mat[i, j] = sm.ratio() 

print(mat)

is there a clever way to speed this up?

Answer 1

Your code seems legit. I did a couple of little tweaks that would shave off a couple of microseconds per loop:

1. No need for the two sorted calls because difflib can calculated an order-indifferent comparison with quick_ratio (Checkout the documentation here for the difference between ratio , quick_ratio , and real_quick_ratio ).
2. No need for the enumerate to access mat by i and j .
3. Removed the access of the list through index first_dict[index] and second_dict[index]

def naive_ratio_comparison(first_dict, second_dict):
    mat = []
    for second in second_dict.values():
        for first in first_dict.values():
            sm = difflib.SequenceMatcher(None, first, second)
            mat.append(sm.quick_ratio())
    result = np.resize(mat, (len(second_dict), len(first_dict)))
    return result

Answer 2

If one dict has M entries and the other N , then you're going to have to do M*N .ratio() calls. There's no way around that, and it's going to be costly.

However, you can easily arrange to do only M+N sorts instead of (as shown) M*N sorts.

For computing .ratio() , the most valuable hint is in the docs:

SequenceMatcher computes and caches detailed information about the second sequence, so if you want to compare one sequence against many sequences, use set_seq2() to set the commonly used sequence once and call set_seq1() repeatedly, once for each of the other sequences.

Putting that all together:

firsts = list(map(sorted, first_dict.values())) # sort these only once

sm = difflib.SequenceMatcher(None)
for i, second in enumerate(second_dict.values()):
    sm.set_seq2(sorted(second))
    for j, first in enumerate(firsts):
        sm.set_seq1(first)
        mat[i, j] = sm.ratio()

That should deliver exactly the same results. To minimize the number of expensive .set_seq2() calls, it would - of course - be best to arrange for the shorter dict to be called "second_dict".