Python difflib 中的 SequenceMatcher 是否有可能提供一种更有效的方法来计算 Levenshtein 距离？

Question

这是计算 Levenshtein 距离的通用算法的教科书示例（我从 Magnus Hetland's webite 中提取）：

def levenshtein(a,b):
    "Calculates the Levenshtein distance between a and b."
    n, m = len(a), len(b)
    if n > m:
        # Make sure n <= m, to use O(min(n,m)) space
        a,b = b,a
        n,m = m,n

    current = range(n+1)
    for i in range(1,m+1):
        previous, current = current, [i]+[0]*n
        for j in range(1,n+1):
            add, delete = previous[j]+1, current[j-1]+1
            change = previous[j-1]
            if a[j-1] != b[i-1]:
                change = change + 1
            current[j] = min(add, delete, change)

    return current[n]

不过，我想知道是否有更高效（并且可能更优雅）的纯 Python 实现，它使用 difflib 的 SequenceManager。在玩了它之后，这就是我想出的：

from difflib import SequenceMatcher as sm

def lev_using_difflib(s1, s2):
    a = b = size = distance = 0
    for m in sm(a=s1, b=s2).get_matching_blocks():
        distance += max(m.a-a, m.b-b) - size
        a, b, size = m
    return distance

我想不出一个失败的测试用例，而且性能似乎明显优于标准算法。

以下是依赖 difflib 的 levenshtein 算法的结果：

>>> from timeit import Timer
>>> setup = """
... from difflib import SequenceMatcher as sm
... 
... def lev_using_difflib(s1, s2):
...     a = b = size = distance = 0
...     for m in sm(a=s1, b=s2).get_matching_blocks():
...         distance += max(m.a-a, m.b-b) - size
...         a, b, size = m
...     return distance
... 
... strings = [('sunday','saturday'),
...            ('fitting','babysitting'),
...            ('rosettacode','raisethysword')]
... """
>>> stmt = """
... for s in strings:
...     lev_using_difflib(*s)
... """
>>> Timer(stmt, setup).timeit(100000)
36.989389181137085

这是标准的纯python实现：

>>> from timeit import Timer
>>> setup2 = """
... def levenshtein(a,b):
...     n, m = len(a), len(b)
...     if n > m:
...         a,b = b,a
...         n,m = m,n
... 
...     current = range(n+1)
...     for i in range(1,m+1):
...         previous, current = current, [i]+[0]*n
...         for j in range(1,n+1):
...             add, delete = previous[j]+1, current[j-1]+1
...             change = previous[j-1]
...             if a[j-1] != b[i-1]:
...                 change = change + 1
...             current[j] = min(add, delete, change)
... 
...     return current[n]
... 
... strings = [('sunday','saturday'),
...            ('fitting','babysitting'),
...            ('rosettacode','raisethysword')]
... """
>>> stmt2 = """
... for s in strings:
...     levenshtein(*s)
... """
>>> Timer(stmt2, setup2).timeit(100000)
55.594768047332764

使用 difflib 的 SequenceMatcher 算法的性能真的更好吗？或者它是否依赖于完全使比较无效的 C 库？如果它依赖于 C 扩展，我如何通过查看 difflib.py 实现来判断？

使用 Python 2.7.3 [GCC 4.2.1 (Apple Inc. build 5666)]

提前感谢您的帮助！

Answer 1

>>> levenshtein('hello', 'world')
4
>>> lev_using_difflib('hello', 'world')
5