Python “编辑距离”(Levenshtein distance)函数的比较

本文搜集了网上比较常用的几种计算Levenshtein distance的函数，

其中函数(1)为调用数学工具包Numpy， 函数(2)和(1)算法类似，都是采用DP, (3)来自wiki(4)是直接调用python的第三方库Levenshtein

源码和结果如下：

import timefrom functools import wrapsimport cProfileimport numpyimport Levenshteindef fn_timer(function):    @wraps(function)    def function_timer(*args, **kwargs):        t0 = time.time()        result = function(*args, **kwargs)        t1 = time.time()        print ("Total time running %s: %s seconds" %                (function.func_name, str(t1-t0))                )        return result    return function_timerdef levenshtein1(source, target):    if len(source) < len(target):        return levenshtein1(target, source)     # So now we have len(source) >= len(target).    if len(target) == 0:        return len(source)     # We call tuple() to force strings to be used as sequences    # ('c', 'a', 't', 's') - numpy uses them as values by default.    source = numpy.array(tuple(source))    target = numpy.array(tuple(target))     # We use a dynamic programming algorithm, but with the    # added optimization that we only need the last two rows    # of the matrix.    previous_row = numpy.arange(target.size + 1)    for s in source:        # Insertion (target grows longer than source):        current_row = previous_row + 1         # Substitution or matching:        # Target and source items are aligned, and either        # are different (cost of 1), or are the same (cost of 0).        current_row[1:] = numpy.minimum(                current_row[1:],                numpy.add(previous_row[:-1], target != s))         # Deletion (target grows shorter than source):        current_row[1:] = numpy.minimum(                current_row[1:],                current_row[0:-1] + 1)         previous_row = current_row     return previous_row[-1]def levenshtein2(s1, s2):    if len(s1) < len(s2):        return levenshtein2(s2, s1)     # len(s1) >= len(s2)    if len(s2) == 0:        return len(s1)     previous_row = range(len(s2) + 1)    for i, c1 in enumerate(s1):        current_row = [i + 1]        for j, c2 in enumerate(s2):            insertions = previous_row[j + 1] + 1 # j+1 instead of j since previous_row and current_row are one character longer            deletions = current_row[j] + 1       # than s2            substitutions = previous_row[j] + (c1 != c2)            current_row.append(min(insertions, deletions, substitutions))        previous_row = current_row     return previous_row[-1]def levenshtein3(s, t):        ''' From Wikipedia article; Iterative with two matrix rows. '''        if s == t: return 0        elif len(s) == 0: return len(t)        elif len(t) == 0: return len(s)        v0 = [None] * (len(t) + 1)        v1 = [None] * (len(t) + 1)        for i in range(len(v0)):            v0[i] = i        for i in range(len(s)):            v1[0] = i + 1            for j in range(len(t)):                cost = 0 if s[i] == t[j] else 1                v1[j + 1] = min(v1[j] + 1, v0[j + 1] + 1, v0[j] + cost)            for j in range(len(v0)):                v0[j] = v1[j]         return v1[len(t)]@fn_timerdef calllevenshtein1(s,t,n):    for i in range(n):        levenshtein3(s,t)@fn_timerdef calllevenshtein2(s,t,n):    for i in range(n):        levenshtein3(s,t)@fn_timerdef calllevenshtein3(s,t,n):    for i in range(n):        levenshtein3(s,t)@fn_timerdef calllevenshtein4(s,t,n):    for i in range(n):        Levenshtein.distance(s,t) if __name__ == "__main__":    n = 50000    a = 'abddcdefdgbd22svb'    b = 'bcdefg34rdyvdfsd'    calllevenshtein1(a, b, n)    calllevenshtein2(a, b, n)    calllevenshtein3(a, b, n)    calllevenshtein4(a, b, n)

结果：

Total time running calllevenshtein1: 16.0750000477 seconds
Total time running calllevenshtein2: 16.4990000725 seconds
Total time running calllevenshtein3: 16.2939999104 seconds
Total time running calllevenshtein4: 0.0629999637604 seconds

从结果来看，调用python第三方包效率最高，原因是其内部调用c库，优化了算法结构