python - 两个长列表之间的最长公共(public)子串

Question

我有两个很长的列表，我想在第二个列表中为第一个列表的每个元素找到最长的公共(public)子字符串。

一个简化的例子是

L1= ["a_b_c","d_e_f"]
L2=["xx""xy_a","xy_b_c","z_d_e","zl_d","z_d_e_y"]

所以我想在 L2 ("xy_b_c") 中找到 "a_b_c"的最佳匹配，然后在 L2("z_d_e_y") 中找到 "d_e_f"的最佳匹配。最适合我的是具有最长公共(public)字符的字符串。在 我查看了 Levenshtein Distance 的示例，它适用于小列表( http://www.stavros.io/posts/finding-the-levenshtein-distance-in-python/ )，但我的列表 L2 有 163531 个元素，并且在过去的 15 分钟内甚至找不到一个匹配项..

我没有 CS 背景，有人能告诉我一些更好的算法(或者更好的算法实现吗？:))非常感谢。

当前代码(从 stackoverflow 复制链接和其他人):

L1= ["a_b_c","d_e_f"]
L2=["xx""xy_a","xy_b_c","z_d_e","zl_d","z_d_e_y"]

def levenshtein_distance(first, second):
    """Find the Levenshtein distance between two strings."""
    if len(first) > len(second):
        first, second = second, first
    if len(second) == 0:
        return len(first)
    first_length = len(first) + 1
    second_length = len(second) + 1
    distance_matrix = [[0] * second_length for x in range(first_length)]
    for i in range(first_length):
       distance_matrix[i][0] = i
    for j in range(second_length):
       distance_matrix[0][j]=j
    for i in xrange(1, first_length):
        for j in range(1, second_length):
            deletion = distance_matrix[i-1][j] + 1
            insertion = distance_matrix[i][j-1] + 1
            substitution = distance_matrix[i-1][j-1]
            if first[i-1] != second[j-1]:
                substitution += 1
            distance_matrix[i][j] = min(insertion, deletion, substitution)
    return distance_matrix[first_length-1][second_length-1]

for string in L1:
    print sorted(L2,key = lambda x:levenshtein_distance(x,string))[0]

编辑 - 只需按下 control+C，它在 15 分钟后给了我一个不正确(但接近)的答案。那只是第一个字符串，还有很多..

Answer 1

使用 difflib模块:

>>> from functools import partial
>>> from difflib import SequenceMatcher
def func(x, y):
    s = SequenceMatcher(None, x, y)
    return s.find_longest_match(0, len(x), 0, len(y)).size
... 
for item in L1:
    f = partial(func, item)
    print max(L2, key=f)
...     
xy_b_c
z_d_e_y