Longest common substring between two long lists

I have two really long lists, and I want to find the longest common sub string for each element of the first list in the second list.

A simplified example is

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

So I want to find the best match for "a_b_c" in L2 ("xy_b_c"), then the best match for "d_e_f" in L2("z_d_e_y"). Best match for me is the string with the longest common characters. In I looked at examples for Levenshtein Distance which works for small lists just fine (http://www.stavros.io/posts/finding-the-levenshtein-distance-in-python/), but my list L2 has 163531 elements and It hasn't been able to find even one match for the last 15 minutes..

I do not have a CS background, can someone point me to some better algorithm (or even better, its implementation? :) ) Thanks a ton.

Current code (copied off the link and someone else from stackoverflow):

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]

edit- just hit control+C and it gave me an incorrect(but close) answer after 15 minutes. Thats only for the first string and there's a lot of them left..