通过 difflib 模块比较 python 中的列表

发布于 2024-10-27 12:44:33 字数 456 浏览 0 评论 0原文

我正在尝试 difflib 库。我有两个列表：L_1 和 L_2 包含字符串。我想知道这些序列是否相似（顺序并不重要）。

L_1 = ["Bob", "Mary", "Hans"]
L_2 = ["Bob", "Marie", "Háns"]

应该没问题。但

L_1 = ["Nirdosch", "Mary", "Rolf"]
L_2 = ["Bob", "Marie", "Háns"]

应该不行。

我提出了迭代第一个列表 L_1 并通过该方法与

difflib.get_close_matches()

第二个列表 L_2 匹配 L_1 的每个元素的想法。如果存在比值更大的匹配项，假设为 0.7，则将其从 L_2 中删除并继续。但我怀疑这是否是一个好的计划。还有更好的吗？

原文

I am trying out the difflib library. I have two lists: L_1 and L_2 containing strings. I want to know, if those sequences are similar (order is not important).

L_1 = ["Bob", "Mary", "Hans"]
L_2 = ["Bob", "Marie", "Háns"]

should be ok. But

L_1 = ["Nirdosch", "Mary", "Rolf"]
L_2 = ["Bob", "Marie", "Háns"]

should not be ok.

I came up with the idea of iterating over the first list L_1 and to match every element of L_1 by the method

difflib.get_close_matches()

against the second list L_2. If there was a match with a ratio bigger then let's say 0.7 remove it from L_2 and continue. But I doubt it is a good plan. Is there a better one?

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

三生路 2024-11-03 12:44:33

我会做类似的事情：

import difflib

L_1 = ["Bob", "Mary", "Hans"]
L_2 = ["Bob", "Marie", "Hans"]

def similiarity(L_1, L_2):
    L_1 = set(intern(w) for w in L_1)
    L_2 = set(intern(w) for w in L_2)

    to_match = L_1.difference( L_2)
    against = L_2.difference(L_1)
    for w in to_match:
        res = difflib.get_close_matches(w, against)
        if len(res):
            against.remove( res[0] )
    return (len(L_2)-len(against)) / (len(L_1))

print similiarity(L_1,L_2)

I would do something like:

import difflib

L_1 = ["Bob", "Mary", "Hans"]
L_2 = ["Bob", "Marie", "Hans"]

def similiarity(L_1, L_2):
    L_1 = set(intern(w) for w in L_1)
    L_2 = set(intern(w) for w in L_2)

    to_match = L_1.difference( L_2)
    against = L_2.difference(L_1)
    for w in to_match:
        res = difflib.get_close_matches(w, against)
        if len(res):
            against.remove( res[0] )
    return (len(L_2)-len(against)) / (len(L_1))

print similiarity(L_1,L_2)

回复收藏 0 原文

~没有更多了~