如何在 Python 中计算包含字符串的两个列表的 Jaccard 相似度？
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How can I calculate the Jaccard Similarity of two lists containing strings in Python?
提问by Aventinus
I have two lists with usernames and I want to calculate the Jaccard similarity. Is it possible?
我有两个带有用户名的列表，我想计算 Jaccard 相似度。是否可以？
Thisthread shows how to calculate the Jaccard Similarity between two strings, however I want to apply this to two lists, where each element is one word (e.g., a username).
该线程显示了如何计算两个字符串之间的 Jaccard 相似度，但是我想将其应用于两个列表，其中每个元素都是一个单词（例如，用户名）。
回答by Aventinus
I ended up writing my own solution after all:
毕竟我最终编写了自己的解决方案：
def jaccard_similarity(list1, list2):
intersection = len(list(set(list1).intersection(list2)))
union = (len(list1) + len(list2)) - intersection
return float(intersection) / union
回答by iamlcc
@aventinus I don't have enough reputation to add a comment to your answer, but just to make things clearer, your solution measures the jaccard_similaritybut the function is misnamed as jaccard_distance, which is actually 1 - jaccard_similarity
@aventinus 我没有足够的声誉来为您的答案添加评论，但只是为了让事情更清楚，您的解决方案测量了jaccard_similarity但函数被错误命名为jaccard_distance，这实际上是1 - jaccard_similarity
回答by w4bo
For Python 3:
对于 Python 3：
def jaccard_similarity(list1, list2):
s1 = set(list1)
s2 = set(list2)
return len(s1.intersection(s2)) / len(s1.union(s2))
list1 = ['dog', 'cat', 'cat', 'rat']
list2 = ['dog', 'cat', 'mouse']
jaccard_similarity(list1, list2)
>>> 0.5
For Python2 use return len(s1.intersection(s2)) / float(len(s1.union(s2)))
对于 Python2 使用 return len(s1.intersection(s2)) / float(len(s1.union(s2)))
回答by klaus
Assuming your usernames don't repeat, you can use the same idea:
假设您的用户名不重复，您可以使用相同的想法：
def jaccard(a, b):
c = a.intersection(b)
return float(len(c)) / (len(a) + len(b) - len(c))
list1 = ['dog', 'cat', 'rat']
list2 = ['dog', 'cat', 'mouse']
# The intersection is ['dog', 'cat']
# union is ['dog', 'cat', 'rat', 'mouse]
words1 = set(list1)
words2 = set(list2)
jaccard(words1, words2)
>>> 0.5
回答by LaSul
回答by Erwin Scholtens
@Aventinus (I also cannot comment): Note that Jaccard similarityis an operation on sets, so in the denominator part it should also use sets (instead of lists). So for example jaccard_similarity('aa', 'ab')should result in 0.5.
@Aventinus（我也无法评论）：请注意，Jaccard相似度是对集合的运算，因此在分母部分它也应该使用集合（而不是列表）。因此，例如jaccard_similarity('aa', 'ab')应该导致0.5.
def jaccard_similarity(list1, list2):
intersection = len(set(list1).intersection(list2))
union = len(set(list1)) + len(set(list2)) - intersection
return intersection / union
Note that in the intersection, there is no need to cast to list first. Also, the cast to float is not needed in Python 3.
注意，在交集的时候，不需要先cast to list。此外，Python 3 中不需要浮点型转换。
回答by kd88
If you'd like to include repeated elements, you can use Counter, which I would imagine is relatively quick since it's just an extended dictunder the hood:
如果您想包含重复的元素，可以使用Counter，我认为它相对较快，因为它只是dict引擎盖下的扩展：
from collections import Counter
def jaccard_repeats(a, b):
"""Jaccard similarity measure between input iterables,
allowing repeated elements"""
_a = Counter(a)
_b = Counter(b)
c = (_a - _b) + (_b - _a)
n = sum(c.values())
return n/(len(a) + len(b) - n)
list1 = ['dog', 'cat', 'rat', 'cat']
list2 = ['dog', 'cat', 'rat']
list3 = ['dog', 'cat', 'mouse']
jaccard_repeats(list1, list3)
>>> 0.75
jaccard_repeats(list1, list2)
>>> 0.16666666666666666
jaccard_repeats(list2, list3)
>>> 0.5
