计算单词列表之间的相似度

Question

我想计算两个单词列表之间的相似度，例如：

['email','user','this','email','address','customer']

类似于此列表：

['email','mail','address','netmail']

例如，我希望比其他列表具有更高的相似性百分比：['address','ip','network']即使 address 该列表中存在相似性
。

Answer 1

由于您实际上还无法演示晶体输出，因此以下是我的最佳镜头：

list_A = ['email','user','this','email','address','customer']
list_B = ['email','mail','address','netmail']

在上面的两个列表中，我们将找到列表中每个元素与其余元素之间的余弦相似度。即email从list_B与每一个元素list_A：

def word2vec(word):
    from collections import Counter
    from math import sqrt

    # count the characters in word
    cw = Counter(word)
    # precomputes a set of the different characters
    sw = set(cw)
    # precomputes the "length" of the word vector
    lw = sqrt(sum(c*c for c in cw.values()))

    # return a tuple
    return cw, sw, lw

def cosdis(v1, v2):
    # which characters are common to the two words?
    common = v1[1].intersection(v2[1])
    # by definition of cosine distance we have
    return sum(v1[0][ch]*v2[0][ch] for ch in common)/v1[2]/v2[2]


list_A = ['email','user','this','email','address','customer']
list_B = ['email','mail','address','netmail']

threshold = 0.80     # if needed
for key in list_A:
    for word in list_B:
        try:
            # print(key)
            # print(word)
            res = cosdis(word2vec(word), word2vec(key))
            # print(res)
            print("The cosine similarity between : {} and : {} is: {}".format(word, key, res*100))
            # if res > threshold:
            #     print("Found a word with cosine distance > 80 : {} with original word: {}".format(word, key))
        except IndexError:
            pass

输出 ：

The cosine similarity between : email and : email is: 100.0
The cosine similarity between : mail and : email is: 89.44271909999159
The cosine similarity between : address and : email is: 26.967994498529684
The cosine similarity between : netmail and : email is: 84.51542547285166
The cosine similarity between : email and : user is: 22.360679774997898
The cosine similarity between : mail and : user is: 0.0
The cosine similarity between : address and : user is: 60.30226891555272
The cosine similarity between : netmail and : user is: 18.89822365046136
The cosine similarity between : email and : this is: 22.360679774997898
The cosine similarity between : mail and : this is: 25.0
The cosine similarity between : address and : this is: 30.15113445777636
The cosine similarity between : netmail and : this is: 37.79644730092272
The cosine similarity between : email and : email is: 100.0
The cosine similarity between : mail and : email is: 89.44271909999159
The cosine similarity between : address and : email is: 26.967994498529684
The cosine similarity between : netmail and : email is: 84.51542547285166
The cosine similarity between : email and : address is: 26.967994498529684
The cosine similarity between : mail and : address is: 15.07556722888818
The cosine similarity between : address and : address is: 100.0
The cosine similarity between : netmail and : address is: 22.79211529192759
The cosine similarity between : email and : customer is: 31.62277660168379
The cosine similarity between : mail and : customer is: 17.677669529663685
The cosine similarity between : address and : customer is: 42.640143271122085
The cosine similarity between : netmail and : customer is: 40.08918628686365

注意：我也已threshold在代码中注释了该部分，以防万一您只需要单词的相似度超过某个阈值（即80％）

编辑 ：

OP ： 但是我想做的不是逐字比较，而是逐个列出

使用Counter和math：

from collections import Counter
import math

counterA = Counter(list_A)
counterB = Counter(list_B)


def counter_cosine_similarity(c1, c2):
    terms = set(c1).union(c2)
    dotprod = sum(c1.get(k, 0) * c2.get(k, 0) for k in terms)
    magA = math.sqrt(sum(c1.get(k, 0)**2 for k in terms))
    magB = math.sqrt(sum(c2.get(k, 0)**2 for k in terms))
    return dotprod / (magA * magB)

print(counter_cosine_similarity(counterA, counterB) * 100)

输出 ：

53.03300858899106