Python 查找两个列表之间的相似性。根据值在列表中的位置给出单独的权重

我试图找到一个列表与另一个列表的相似性值。比如找到一个句子的jaccard相似度值。但这里唯一的区别是，如果值在两个列表的同一个索引中，那么它得到的是一个静态权重，否则它的权重将根据它离该索引的位置的多少而惩罚

a=["are","you","are","you","why"]
b=['you',"are","you",'are',"why"]
li=[]
va=[]
fi=[]
weightOfStatic=1/len(a)
for i in range(len(a)):    
    if a[i]==b[i]:
    print("true1", weightOfStatic,a[i],b[i])
    fi.append({"static":i, "dynamic":i,"Weight":weightOfStatic})
    li.append([weightOfStatic,a[i],b[i]])
    va.append(li)
else:
     for j in range(len(b)):
         if a[i]==b[j]:
         weightOfDynamic = weightOfStatic*(1-(1/len(b))*abs(i-j))
         fi.append({"static":i, "dynamic":j,"Weight":weightOfDynamic})
         print("true2 and index diiference between words =%d"% abs(i-j),weightOfDynamic, i,j)
         li.append([weightOfDynamic,a[i],b[j]])
         va.append(weightOfDynamic)

sim_value=sum(va)
print("The similarity value is = %f" %(sim_value))

以下代码在没有重复单词的情况下运行良好。
比如a=[“你好”，“你在吗”，“你”] b=[“你”、“是”、“如何”]。 对于这个意义，它给出了0.5的相似性值

上述示例的预期结果将介于列表A和B之间。如果列表A中有重复的单词，则列表A中的值应取其在B中最近的索引。这就是在给定代码的情况下对aboe示例进行匹配的方式

      {'static': 0, 'dynamic': 1, 'Weight': 0.160}
 here 0 should not match with 3 again
      {'static': 0, 'dynamic': 3, 'Weight': 0.079}
      {'static': 1, 'dynamic': 0, 'Weight': 0.160}
 same for 1 and 2
      {'static': 1, 'dynamic': 2, 'Weight': 0.160}
 dynamic 1 is already overhere 
      {'static': 2, 'dynamic': 1, 'Weight': 0.160}
      {'static': 2, 'dynamic': 3, 'Weight': 0.160}
 dynamic 0 is already over
      {'static': 3, 'dynamic': 0, 'Weight': 0.079}
      {'static': 3, 'dynamic': 2, 'Weight': 0.160}
      [0.2, 'why', 'why'] 

此处的重量为1.3200（重量将从0到1）

结果应该是

      {'static': 0, 'dynamic': 1, 'Weight': 0.160}
      {'static': 1, 'dynamic': 0, 'Weight': 0.160}
      {'static': 2, 'dynamic': 3, 'Weight': 0.160}
      {'static': 3, 'dynamic': 2, 'Weight': 0.160}
      [0.2, 'why', 'why'] 

总重量为0.84

首先，我“美化”了您的代码，使其看起来更像蟒蛇我觉得你把事情复杂化了一点。事实上，它甚至没有为我运行，因为你试图对一个包含int和list的列表求和

a = ['are','you','are','you','why']
b = ['you','are','you','are','why']

total_weight = 0
weight_of_static = 1/len(a)
for i, a_word in enumerate(a):
    if a_word == b[i]:
        print('{0} <-> {1} => static\t\t// weight: {2:.2f}'.format(a_word, b[i], weight_of_static))
        total_weight += weight_of_static
    else:
        distances = []
        for j, b_word in enumerate(b):
            if a_word == b_word:
                distances.append(abs(i - j))

        dynamic_weight = weight_of_static*(1 - ( 1 / len(b)) * min(distances))
        total_weight += dynamic_weight
        print('{0} <-> {1} => not static\t// weight: {2:.2f}'.format(a_word, b[i], dynamic_weight))

print('The similarity value is = {0:.2f}'.format(total_weight))

---

我希望这能有所帮助。

嘿，非常感谢。。！

$ python similarity.py
are <-> you => not static       // weight: 0.16
you <-> are => not static       // weight: 0.16
are <-> you => not static       // weight: 0.16
you <-> are => not static       // weight: 0.16
why <-> why => static           // weight: 0.20
The similarity value is = 0.84