2个不同类型的对象的交点

发布于 2025-01-21 00:42:30 字数 516 浏览 2 评论 0原文

我想进行：：：：但是这2个类型的交叉点

(('e', 13.02338360095244), 
 ('a', 11.820318700775383), 
 ('o', 9.20172171683253), 
 ('s', 7.635081506807498), 
 ('n', 7.547469320471335), 
 ('i', 7.219915745772025), 
 ('r', 6.704927040722877), 
 ('l', 5.650833384211491), 
 ('d', 5.098296599303987), 
 ('t', 4.7109103119848585))

，

[('e', 1636), ('n', 991), ('a', 930), ('t', 678), ('o', 677), ('r', 612), ('d', 581), ('i', 507), ('l', 405), ('s', 399)]

我该如何进行十字路口并显示交集中的元素量？

原文

i want to take the intersection of:

(('e', 13.02338360095244), 
 ('a', 11.820318700775383), 
 ('o', 9.20172171683253), 
 ('s', 7.635081506807498), 
 ('n', 7.547469320471335), 
 ('i', 7.219915745772025), 
 ('r', 6.704927040722877), 
 ('l', 5.650833384211491), 
 ('d', 5.098296599303987), 
 ('t', 4.7109103119848585))

and:

[('e', 1636), ('n', 991), ('a', 930), ('t', 678), ('o', 677), ('r', 612), ('d', 581), ('i', 507), ('l', 405), ('s', 399)]

but these 2 have different types, how do i go about taking the intersection and displaying the amount of elements in the intersection?

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橘和柠 2025-01-28 00:42:30

您可以集合的交集。例如：

a = (('e', 13.02338360095244), 
 ('a', 11.820318700775383), 
 ('o', 9.20172171683253), 
 ('s', 7.635081506807498), 
 ('n', 7.547469320471335), 
 ('i', 7.219915745772025), 
 ('r', 6.704927040722877), 
 ('l', 5.650833384211491), 
 ('d', 5.098296599303987), 
 ('t', 4.7109103119848585))
b = [('e', 1636), ('n', 991), ('a', 930), ('t', 678), ('o', 677), ('r', 612), ('d', 581), ('i', 507), ('l', 405), ('s', 399)]

set(k for k,_ in a).intersection(k for k,_ in b)

输出将为 {'i', 'a', 's', 'n', 'r', 'l', 't', 'o', 'd', 'e'}。

You can you the intersection of set. For example:

a = (('e', 13.02338360095244), 
 ('a', 11.820318700775383), 
 ('o', 9.20172171683253), 
 ('s', 7.635081506807498), 
 ('n', 7.547469320471335), 
 ('i', 7.219915745772025), 
 ('r', 6.704927040722877), 
 ('l', 5.650833384211491), 
 ('d', 5.098296599303987), 
 ('t', 4.7109103119848585))
b = [('e', 1636), ('n', 991), ('a', 930), ('t', 678), ('o', 677), ('r', 612), ('d', 581), ('i', 507), ('l', 405), ('s', 399)]

set(k for k,_ in a).intersection(k for k,_ in b)

Output will be {'i', 'a', 's', 'n', 'r', 'l', 't', 'o', 'd', 'e'}.

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