从元组列表中获取交集 [英] Get intersection from list of tuples

问题描述

我有两个元组列表

a = [('head1','a'),('head2','b'),('head3','x'),('head4','z')]
b = [('head5','u'),('head6','w'),('head7','x'),('head8','y'),('head9','z')]

我想使用每个元组的第二个元素的交集，例如set {a[0][0],a[0][1],a[0][2],a[0][3]}与列表a和b中的set {b[0][0],b[0][1],b[0][2],b[0][3],b[0][4]}的交集，以便如果交集值返回元组的第一个元素映射存在.结果输出应如下所示:

I want to take the intersection of 2nd element of each tuple for example set {a[0][0],a[0][1],a[0][2],a[0][3]} intersection with set {b[0][0],b[0][1],b[0][2],b[0][3],b[0][4]} from list a and b such that it returns the first element mapping of the tuple if intersection value exist. The resulting output should be like following:

res = [('head3','head7'),('head4','head9')]

到目前为止，我已经尝试过:

So far I have tried this:

x = [(a[i][0],b[j][0]) for i in range(len(a)) for j in range(len(b)) if a[0][i] == b[0][j]]

但出现错误IndexError: tuple index out of range

这样做的正确，最快的方法是什么?

What could be the correct and fastest way of doing this ?

推荐答案

您可以在Python 3中执行以下操作.从列表中创建字典，并从两个字典中获取键的交集，然后从键中获取相应的值:

You can do the following in Python 3. Create dicts from your lists, taking the intersection of keys from both dicts, fetch the corresponding values at the key:

>>> da = {k:v for v, k in a}
>>> db = {k:v for v, k in b}
>>> [(da[k], db[k])  for k in da.keys()&db.keys()]
[('head4', 'head9'), ('head3', 'head7')]

在Python 2中，可以使用set(da).intersection(db)代替da.keys()&db.keys().

In Python 2, you can use set(da).intersection(db) in place of da.keys()&db.keys().

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