更新时间:2023-10-22 14:05:58
对于2.4,你可以只定义一个交集函数.
def intersect(*d):集合 = iter(map(set, d))结果 = set.next()对于 s 集合:结果 = result.intersection(s)返回结果
对于较新版本的python:
交集方法接受任意数量的参数
result = set(d[0]).intersection(*d[1:])
或者,您可以将第一个集合与其自身相交以避免对列表进行切片和复制:
result = set(d[0]).intersection(*d)
我不确定哪个更有效,并且感觉这取决于 d[0]
的大小和列表的大小,除非 python 有内置检查因为喜欢
如果 s1 是 s2:返回 s1
在交集方法中.
I am playing with python and am able to get the intersection of two lists:
result = set(a).intersection(b)
Now if d
is a list containing a
and b
and a third element c
, is there an built-in function for finding the intersection of all the three lists inside d
? So for instance,
d = [[1,2,3,4], [2,3,4], [3,4,5,6,7]]
then the result should be
[3,4]
for 2.4, you can just define an intersection function.
def intersect(*d):
sets = iter(map(set, d))
result = sets.next()
for s in sets:
result = result.intersection(s)
return result
for newer versions of python:
the intersection method takes an arbitrary amount of arguments
result = set(d[0]).intersection(*d[1:])
alternatively, you can intersect the first set with itself to avoid slicing the list and making a copy:
result = set(d[0]).intersection(*d)
I'm not really sure which would be more efficient and have a feeling that it would depend on the size of the d[0]
and the size of the list unless python has an inbuilt check for it like
if s1 is s2:
return s1
in the intersection method.
>>> d = [[1,2,3,4], [2,3,4], [3,4,5,6,7]]
>>> set(d[0]).intersection(*d)
set([3, 4])
>>> set(d[0]).intersection(*d[1:])
set([3, 4])
>>>