问题：如何在不使用“ |”的情况下将两组连接在一起
假定S和T被分配了集合。不使用join运算符|，如何找到两个集合的并集？例如，找到交叉点：
S = {1, 2, 3, 4}
T = {3, 4, 5, 6}
S_intersect_T = { i for i in S if i in T }
那么如何在不使用的情况下在一行中找到两个集合的并集|呢？
Assume that S and T are assigned sets. Without using the join operator |, how can I find the union of the two sets? This, for example, finds the intersection:
S = {1, 2, 3, 4}
T = {3, 4, 5, 6}
S_intersect_T = { i for i in S if i in T }
So how can I find the union of two sets in one line without using |?
回答 0
您可以对集合使用联合方法： set.union(other_set)
请注意，它会返回一个新集合，即不会对其自身进行修改。
You can use union method for sets: set.union(other_set)
Note that it returns a new set i.e it doesn’t modify itself.
回答 1
您可以使用or_别名：
>>> from operator import or_
>>> from functools import reduce # python3 required
>>> reduce(or_, [{1, 2, 3, 4}, {3, 4, 5, 6}])
set([1, 2, 3, 4, 5, 6])
You could use or_ alias:
>>> from operator import or_
>>> from functools import reduce # python3 required
>>> reduce(or_, [{1, 2, 3, 4}, {3, 4, 5, 6}])
set([1, 2, 3, 4, 5, 6])
回答 2
如果您可以修改原始集（在某些情况下可能需要这样做），可以使用set.update()：
S.update(T)
返回值是None，但S将更新为原始S和的并集T。
If you are fine with modifying the original set (which you may want to do in some cases), you can use set.update():
S.update(T)
The return value is None, but S will be updated to be the union of the original S and T.
回答 3
假设您也无法使用s.union(t)，等于s | t，您可以尝试
>>> from itertools import chain
>>> set(chain(s,t))
set([1, 2, 3, 4, 5, 6])
或者，如果您想理解，
>>> {i for j in (s,t) for i in j}
set([1, 2, 3, 4, 5, 6])
Assuming you also can’t use s.union(t), which is equivalent to s | t, you could try
>>> from itertools import chain
>>> set(chain(s,t))
set([1, 2, 3, 4, 5, 6])
Or, if you want a comprehension,
>>> {i for j in (s,t) for i in j}
set([1, 2, 3, 4, 5, 6])
回答 4
如果加入表示您是工会，请尝试以下方法：
set(list(s) + list(t))
这有点骇人听闻，但我想不出更好的衬里来做。
If by join you mean union, try this:
set(list(s) + list(t))
It’s a bit of a hack, but I can’t think of a better one liner to do it.
回答 5
假设您有2个清单
A = [1,2,3,4]
B = [3,4,5,6]
所以你可以找到A联盟B如下
union = set(A).union(set(B))
另外，如果要查找相交和非相交，请按照以下步骤进行操作
intersection = set(A).intersection(set(B))
non_intersection = union - intersection
Suppose you have 2 lists
A = [1,2,3,4]
B = [3,4,5,6]
so you can find A Union B as follow
union = set(A).union(set(B))
also if you want to find intersection and non-intersection you do that as follow
intersection = set(A).intersection(set(B))
non_intersection = union - intersection
回答 6
您可以像这样将两个集合解压缩成一个集合：
>>> set_1 = {1, 2, 3, 4}
>>> set_2 = {3, 4, 5, 6}
>>> union = {*set_1, *set_2}
>>> union
{1, 2, 3, 4, 5, 6}
在*解包集。拆包是将可迭代项（例如集合或列表）表示为它产生的每个项目的地方。这意味着上面的示例简化为{1, 2, 3, 4, 3, 4, 5, 6}，然后简化为，{1, 2, 3, 4, 5, 6}因为该集合只能包含唯一项。
You can just unpack both sets into one like this:
>>> set_1 = {1, 2, 3, 4}
>>> set_2 = {3, 4, 5, 6}
>>> union = {*set_1, *set_2}
>>> union
{1, 2, 3, 4, 5, 6}
The * unpacks the set. Unpacking is where an iterable (e.g. a set or list) is represented as every item it yields. This means the above example simplifies to {1, 2, 3, 4, 3, 4, 5, 6} which then simplifies to {1, 2, 3, 4, 5, 6} because the set can only contain unique items.
回答 7
您可以做union或简单的列表理解
[A.add(_) for _ in B]
A将拥有B的所有元素
You can do union or simple list comprehension
[A.add(_) for _ in B]
A would have all the elements of B
