如何在不使用“ |”的情况下将两组连接在一起

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 |?

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.

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])

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.

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])

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.

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

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.

You can do union or simple list comprehension

[A.add(_) for _ in B]

A would have all the elements of B