Python：元组集合的联合

Question

Let's say we have two sets:

t = {('b', 3), ('a', 2)}
r = {('b', 4), ('c', 6)}

I want a union on 1st element to result in

u = {('b', 3), ('a', 2), ('c', 6)}

if duplicate symbol is present in both place (example 'b' in the above) then the element of the first list should be retained. Thanks.

Answer 1

Just do:

t = {('b', 3), ('a', 2)}
r = {('b', 4), ('c', 6)}
d = dict(r)
d.update(t)
u = set(d.items())
print(u)

Output:

{('c', 6), ('a', 2), ('b', 3)}

Answer 2

A little bit shorter version:

s = dict((*r, *t))
set(s.items())

Output:

{('a', 2), ('b', 3), ('c', 6)}

Answer 3

for el in r:
    if not el[0] in [x[0] for x in t]:
        t.add(el)

t 

{('a', 2), ('b', 3), ('c', 6)}

Answer 4

You can't do that with set intersecion. Two objects are either equal or they are not. Since your objects are tuples, (b, 3) and (b, 4) are not equal, and you don't get to change that.

The obvious way would be to create your own class and redefine equality, something like

class MyTuple:
    def __init__(self, values):
         self.values = values

    def __eq__(self, other):
        return self.values[0] == other[0]

and create sets of such objects.

Answer 5

Here is my one-line style solution based on comprehensions:

t = {('b', 3), ('a', 2)}
r = {('b', 4), ('c', 6)}

result = {*t, *{i for i in r if i[0] not in {j[0] for j in t}}}

print(result)  # {('b', 3), ('a', 2), ('c', 6)}

Using conversion to dictionary to eliminate the duplicates, you can also do that, which is a quite smart solution IMHO:

t = {('b', 3), ('a', 2)}
r = {('b', 4), ('c', 6)}

result = {(k,v) for k,v in dict((*r,*t)).items()}

print(result)  # {('b', 3), ('a', 2), ('c', 6)}

Answer 6

An alternative using chain :

from itertools import chain

t = {('b', 3), ('a', 2)}
r = {('b', 4), ('c', 6)}

result = set({k: v for k, v in chain(r, t)}.items())

Output

{('b', 3), ('a', 2), ('c', 6)}