CODEWITHSUNDEEP
pythonpython-3.x
Shh
Shh

Reputation: 1006

Python: union of set of tuples

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.

Upvotes: 2

Views: 1646

Answers (6)

cyberspace
cyberspace

Reputation: 211

A little bit shorter version: 

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

Output: 

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

Upvotes: 2

Laurent H.
Laurent H.

Reputation: 6536

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)}

Upvotes: 0

Dani Mesejo
Dani Mesejo

Reputation: 61930

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)}

Upvotes: 0

blue note
blue note

Reputation: 29099

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.

Upvotes: 0

LetzerWille
LetzerWille

Reputation: 5668

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)}

Upvotes: 1

javidcf
javidcf

Reputation: 59741

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)}

Upvotes: 6

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