如何将元组列表中的唯一元素作为另一个元组返回

Question

Assume I have list of tuples a=[(0,1),(2,0),(1,4)] and I want to return unique element of each two tuples as a new tuple. For example (0,1) and (2,0) returns (1,2) . Also (0,1) and (1,4) return (0,4) .

Therefore, output is unique=[(1,2),(0,4)]

I tried code below but it seems I am not in correct path:

from itertools import combinations
a=[(0,1),(2,0),(1,4)]

b=list(combinations(a,2))
def myComp(pair1, pair2):
    if any(x == y for x, y in zip(pair1, pair2)):
        return(pair1,pair2)

d=[]
for i in range(len(b)):
    c=myComp(b[i][0],b[i][1])
    d.append(c) 

Answer 1

First, convert all the tuples in a to sets.

a2 = [set(i) for i in a]

Then, take combinations of two elements of a2 .

b2 = itertools.combinations(a2, 2)
# print(list(b2)) gives:
# [({0, 1}, {0, 2}), ({0, 1}, {1, 4}), ({0, 2}, {1, 4})]

Then, for every pair in b2 , find the symmetric difference.

answer = [tuple(x ^ y) for x, y in b2]
# answer: [(1, 2), (0, 4), (0, 1, 2, 4)]

Like Ironkey mentions in the comments, you get (0, 1, 2, 4) in the last element because you're comparing (0, 2) and (1, 4) , and all elements in these tuples are mutually exclusive.

To filter out the four-element tuples, you can simply add that condition:

answer_f = [x for x in answer if len(x) == 2]
# answer_f: [(1, 2), (0, 4)]

Answer 2

crude answer without any list comps:

from itertools import combinations

a = [(0,1),(0,2),(0,3),(1,2),(1,3),(1,4),(2,3)]

uniques = []
for x, y in combinations(a,2):
    z = []
    for i in x:
        if i not in y:
           z.append(i)
    for i in y:
        if i not in x:
           z.append(i)
    if len(z) == 2:
        uniques.append(tuple(z))

print(list(set(uniques)))

[(0, 1), (2, 4), (1, 2), (0, 4), (3, 4), (0, 3), (2, 3), (0, 2), (1, 3)]

Answer 3

This doesn't use packages.

The conditions if i.= x and a.index(i)<a.index(x) make sure we aren't comparing identical tuples and also we are comparing the same pair of tuples once

Then we just grab the element of they are not in the other tuple

The condition if len(m) == 2 and sorted(m) in uniq makes sure that the list is only 2 elements long, and also solves your comment about having (2,1) & (1,2)

a=[(0,1),(2,0),(1,4),(0,2)]

uniq = []
for i in a:
  for x in a:
    if i != x and a.index(i)<a.index(x):
      m = [y for y in i if y not in x] + [z for z in x if z not in i]
      if len(m) == 2 and tuple(sorted(m)) not in uniq:
        uniq.append(tuple(m))

print(uniq)

>>> [(1, 2), (0, 4)]