List Tuple Representing

Question

Let's say , I have following even number names list and should return representing pairs as result :

['A', 'B', 'C', 'D']

>>> [[('B', 'C'), ('A', 'D')],
     [('A', 'B'), ('C', 'D')],
     [('A', 'C'), ('B', 'D')]]

I wrote the following code :

import itertools
combinations = list(itertools.combinations(['A', 'B', 'C', 'D'], 2))
result = []

for i in range( 0 , len(combinations) ):
    if (combinations[i-1][0] != combinations[i][0]) & (combinations[i-1][0] != combinations[i][1])  :
        if (combinations[i-1][1] != combinations[i][0]) & (combinations[i-1][1] != combinations[i][1]) :
            zipped = zip(combinations[i], combinations[i-1])
            result.append(list(zipped))
result

But it gives following as a result ;

[[('A', 'C'), ('B', 'D')],
 [('B', 'A'), ('C', 'D')]]

What is the missing point in my code ?

Isn't the original `combinations` what you want? I don't see what the `result` filtering does for you. — hpaulj, Oct 17 '21 at 23:22
No, `combinations` just delivers two at a time. He needs the others as well. And the problem gets trickier when extended to more than 4 names. — Tim Roberts, Oct 17 '21 at 23:30
Does this answer your question? [How to split a list into pairs in all possible ways](https://stackoverflow.com/questions/5360220/how-to-split-a-list-into-pairs-in-all-possible-ways) — hilberts_drinking_problem, Oct 18 '21 at 07:37

score 2 · Accepted Answer · answered Oct 17 '21 at 23:37

This is a tricky little problem, especially when extended to more than four names. What I've done here is to create all the permutations of the four names. Then, to determine uniqueness, I sort the pairs, and sort the list of pairs. I then keep the unique ones.

import itertools
gather = []
for sets in itertools.permutations(['Andrea', 'Bob', 'Cassandra', 'Doug']):
    pairs = sorted([sorted(p) for p in zip(sets[0::2],sets[1::2])])
    if pairs not in gather:
        gather.append(pairs)
from pprint import pprint
pprint(gather)

Output for four names:

[[['Andrea', 'Bob'], ['Cassandra', 'Doug']],
 [['Andrea', 'Cassandra'], ['Bob', 'Doug']],
 [['Andrea', 'Doug'], ['Bob', 'Cassandra']],

Output for six names:

[[['Andrea', 'Bob'], ['Cassandra', 'Doug'], ['Ethel', 'Fred']],
 [['Andrea', 'Bob'], ['Cassandra', 'Ethel'], ['Doug', 'Fred']],
 [['Andrea', 'Bob'], ['Cassandra', 'Fred'], ['Doug', 'Ethel']],
 [['Andrea', 'Cassandra'], ['Bob', 'Doug'], ['Ethel', 'Fred']],
 [['Andrea', 'Cassandra'], ['Bob', 'Ethel'], ['Doug', 'Fred']],
 [['Andrea', 'Cassandra'], ['Bob', 'Fred'], ['Doug', 'Ethel']],
 [['Andrea', 'Doug'], ['Bob', 'Cassandra'], ['Ethel', 'Fred']],
 [['Andrea', 'Doug'], ['Bob', 'Ethel'], ['Cassandra', 'Fred']],
 [['Andrea', 'Doug'], ['Bob', 'Fred'], ['Cassandra', 'Ethel']],
 [['Andrea', 'Ethel'], ['Bob', 'Cassandra'], ['Doug', 'Fred']],
 [['Andrea', 'Ethel'], ['Bob', 'Doug'], ['Cassandra', 'Fred']],
 [['Andrea', 'Ethel'], ['Bob', 'Fred'], ['Cassandra', 'Doug']],
 [['Andrea', 'Fred'], ['Bob', 'Cassandra'], ['Doug', 'Ethel']],
 [['Andrea', 'Fred'], ['Bob', 'Doug'], ['Cassandra', 'Ethel']],
 [['Andrea', 'Fred'], ['Bob', 'Ethel'], ['Cassandra', 'Doug']]]

List Tuple Representing

1 Answers1