Intersection and difference of multiple sets

Question

I have multiple lists of "associations" as input:

input1 = [
    ('id1', 'id2', 'id3', 'id4'),
    ('id5',),
    ('id6', 'id7', 'id8')
]
input2 = [
    ('id1', 'id2', 'id4'),
    ('id3',),
    ('id5',),
    ('id7', 'id6', 'id8')
]
input3 = [
    ('id1', 'id2'),
    ('id3', 'id4'),
    ('id5',),
    ('id8', 'id7', 'id6')
]

Each input contains all the ids.
Each id appears a single time in an input.
The number of inputs can vary from 1 to 4.

I would like to process those inputs and generate something like this:

assocs = {
    # never associated
    0: [
        ('id5',),
    ],
    # associated a single time
    1: [
        ('id1', 'id3'),
        ('id2', 'id3'),
    ],
    # associated twice
    2: [ # 
        ('id1', 'id4'),
        ('id2', 'id4'),
        ('id3', 'id4'),
    ],
    # associated everytime
    3: [
        ('id1', 'id2'),
        ('id6', 'id7', 'id8'),
    ]
}

I can somehow count the number of times that each association is encountered with:

temp = defaultdict(int)

for ids in map(sorted, itertools.chain(input1, input2, input3)):
    for i in range(1, len(ids)+1):
        for comb in itertools.combinations(ids, i):
            temp[comb] += 1

But now I’m stuck with the following step, which would be cleaning the temp dict; not to mention that I’m not sure that I chose the right strategy in the first place.

Asked By: Fravadona

||

Source

Answer 1

The logic is not fully clear to me, but it looks like you could generate the powersets of each tuple and count them, handling the loners specifically:

from itertools import chain, combinations
from collections import Counter

inputs = [input1, input2, input3]

def powerset(iterable, min=2):
    # modified powerset recipe to yield at least pairs
    s = list(iterable)
    return chain.from_iterable(combinations(s, r) for r in range(min, len(s)+1))

cnt = Counter(tuple(sorted(s)) for lst in inputs for t in lst
              for s in (powerset(t) if len(t)>1 else (t,)))
out = {}
for k, c in cnt.items():
    out.setdefault(c if len(k)>1 else 0, []).append(k)

Output:

{0: [('id5',),
     ('id3',)],
 1: [('id1', 'id3'),
     ('id2', 'id3'),
     ('id1', 'id2', 'id3'),
     ('id1', 'id3', 'id4'),
     ('id2', 'id3', 'id4'),
     ('id1', 'id2', 'id3', 'id4')],
 2: [('id1', 'id4'),
     ('id2', 'id4'),
     ('id3', 'id4'),
     ('id1', 'id2', 'id4')],
 3: [('id1', 'id2'),
     ('id6', 'id7'),
     ('id6', 'id8'),
     ('id7', 'id8'),
     ('id6', 'id7', 'id8')]}

Different way of handling the loners:

cnt = Counter(tuple(sorted(s)) for lst in inputs for t in lst
              for s in powerset(t))

out = {0: [(x,) for x in set().union(*input1) - set().union(*cnt.keys())]}
for k, c in cnt.items():
    out.setdefault(c if len(k)>1 else 0, []).append(k)

output:

{0: [('id5',)],
 1: [('id1', 'id3'),
     ('id2', 'id3'),
     ('id1', 'id2', 'id3'),
     ('id1', 'id3', 'id4'),
     ('id2', 'id3', 'id4'),
     ('id1', 'id2', 'id3', 'id4')],
 2: [('id1', 'id4'),
     ('id2', 'id4'),
     ('id3', 'id4'),
     ('id1', 'id2', 'id4')],
 3: [('id1', 'id2'),
     ('id6', 'id7'),
     ('id6', 'id8'),
     ('id7', 'id8'),
     ('id6', 'id7', 'id8')]}

Answered By: mozway

Intersection and difference of multiple sets

Question:

Answers: