Unique combinations of a list of tuples
You can use recursion with a generator:
data = [(1,2,3), (4,5,6), (7,8,9)]
def combos(d, c = []):
if len(c) == len(d):
yield c
else:
for i in d:
if i not in c:
yield from combos(d, c+[i])
def product(d, c = []):
if c:
yield tuple(c)
if d:
for i in d[0]:
yield from product(d[1:], c+[i])
result = sorted({i for b in combos(data) for i in product(b)})
final_result = [a for i, a in enumerate(result) if all(len(c) != len(a) or len(set(c)&set(a)) != len(a) for c in result[:i])]
Output:
[(1,), (1, 4), (1, 4, 7), (1, 4, 8), (1, 4, 9), (1, 5), (1, 5, 7), (1, 5, 8), (1, 5, 9), (1, 6), (1, 6, 7), (1, 6, 8), (1, 6, 9), (1, 7), (1, 8), (1, 9), (2,), (2, 4), (2, 4, 7), (2, 4, 8), (2, 4, 9), (2, 5), (2, 5, 7), (2, 5, 8), (2, 5, 9), (2, 6), (2, 6, 7), (2, 6, 8), (2, 6, 9), (2, 7), (2, 8), (2, 9), (3,), (3, 4), (3, 4, 7), (3, 4, 8), (3, 4, 9), (3, 5), (3, 5, 7), (3, 5, 8), (3, 5, 9), (3, 6), (3, 6, 7), (3, 6, 8), (3, 6, 9), (3, 7), (3, 8), (3, 9), (4,), (4, 7), (4, 8), (4, 9), (5,), (5, 7), (5, 8), (5, 9), (6,), (6, 7), (6, 8), (6, 9), (7,), (8,), (9,)]
Here is a non-recursive solution with a simple for
loop. Uniqueness is enforced by applying set
to the list of the outputted tuples.
lsts = [(1,2,3), (4,5,6), (7,8,9)]
res = [[]]
for lst in lsts:
res += [(*r, x) for r in res for x in lst]
# print({tuple(lst) for lst in res[1:]})
# {(5, 9), (4, 7), (6, 9), (1, 4, 7), (2, 6, 9), (4, 8), (3, 4, 7), (2,
# 8), (2, 6, 8), (9,), (2, 5, 8), (1, 6), (3, 6, 8), (2, 5, 9), (3, 5,
# 9), (3, 7), (2, 5), (3, 6, 9), (5, 8), (1, 6, 8), (3, 5, 8), (2, 6,
# 7), (4, 9), (6, 7), (1,), (2, 9), (1, 6, 9), (3,), (1, 5), (5,), (3,
# 6), (7,), (3, 6, 7), (1, 5, 9), (2, 6), (2, 4, 7), (1, 5, 8), (3, 4,
# 8), (8,), (3, 4, 9), (1, 4), (1, 6, 7), (3, 9), (1, 9), (2, 5, 7), (3,
# 5), (2, 7), (2, 4, 9), (6, 8), (1, 5, 7), (2,), (2, 4, 8), (5, 7), (1,
# 4, 8), (3, 5, 7), (4,), (3, 8), (1, 8), (1, 4, 9), (6,), (1, 7), (3,
# 4), (2, 4)}
Using itertools:
import itertools as it
def all_combinations(groups):
result = set()
for prod in it.product(*groups):
for length in range(1, len(groups) + 1):
result.update(it.combinations(prod, length))
return result
all_combinations([(1,2,3), (4,5,6), (7,8,9)])