let say I have a list of lists
[[a1, a2, a3], [b1, b2], [c1, c2, c3, c4]]
The number of lists in the list is not known in advance.
I want to have all combinations of elements from the different list, so
[a1, b1, c1], [a1, b1, c2], ..., [a3, b2, c4]
but if there common elements in the different list, all these combinations need to be deleted. So if for example, a1 = c2
, then the combinations [a1, b1, c2], [a1, b2, c2]
need to be deleted in the resulting list.
To get all possible combinations, you can use the answer on All possible permutations of a set of lists in Python , but can you automaticaly delete all combinations with common elements?
You are looking for the Cartesian Product of your lists. Use itertools.product()
, and filter the elements to make sure none are equal:
from itertools import product
for combo in product(*input_lists):
if len(set(combo)) != len(combo): # not all values unique
continue
print(*combo)
I'm assuming that by a1 = c2
you mean that all values in the combination need to be unique , the above tests for this by creating a set from the combination. If the set length is smaller than the combination length, you had repeated values.
You can put this filter into a generator function:
def unique_product(*l, repeat=None):
for combo in product(*l, repeat=repeat):
if len(set(combo)) == len(combo): # all values unique
yield combo
then use for unique in unique_product(*input_lists):
You can also use the filter()
function to achieve the same, but that incurs a function call for each combination produced.
As the other guys said, you can do with itertools but you may need to drop duplicates:
import itertools
L = [1,2,3,4]
combos = list(itertools.combinations(L, 2))
pairs = [[x, y] for x in combos for y in combos if not set(x).intersection(set(y))]
list(set(sum(pairs, [])))
And then the you will have this as output:
[(1, 2), (1, 3), (1, 4), (2, 3), (3, 4), (2, 4)]
[EDIT]
Inspired on the answer provided here: https://stackoverflow.com/a/42924469/8357763
1) itertools.product
all_combinations = itertools.product(elements)
2) filter
with lambda
filtered_combinations = filter(lambda x: len(x) != len(set(x)), all_combinations)
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