Reputation: 4130
I want to iterate over a list of lists, and pull all the combinations from the lists.
I can do it in the following way:
list = [['1','2','3'],['a','b','c'],['x','y','z']]
for itemi in list[0]:
print itemi
for itemj in list[1]:
print itemi+itemj
for itemk in list[2]:
print itemi+itemj+itemk
My problem is that I want to do it over a varying number of lists in the list o' lists. There will be at first just one list, and (for now) end with 64 lists.
The above example is for 3 lists, and in reality all the lists contain the same values (zero to 255). I used the above examples just so you can see what the output should look like.
I figured there must be a better way to this, without having to build the nested for loops for each size of list of lists.
Upvotes: 2
Views: 916
Reputation: 336258
Use itertools.product()
:
>>> l = [['1','2','3'],['a','b','c'],['x','y','z']]
>>> import itertools
>>> list(itertools.product(*l))
[('1', 'a', 'x'), ('1', 'a', 'y'), ('1', 'a', 'z'), ('1', 'b', 'x'),
('1', 'b', 'y'), ('1', 'b', 'z'), ('1', 'c', 'x'), ('1', 'c', 'y'),
('1', 'c', 'z'), ('2', 'a', 'x'), ('2', 'a', 'y'), ('2', 'a', 'z'),
('2', 'b', 'x'), ('2', 'b', 'y'), ('2', 'b', 'z'), ('2', 'c', 'x'),
('2', 'c', 'y'), ('2', 'c', 'z'), ('3', 'a', 'x'), ('3', 'a', 'y'),
('3', 'a', 'z'), ('3', 'b', 'x'), ('3', 'b', 'y'), ('3', 'b', 'z'),
('3', 'c', 'x'), ('3', 'c', 'y'), ('3', 'c', 'z')]
This is not quite what you want yet, but it's easy to get there:
>>> for i in range(len(l)):
... print(list(itertools.product(*l[:i+1])))
...
[('1',), ('2',), ('3',)]
[('1', 'a'), ('1', 'b'), ('1', 'c'), ('2', 'a'), ('2', 'b'), ('2', 'c'),
('3', 'a'), ('3', 'b'), ('3', 'c')]
[('1', 'a', 'x'), ('1', 'a', 'y'), ('1', 'a', 'z'), ('1', 'b', 'x'),
('1', 'b', 'y'), ('1', 'b', 'z'), ('1', 'c', 'x'), ('1', 'c', 'y'),
('1', 'c', 'z'), ('2', 'a', 'x'), ('2', 'a', 'y'), ('2', 'a', 'z'),
('2', 'b', 'x'), ('2', 'b', 'y'), ('2', 'b', 'z'), ('2', 'c', 'x'),
('2', 'c', 'y'), ('2', 'c', 'z'), ('3', 'a', 'x'), ('3', 'a', 'y'),
('3', 'a', 'z'), ('3', 'b', 'x'), ('3', 'b', 'y'), ('3', 'b', 'z'),
('3', 'c', 'x'), ('3', 'c', 'y'), ('3', 'c', 'z')]
Get everything in one list:
>>> result = []
>>> for i in range(len(l)):
... result.extend(list(itertools.product(*l[:i+1])))
...
>>> result
[('1',), ('2',), ('3',), ('1', 'a'), ('1', 'b'), ('1', 'c'), ('2', 'a'),
('2', 'b'), ('2', 'c'), ('3', 'a'), ('3', 'b'), ('3', 'c'), ('1', 'a', 'x'),
('1', 'a', 'y'), ('1', 'a', 'z'), ('1', 'b', 'x'), ('1', 'b', 'y'),
('1', 'b', 'z'), ('1', 'c', 'x'), ('1', 'c', 'y'), ('1', 'c', 'z'),
('2', 'a', 'x'), ('2', 'a', 'y'), ('2', 'a', 'z'), ('2', 'b', 'x'),
('2', 'b', 'y'), ('2', 'b', 'z'), ('2', 'c', 'x'), ('2', 'c', 'y'),
('2', 'c', 'z'), ('3', 'a', 'x'), ('3', 'a', 'y'), ('3', 'a', 'z'),
('3', 'b', 'x'), ('3', 'b', 'y'), ('3', 'b', 'z'), ('3', 'c', 'x'),
('3', 'c', 'y'), ('3', 'c', 'z')]
Get it in the exact shape you want:
>>> sorted(result)
[('1',), ('1', 'a'), ('1', 'a', 'x'), ('1', 'a', 'y'), ('1', 'a', 'z'),
('1', 'b'), ('1', 'b', 'x'), ('1', 'b', 'y'), ('1', 'b', 'z'), ('1', 'c'),
('1', 'c', 'x'), ('1', 'c', 'y'), ('1', 'c', 'z'), ('2',), ('2', 'a'),
('2', 'a', 'x'), ('2', 'a', 'y'), ('2', 'a', 'z'), ('2', 'b'), ('2', 'b', 'x'),
('2', 'b', 'y'), ('2', 'b', 'z'), ('2', 'c'), ('2', 'c', 'x'), ('2', 'c', 'y'),
('2', 'c', 'z'), ('3',), ('3', 'a'), ('3', 'a', 'x'), ('3', 'a', 'y'),
('3', 'a', 'z'), ('3', 'b'), ('3', 'b', 'x'), ('3', 'b', 'y'), ('3', 'b', 'z'),
('3', 'c'), ('3', 'c', 'x'), ('3', 'c', 'y'), ('3', 'c', 'z')]
Upvotes: 11