Given a list l=[1,2] I am trying to create a product of permutations. In other words, I am trying to create the following.
from itertools import permutations
l=[1,2]
print([(e1, e2) for e1 in permutations(l) for e2 in permutations(l)])
The above code prints [((1, 2), (1, 2)), ((1, 2), (2, 1)), ((2, 1), (1, 2)), ((2, 1), (2, 1))] as expected.
However, if I use the code below,
from itertools import permutations
l=[1,2]
lp1 = permutations(l)
lp2 = permutations(l)
print([(e1, e2) for e1 in lp1 for e2 in lp2])
The code prints [((1, 2), (1, 2)), ((1, 2), (2, 1))] .
I guess this is due to lp1 and lp2 pointing to the same iterator. But I do not understand why this is the case. Is this a intended behavior or is this a bug?
>Solution :
Yes. permutation is a generator. You can use it only once.
Just to illustrate with an easier example, it is exactly as if you
tried
for i in range(3):
for j in enumerate([1,2,3]):
print(i,j)
To get
0 (0, 1)
0 (1, 2)
0 (2, 3)
1 (0, 1)
1 (1, 2)
1 (2, 3)
2 (0, 1)
2 (1, 2)
2 (2, 3)
And then were surprised that
range1=range(3)
range2=enumerate([1,2,3])
for i in range1:
for j in range2:
print(i,j)
was not working as expected, and gives:
0 (0, 1)
0 (1, 2)
0 (2, 3)
Because you need to recreate range2 for each i iteration. Otherwise, j with iterates only once. The 2 other times, the iterator is over.
(Edit note: initialy I used range(3) in my example for range2. But that is not a good example, since a range is a range, not a generator. So you can use it several times)