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What happens to a <itertools.permutations object at 0x7fe3cc66af68> after it is read?
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What happens to a <itertools.permutations object at 0x7fe3cc66af68> after it is read?
I want to get all permutations of 2 vowels

from itertools import permutations

doubles = permutations (['a', 'e', 'i', 'o', 'u'],2) 
for i in doubles: 
This prints all the permutations. I want to save them in a list.

I tried:

perm = ('a', 'e')
perm[0] gives me 'a'
perm[1] gives me 'e'
Now I find, if I run:

doubles = permutations (['a', 'e', 'i', 'o', 'u'],2)
for i in doubles:
I see my permutations

If I run

for i in doubles:
again, I get nothing. No error. Nothing.

Where has doubles gone? Seems it can only be read 1 time. Why is that? Read it and it is gone!

This gets me what I want:

doubles = permutations (['a', 'e', 'i', 'o', 'u'],2) 
perms = []
for i in doubles:
    perm = i[0] + i[1]
once you iterate over it, the itertools.permutations object is exhausted and there is nothing to return on second attempt to use/iterate over it
from itertools import permutations
doubles = permutations(['a', 'e', 'i', 'o', 'u'], 2)
doubles = list(permutations(['a', 'e', 'i', 'o', 'u'], 2))
<class 'itertools.permutations'> <class 'list
Pedroski55 likes this post
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Correct. The permutations() function returns an iterator. You can only consume an individual iterator once.

If you need the data again, you need to either recreate the iterator (call permutations() again), or save the information in another data structure. Memory requirements and access patterns may make one or the other a better choice.
buran and Pedroski55 like this post
Just hold a reference to the list/tuple and create new permutations and process them in the for-loop.
I made some other improvements.

# definition somewhere
doubles = ['a', 'e', 'i', 'o', 'u'] 

# if this never will change, choose a tuple
# this will prevent the ability to assign new objects to it
# doubles = ('a', 'e', 'i', 'o', 'u')
# then doubles[0] = "A" will raise an TypeError

perms = []
# creating always a new permutations object from doubles
for p in permutations(doubles, 2):
    # p is a tuple with permutations
    # we want to join them into a string
    p = "".join(p)
    # printing and appending
Pedroski55 likes this post
My code examples are always for Python >=3.6.0
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