Large list generation memory managment

Question

import itertools
Num = 11

base = list(range(1,Num+1))

Permutations = list(itertools.permutations(base))

I'm getting a memory error trying to run this. In reality I only need to generate the 1st (Num-1)! permutations but I'm not sure how to (so if Num = 7 i would need to generate the first 6! = 720 permutations). But I would ideally like to be able to generate permutations for significantly higher values of Num so any suggestions would be great

John Kugelman · Accepted Answer · 2020-03-26T15:23:49.350

range() and permutation() both return generators that generate items on demand. You don't need to call list() and turn them into lists. Just iterate over them directly and access the items one by one.

num = 11
base = range(1, num+1)
permutations = itertools.permutations(base)

for permutation in permutations:
    # Do something with `permutation`.

(Note that a generator can only be used once. If you want to iterate over the permutations more than once you'll need to call itertools.permutations() multiple times.)

To stop after n items use itertools.islice():

for permutation in itertools.islice(permutations, n):
    # Do something with `permutation`.

You can skip items at the beginning, too. This would skip the first five permutations:

for permutation in itertools.islice(permutations, 5, n):
    # Do something with `permutation`.

If you want to count the permutations you can add enumerate(), which attaches an index to every entry:

for i, permutation in enumerate(itertools.islice(permutations, n)):
    # Skip the fifth permutation.
    if i == 4:
        continue

    # Do something with `permutation`.

By the way, please use lowercase for variable names. Only class names should be capitalized.

Large list generation memory managment

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