Nested loop through list avoiding same element

Question

I can't explain the concept well at all, but I am trying to loop through a list using a nested loop, and I can't figure out how to avoid them using the same element.

list = [1, 2, 2, 4]
for i in list:
    for j in list:
        print(i, j) # But only if they are not the same element

So the output should be:

Edit as the solutions don't work in all scenarios:

The if i != j solution only works if all elements in the list are different, I clearly chose a poor example, but I meant same element rather than the same number; I have changed the example

I disagree, I said explicitly asked about them being the same element rather than just the same number, how else would you have worded it? — user11035198
– user11035198, Commented Mar 16, 2020 at 14:19
@JonathanDyke What if your list is [2, 2, 2]. What is the expected output? — a_guest
– a_guest, Commented Mar 16, 2020 at 14:22
@JonathanDyke I didnot see that you changed the original list too. ;) I apologize. — Ch3steR
– Ch3steR, Commented Mar 16, 2020 at 14:23
@a_guest regardless of the numbers it would be all permutations of list[i] list[j] except where i == j, so in the case of [2, 2, 2] it would be 2 2 six times — user11035198
– user11035198, Commented Mar 16, 2020 at 14:24

blhsing · Accepted Answer · 2020-03-16 14:16:24Z

7

You can compare the indices of the two iterations instead:

lst = [1, 2, 2, 4]
for i, a in enumerate(lst):
    for j, b in enumerate(lst):
        if i != j:
            print(a, b)

You can also consider using itertools.permutations for your purpose:

lst = [1, 2, 2, 4]
from itertools import permutations
for i, j in permutations(lst, 2):
    print(i, j)

Both would output:

answered Mar 16, 2020 at 14:16

blhsing

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Comments

Petr Blahos · Accepted Answer · 2020-03-16 14:06:52Z

3

Simply:

if i != j:
    print(i, j)

answered Mar 16, 2020 at 14:06

Petr Blahos

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