I am trying to implement an "i not equal to j" (i<j) loop, which skips cases where i = j, but I would further like to make the additional requirement that the loop does not repeat the permutation of (j,i), if (i,j) has already been done (since, due to symmetry, these two cases give the same solution).

First Attempt

In the code to follow, I make the i<j loop by iterating through the following lists, where the second list is just the first list rolled ahead 1:

mylist = ['a', 'b', 'c']
np.roll(mylist,2).tolist() = ['b', 'c', 'a']

The sequence generated by the code below turns out to not be what I want:

import numpy as npmylist = ['a', 'b', 'c']
for i in mylist:for j in np.roll(mylist,2).tolist():print(i,j)

since it returns a duplicate a a and has repeated permutations a b and b a:

a b
a c
a a
b b
b c
b a
c b
c c
c a

The desired sequence should instead be the pair-wise combinations of the elements in mylist, since for N=3 elements, there should only be N*(N-1)/2 = 3 pairs to loop through:

a b
a c
b c

You can do this using quite a hacky method just by removing the first element and appending it:

mylist.append(mylist.pop(0))

Where .append(...) will append an element to the end of a list, and .pop(...) will remove an element from a given index and return it.

• You can read up about the builtin data structure functions here