Question

What is the equivalent of the following in python?

for (i=0; i<n; i++)
    for (j=i+1; j<n; j++)
        //do stuff with A[i], A[j]

Or in some sense, the following. It should also remove the element from A at the completion of each round of the loop.

for a in A:
    for a' in A/{a}: #i.e. rest of the elements of A
        #do something with a,a'
    #remove a from A

Is there a pythonic way of doing this without using enumerate()?

Edits:

Sorry for the bad description.

1. In the first example, I mean to use i & j only as indices. Their values do not matter. Its just a rough c++ equivalent of the latter.

2. The outer loop is executed n times. The inner loop is executed (n-1), (n-2)…0 times for each iteration of the outer loop.

Maybe this might help (pseudocode):

function next_iteration(list):
    head = first element
    tail = remaining elements #list
    each element in tail interacts with head one by one
    next_iteration(tail)

PS: All code samples above are pseudocodes. I’m trying to express something that is still a bit vague in my mind.

Answer 1

Since your two questions are different, here is solution for your second problem:

for i in xrange(len(A)):
    for j in xrange(len(A)):
        if i != j:
            do_stuff(A[i], A[j])

or using itertools (I think using the included batteries is very pythonic!):

import itertools

for a, b in itertools.permutations(A, 2):
    do_stuff(a, b)

This applies do_stuff to all combinations of 2 different elements from A. I you want to store the result just use:

[do_stuff(a, b) for a, b in itertools.permutations(A, 2)]