Question

I understand the gist of the code, that it forms permutations; however, I was wondering if someone could explain exactly what is going on in the return statement.

def perm(l):
    sz = len(l)
    print (l)
    if sz <= 1:
        print ('sz <= 1')
        return [l]
    return [p[:i]+[l[0]]+p[i:] for i in range(sz) for p in perm(l[1:])]

Answer 1

This return is returning a list comprehension whose items are made by inserting the first item of l into each position of p, from the first to the last — p in turn is a list of lists, obtained by a recursive call to perm which excludes the first item of l (and thus permutes all other items in all possible ways).

If you don’t understand recursion, it’s not really trivial to explain;-). If you don’t understand list comprehensions, they are trivial to explain — that return is semantically equivalent to

result = []
for i in range(sz):
  for p in perm(l[1:]):
    result.append(p[:i]+[l[0]]+p[i:])
return result

this also shows how inefficient this code is: it’s calling perm recursively sz times, and obviously there’s no need for it. Much better would be to simply swap the two for loops:

result = []
for p in perm(l[1:]):
  for i in range(sz):
    result.append(p[:i]+[l[0]]+p[i:])
return result

and the equivalent of this, much better code, is a list comprehension with the two for clauses swapped:

return [p[:i]+[l[0]]+p[i:] for p in perm(l[1:]) for i in range(sz)]