Stopping a recursive generator & permutations

Question

As an exercise, I've been trying out various ways of generating all permutations of a list in Python -- recursive, non-recursive... -- and comparing the performance with itertools.permutations(). But I'm having trouble with the generator version of the recursive method, which doesn't finish cleanly with a StopIteration exception, but instead throws an IndexError:

def spawnperms(alist):
    """same algorithm as recursive option, but a generator"""
    if (alist == []):
        yield []
    for perm in spawnperms(alist[:-1]):
        for i in range(len(perm)+1):
            yield perm[:i] + [alist[-1]] + perm[i:]

Calling this from the Python interpreter:

>>> for i in spawnperms(range(3)):
...     print i
... 
[2, 1, 0]
[1, 2, 0]
[1, 0, 2]
[2, 0, 1]
[0, 2, 1]
[0, 1, 2]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 5, in spawnperms
  File "<stdin>", line 5, in spawnperms
  File "<stdin>", line 5, in spawnperms
  File "<stdin>", line 7, in spawnperms
IndexError: list index out of range

Ouch. I tried stepping through it with pdb, which nearly created a stack overflow in my brain, but what I understood is that the recursion "reaches down" to the empty list, and then the outer (I think) for loop runs out of indices.

How can I correct my code?

EDIT: One learning from Mark Byers' deceptively simple correct answer is that clean coding practices can prevent mistakes. Had I used an else systematically after if, regardless of whether I thought the condition could ever be revisited, this wouldn't have happened. And it still feels very stupid!

Answer 1

You are missing an else:

if (alist == []):
    yield []
else:
    for ...

This is because yield does not behave in the same way as return. Execution continues after the yield statement when you request the next value.