I think it's worth putting a few timings up here for some perspective.
All timings done on OS-X 10.5.8 with python2.7
John Clement's answer:
python -m timeit -s 'my_list = range(1000)[::-1]; from operator import itemgetter' 'min(enumerate(my_list),key=itemgetter(1))'
1000 loops, best of 3: 239 usec per loop
David Wolever's answer:
python -m timeit -s 'my_list = range(1000)[::-1]' 'min((val, idx) for (idx, val) in enumerate(my_list))
1000 loops, best of 3: 345 usec per loop
OP's answer:
python -m timeit -s 'my_list = range(1000)[::-1]' 'my_list.index(min(my_list))'
10000 loops, best of 3: 96.8 usec per loop
Note that I'm purposefully putting the smallest item last in the list to make .index as slow as it could possibly be. It would be interesting to see at what N the iterate once answers would become competitive with the iterate twice answer we have here.
Of course, speed isn't everything and most of the time, it's not even worth worrying about ... choose the one that is easiest to read unless this is a performance bottleneck in your code (and then profile on your typical real-world data -- preferably on your target machines).