This way is probably slowe (two scans of the list for l1, and even more work for l2), but for small lists it's probably simple enough to be considered:
For a simple list: >>> l1 = [5, 3, 2, 1, 4] >>> l1.index(min(l1)) 3 For a list of lists: >>> l2 = [[3, 3, 3, 3], [6], [10], [3, 3, 3, 1, 4], [3, 0, 3, 3]] >>> mins = map(min, l2) >>> mins [3, 6, 10, 1, 0] >>> pos1 = mins.index(min(mins)) >>> pos1 4 >>> subl = l2[pos1] >>> subl.index(min(subl)) 1 This solution is also fragile: >>> l3 = [[3], []] >>> mins = map(min, l3) Traceback (most recent call last): File "<interactive input>", line 1, in ? ValueError: min() arg is an empty sequence Bye, bearophile -- http://mail.python.org/mailman/listinfo/python-list
