David Isaac wrote: > 2. Is this a good argmax (as long as I know the iterable is finite)? > def argmax(iterable): return max(izip( iterable, count() ))[1]
There's a subtle difference to the builtin: argmax() gives you the (index of the) last maximum while max() returns the (value of the) first maximum: >>> from itertools import count, izip >>> def argmax(iterable): ... return max(izip(iterable, count()))[1] ... >>> class Int(int): pass ... >>> type(max([Int(0), 0])) <class '__main__.Int'> # must be the first item then >>> argmax([Int(0), 0]) 1 If you care, here's the fix building on George's implementation: >>> def argmax2(iterable): ... return -max((v, -i) for i, v in enumerate(iterable))[1] ... >>> argmax2([Int(0), 0]) 0 Peter -- http://mail.python.org/mailman/listinfo/python-list
