implementing argmax in Python

Question

How should argmax be implemented in Python? It should be as efficient as possible, so it should work with iterables.

Three ways it could be implemented:

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Answer 1

Based on Neil's answer, but specialized for functions that take multiple arguments.

argmax = lambda keys, func: max(imap(lambda key: (func(*key), key), keys))[1]

For example:

argmax([(5, 2), (3, 3), (2, 5)], pow)
# (2, 5)

Answer 2

Is the following code a fast and pythonic way?

idx_max = max(enumerate(x), key=lambda x:x[1])[0]

Answer 3

I found this way easier to think about argmax: say we want to calculate argmax(f(y)) where y is an item from Y. So for each y we want to calculate f(y) and get y with maximum f(y).

This definition of argmax is general, unlike "given an iterable of values return the index of the greatest value" (and it is also quite natural IMHO).

And ..drumroll.. Python allows to do exactly this using a built-in max:

best_y = max(Y, key=f)

So argmax_f (from the accepted answer) is unnecesary complicated and inefficient IMHO - it is a complicated version of built-in max. All other argmax-like tasks should become clear at this point: just define a proper function f.

Answer 4

I modified the best solution I found:

# given an iterable of pairs return the key corresponding to the greatest value
def argmax(pairs):
    return max(pairs, key=lambda x: x[1])[0]

# given an iterable of values return the index of the greatest value
def argmax_index(values):
    return argmax(enumerate(values))

# given an iterable of keys and a function f, return the key with largest f(key)
def argmax_f(keys, f):
    return max(keys, key=f)

Answer 5

def argmax(lst):
     return lst.index(max(lst))

or analogously:

argmax = lambda lst: lst.index(max(lst)