Getting the index of the returned max or min item using max()/min() on a list

Say that you have a list values = [3,6,1,5], and need the index of the smallest element, i.e. index_min = 2 in this case.

Avoid the solution with itemgetter() presented in the other answers, and use instead

index_min = min(range(len(values)), key=values.__getitem__)

because it doesn't require to import operator nor to use enumerate, and it is always faster(benchmark below) than a solution using itemgetter().

If you are dealing with numpy arrays or can afford numpy as a dependency, consider also using

import numpy as npindex_min = np.argmin(values)

This will be faster than the first solution even if you apply it to a pure Python list if:

• it is larger than a few elements (about 2**4 elements on my machine)
• you can afford the memory copy from a pure list to a numpy array

as this benchmark points out:

I have run the benchmark on my machine with python 2.7 for the two solutions above (blue: pure python, first solution) (red, numpy solution) and for the standard solution based on itemgetter() (black, reference solution).The same benchmark with python 3.5 showed that the methods compare exactly the same of the python 2.7 case presented above

if is_min_level:    return values.index(min(values))else:    return values.index(max(values))

You can find the min/max index and value at the same time if you enumerate the items in the list, but perform min/max on the original values of the list. Like so:

import operatormin_index, min_value = min(enumerate(values), key=operator.itemgetter(1))max_index, max_value = max(enumerate(values), key=operator.itemgetter(1))

This way the list will only be traversed once for min (or max).