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Suppose I want to find the index of the maximal value of a given sequence. In Matlab, this is:

[value, index] = max(x)

In Python, I would abstract the following pattern:

def find_best_element_index(elements, is_better):
    best_so_far = elements[0]
    best_so_far_index = 0
    for (index, element) in enumerate(elements):
        if is_better(element, best_so_far):
            best_so_far = element
            best_so_far_index = index
    return best_so_far_index

Which I could use in the following way:

import operator
assert find_best_element_index([6, 3, 9, 8, 0, 9, 7, 9, 7, 8], operator.gt) == 2

But I wonder if there is a more pythonic way to go when I just want to find the index of the "best" item.

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1 Answer 1

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  1. I just want to find the index of the "best" item.

This can be written very shortly and readably using enumerate, first class functions and list indexing.

def index_of_max_value(items):
    return max(enumerate(items), key = lambda x: x[1])[0]

Explanation:

  1. I create a list that contains the numbers and their indexes, the index is first, the number second.
>>> list(enumerate([4,7,4,8]))
[(0, 4), (1, 7), (2, 4), (3, 8)]
  1. I ask the item that has the biggest value when a function I decide is applied to it, I do so by passing the function as the key argument, when functions are passed around like this they are first class citizens
>>> max(["rg","gu","sd"], key = lambda s: sum(map(ord,s)))
'gu'

For example in the above I asked which of the strings had the most ASCII value.

key = lambda x: x[1]

Means that I want the pair (index,number) that has the biggest 'second' item (remember zero indexing).

  1. [n] gives you the n-th item of a list, I ask the '0-th' (first) item that is the index.

I hope that you like my alternative shorter solution, if you have any doubts, fell free to ask.

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  • \$\begingroup\$ Great one-liner! I didn't know that I could provide an optional key argument to max. In this way, I can use max to calculate the index of the min value too! ;) \$\endgroup\$ Commented Mar 8, 2015 at 10:33
  • \$\begingroup\$ BTW, can you confirm that the required size is constant, i.e., that max does not construct the whole list of couples? \$\endgroup\$ Commented Mar 8, 2015 at 10:43
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    \$\begingroup\$ @Aristide enumerate yields an iterator, that is like a promise or a lazy list, it is consumed value after value, so the space requirement are O(1), that is only two items in memory at any given time: the current max and the item to compare. \$\endgroup\$ Commented Mar 8, 2015 at 11:20
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    \$\begingroup\$ Sorry, my question was too broad, I have edited it in order to suppress the second part, since your answer is a perfect fit for the first one. I'll ask the second question separately. Thanks for your explanations! \$\endgroup\$ Commented Mar 8, 2015 at 12:12
  • \$\begingroup\$ @Aristide you are welcome! \$\endgroup\$ Commented Mar 8, 2015 at 19:20

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