Finding the index of the value which is the min or max in Python

Question

I\'ve got a structure of the form:

>>> items
[([[0, 1], [2, 20]], \'zz\', \'\'), ([[1, 3], [5, 29], [50, 500]], \'a\', \'b\')]

The

Answer 1

This works:

by_index = ([sub_index, list_index] for list_index, list_item in
             enumerate(items) for sub_index in list_item[0])
[item[1] for item in sorted(by_index)]

Gives:

[0, 1, 0, 1, 1]

In detail. The generator:

by_index = ([sub_index, list_index] for list_index, list_item in
             enumerate(items) for sub_index in list_item[0])
list(by_index)    
[[[0, 1], 0], [[2, 20], 0], [[1, 3], 1], [[5, 29], 1], [[50, 500], 1]]

So the only thing needed is sorting and getting only the desired index:

[item[1] for item in sorted(by_index)]

Answer 2

 from operator import itemgetter
 index, element = max(enumerate(items), key=itemgetter(1))

Return the index of the biggest element in items and the element itself.

Answer 3

Yet another way to get the argmax is:

def argmax(lst):
    return max(range(len(lst)), key=lst.__getitem__)

Answer 4

This method finds the index of the maximum element of any iterable and does not require any external imports:

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

Answer 5

The index of the max of a list:

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

If there are duplicate max values in lst, this will return the index of the first maximum value found.

Answer 6

It's easy if you don't try to use the fact that the internal range lists are sorted

sorted(sum([ [(rng,) + i[1:] for rng in i[0]] for i in items ], []), lambda i: i[0][0])

It sounds like you want a function that returns the index of the smallest value though

def min_idx(l, key=lambda x: x):
    min_i, min_key = None, float('inf')
    for i, v in enumerate(l):
        key_v = key(v)
        if key_v < min_key:
            mini_i = i
            min_key = key_v
    return min_i

def merge_items(items):
    res = []
    while True:
        i = min_idx(items, key=lambda i: i[0][0][0])
        item = items[i]
        res.append((item[0][0],) + item[1:])
    return res