How to find index of item with shortest list inside?

Question

So I have a list in this format:

lst = [[x1, y1, [1,3,4]], [x2, y2, [6,3]], [x3, y3, [8,6,3,9]]]

I want to find the index of item with the shortest list inside . ie In this case it would return [x2, y2, [6,3]] since [6,3] is the shortest list in this case.

Of course I can iterate through to find the shortest item but is there better/faster way to do this in python?

Answer 1

Use the min , providing a key function that returns the length of the sublist:

min(lst, key=lambda item: len(item[2]))

Note that iteration over lst still takes place implicitly when using min . There is no other way to check every item than to iterate.

If you really need the index do:

index = min(range(len(lst)), key=lambda dx: len(lst[dx][2])

Answer 2

In [33]: lst = [['x1', 'y1', [1,3,4]], ['x2', 'y2', [6,3]], ['x3', 'y3', [8,6,3,9]]]

In [34]: min(enumerate(lst), key=lambda t : min(len(s) for s in t[1] if isinstance(s,list)))[0]
Out[34]: 1

Answer 3

确实，有：

min(lst, key=lambda x: len(x[-1]))

Answer 4

If inner list is always the last one item

from itertools import imap

your_min = min(imap(len, (item[-1] for item in your_list)))

or if you want index:

def find_index(lst):
    min_len = 1e10
    min_idx = None
    for i, v in enumerate(your_list):
        curr_len = len(v[-1])
        if curr_len < min_len:
            min_len = curr_len
            min_idx = i
    return i