如何获取字典键列表中第二项的最低值？

Question

Suppose we have a dictionary like this:

Dict={'2020-12-23_36.JPG': [2959, 4690],
 '2020-12-28_106.JPG': [2956, 4592],
 '2020-12-28_117.JPG': [2993, 4752],
 '2020-12-28_13.JPG': [2919, 4921],
 '2020-12-28_141.JPG': [2984, 4944]}

How do we retrieve from this the key with the lowest value of the second item in the list?:

import operator
min(Dict.items(), key=operator.itemgetter(1))[0]
"2020-12-28_13.JPG"

This makes sense as "2020-12-28_13.JPG" does indeed have the lowest value in the first item of the list: 2919

I attempted the following but with no success:

min(Dict.items(), key=operator.itemgetter(1))[1][1]
"4921"

It only prints the second item of the list, but based on the first item of the list.

I also tried:

min(Dict.items(), key=operator.itemgetter(1))[1])[0]
"TypeError: 'operator.itemgetter' object is not subscriptable"

Expected result

min(Dict.items(), key=operator.itemgetter(1))[1][1]
'2020-12-28_106.JPG' #(because 4592 is the lowest value of the -1 in list)

Answer 1

This might actually be very simple. sorted sorts based on any "key" for any iterable. In the instant case, the key is the 2nd item in the value of the dictionary.

sorted(Dict, key = lambda x: Dict[x][1])[0]

Answer 2

You have small typo bug on you code, compare this:

Your line

min(Dict.items(), key=operator.itemgetter(1)[1])[0]

Correct one

min(Dict.items(), key=operator.itemgetter(1))[1][0]