当值为列表时，dict(sorted(dictionary.items(), key=operator.itemgetter(1)) 并不总是返回有序的dict

Question

I have a dict:

count2:defaultdict(<class 'list'>, {'i': [3, 2, 2, 1], 'w': [2, 2], 'p': [2, 2], 'd': [2, 2], 'm': [2, 2], 'y': [2, 2, 2, 1], 'x': [2, 2, 4, 1], 'j': [2, 2], 'o': [2, 1], 'r': [2, 1]})

when I try to sort it by using

ordered = dict(sorted(count2.items(), key=operator.itemgetter(1), reverse=True)) 

It not always sorts it like I want it to sort. (the value must have the biggest number and then descend) so it returns this:

orderedStart:{'i': [3, 2, 2, 1], 'x': [2, 2, 4, 1], 'y': [2, 2, 2, 1], 'w': [2, 2], 'p': [2, 2], 'd': [2, 2], 'm': [2, 2], 'j': [2, 2], 'o': [2, 1], 'r': [2, 1]}

everything is right except for that x should be in front of i since 4 > 3 . Are some indexes more prioritized?

To facilitate the users, here is an Example of a well sorted list using the same code.

Before:

count2:defaultdict(<class 'list'>, {'r': [2, 2, 2, 1], 'g': [2, 1], 'e': [3, 1], 'n': [5, 1], 't': [4, 1], 'i': [2, 1], 'o': [5, 1], 'm': [2, 1]})

After:

{'n': [5, 1], 'o': [5, 1], 't': [4, 1], 'e': [3, 1], 'r': [2, 2, 2, 1], 'g': [2, 1], 'i': [2, 1], 'm': [2, 1]}

Answer 1

You misunderstand how python compares tuples. You have asked it to compare [3, 2, 2, 1] and [2, 2, 4, 1] . Since 3 > 2, it comes first.

Python uses "lexicographic comparison", which is identical to the way you look up words in a dictionary. First you compare the first letters. If they are different, you're done; if they are different, you look at the second letter. And so forth.

Answer 2

When you compare lists, they're compared lexicographically, so the first element takes precedence, then the second, and so on. If you want the largest list element to take precedence, compare sorted lists.

ordered = dict(sorted(count2.items(), key=lambda item: sorted(item[1], reverse=True), reverse=True))