按值对 Python 字典进行排序； value 是一个包含两项的列表； 对第一个列表项进行常规排序，对第二项进行反向排序

Question

I have a dictionary whose values are all two-item lists (the key is a six-item tuple, but we won't get into that).

Here's a visualization:

my_ridiculous_dict = {
    ('0', '2', '0', '4', '0', '1'): [1, 1000], 
    ('5', '3', '0', '2', '0', '0'): [1, 400], 
    ('3', '6', '0', '2', '0', '0'): [2, 3000], 
    ('4', '3', '0', '2', '0', '0'): [2, 80], 
    ('0', '0', '3', '9', '0', '3'): [3, 0]
}

I need to sort it first by the first item in the value's list, then with a reverse sort on the second item. So what you see above would be sorted correctly

Apologies if this has been addressed, but all the questions I've seen on sorting dictionaries by value seem to assume a simple int or str as the value, & I'm having trouble wrapping my head around the use of lambdas or itemgetters given this added complexity.

Should mention I don't mind doing this in two sort passes, just would like to git'r'done!

Many thanks in advance.

Answer 1

Dictionaries aren't meant to be sorted but if you copy the items to a list, you can sort the list.

mrd = {
('4', '3', '0', '2', '0', '0'): [2, 80], 
('0', '2', '0', '4', '0', '1'): [1, 1000], 
('3', '6', '0', '2', '0', '0'): [2, 3000], 
('5', '3', '0', '2', '0', '0'): [1, 400], 
('0', '0', '3', '9', '0', '3'): [3, 0]
}

lst = list(mrd)
lst.sort(key=lambda x:[mrd[x][0], -mrd[x][1]])
for i in lst: print(i, mrd[i])

Output:

('0', '2', '0', '4', '0', '1') [1, 1000]
('5', '3', '0', '2', '0', '0') [1, 400]
('3', '6', '0', '2', '0', '0') [2, 3000]
('4', '3', '0', '2', '0', '0') [2, 80]
('0', '0', '3', '9', '0', '3') [3, 0]

Answer 2

Python uses a stable sort

A more general solution relies on the fact that the sorts are stable in Python. That means that if two items sort the same, then they retain their original order.

To get what you want, do a reversed sort by the second value. Use my_dict.items() to get (key, value) tuples. t[1][1] is the second value.

items = sorted(my_dict.items(), key=lambda t:t[1][1], reverse=True)

Then sort by the first value.

items.sort(key=lambda t:t[1][0])

In current versions of Python, dicts maintain items in key-insertion-order. So you could create a sorted dict by:

my_sorted_dict = dict(items)

using operator.itemgetter()

I don't see a way to use itemgetter on a nested data structure, such as the (key, [value1, value2]) that my_dict.items() provides. So, to use operator.itemgetter() the structure would need to be flattened.

import operator

items = [(k, *v) for k,v in my_dict.items()]

This turns it into a tuple of elements like: (k, value1, value2).

Now sort the list like before; second element, then first element. operator.itemgetter(n) creates and returns a function that takes an object and calls __getitem__(n) on that object. Thus, these lines do a reverse sort keyed by item[2] for each item in items, then a sort keyed by item[1].

items.sort(key=operator.itemgetter(2), reverse=True)
items.sort(key=operator.itemgetter(1))

Then a dict comprehension builds the sorted dict:

my_sorted_dict = {k:[v1,v2] for k,v1,v2 in items}

my_sorted_dict