Sort python dictionary by value and the value is a tuple

Question

I need to sort this dictionary by points:

key: [(attacker, points), (attacker, points)]
key2: [(attacker, points), (attacker, points)]

so if I have this:

key: [(attacker, 20), (attacker, 25)]
key2: [(attacker, 5), (attacker, 10)]

I will get this:

key2: [(attacker, 5), (attacker, 10)]
key: [(attacker, 20), (attacker, 25)]

1. dictionaries are not ordered/cannot be sorted. Do you mean you want to get maybe a sorted list? 2. Each dictionary item contains two `points` variables. Which one you like to sort by? — kyriakosSt, Feb 27 '20 at 15:48
Assuming `attacker = 'attacker'`, and you are properly [sorting by value](https://stackoverflow.com/q/613183/4518341), it already sorts like that. You need to provide a [mre]. — wjandrea, Feb 27 '20 at 15:49

alikhtag · Answer 1 · 2020-02-27T15:57:21.593

0

This should work, but dictionary keys will not be sorted (because it is unsorted collection) only what is inside (value) the key will be sorted because you are using the list of tuples.

for k,v in your_dict.items():
    v.sort(key=lambda i: i[1])

edited Feb 27 '20 at 15:57

answered Feb 27 '20 at 15:49

alikhtag

316
1
6

score 0 · Accepted Answer · answered Feb 27 '20 at 16:01

Here goes:

from collections import OrderedDict
ordered_keys = sorted(list(my_dict.keys()), key=lambda x: x[1])
my_ordered_dict = OrderedDict((key, my_dict[key]) for key in ordered_keys)

Example input:

my_dict = {("a", 3): "toto", ("b", 2): "titi", ("c", 1): "tata"}

Output:

OrderedDict([(('c', 1), 'tata'), (('b', 2), 'titi'), (('a', 3), 'toto')])

Sort python dictionary by value and the value is a tuple

2 Answers2