How to properly sort a dictionary by value of a key?

Question

I would like to sort a dictionary by value of specific key ("foo" in the example)

mydict = dict()
mydict["a"] = {"foo": 0.3, "boo": 0.1}
mydict["b"] = {"foo": 0.1, "boo": 0.2}
mydict["c"] = {"foo": 0.8, "boo": 0.3}
mydict["d"] = {"foo": 0.5, "boo": 0.4}
mydict["e"] = {"foo": 0.2, "boo": 0.5}

# sort by "foo" in descending order
# expected result: [{"foo": 0.8, "boo": 0.3}, {"foo": 0.5, "boo": 0.4}, {"foo": 0.3, "boo": 0.1}, {"foo": 0.2, "boo": 0.5}, {"foo": 0.1, "boo": 0.2}]

How can I get the result I want like shown in the comment above?

Andrej Kesely · Accepted Answer · 2021-05-09T16:39:31.973

11

print(sorted(mydict.values(), key=lambda k: k["foo"], reverse=True))

Prints:

[{'foo': 0.8, 'boo': 0.3}, {'foo': 0.5, 'boo': 0.4}, {'foo': 0.3, 'boo': 0.1}, {'foo': 0.2, 'boo': 0.5}, {'foo': 0.1, 'boo': 0.2}]

EDIT: To include "key":

print(
    sorted(
        [{"key": k, **v} for k, v in mydict.items()],
        key=lambda k: k["foo"],
        reverse=True,
    )
)

Prints:

[{'key': 'c', 'foo': 0.8, 'boo': 0.3}, 
 {'key': 'd', 'foo': 0.5, 'boo': 0.4}, 
 {'key': 'a', 'foo': 0.3, 'boo': 0.1}, 
 {'key': 'e', 'foo': 0.2, 'boo': 0.5}, 
 {'key': 'b', 'foo': 0.1, 'boo': 0.2}]

edited May 09 '21 at 16:39

answered May 09 '21 at 15:55

Andrej Kesely

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1

You probably want to sort by foo and boo to have a tie breaker imo. Like key=lambda k: (k["foo"], k["boo"]) – Cedric May 09 '21 at 15:57
Thank you so much. That worked. May I ask you if it's possible to also include keys in each element? For example, `{"key": "c", "foo": 0.8, "boo": 0.3}` – Zack Lee May 09 '21 at 16:35
1

@ZackLee See my edit. – Andrej Kesely May 09 '21 at 16:39
1

@AndrejKesely Thank you so much you are my hero! – Zack Lee May 09 '21 at 16:41

How to properly sort a dictionary by value of a key?

1 Answers1