Sort list of dictionaries with inconsistent keys

Question

I want to sort a list of dictionaries. The problem is that key in the dictionaries are not same, but every dictionary will have only one item for sure. For example, [{'foo':39}, {'bar':7}, {'spam':35}, {'buzz':4}] Here, key is the name of the person and value is the age. I want result as [{'buzz': 4}, {'bar': 7}, {'spam': 35}, {'foo': 39}] What I am doing is :

def get_val(d):
    for k, v in d.items():
        return v

sorted_lst = sorted(lst, key=lambda d: get_val(d))

Is there any better solution?

I believe this question has already been asked here: https://stackoverflow.com/questions/9001509/how-can-i-sort-a-dictionary-by-key — ramzeek, Oct 24 '21 at 04:38
@rv.kvetch. Actually `next(iter(d.values())`. (Your solution was the first thing I tried and I got an error.) — Frank Yellin, Oct 24 '21 at 04:40
@wikikikitiki I have gone through the link. That was a different problem. — soumya-kole, Oct 24 '21 at 04:41
@FrankYellin thanks, I was looking for something like that only to avoid a function just to return the value — soumya-kole, Oct 24 '21 at 04:45

I'mahdi · Accepted Answer · 2021-10-24T04:48:36.097

You can use values of dict in lambda like below :

>>> lst_dct = [{'foo':39}, {'bar':7}, {'spam':35}, {'buzz':4}]

>>> sorted(lst_dct, key=lambda x: list(x.values()))
[{'buzz': 4}, {'bar': 7}, {'spam': 35}, {'foo': 39}]

You can extent this for use list with multi elements:

>>> lst_dct = [{'foo':[39, 40]}, {'bar':[7,8]}, {'spam':[4,5]}, {'buzz':[4,6]}]

>>> sorted(lst_dct, key=lambda x: sorted(x.values()))
[{'spam': [4, 5]}, {'buzz': [4, 6]}, {'bar': [7, 8]}, {'foo': [39, 40]}]

Sort list of dictionaries with inconsistent keys

1 Answers1