How to sort a list of nested dictionaries with different keys

Question

I have the following list:

mylist = [
   { 'frame': { 'aaa': 50 } },
   { 'frame': { 'bbb': 12 } },
   { 'frame': { 'ccc': 23 } },
   { 'frame': { 'ddd': 72 } }
]

I was hoping to sort the list based on those integer values with different keys 'aaa', bbb', 'ccc', etc.

I have seen this question but the keys are consistent to this problem.

Any help is highly appreciated!

EDIT: My desired output is:

sorted_list = [
       { 'frame': { 'ddd': 72 } },
       { 'frame': { 'aaa': 50 } },
       { 'frame': { 'ccc': 23 } },
       { 'frame': { 'bbb': 12 } }
]

Answer 1

mylist = sorted(mylist, key=lambda k: -list(k["frame"].values())[0])
print(mylist)

Prints:

[{'frame': {'ddd': 72}}, 
 {'frame': {'aaa': 50}}, 
 {'frame': {'ccc': 23}}, 
 {'frame': {'bbb': 12}}]

Answer 2

You can usesorted with a lambda for the key which just returns the first value stored in the dict under the 'frame' key.

We get the first value by using next with iter . This avoids creating a new list object.

We also pass the reverse parameter as True , so we get biggest numbers first. :

>>> mylist = [
...    { 'frame': { 'aaa': 50 } },
...    { 'frame': { 'bbb': 12 } },
...    { 'frame': { 'ccc': 23 } },
...    { 'frame': { 'ddd': 72 } }
... ]

>>> sorted(mylist, key=lambda d: next(iter(d['frame'].values())), reverse=True)
[{'frame': {'ddd': 72}},
 {'frame': {'aaa': 50}},
 {'frame': {'ccc': 23}},
 {'frame': {'bbb': 12}}]

Answer 3

A slight tweak to the answer in that question gets you this which works.

sorted_list = sorted(mylist, key=lambda k: k['frame'][list(k['frame'].keys())[0]], reverse=True)

Essentially it just goes down into the list, To then sort by the values. Set reverse to True to get it to reverse the list.