Sort list of single item dictionaries?

Question

I have a list with individual items as dictionaries which might have different keys. I want to sort them based on the values. E.g.

Lets say,

unsorted_list = [{'a': 23}, {'b': 34}, {'c': 2}]

After sort, (descending)

sorted_list = [{'b': 34}, {'a': 23}, {'c': 2}]

Please let me how to do it in python

Folks: This is not a duplicate of any of those. The question is about sorting a list that contains dictionaries (by the contents of the dictionaries). — martineau, Oct 13 '17 at 14:56
This question is slightly different in the case that here each item in the list is a dict. Also keys are different for different dicts. — Kanhaiya Choudhary, Oct 13 '17 at 14:56

score 2 · Accepted Answer · answered Oct 13 '17 at 14:59

you need to sort the elements according to the dict values (there is only one value anyway), in reverse:

unsorted_list = [{'a': 23}, {'b': 34}, {'c': 2}]

sorted_list = sorted(unsorted_list, key = lambda d : list(d.values()), reverse=True)

result:

[{'b': 34}, {'a': 23}, {'c': 2}]

score 1 · Answer 2 · answered Oct 13 '17 at 15:03

1

You can try this:

unsorted_list = [{'a': 23}, {'b': 34}, {'c': 2}]
final_data = sorted(unsorted_list, key=lambda x:x.values()[0])[::-1]

Output:

[{'b': 34}, {'a': 23}, {'c': 2}]

answered Oct 13 '17 at 15:03

Ajax1234

1

You should pass the `reverse=True` keyword to `sorted()` instead of the having the `[::-1]` at the end. – martineau Oct 13 '17 at 15:11

score 0 · Answer 3 · answered Oct 13 '17 at 14:59

0

This should do what you need:

sorted_list = sorted(unsorted_list, key=lambda x: list(x.values())[0]*-1)

or

sorted_list = sorted(unsorted_list, key=lambda x: list(x.values())[0], reverse=True)

answered Oct 13 '17 at 14:59

zipa

3 Answers3