Sort dictionary of lists by the second element in each list

Question

I have a dictionary of lists such as:

test_dict = { 'a' : [[1, 6, 8], [2, 5, 9], [54, 1, 34]],
              'b' : [[1, 3, 8], [2, 1, 9], [54, 2, 34]],
              'c' : [[1, 1, 8], [2, 9, 9], [54, 7, 34]]
            }

and I want to sort (ascending) each value list by the second element in every sublist. The desired output dictionary would be:

output_dict = { 'a' : [[54, 1, 34], [2, 5, 9], [1, 6, 8]],
                'b' : [[2, 1, 9], [54, 2, 34], [1, 3, 8]],
                'c' : [[1, 1, 8], [54, 7, 34], [2, 9, 9]]
              }

I am trying:

sorted_dict = dict(sorted(test_dict.items(), key=lambda e: e[1][1]))
sorted_dict.items()

bit this does not seem to do anything.

Answer 1

You are looking to sort lists of lists in your dictionary values, not the order of dictionary keys . For this, you can use a dictionary comprehension:

res = {k: sorted(v, key=lambda x: x[1]) for k, v in test_dict.items()}

{'a': [[54, 1, 34], [2, 5, 9], [1, 6, 8]],
 'b': [[2, 1, 9], [54, 2, 34], [1, 3, 8]],
 'c': [[1, 1, 8], [54, 7, 34], [2, 9, 9]]}

For the functional equivalent, you can use key=operator.itemgetter(1) . In Python 3.6+, your dictionary order should be maintained. Prior to 3.6 dictionaries are unordered and you should not expect any particular ordering of keys.

To order by key, you can use collections.OrderedDict :

from collections import OrderedDict

res_ordered = OrderedDict(sorted(res.items(), key=lambda x: x[0]))

OrderedDict([('a', [[54, 1, 34], [2, 5, 9], [1, 6, 8]]),
             ('b', [[2, 1, 9], [54, 2, 34], [1, 3, 8]]),
             ('c', [[1, 1, 8], [54, 7, 34], [2, 9, 9]])])

Answer 2

You can do like below, it will update the existing dictionary

test_dict = { 'a' : [[1, 6, 8], [2, 5, 9], [54, 1, 34]],
              'b' : [[1, 3, 8], [2, 1, 9], [54, 2, 34]],
              'c' : [[1, 1, 8], [2, 9, 9], [54, 7, 34]]
            }

for k, v in test_dict.items():
    test_dict[k] = sorted(v, key=lambda e: e[1])

print(test_dict)

to create a new dictionary

test_dict = { 'a' : [[1, 6, 8], [2, 5, 9], [54, 1, 34]],
              'b' : [[1, 3, 8], [2, 1, 9], [54, 2, 34]],
              'c' : [[1, 1, 8], [2, 9, 9], [54, 7, 34]]
            }

new_dict = {k:sorted(v, key=lambda e: e[1]) for k, v in test_dict.items()}

print(new_dict)