Sorting Dictionaries by Internal State of Value in Python

Question

How can we sort a dictionary in python using the internal state of the value?

For example:

myDict = {'a': {'day': [0, 1, 2, 5], 'week': [4, 3, 1, 2, 3]},
          'b': {'day': [0, 6, 2, 5], 'week': [4, 3, 1, 2, 3]}}

I want to sort the dictionary keys by the second value of the day list for their corresponding value dictionaries.

Dictionaries can't be sorted, what was the output you had in mind? — thefourtheye, May 12 '14 at 10:45
Yes, but sorted based on a value in their day lists in their value objects. — Moriarty, May 12 '14 at 10:48
Checked http://stackoverflow.com/questions/613183/python-sort-a-dictionary-by-value, but its not clear how to sort based on a more complex value object (like a dict in this case). itemgetter does not work with str — Moriarty, May 12 '14 at 10:50

score 3 · Accepted Answer · answered May 12 '14 at 10:53

You can sort that like this

print sorted(myDict, key=lambda v: myDict[v]['day'][1])
# ['a', 'b']

The key parameter accepts a function which will be used to determine the value to be used for the current item during the sorting.

As sorted iterates over the myDict, we get each and every key. So, for each key, we pick the second element in the day dictionary and use it as the value which represents the key during sorting.

What we have done is, when sorting function looks for a value of a we return 1 and when it looks for b we return 6.

This is what I needed. The question marked as a duplicate does not address this specific case. — Moriarty, May 13 '14 at 07:27

Sorting Dictionaries by Internal State of Value in Python

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