How can I sort a dictionary in Python by the value of a specific sub-key in descending order?

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1

How can I sort a dictionary in Python by the value of a specific sub-key in descending order?

I have the following dictionary d and I want to sort it based on the value of key3 in descending order:

d = { '123': { 'key1': 3, 'key2': 11, 'key3': 3 },
      '124': { 'key1': 6, 'key2': 56, 'key3': 6 },
      '125': { 'key1': 7, 'key2': 44, 'key3': 9 },
    }

After sorting, the dictionary should look like this:

d = { '125': { 'key1': 7, 'key2': 44, 'key3': 9 },
      '124': { 'key1': 6, 'key2': 56, 'key3': 6 },
      '123': { 'key1': 3, 'key2': 11, 'key3': 3 },
    }

My approach was to create another dictionary e from d, using the value of key3 as the key in e and then sorting the dictionary. However, since the value of key3 can be the same for multiple entries, I lost some keys and their values.

How can I properly sort this dictionary by the value of key3 in descending order without losing any keys or values?

2

Using OrderedDict: This one’s my go-to approach when I want to preserve order in the sorted dictionary.

from collections import OrderedDict

d = { '123': { 'key1': 3, 'key2': 11, 'key3': 3 },
      '124': { 'key1': 6, 'key2': 56, 'key3': 6 },
      '125': { 'key1': 7, 'key2': 44, 'key3': 9 },
    }

# Sort dictionary in descending order based on 'key3'
d_descending = OrderedDict(sorted(d.items(), key=lambda kv: kv[1]['key3'], reverse=True))

In this solution, d_descending will contain the dictionary sorted by the values of key3 in descending order. The OrderedDict ensures that the order is preserved. This method works perfectly when order retention matters.

3

That’s a solid way, Charity! Another neat trick is to avoid creating a new dictionary altogether and just use sorted() when iterating. This is handy when you want to process the items directly without extra overhead.

d = { '123': { 'key1': 3, 'key2': 11, 'key3': 3 },
      '124': { 'key1': 6, 'key2': 56, 'key3': 6 },
      '125': { 'key1': 7, 'key2': 44, 'key3': 9 },
    }

# Iterate over sorted dictionary based on 'key3' in descending order
for key, value in sorted(d.items(), key=lambda kv: kv[1]['key3'], reverse=True):
    do_something_with(key, value)

This approach skips modifying the dictionary itself and gets the job done right in the loop. Clean, efficient, and no need for an additional data structure.

4

Great points, Charity and Devan! Let me add another angle to this. If you’re looking for the top n elements based on key3 in descending order, you can use heapq.nlargest(). It’s incredibly efficient for that specific purpose.

import heapq

d = { '123': { 'key1': 3, 'key2': 11, 'key3': 3 },
      '124': { 'key1': 6, 'key2': 56, 'key3': 6 },
      '125': { 'key1': 7, 'key2': 44, 'key3': 9 },
    }

# Get the top 2 items sorted by 'key3' in descending order
top_n = heapq.nlargest(2, d.items(), key=lambda kv: kv[1]['key3'])

This solution shines when you’re dealing with a large dictionary but only care about a small subset of the highest-ranked items. Why sort the entire dictionary when you only need the top n? heapq.nlargest() nails that efficiency!