Sorting Python dictionary based on number of items in list

Question

I have a small script in Python with a dictionary same as follows:

d = {'category_1' : ['a', 'b'], 
'category_2' : ['c', 'd', 'e'],
'category_3' : ['z']}

How can I sort it based on number of values in list? I want it looks like:

d = {'category_3' : ['z'], 
'category_1' : ['a', 'b'], 
'category_2' : ['c', 'd', 'e']}

score 7 · Accepted Answer · answered Jun 28 '13 at 10:47

7

Dictionaries in Python are orderless.

In order to actually store ordering, you will need to either have a list of tuples, or use a collections.OrderedDict().

>>> from collections import OrderedDict
>>> OrderedDict(sorted(d.items(), key=lambda item: len(item[1])))
OrderedDict([('category_3', ['z']), ('category_1', ['a', 'b']), ('category_2', ['c', 'd', 'e'])])

The ordering is achieved here by using the sorted() built-in, with a simple key function.

answered Jun 28 '13 at 10:47

Gareth Latty

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or `key=lambda (k,v): len(v)` (for Python2) – John La Rooy Jun 28 '13 at 11:11
@gnibbler Only in 2.x though. – Gareth Latty Jun 28 '13 at 11:12
@Lattyware, unfortunately it's one use case I miss with 3, although 2to3 should rewrite it ok to `key=lambda k_v: len(k_v[1])`, `item` is more readable than `k_v` – John La Rooy Jun 28 '13 at 11:27

Sorting Python dictionary based on number of items in list

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