python: sorting a dict of dicts on a key

python

Solution

d = {'ford': {'count': 3},
     'mazda': {'count': 0},
     'toyota': {'count': 1}}

>>> sorted(d.items(), key=lambda (k, v): v['count'])
[('mazda', {'count': 0}), ('toyota', {'count': 1}), ('ford', {'count': 3})]

To keep the result as a dictionary, you can used `collections.OrderedDict`:

>>> from collections import OrderedDict
>>> ordered = OrderedDict(sorted(d.items(), key=lambda (k, v): v['count']))
>>> ordered
OrderedDict([('mazda', {'count': 0}), ('toyota', {'count': 1}), ('ford', {'count': 3})])
>>> ordered.keys()          # this is guaranteed to come back in the sorted order
['mazda', 'toyota', 'ford']
>>> ordered['mazda']        # still a dictionary
{'count': 0}

Version concerns:

- On Python 2.x you could use `d.iteritems()` instead of `d.items()` for better memory efficiency

- `collections.OrderedDict` is only available on Python 2.7 and Python 3.2 (and higher)

Problem

Original source