Python 3 sort a dict by its values

from collections import OrderedDict
from operator import itemgetter    

d = {"aa": 3, "bb": 4, "cc": 2, "dd": 1}
print(OrderedDict(sorted(d.items(), key = itemgetter(1), reverse = True)))

prints

OrderedDict([('bb', 4), ('aa', 3), ('cc', 2), ('dd', 1)])

Though from your last sentence, it appears that a list of tuples would work just fine, e.g.

from operator import itemgetter  

d = {"aa": 3, "bb": 4, "cc": 2, "dd": 1}
for key, value in sorted(d.items(), key = itemgetter(1), reverse = True):
    print(key, value)

which prints

bb 4
aa 3
cc 2
dd 1

You can sort by values in reverse order (largest to smallest) using a dictionary comprehension:

{k: d[k] for k in sorted(d, key=d.get, reverse=True)}
# {'b': 4, 'a': 3, 'c': 2, 'd': 1}

If you want to sort by values in ascending order (smallest to largest)

{k: d[k] for k in sorted(d, key=d.get)}
# {'d': 1, 'c': 2, 'a': 3, 'b': 4}

If you want to sort by the keys in ascending order

{k: d[k] for k in sorted(d)}
# {'a': 3, 'b': 4, 'c': 2, 'd': 1}

This works on CPython 3.6+ and any implementation of Python 3.7+ because dictionaries keep insertion order.

itemgetter (see other answers) is (as I know) more efficient for large dictionaries but for the common case, I believe that d.get wins. And it does not require an extra import.

>>> d = {"aa": 3, "bb": 4, "cc": 2, "dd": 1}
>>> for k in sorted(d, key=d.get, reverse=True):
...     k, d[k]
...
('bb', 4)
('aa', 3)
('cc', 2)
('dd', 1)

Note that alternatively you can set d.__getitem__ as key function which may provide a small performance boost over d.get.