Python3按其值对一个 dict 进行排序

To sort a dictionary and keep it functioning as a dictionary afterwards, you could use OrderedDict from the standard library.

If that's not what you need, then I encourage you to reconsider the sort functions that leave you with a list of tuples. What output did you want, if not an ordered list of key-value pairs (tuples)?

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

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.

To sort dictionary, we could make use of operator module. Here is the operator module documentation.

import operator                             #Importing operator module
dc =  {"aa": 3, "bb": 4, "cc": 2, "dd": 1}  #Dictionary to be sorted


dc_sort = sorted(dc.items(),key = operator.itemgetter(1),reverse = True)
print dc_sort

Output sequence will be a sorted list :

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

If we want to sort with respect to keys, we can make use of

dc_sort = sorted(dc.items(),key = operator.itemgetter(0),reverse = True)

Output sequence will be :

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

Another way is to use a lambda expression. Depending on interpreter version and whether you wish to create a sorted dictionary or sorted key-value tuples (as the OP does), this may even be faster than the accepted answer.

d = {'aa': 3, 'bb': 4, 'cc': 2, 'dd': 1}
s = sorted(d.items(), key=lambda x: x[1], reverse=True)


for k, v in s:
print(k, v)

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.