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First, we use the sorted() function to order the values of the dictionary. We then loop through the sorted values, finding the keys for each value. We add these keys-value pairs in the sorted order into a new dictionary. Note: Sorting does not allow you to re-order the dictionary in-place.
We can sort lists, tuples, strings, and other iterable objects in python since they are all ordered objects. Well, as of python 3.7, dictionaries remember the order of items inserted as well. Thus we are also able to sort dictionaries using python's built-in sorted() function.
Use dict. items() to get a list of tuple pairs from d and sort it using a lambda function and sorted(). Use dict() to convert the sorted list back to a dictionary. Use the reverse parameter in sorted() to sort the dictionary in reverse order, based on the second argument.
The key=lambda x: (x[1],x[0]) tells sorted that for each item x in y. items() , use (x[1],x[0]) as the proxy value to be sorted. Since x is of the form (key,value) , (x[1],x[0]) yields (value,key) . This causes sorted to sort by value first, then by key for tie-breakers.
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.
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.
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)
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)]
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