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order dictionary by value python

Holteeer 
import operator
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=operator.itemgetter(1))


# Sort by key
import operator
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=operator.itemgetter(0))
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3.78
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Stranger 95 points 

                                    dict(sorted(x.items(), key=lambda item: item[1]))

3.78 (9 Votes)
0
3.78
9
George Feb 135 points 

                                    x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_dict = {k: v for k, v in sorted(x.items(), key=lambda item: item[1])}
print(sorted_dict)
#{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}

3.78 (9 Votes)
0
3.89
9
Satyam 85 points 

                                    x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
{k: v for k, v in sorted(x.items(), key=lambda item: item[1])}
{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}

3.89 (9 Votes)
0
0
0
Manny265 100 points 

                                    >>> 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)

0
0
0
1
Manjul Rathee 130 points 

                                    from operator import itemgetter
new_dict = sorted(data.items(), key=itemgetter(1))

0
0
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