how can I sort a dictionary in python according to its values?
s = {1: 1, 7: 2, 4: 2, 3: 1, 8: 1}
k = dict(sorted(s.items(),key=lambda x:x[0],reverse = True))
print(k)how can I sort a dictionary in python according to its values?
s = {1: 1, 7: 2, 4: 2, 3: 1, 8: 1}
k = dict(sorted(s.items(),key=lambda x:x[0],reverse = True))
print(k)how to sort a dictionary by value in python
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))python - sort dictionary by value
d = {'one':1,'three':3,'five':5,'two':2,'four':4}
# Sort
a = sorted(d.items(), key=lambda x: x[1])
# Reverse sort
r = sorted(d.items(), key=lambda x: x[1], reverse=True)sort dict by value
dict(sorted(x.items(), key=lambda item: item[1]))sort dict by value
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}python sort dict by value
A={1:2, -1:4, 4:-20}
{k:A[k] for k in sorted(A, key=A.get)}
output:
{4: -20, 1: 2, -1: 4}Copyright © 2021 Codeinu
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