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)python sort dictionary alphabetically by key
sortednames=sorted(dictUsers.keys(), key=lambda x:x.lower())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 dict by key
A={1:2, -1:4, 4:-20}
{k:A[k] for k in sorted(A)}
output:
{-1: 4, 1: 2, 4: -20}sort dictionary python
l = {1: 40, 2: 60, 3: 50, 4: 30, 5: 20}
d1 = dict(sorted(l.items(),key=lambda x:x[1],reverse=True))
print(d1) #output : {2: 60, 3: 50, 1: 40, 4: 30, 5: 20}
d2 = dict(sorted(l.items(),key=lambda x:x[1],reverse=False))
print(d2) #output : {5: 20, 4: 30, 1: 40, 3: 50, 2: 60}sort the dictionary in python
d = {2: 3, 1: 89, 4: 5, 3: 0}
od = sorted(d.items())
print(od)Copyright © 2021 Codeinu
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