python sort a dictionary by values
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sort_by_key = dict(sorted(x.items(),key=lambda item:item[0]))
sort_by_value = dict(sorted(x.items(), key=lambda item: item[1]))
print("sort_by_key:", sort_by_key)
print("sort_by_value:", sort_by_value)
# sort_by_key: {0: 0, 1: 2, 2: 1, 3: 4, 4: 3}
# sort_by_value: {0: 0, 2: 1, 1: 2, 4: 3, 3: 4}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)sort dictionary by values
from collections import OrderedDict
dd = OrderedDict(sorted(d.items(), key=lambda x: x[1]))
print(dd)sorting values in dictionary in python
#instead of using python inbuilt function we can it compute directly.
#here iam sorting the values in descending order..
d = {1: 1, 7: 2, 4: 2, 3: 1, 8: 1}
s=[]
for i in d.items():
s.append(i)
for i in range(0,len(s)):
for j in range(i+1,len(s)):
if s[i][1]<s[j][1]:
s[i],s[j]=s[j],s[i]
print(dict(s))sorting-a-dictionary-by-value-then-by-key
[(14,2), (12,2), (9,1)] # output from print
[(90,4), (99,3), (101,1), (100,1), (92,1)]Copyright © 2021 Codeinu
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