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)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)how to sort dict by value
dict1 = {1: 1, 2: 9, 3: 4}
sorted_dict = {}
sorted_keys = sorted(dict1, key=dict1.get) # [1, 3, 2]
for w in sorted_keys:
sorted_dict[w] = dict1[w]
print(sorted_dict) # {1: 1, 3: 4, 2: 9}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 dictionary by value
from operator import itemgetter
new_dict = sorted(data.items(), key=itemgetter(1))Copyright © 2021 Codeinu
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