Answered by : mobile-star
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))
Source : | Last Update : Wed, 04 Dec 19
Answered by : aman-kumar-verma
# your dictionary
a = {'a':4, 'c':5, 'b':3, 'd':0}
# sort x by keys
a_keys = dict(sorted(a.items(),key=lambda x:x[0],reverse = False)) # ascending order
# output: {'a': 4, 'b': 3, 'c': 5, 'd': 0}
# # sort x by values
a_values = dict(sorted(a.items(),key=lambda x:x[1],reverse = False)) # ascending order
# output: {'d': 0, 'b': 3, 'a': 4, 'c': 5}
Source : | Last Update : Tue, 31 Aug 21
Answered by : razvan
from operator import itemgetter
new_dict = sorted(data.items(), key=itemgetter(1))
Source : | Last Update : Thu, 01 Oct 20
Answered by : powerful-porpoise-5vfq6nc5275i
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
Source : | Last Update : Fri, 27 May 22
Answered by : crowded-crossbill-m7g8nzp5m881
sorted(y.items(), key=lambda x: (x[1],x[0]))
Source : https://stackoverflow.com/questions/7742752/sorting-a-dictionary-by-value-then-by-key | Last Update : Wed, 11 May 22
Answered by : kiril-klein
>>> 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}
Source : https://stackoverflow.com/questions/613183/how-do-i-sort-a-dictionary-by-value | Last Update : Mon, 28 Feb 22
Answered by : jinn-world
dt = {5:4, 1:6, 6:3}
sorted_dt = {key: value for key, value in sorted(dt.items(), key=lambda item: item[1])}
print(sorted_dt)
Source : https://www.programiz.com/python-programming/examples/sort-dictionary-value | Last Update : Tue, 02 Aug 22
Answered by : aldo-cauvi
d = {'apple': 2, 'banana': 3, 'almond':2 , 'beetroot': 3, 'peach': 4}
[v[0] for v in sorted(d.items(), key=lambda kv: (-kv[1], kv[0]))]
Source : https://stackoverflow.com/questions/9919342/sorting-a-dictionary-by-value-then-key | Last Update : Sat, 13 Mar 21
Answered by : jinn-world
dt = {5:4, 1:6, 6:3}
sorted_dt_value = sorted(dt.values())
print(sorted_dt_value)
Source : https://www.programiz.com/python-programming/examples/sort-dictionary-value | Last Update : Tue, 02 Aug 22