Answers for "sorted dictionary python by value"

12

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))
Posted by: Guest on November-27-2019
1

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)
Posted by: Guest on February-25-2021
1

python get dictionary keys sorted by value

sorted(A, key=A.get)
Posted by: Guest on October-10-2020
3

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}
Posted by: Guest on April-22-2020
0

order dictionary by value python

x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_dict = {k: v for k, v in sorted(x.items(), key=lambda item: item[1])}
print(sorted_dict)
#{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}
Posted by: Guest on June-28-2021
0

sort a dictionary by value then key

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]))]
Posted by: Guest on March-06-2021

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