Answers for "sort dictionary based on values"

5

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}
Posted by: Guest on July-12-2021
35

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)
Posted by: Guest on November-23-2020
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

sort dict by value

dict(sorted(x.items(), key=lambda item: item[1]))
Posted by: Guest on March-17-2021
1

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}
Posted by: Guest on August-08-2021

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