Answers for "sort by dictionary value python"

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
4

python sort dictionary by value descending

Python Code:
import operator
d = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
print('Original dictionary : ',d)
sorted_d = dict(sorted(d.items(), key=operator.itemgetter(1)))
print('Dictionary in ascending order by value : ',sorted_d)
sorted_d = dict(sorted(d.items(), key=operator.itemgetter(1),reverse=True))
print('Dictionary in descending order by value : ',sorted_d)

Sample Output:
Original dictionary :  {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
Dictionary in ascending order by value :  {0: 0, 2: 1, 1: 2, 4: 3, 3: 4}
Dictionary in descending order by value :  {3: 4, 4: 3, 1: 2, 2: 1, 0: 0}
Posted by: Guest on June-06-2020
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
2

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 November-25-2020
1

sort dict by value

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

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