Answers for "order dict by value python 3"

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
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
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
0

how to sort dictionary in python by value

# 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}
Posted by: Guest on August-24-2021

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