Python sort dict by value

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)

Sort list of dictionaries python by value

newlist = sorted(list_to_be_sorted, key=lambda k: k['name'])

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))

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}

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)

Source: stackoverflow.com

Python 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])}

Source: stackoverflow.com

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