Sort dictionary by value python

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}

Sort dictionary by value python

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}

Source: stackoverflow.com

Sort dictionary by value python

orders = {
	'Pizza': 33,
	'Burger': 45,
	'Sandwich': 67,
	'Latte': 39,
	'Snickers': 48
}

sort_orders = sorted(orders.items(), key=lambda x: x[1], reverse=True)

for i in sort_orders:
	print(i[0], i[1])

Source: itsmycode.com

Similar codes

0 vote116 views1 answer
Python running mean pandas
Python
1 vote104 views1 answer
Mypy ignore line
Python
13 votes100 views1 answer
Python read file to variable
Python
0 vote107 views1 answer
Ttk button
Python
12 votes109 views4 answers
Dictionary append value python
Python
3 votes133 views2 answers
How to restart program in python
Python
26 votes116 views3 answers
Python split string in characters
Python
18 votes149 views4 answers
Remove n characters from string python
Python
0 vote123 views1 answer
Zen of python source code
Python
3 votes105 views2 answers
Creating new virtual environment in python
Python