Sort a dictionary

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)

Sorting a dictionary in 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

Python sort dict by key

A={1:2, -1:4, 4:-20}
{k:A[k] for k in sorted(A)}

output:
{-1: 4, 1: 2, 4: -20}

Python sort dictionary by key

In [1]: import collections

In [2]: d = {2:3, 1:89, 4:5, 3:0}

In [3]: od = collections.OrderedDict(sorted(d.items()))

In [4]: od
Out[4]: OrderedDict([(1, 89), (2, 3), (3, 0), (4, 5)])

Source: stackoverflow.com

Sort a dictionary

from operator import itemgetter
new_dict = sorted(data.items(), key=itemgetter(1))

How to sort a dictionary py

orders = {
	'cappuccino': 54,
	'latte': 56,
	'espresso': 72,
	'americano': 48,
	'cortado': 41
}

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

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

Source: careerkarma.com

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