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)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)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))sort the dictionary in python
d = {2: 3, 1: 89, 4: 5, 3: 0}
od = sorted(d.items())
print(od)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}sort dictionary
#for dictionary d
sorted(d.items(), key=lambda x: x[1]) #for inceasing order
sorted(d.items(), key=lambda x: x[1], reverse=True) # for decreasing order
#it will return list of key value pair tupleshow 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])Copyright © 2021 Codeinu
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