0
Q:

python sort a dictionary

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))
8
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}
6
from operator import itemgetter
new_dict = sorted(data.items(), key=itemgetter(1))
0
#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 tuples
1
d = {2: 3, 1: 89, 4: 5, 3: 0}
od = sorted(d.items())
print(od)
1

 print(sorted(key_value.items(), key = lambda kv:(kv[1], kv[0])))     

2

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