按照字典的值进行排序

Posted 2020-10-06

按照字典值的大小对字典进行排序

1. 利用zip函数将key值和value值位置进行颠倒，再使用sorted进行排序

例如：

from random import  randint
d= {k:randint(0,20) for k in ‘abcdefg‘}
#{‘f‘: 10, ‘g‘: 16, ‘a‘: 18, ‘c‘: 12, ‘e‘: 16, ‘d‘: 7, ‘b‘: 16}
z = list(zip(d.values(),d.keys()))
# zip函数就是将输入的序列重新组合成每个元素都是元组的列表，其中每个元素由输入序列相同位置的元素组成
#[(10, ‘f‘), (16, ‘g‘), (18, ‘a‘), (12, ‘c‘), (16, ‘e‘), (7, ‘d‘), (16, ‘b‘)]
print(sorted(zip(d.values(),d.keys())))
#[(7, ‘d‘), (10, ‘f‘), (12, ‘c‘), (16, ‘b‘), (16, ‘e‘), (16, ‘g‘), (18, ‘a‘)]

2.利用匿名函数进行排序

print(sorted(d.items(),key = lambda k : k[1]))
#[(‘d‘, 7), (‘f‘, 10), (‘c‘, 12), (‘g‘, 16), (‘e‘, 16), (‘b‘, 16), (‘a‘, 18)]

3.　使用operator.itemgetter函数

import operator
print(sorted(d.items(), key=operator.itemgetter(1)))
#[(‘d‘, 7), (‘f‘, 10), (‘c‘, 12), (‘g‘, 16), (‘e‘, 16), (‘b‘, 16), (‘a‘, 18)]