python字典排序总结

dic = {"abc":18,"adc":19,"abe":20}
# 默认对键排序，从小到大，返回排序后键组成的列表
zidian = sorted(dic)#[‘abc‘, ‘abe‘, ‘adc‘]
print(zidian)
# 对键进行反向（从大到小）排序
zidian = sorted(dic,reverse=True)#[‘adc‘, ‘abe‘, ‘abc‘]
print(zidian)
# 拿到所有的key，然后再对key排序
zidian = sorted(dic.keys(),reverse=True)#[‘adc‘, ‘abe‘, ‘abc‘]
print(zidian)
# 对值排序，从小到大
print(dic)
zidian = sorted(dic.values())#[18, 19, 20]
print(zidian)
# 对值排序，从大到小
zidian = sorted(dic.values(),reverse=True)#[20, 19, 18]
print(zidian)
#可以用dict1.items()，得到包含键，值的元组,
# 由于迭代对象是元组，返回值自然是元组组成的列表,x指元组，x[1]是值，x[0]是键

# 键由小到大排序
zidian = sorted(dic.items(),key=lambda x:x[0])
print(zidian)

# 键由大到小排序
zidian = sorted(dic.items(),key=lambda x:x[0],reverse=True)
print(zidian)

# 值由小到大排序
zidian = sorted(dic.items(),key=lambda x:x[1])
print(zidian)
# 值由大到小排序
zidian = sorted(dic.items(),key=lambda x:x[1],reverse=True)
print(zidian)

#itemgetter(0)，获取key
# itemgetter(1)，获取value
from operator import itemgetter
d = {"a":8,"b":4,"c":12,"a":10,"b":1,"e":10}
# 键由大到小
print(sorted(d.items(),key=itemgetter(0),reverse=True))#[(‘e‘, 10), (‘c‘, 12), (‘b‘, 1), (‘a‘, 10)]
# 值由大到小
print(sorted(d.items(),key=itemgetter(1),reverse=True))#[(‘c‘, 12), (‘a‘, 10), (‘e‘, 10), (‘b‘, 1)]

原文地址：https://www.cnblogs.com/songdanlee/p/11135282.html