一、列表排序

demoList = [1, 3, 2, 4, 9 ,7]

res = sorted(demoList)  # 默认升序

# 降序
# res = sorted(demoList, reverse=True)

print(res)

二、字典排序

demoDict = {"篮球": 5, "排球": 9, "网球": 6, "足球": 3}

# sorted排序
res = sorted(demoDict.items(),key=lambda x:x[1])
print(res)
# sort排序，其实将字典转成<<列表+元组>>格式排序
newList = list(demoDict.items())  # [('篮球', 5), ('排球', 9), ('网球', 6), ('足球', 3)]
newList.sort(key=lambda x:x[1])
print(newList)
print(dict(newList))

三、<<列表+字典>>排序

demoList = [
	{"name": "张三", "age": 16}, {"name": "李四", "age": 24}, 
	{"name": "王五", "age": 20}, {"name": "赵六", "age": 18}
]

sorted(demoList, key=lambda x:x["age"])

四、<<列表+元组>>排序

• 在 (二、字典排序)中有介绍

五、<<列表+列表>>排序

demoList = [
	["张三", 16, "北京市"], ["李四", 24, "上海市"], 
	["王五", 20, "广州市"], ["赵六", 18, "深圳市"]
]

sorted(demoList, key=lambda x:x[1])

六、<<列表+对象>>排序

class Student:
    def __init__(self, name, grade, age):
        self.name = name
        self.grade = grade
        self.age = age
    def __repr__(self):
        return repr((self.name, self.grade, self.age))

stuObj = [
        Student('john', 'A', 15),
        Student('jane', 'B', 12),
        Student('dave', 'B', 10),
]

sorted(stuObj, key=lambda x: x.age)

七、<<列表+字典运算>>排序

demoList = [
    {'key1': 1, 'key2': 1}, {'key1': 5, 'key2': 1}, {'key1': 3, 'key2': 6}, 
    {'key1': 2, 'key2': 3}, {'key1': 4, 'key2': 9}
]

demoList.sort(key=lambda x:x["key1"]+x["key2"])
print(demoList)

八、高级拓展

8.1、None值无法被排序

demoList = [
	{"name": "张三", "age": 16}, {"name": "李四", "age": 24}, 
	{"name": "王五", "age": 20}, {"name": "赵六", "age": 18},
	{"name": "孙七", "age": None}
]

• 解决方式

demoList.sort(key=lambda x:(x["age"] is None, x["age"] == "", x["age"]))
print(demoList)

8.2、字典取最大|小值

demoDict = {"篮球": 5, "排球": 9, "网球": 6, "足球": 3}

# 最大值
maxData = max(demoDict.items(), key=lambda x:x[1])
print(maxData)
# 最小值
minData = min(demoDict.items(), key=lambda x:x[1])
print(minData)

8.3、多重条件排序

demoList = [
    {"name": "诺基亚K", "price": 2399, "level": 1}, {"name":"魅族", "price": 2499, "level": 3},
    {"name": "中兴E", "price": 2399, "level": 3}, {"name": "vivoZ", "price": 2799, "level": 2}, 
    {"name": "oppoT", "price": 2699, "level": 2}, {"name": "小米Y", "price": 2999, "level": 2}, 
    {"name": "华为X", "price": 2999, "level": 1}, {"name": "苹果P", "price": 5799, "level": 1}, 
    {"name": "三星W", "price": 5799, "level": 1}
]

# 两重查询条件，价格(默认升序)、级别(默认升序)
sorted(demoList, key=lambda x:(x["price"], x["level"]))
# 两重查询条件，价格(默认升序)、级别(降序可以用-)
sorted(demoList, key=lambda x:(x["price"], -x["level"]))

• 价格⬆级别⬆
  
• 价格⬆级别⬇
  

8.4、operator实现排序

• 排序的思想同匿名函数lambda一样

# 举个字典排序
demoDict = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}

# lambda排序
sorted(demoDict.items(), key=x: x[1])

# operator排序
import operator
sorted(demoDict.items(), key=operator.itemgetter(1))

# 仔细看其实排序都是一样的，无非是换种写法

8.5、多层列表字典排序

demoDict = {
    "a": [{"key": 2}], "c": [{"key": 1}], "b": [{"key": 3}]
}

sorted(demoDict.items(), key=lambda x:x[1][0]["key"])
print(demoDict)
print(dict(demoDict))

九、结束！