Python 列表內建 list.sort() 方法實現原地排序，同時 Python 還提供 sorted() 函式返回對可迭代物件排序後的新列表。

首先匯入可能需要的模組：

# coding:utf-8
import operator
import collections

data_list = []
data_list.append({'id':'1','name':'zhangsan','age':'22'})
data_list.append({'id':'2','name':'wangwu','age':'23'})
data_list.append({'id':'7','name':'張三','age':'22'})
data_list.append({'id':'4','name':'李四','age':'21'})
data_list.append({'id':'3','name':'肖戰','age':'23'})
data_list.append({'id':'5','name':'無視','age':'21'})
data_list.append({'id':'6','name':'排查','age':'23'})
'''先按照id排序，再按照age排序'''
data_list.sort(key=lambda obj:(obj.get('id'),obj.get('age')), reverse=False)
for s in data_list:
    print(s)
 

L = [('b', 6), ('a', 1), ('c', 3), ('d', 4)]
L.sort(key=lambda x:x[1])
print(L)

L.sort(key=lambda x: (x[1], x[0]))
print(L)

aa=sorted(L,key=lambda x: (x[1], x[0]))
print(aa)

L.sort(key=operator.itemgetter(1, 0))
print(L)

class Student(object):
    def __init__(self,id,name,age):
        self.id=id
        self.name=name
        self.age=age

list=[]
s1=Student(1,'zhangsan',23)
s2=Student(2,'wangwu',32)
s3=Student(3,'lisi',21)
s4=Student(3,'lisi',20)
list.append(s1)
list.append(s2)
list.append(s3)
list.append(s4)
list.sort(key=operator.attrgetter('id','age'))
for ssss in list:
    print('id:  '+str(ssss.id)+'  name:  '+ssss.name+"  age:  "+str(ssss.age))
 

字典排序，按照值排序注意值得型別。

dict2={'id':'1','name':'zhangsan','age':'23'}
'''按鍵排序：sorted(dict2.items(), key=lambda e:e[0], reverse=True)
   按鍵排序：sorted(dict2.items(), key=lambda e:e[1], reverse=True)
'''
sorted(dict2.items(), key=lambda e:e[1], reverse=True)
print(dict2)