#Auther Duncan
#--*--coding=utf-8--*--
# name1 = ‘zidane‘
# name2 = "zidane"
# print(type(name1),type(name2))
# username = raw_input("Input your name:")
# if username.strip() == "root":
# print("welcome to login")
# else:
# print("Invalid username")
# name = ‘hello,world‘
# print("h" in name)
# print(‘,‘ in name)
# print(‘a‘ in name)
# name = ‘root‘
# print(name.capitalize())
# name = ‘administrator‘
# print(name[0:1])
# print(name[2:9])
# name = ‘administrator‘
# print(name.center(40,‘-‘))
# name = ‘administrator‘
# print(name.find("a"))
# print(name.find("b"))
# age = raw_input("input age")
#
# age = ‘adddddddddddddddd‘
# age = age.replace(‘d‘,‘D‘)
# print(age)
# print(ord(‘s‘))
# print(chr(116))
# dict = {‘name‘:‘Bob‘,‘age‘:17,‘job‘:‘IT‘}
# print(type(dict))
# dict.clear()
# print(dict)
# dict1 = dict.copy()
# print(dict1)
# print(dict.get(‘name‘))
# print(dict.get(‘NAME‘))
# print(dict.has_key(‘name‘))
# print(dict.has_key(‘NAME‘))
# print(dict.items())
# print(dict.iterkeys())
# print(dict.values())
#
# print(dict.keys())
# del dict[‘name‘]
# print(dict)
# print(dict.pop("name"))
# print(dict.pop(‘age‘))
# print(dict.pop(‘agE‘))
# print(dict.popitem())
# print(dict)
# print(dict.popitem())
# print(dict)
# print(dict.popitem())
# print(dict)
# dict.setdefault(‘name‘,‘shenzhen‘)
# print(dict)
#
# 如果是一个已经存在的键,不会修改已有的键值
# {‘job‘: ‘IT‘, ‘age‘: 17, ‘name‘: ‘Bob‘}
# dict.setdefault(‘add‘,‘shenzhen‘)
# print(dict)
#
# #如果是一个不存在的键值,则会添加到dict中
# {‘job‘: ‘IT‘, ‘age‘: 17, ‘add‘: ‘shenzhen‘, ‘name‘: ‘Bob‘}
# list = ["name","age","job"]
# dict1 = dict.fromkeys(list)
# print(dict1)
#结果,把一个列表中的元素当做新的dict的键值,value是可选的,如果不写,则默认是None
# {‘job‘: None, ‘age‘: None, ‘name‘: None}
#
#
# tuple = ("name","age","job")
# print(type(tuple))
# dict2 = dict.fromkeys(tuple,‘abc‘)
# print(dict2)
#
# 结果,结果,把一个元组中的元素当做新的dict的键值,value是可选的,如果写,则只能写一个value的值
# {‘job‘: ‘abc‘, ‘age‘: ‘abc‘, ‘name‘: ‘abc‘}
# 查看k值和v值
# print(dict.viewitems())
# 查看k值
# print(dict.viewkeys())
# 查看v值
# print(dict.viewvalues())
# dict1 = {‘name‘:‘Bob‘,‘age‘:17,‘job‘:‘IT‘}
# dict2 = {‘name‘:‘Cui‘,‘age‘:19,‘job‘:‘teacher‘}
# dict1.update(dict2)
# print(dict1)
#
# #结果,把dict2中的k值和v值update到dict1中,如果有重复的键值,会被更新
# {‘job‘: ‘teacher‘, ‘name‘: ‘Wang‘, ‘age‘: 19}
# dict1 = {‘name‘:‘Bob‘,‘age‘:17,‘job‘:‘IT‘}
# dict3 = {‘add‘:‘shenzhen‘}
# dict1.update(dict3)
# print(dict1)
#
# #结果,把dict3中的k和v值更新到dict1中,如果dict3中的k值在不dict1中不存在,则会追加到dict1中
# {‘job‘: ‘IT‘, ‘age‘: 17, ‘add‘: ‘shenzhen‘, ‘name‘: ‘Bob‘}
原文:http://www.cnblogs.com/bainianminguo/p/6412978.html