python基础—字典

Posted 2020-11-02 badweather

#########################################################################
# dict
# dict
# dic = {
# "k1": ‘v1‘,
# "k2": ‘v2‘
# }
# 1 根据序列，创建字典，并指定统一的值
　　# v = dict.fromkeys(["k1",123,"999"],123)
　　# print(v)

# 2 根据Key获取值，key不存在时，可以指定默认值（None）
　　# v = dic[‘k11111‘]
　　# print(v)
　　# v = dic.get(‘k1‘,111111)
　　# print(v)

# 3 删除并获取值
　　# dic = {
　　# "k1": ‘v1‘,
　　# "k2": ‘v2‘
　　# }
　　# v = dic.pop(‘k1‘,90)
　　# print(dic,v)
　　# k,v = dic.popitem()
　　# print(dic,k,v)

# 4 设置值，
　　# 已存在，不设置，获取当前key对应的值
　　# 不存在，设置，获取当前key对应的值
　　# dic = {
　　# "k1": ‘v1‘,
　　# "k2": ‘v2‘
　　# }
　　# v = dic.setdefault(‘k1111‘,‘123‘)
　　# print(dic,v)

# 5 更新
　　# dic = {
　　# "k1": ‘v1‘,
　　# "k2": ‘v2‘
　　# }
　　# dic.update({‘k1‘: ‘111111‘,‘k3‘: 123})
　　# print(dic)
　　# dic.update(k1=123,k3=345,k5="asdf")
　　# print(dic)

# 6 keys()

7 values()

8 items()

9 get

10update
##########################################################################

 

#1、基本机构
# info = {
# "k1": "v1", # 键值对
# "k2": "v2"
# }

#2 字典的value可以是任何值
# info = {
# "k1": 18,
# "k2": True,
# "k3": [
# 11,
# [],
# (),
# 22,
# 33,
# {
# ‘kk1‘: ‘vv1‘,
# ‘kk2‘: ‘vv2‘,
# ‘kk3‘: (11,22),
# }
# ],
# "k4": (11,22,33,44)
# }
# print(info)

# 3 布尔值(1,0)、列表、字典不能作为字典的key
# info ={
# 1: ‘asdf‘,
# "k1": ‘asdf‘,
# True: "123",
# # [11,22]: 123
# (11,22): 123,
# # {‘k1‘:‘v1‘}: 123
#
# }
# print(info)

#4 字典无序

# info = {
# "k1": 18,
# "k2": True,
# "k3": [
# 11,
# [],
# (),
# 22,
# 33,
# {
# ‘kk1‘: ‘vv1‘,
# ‘kk2‘: ‘vv2‘,
# ‘kk3‘: (11,22),
# }
# ],
# "k4": (11,22,33,44)
# }
# print(info)

# 5、索引方式找到指定元素
# info = {
# "k1": 18,
# 2: True,
# "k3": [
# 11,
# [],
# (),
# 22,
# 33,
# {
# ‘kk1‘: ‘vv1‘,
# ‘kk2‘: ‘vv2‘,
# ‘kk3‘: (11,22),
# }
# ],
# "k4": (11,22,33,44)
# }
# # v = info[‘k1‘]
# # print(v)
# # v = info[2]
# # print(v)
# v = info[‘k3‘][5][‘kk3‘][0]
# print(v)

# 6 字典支持 del 删除
# info = {
# "k1": 18,
# 2: True,
# "k3": [
# 11,
# [],
# (),
# 22,
# 33,
# {
# ‘kk1‘: ‘vv1‘,
# ‘kk2‘: ‘vv2‘,
# ‘kk3‘: (11,22),
# }
# ],
# "k4": (11,22,33,44)
# }
# del info[‘k1‘]
#
# del info[‘k3‘][5][‘kk1‘]
# print(info)

# 7 for循环
# dict
# info = {
# "k1": 18,
# 2: True,
# "k3": [
# 11,
# [],
# (),
# 22,
# 33,
# {
# ‘kk1‘: ‘vv1‘,
# ‘kk2‘: ‘vv2‘,
# ‘kk3‘: (11,22),
# }
# ],
# "k4": (11,22,33,44)
# }
# for item in info:
# print(item)
#
# for item in info.keys():
# print(item)

# for item in info.values():
# print(item)

# for item in info.keys():
# print(item,info[item])

# for k,v in info.items():
# print(k,v)

# True 1 False 0
# info ={
# "k1": ‘asdf‘,
# True: "123",
# # [11,22]: 123
# (11,22): 123,
# # {‘k1‘:‘ v1‘}: 123
#
# }
# print(info)