day12-python之深灰魔法

#######################################灰魔法: list类中提供的方法列表#######################################

 #!/usr/bin/env python

 # -*- coding:utf-8 -*-

 # 参数

 # 1. 原来值最后追加

 # 对象.方法(..)   # li对象调用append方法

 li = [11, 22, 33, 22, 44]

 # li.append(5)

 # li.append("alex")

 # li.append([1234,2323])

 # print(li)

 # 2 清空列表

 # li.clear()

 # print(li)

 # 3 拷贝，浅拷贝

 # v = li.copy()

 # print(v)

 # 4. 计算元素出现的次数

 # v = li.count(22)

 # print(v)

 # 5. 扩展原列表，参数：可迭代对象

 # li = [11, 22, 33, 22, 44]

 # li.append([9898,"不得了"])

 # print(li)

 #

 # li.extend([9898,"不得了"])

 # for i in [9898,"不得了"]:

 #     li.append(i)

 #

 # li.extend("不得了")

 # print(li)

 # 6. 根据值获取当前值索引位置（左边优先）

 # li = [11, 22, 33, 22, 44]

 # v= li.index(22)

 # print(v)

 # 7. 在指定索引位置插入元素

 # li = [11, 22, 33, 22, 44]

 # li.insert(0,99)

 # print(li)

 # 8、 删除某个值(1.指定索引；2. 默认最后一个)，并获取删除的值

 # li = [11, 22, 33, 22, 44]

 # v = li.pop()

 # print(li)

 # print(v)

 # li = [11, 22, 33, 22, 44]

 # v = li.pop(1)

 # print(li)

 # print(v)

 # 9. 删除列表中的指定值，左边优先

 # li = [11, 22, 33, 22, 44]

 # li.remove(22)

 # print(li)

 # PS: pop remove del li[0]    del li[7:9]   clear

 # 10 将当前列表进行翻转

 # li = [11, 22, 33, 22, 44]

 # li.reverse()

 # print(li)

 # 11 列表的排序

 # li = [11,44, 22, 33, 22]

 # li.sort()

 # li.sort(reverse=True)

 # print(li)

 ### 欠

 # cmp

 # key

 # sorted

# 1. 列表格式
# 2. 列表中可以嵌套任何类型
# 中括号括起来
# ，分割每个元素
# 列表中的元素可以是数字，字符串,列表，布尔值..所有的都能放进去
# “集合”，内部放置任何东西

 #!/usr/bin/env python

 # -*- coding:utf-8 -*-

 # 3.

 # 索引取值

 li = [1,23,212,45,67,89]

 # print(li[3])

 # 4 切片，切片结果也是列表

 # print(li[3:-1])

 # 5 for循环

 # while循环

 # for item in li:

 #     print(item)

 # 列表元素，可以被修改

 # li = [1, 12, 9, "age", ["石振文", ["19", 10], "庞麦郎"], "alex", True]

 # print(li)

 ############## 6 索引

 # 修改

 # li[1] = 120

 # print(li)

 # li[1] = [11,22,33,44]

 # print(li)

 # 删除,第一种方式

 # print(li)

 # del li[1]

 # print(li)

 ############## 7 切片

 # 修改

 # print(li)

 # li[1:3] = [120,90]

 # print(li)

 # 删除

 # print(li)

 # del li[2:6]

 # print(li)

 # 8 in 操作

 # li = [1, 12, 9, "age", ["石振文", ["19", 10], "庞麦郎"], "alex", True]

 # v1 = "石振文" in li

 # print(v1)

 # v2 = "age" in li

 # print(v2)

 ###### 列表中的元素，

 # 9 操作

 # li = [1, 12, 9, "age", ["石振文", ["19", 10], "庞麦郎"], "alex", True]

 # print(li[4][1][0])

 # 10 转换

 # 字符串转换列表   li =  list("asdfasdfasdf"), 内部使用for循环

 # s = "pouaskdfauspdfiajsdkfj"

 # new_li = list(s)

 # print(new_li)

 # 列表转换成字符串，

 # 需要自己写for循环一个一个处理： 既有数字又有字符串

 # li = [11,22,33,"123","alex"]

 # r = str(li) # '[11,22,33,"123","alex"]'

 # print(r)

 # s = ""

 # for i in li:

 #     s = s + str(i)

 # print(s)

 # 直接使用字符串join方法：列表中的元素只有字符串

 # li = ["123","alex"]

 # v = "".join(li)

 # print(v)

 ### 补充：字符串创建后，不可修改

 # v = "alex"

 # v = v.replace('l','el')

 # print(v)

 # li = [11,22,33,44]

 # li[0]

 # li[0] = 999

 # s = "alex"

 # li[0]

 # s[0] = "E"

 # print(s)

 # 列表，有序；元素可以被修改

#########################################元组###########################################################

 #!/usr/bin/env python

 # -*- coding:utf-8 -*-

 # 元组，元素不可被修改，不能被增加或者删除

 # tuple

 # tu = (11,22,33,44,22,22,22)

 # print(tu.count(22))#获取指定元素在元组中出现的次数

 # print(tu.index(22))

 # 1. 书写格式

 tu = (111,"alex",(11,22),[(33,44)],True,33,44,)

 # 一般写元组的时候，推荐在最后加入 ,

 # 元素不可被修改，不能被增加或者删除

 # 2. 索引

 # v = tu[0]

 # print(v)

 # 3. 切片

 # v = tu[0:2]

 # print(v)

 # 4. 可以被for循环，可迭代对象

 # for item in tu:

 #     print(item)

 # 5. 转换

 s = "asdfasdf0"

 li = ["asdf","asdfasdf"]

 tu = ("asdf","asdf")

 # v = tuple(s)

 # print(v)

 # v = tuple(li)

 # print(v)

 # v = list(tu)

 # print(v)

 # v = "_".join(tu)

 # print(v)

 # li = ["asdf","asdfasdf"]

 # li.extend((11,22,33,))

 # print(li)

 # 6.元组的一级元素不可修改/删除/增加

 tu = (111,"alex",(11,22),[(33,44)],True,33,44,)

 # # 元组，有序。

 # v = tu[3][0][0]

 # print(v)

 # v=tu[3]

 # print(v)

 # tu[3][0] = 567

 # print(tu)

###################字典######################

 #!/usr/bin/env python

 # -*- coding:utf-8 -*-

 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()   get   update

 ##########

 # 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 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)

#######################　整理 #################

# 一、数字
# int(..)
# 二、字符串
# replace/find/join/strip/startswith/split/upper/lower/format
# tempalte = "i am {name}, age : {age}"
# # v = tempalte.format(name='alex',age=19)
# v = tempalte.format(**{"name": 'alex','age': 19})
# print(v)
# 三、列表
# append、extend、insert
# 索引、切片、循环
# 四、元组
# 忽略
# 索引、切片、循环以及元素不能被修改
# 五、字典
# get/update/keys/values/items
# for,索引

# dic = {
# "k1": 'v1'
# }

# v = "k1" in dic
# print(v)

# v = "v1" in dic.values()
# print(v)
# 六、布尔值
# 0 1
# bool(...)
# None "" () [] {} 0 ==> False

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