python 学习第二天 （下）

Posted 2020-08-09

继续上次的笔记

 

####判断一个元素是否在列表中

9 in name

print(9 in name) 会返回一个True 或 False 的结果

 

if 9 in name: #判断一个元素是否在列表中
    print("9 is in name")

####判断一个元素出现的次数 count()方法

 

name =["Alex","Jack","Rain",9,4,3,5,634,34,89,"Eric","Monica","Fiona",3,65,3,2,6,8,2,4,7]
if 9 in name:
    print("9 is in name ")
    if 9 in name:
        num_of_ele = name.count(9)
        print("[%s] 3 is/are in name" % num_of_ele)

 

 

####找到某个元素的索引 index() 方法

 

name = ["Alex","Jack","Rain",9,4,3,5,634,34,89,"Eric","Monica","Fiona",3,65,3,2,6,8,2,4,7]

if 9 in name:
    num_of_ele = name.count(9)
    posistion_of_ele = name.index(9)   #只匹配查找到的一个元素就返回
    name[posistion_of_ele] = 999
    print("[%s] 9 is/are in name,posistion:[%s]" % (num_of_ele,posistion_of_ele))

for i in range(name.count(9)):
    ele_index = name.index(9)
    name[ele_index] = 999
print(name)

 

####列表扩展 extend

name.extend(name2) 将列表name2 复制到列表name1 后面，可以有重复的元素，复制后列表name2 依然存在。

name = ["Alex","Jack","Rain",9,4,3,5,634,34,89,"Eric","Monica","Fiona",3,65,3,2,6,8,2,4,7]
name2 = [‘zhangsan‘,‘LiSi‘,‘ZhaoSha‘,634]

name.extend(name2)
print(name)
print(name2)

执行结果：

[‘Alex‘, ‘Jack‘, ‘Rain‘, 999, 4, 3, 5, 634, 34, 89, ‘Eric‘, ‘Monica‘, ‘Fiona‘, 3, 65, 3, 2, 6, 8, 2, 4, 7, ‘zhangsan‘, ‘LiSi‘, ‘ZhaoSha‘, 634]
[‘zhangsan‘,‘LiSi‘,‘ZhaoSha‘,634]

 

####列表反转 reverse() 方法

name2 = [‘zhangsan‘,‘LiSi‘,‘ZhaoSha‘,‘634‘]
name2.reverse()
print(name2)

*执行结果*

[‘634‘, ‘ZhaoSha‘, ‘LiSi‘, ‘zhangsan‘]

 

####列表排序 sort()

**python 2.7 排序。3.0中字符串不能和数字一起排序会报错。**

>>> name = ["Alex","Jack","Rain",9,4,3,5,634,34,89,"Eric","Monica","Fiona",3,65,3,2,6,8,2,4,7]
>>> name.sort()
>>> print(name)
[2, 2, 3, 3, 3, 4, 4, 5, 6, 7, 8, 9, 34, 65, 89, 634, ‘Alex‘, ‘Eric‘, ‘Fiona‘, ‘Jack‘, ‘Monica‘, ‘Rain‘]

 

**3.0 排序** 将数字转换成字符

name2 = [‘zhangsan‘,‘LiSi‘,‘ZhaoSha‘,‘634‘]
name2.sort()
print(name2)

 

*执行结果*

[‘634‘, ‘LiSi‘, ‘ZhaoSha‘, ‘zhangsan‘]

 

#### 按下标删除元素，并返回 pop()

 

####列表拷贝 copy()方法

name = [‘zhangsan‘,‘LiSi‘,‘ZhaoSha‘,[1,2,3,4],‘634‘]
name2 = name.copy()
print(name)
print(name2)

 

*执行结果：*

[‘zhangsan‘, ‘LiSi‘, ‘ZhaoSha‘, [1, 2, 3, 4], ‘634‘]
[‘zhangsan‘, ‘LiSi‘, ‘ZhaoSha‘, [1, 2, 3, 4], ‘634‘]

 

####列表嵌套

**注意：嵌套的列表只是保存了列表的内存入口地址在外层列表中**

如果使用copy()方法 默认只COPY了列表的第一层

name = [‘zhangsan‘,‘LiSi‘,‘ZhaoSha‘,[1,2,3,4],‘634‘]
name2 = name.copy()
name[0] = "Zhangsan"
name[3][2] = 33333333333333
print(name)
print(name2)

 

*执行结果：*

[‘Zhangsan‘, ‘LiSi‘, ‘ZhaoSha‘, [1, 2, 33333333333333, 4], ‘634‘]
[‘zhangsan‘, ‘LiSi‘, ‘ZhaoSha‘, [1, 2, 33333333333333, 4], ‘634‘]

 

**需要完全克隆，需要借助一个标准库copy库来实现**

import copy
name = [‘zhangsan‘,‘LiSi‘,‘ZhaoSha‘,[1,2,3,4],‘634‘]
name2 = name.copy()
name3 = copy.deepcopy(name)
name[3][2] = 33333333333333
print(name)
print(name2)
print("name3",name3)

 

*执行结果：*

[‘zhangsan‘, ‘LiSi‘, ‘ZhaoSha‘, [1, 2, 33333333333333, 4], ‘634‘]
[‘zhangsan‘, ‘LiSi‘, ‘ZhaoSha‘, [1, 2, 33333333333333, 4], ‘634‘]
name3 [‘zhangsan‘, ‘LiSi‘, ‘ZhaoSha‘, [1, 2, 3, 4], ‘634‘]

 

####查看列表长度

len(name4)

**课堂练习**

name = [‘Alex‘,‘jack‘,‘Rain‘,[9,4,3,5,9],634,34,89,9,34]

找出有多少个9，把它改成9999
同时找出所有的34，把它删除

##元租

元组可以理解为只读列表

创建元组

>>> r = (1,2,3,4,5)
>>> type(r)
<type ‘tuple‘>

 

两个方法count() index()

##字符串

**常用功能：**

* 移除空白
* 分割
* 长度
* 索引
* 切片

####1 移除空白 strip()

username = input("user:")
if username.strip() == ‘alex‘:
print("Welcome")

 

####2 分割

names = "alex,jack,rain"
name2 = names.split(",")
print(name2)

 

*执行结果：*

[‘alex‘, ‘jack‘, ‘rain‘]

####字符串合并

names = "alex,jack,rain"
name2 = names.split(",")
print("|".jion(name2))

 

*执行结果：*

alex|jack|rain

 

####判断空格

name = "Alex Li"

print(‘‘ in name)

首字母大写

name = "alex Li"

print(name.capitalize()) #首字母大写

字符串格式化

msg = "Hello, {name},it‘s been along {age} since last time sopke..."

msg2 = msg.format(name=‘MingHu‘,age=333)
print(msg2)

 

*执行结果：*

Hello, MingHu,it‘s been along 333 since last time sopke...

 

 

另一写法

msg2 = "hahah{0},dddd{1}"
print(msg2.format(‘Alex‘,33))

 

切片

name = "alex li"
print(name[2:4])

 

*执行结果：*

ex

 

 

center用法

print(name.center(40,‘-‘))

 

*执行结果：*

----------------alex li-----------------

 

find() 方法

name = "alex li"
print(name .find(‘l‘)) #找到值返回下标
print(name .find(‘sdfs‘)) #没找到对应值 返回 -1

 

*执行结果：*

1
-1

 

判断用户输入是否为数字

age = input("your age:")
if age.isdigit():
age = int(age)
print(age)
else:
print("invalid data type")

 

*执行结果：*

your age:33
33
your age:a
invalid data type

 

**判断字符串中是否有特殊字符 isalnum()**

ame = ‘alex3*sdf‘
print(name.isalnum())


False

 

name = ‘alex3sdf‘
print(name.isalnum())

True

 

**判断字符串以什么开始和结束**

startswith()

endswith()

**大小写转换**

upper() 大写

lower() 小写

 

**数字运算**




% 返回余数的用法：最简单的求奇偶数的一个用法

**比较运算符**

 

**赋值运算**


**逻辑运算**




**成员运算**

**身份运算符**




**is 用法**

 >>> a = [1,2,3,4]
 >>> type(a)
 <type ‘list‘>
 >>> type(a) is list
 True

 

**位运算符**

###优先级

###死循环

count =0
while True:
    print("你是风儿我是沙，缠缠绵绵到天涯"，count)
    count +=1
    if count == 100：
        print("去你妈的风和沙，你们这些脱了裤子是人，穿上裤子是鬼的臭男人..")
        break

 

 

在第50次到60次循环之间不打印

 

count =0
while True:
    count +=1
    if count > 50 and count <60:
        continue
    print("你是风儿我是沙，缠缠绵绵到天涯"，count)

    if count == 100：
        print("去你妈的风和沙，你们这些脱了裤子是人，穿上裤子是鬼的臭男人..")
        break

 

 

##字典

列表不具备去重的功能，而且操作复杂

字典格式 key：value 结构

id_db = {
371471199306143632:{
‘name‘:"Alex LI",
‘age‘:20,
‘addr‘:‘shanDong‘,
},
220435493306143632: {
‘name‘: "ShanPao",
‘age‘: 24,
‘addr‘: ‘DongBei‘,
},
220435493306143632: {
‘name‘: "DaShanPao",
‘age‘: 24,
‘addr‘: ‘DongBei‘,
},
}

 

显示字典内容

print(id_db)

 

打印某条记录

print(id_db[371471199306143632] )

 

修改key值

id_db[371471199306143632][‘name‘] = "WangMingHu" #修改已经存在的值
id_db[371471199306143632][‘qq_of_wife‘] = 3823354 #添加一个没有的 key-value
id_db[371471199306143632].pop("qq_of_wife") #删除一个key-value

 

###get() 方法 获取一个key的值

v = id_db .get(371471199306143632)
print(v)

 

*执行结果：*

{‘age‘: 20, ‘addr‘: ‘shanDong‘, ‘name‘: ‘Alex LI‘}

 

**注意**

print(id_db[371471199306143632])和 id_db .get(371471199306143632) 如果该键值不存在 直接打印会导致程序报错，而get方法会返回 None

###update 方法

用另一个字段覆盖一个字典

dic2 = { 
‘name‘: ‘adasdasdasd‘,
220435493306143632: {
‘name‘: "WangWang",

},
}

id_db.update(dic2) #用dic2 覆盖 id_db 覆盖已经存在的key 添加不存在的key
print(id_db)

 

*执行结果：*

{220435493306143631: {‘age‘: 24, ‘addr‘: ‘DongBei‘, ‘name‘: ‘ShanPao‘}, 220435493306143632: {‘name‘: ‘WangWang‘}, 371471199306143632: {‘age‘: 20, ‘addr‘: ‘shanDong‘, ‘name‘: ‘Alex LI‘}, ‘name‘: ‘adasdasdasd‘}

 

 items()方法 将字段转换成元组 或 列表 在字典数据量很多的情况下不要这么操作 很耗时

###setdefault()方法

id_db .setdefault() 方法 判断一个key 是否存在于字典中，如过存在就返回，如果不存在就在字典里生成这个key，默认值为None(可以自己指定语法为setdefault(key,"value")

print(id_db .setdefault(371471199306143632))

 

*执行结果：*

{‘age‘: 20, ‘addr‘: ‘shanDong‘, ‘name‘: ‘Alex LI‘}

 

print(id_db .setdefault(37147119930614363200))
print(id_db)

*执行结果：*

None
{220435493306143631: {‘addr‘: ‘DongBei‘, ‘age‘: 24, ‘name‘: ‘ShanPao‘}, 220435493306143632: {‘addr‘: ‘DongBei‘, ‘age‘: 24, ‘name‘: ‘DaShanPao‘}, 371471199306143632: {‘addr‘: ‘shanDong‘, ‘age‘: 20, ‘name‘: ‘Alex LI‘}, 37147119930614363200: None}

 

###fromkeys() 方法

把列表的里每个值拿出来当成字典里的一个key

print(dict .fromkeys([1,2,34,4,5,6],‘dddd‘))

*执行结果：*

{1: ‘dddd‘, 2: ‘dddd‘, 4: ‘dddd‘, 5: ‘dddd‘, 6: ‘dddd‘, 34: ‘dddd‘}

 

###popitem()方法

随机删掉个key 并返回这个key的值

print(id_db.popitem())

*执行结果：*

(220435493306143631, {‘addr‘: ‘DongBei‘, ‘age‘: 24, ‘name‘: ‘ShanPao‘})

 

###字典用于循环

方法1：

for k,v in id_db .items(): #效率低，因为要有一个dict to list的转换过程
print(k,v)

方法2:
for key in id_db: #建议使用
print(key,id_db[key])

###购物车练习 

# -*- coding:utf-8 -*-
#Author:Koala W

#shop mall program exercises

salary = input("Please input your salary:")
if salary.isdigit():
    salary = int(salary)
else:
    print("invald data!")
    exit()
welcome_msg = ‘Welcome to Alex Shopping mall‘.center(50,‘-‘)

exit_flag = False

print(welcome_msg)
product_list = [
    (‘iphone7‘,5888),
    (‘MAC pro‘,9888),
    (‘MAC air‘,7888),
    (‘XIAOMI 5‘,2499),
    (‘可口可乐‘,2),
    (‘IPAD mini‘,3299),
    (‘Alex Li‘,19.9),
    (‘Bike‘,700),
    (‘TOMMIY‘,200),
    (‘宝马汽车‘,316000),]
shop_car = []
while exit_flag is not True:
    print("product list".center(50,‘-‘))
    for item in enumerate(product_list):
        index = item[0]
        p_name = item[1][0]
        p_price = item[1][1]
        print(index, ‘.‘, p_name, p_price)
    user_choice = input("[q=quit,c=check]What do you want to buy?:")
    if user_choice.isdigit():  #输入数字字符必定是选择商品
        user_choice = int(user_choice)
        if user_choice < len(product_list):
            p_item = product_list[user_choice]
            if p_item[1]<= salary: #余额是否买得起商品
                shop_car.append(p_item) # 放入购物车
                salary -= p_item[1]   #从余额中扣除商品费用
                print("Add [%s] into shop car,you current balance is \033[31;1m[%s]\033[0m" %(p_item,salary))
            else:
                print("Your balance is [%s],cannot afford this.." % salary)
    else:
        if user_choice == ‘q‘ or user_choice == ‘quit‘:
            print("purchased products as below".center(40, ‘*‘))
            for item in shop_car:
                print(item)
                print("END".center(40, ‘*‘))
                print("Your balance is [%s]" % salary)
                print("Bye!")
                exit_flag = True
        elif user_choice == ‘c‘ or user_choice == ‘check‘:
            print("purchased products as below".center(40, ‘*‘))
            for item in shop_car:
                print(item)
                print("END".center(40, ‘*‘))
                print("Your balance is \033[41;1m[%s]\033[0m" % salary)