基础数据类型补充 以后会遇见的坑 二次编码

1.基础数据类型补充

str:

不可变数据

1.首字母大写:

name = "alex"
name1 = name.capitalize()
print(name1)

2.每个单词首字母大写:

name = "alex wusir"
print(name.title())

3.大小写反转

name = "Alex"
print(name.swapcase())

4.居中 -- 填充

name = "alex"
print(name.center(20,"-"))

5.查找

从左向右 只查找一个
name = "alelx"
print(name.find("b"))  #find查找不存在的返回-1 存在返回"0"
print(name.index("b")) #index查找不存在的就报错,,存在的返回索引值

6.拼接 ***

name = "al3x"
print("_".join(name))  

7.格式化

① %s
② f''
③ name.format()
name = "alex{},{},{}"
print(name.format(1,2,3)) # 按照顺序位置进行填充
name = "alex{2},{0},{1}"
print(name.format("a","b","c")) # 按照索引值进行填充
name = "alex{a},{b},{c}"
print(name.format(a=1,c=11,b=67)) # 按照关键字进行填充

8.字符串的 加 和 乘

开辟新的内存空间
name = "alex"
name1 = "wusir"
print(name)
print(name1)
print(name + name1)

list

1.定义方式

list("123")

2.其他方法

2.1 排序 (默认升序)

lst = [1,2,23,234,435,36,23,213421,421,4231,534,65]
lst.sort()  
print(lst)

lst = ["你好","我好"]
lst.sort()  
print(lst)

2.2 反转

lst = [1,2,3,4453,5,6,7]
print(lst[::-1])
lst.reverse()    
print(lst)

lst = [1,2,3,4,5123,21345,231123,4,1235,234,123]
lst.sort()
lst.reverse()
print(lst)

面试题:

拼接两个列表

方式一:
lst.extend(lst1)
print(lst)
方式二:
print(lst+lst1)

正题

lst = [1,2,3,4,5]
lst1=  lst * 5
new_lst = lst * 5
print(new_lst[0] is new_lst[-5])        # True

扩展

lst = [[]]
new_lst = lst * 5
new_lst[0].append(10)
print(new_lst)
# [[10], [10], [10], [10], [10]]

lst = [1,[]]
new_lst = lst * 5
new_lst[0] = 10
print(new_lst)
# [10, [], 1, [], 1, [], 1, [], 1, []]

lst = [1,[]]
new_lst = lst * 5
new_lst[1] = 10
print(new_lst)
# [1, 10, 1, [], 1, [], 1, [], 1, []]

lst = [[]]
new_lst = lst * 5
new_lst[0].append(10)
print(new_lst)
# [[10], [10], [10], [10], [10]]

tuple

tu = ("12")      # 数据类型是( )中数据本身
print(type(tu))
# <class 'str'>

tu = (1,)        # (1,) 是元组
print(type(tu))  
# <class 'tuple'>

元组 加 乘 不可变共用,可变也共用

dict

定义一个字典:

print(dict(k = 1,k = 2))

随机删除:

dic = {"key":1,"key2":2,"key3":56}
print(dic.popitem())         # 返回的是被删除的键值对(键,值)
print(dic)
# python3.6 默认删除最后一个(bug)

批量添加:

dic = {}
dic.fromkeys("123",[23]) 
# 批量添加键值对{"1":[23],"2":[23],"3":[23]}
print(dic)

dic = dict.fromkeys("123456789",1) 
# 批量添加键值对"键是可迭代对象",值 -- 会被共用
dic["1"] = 18
print(dic)

set

set() -- 空集合

{} -- 空字典

定义集合:set("alex")  # 迭代添加的

bool

True   False
数字: 0
字符串: ""
列表:[]
元组:()
字典:{}
集合: set()
其他: None

数据类型之间转换

list  tuple
tuple list

str转 list的方法:
name = "alex"  
print(name.split())

list转 str的方法:
lst = ["1","2","3"] 
print(''.join(lst))

set - list
list - set

python的数据类型

可变:
list ,dict ,set
不可变:
int bool str tuple
有序:
list,tuple,str,int,bool
无序:
dict,set
取值方式:
索引: str list tuple
直接: set ,int ,bool
  键: dict

2.以后会遇见的坑

for的死循环

lst = [1,2]
for i in lst:
    lst.append(3)
print(lst)
# 打印不出内容

删除列表的坑

lst = [1,2,3,4]
for i in lst:
    lst.pop()
print(lst)
# [1, 2]

lst = [1,2,3,4]
for i in lst:
    lst.pop(0)
print(lst)
# [3, 4]

lst = [1,2,3,4]
for i in lst:
    lst.remove(i)
print(lst)
# [2, 4]

成功删除的方式:

lst = [1,2,3,4,5]
for i in range(len(lst)):
    lst.pop()
print(lst)

lst = [1,2,3,4,5]
for i in range(len(lst)-1,-1,-1):
    del lst[i]
print(lst)

lst = [1,2,3,4,5]
for i in range(len(lst)):
    del lst[-1]
print(lst)

lst = [1,2,3,4,5]
lst1 = lst.copy()
for i in lst1:
    lst.remove(i)
print(lst)

删除字典的坑

dic = dict.fromkeys("12345",1)
for i in dic:
    dic.pop(i)
print(dic)
# 报错
# 字典的迭代的时候改变了原来的大小(不能加不能删)

正确删除的方式

dic = dict.fromkeys("12345",1)
dic1 = dic.copy()
for i in dic1:
    dic.pop(i)
print(dic)
# {}

# 集合和字典都是迭代的时候不能改变原来的大小

3.二次编码

密码本:

ascii -- 没有中文

gbk -- 英文 8b(位) 1B(字节) 中文 16b 2B

unicode -- 英文16b 2B 中文32b 4B

utf-8 -- 英文8b 1B 欧洲16b 2B 亚洲24b 3B

name = "你好啊"

s1 = name.encode("utf-8") # 编码 

s2 = name.encode("gbk") # 编码 

s2 = s1.decode("utf-8") # 解码

print(s2.encode("gbk"))

以什么编码集(密码本)进行编码就要用什么编码集(密码本)解码