5.遍历字符串

container = “有时候你的一个微笑,会改变你的一生”

for i in container:

print(i)

6.遍历字典

container = {1:“a”,2:“b”,3:“c”}

for i in container:

print(i)

7.变量的解包[个数要匹配]

a,b = 1,2
a,b = (1,2)
a,b = [1,2]
#a,b = “567” #字符串有局限性,个数是2个,只能用2个变量收;
#a,b = {“a”,“b”} 集合有局限性,因为无序;
a,b ={“a”:1,“b”:2}
print(a,b)

8.遍历等长的二级容器

lst = [(“a”,“b”,“c”),[“d”,“e”,“f”],(“q”,“w”,“e”)]
“”"
#方法一
for i in lst:
for en in i:
print(en)
“”"
for en,line,i in lst:
print(en,line,i)
#en,line,i (“a”,“b”,“c”)

9.遍历不等长的二级容器

lst = lst = [(“a”,“b”,“c”),[“d”,“e”],(“q”,)]
for i in lst:
for en in i:
print(en)
#range 对象
“”"
range(start,end,step)
start:开始值
end:结束值
step: 步长
最大值end 取不到的,渠道end这个数之前的那个值
“”"

range()作用

res = range(10)

print(res)

1)range当中只有1个值

for i in range(10):

print(i)

2)range当中二个值的情况

for i in range(5,10):

print(i)

3)range当中三个值得情况下

for i in range(1,15,2):

print(i)

4)range倒着打印

for i in range(9,0,-1):

print(i)

5）99乘法表for改写

#while
i = 1
while i<=9:
j = 1
while j <= i:
print(“%s*%s=%2s”% (i,j,i*j),end=" ")