for循环

在Python中，for循环是一种迭代结构，用于遍历可迭代对象（如列表、元组、字符串、字典、集合等）中的元素。

for循环可以解决while循环的局限性
#while循环在遍历数据时的局限性

如下for循环可以实现

lst = [1,2,3,4,5] # ok
i = 0
while i < len(lst):
print(lst[i])
i+=1

用while循环无法实现：

setvar = {“a”,“b”,“c”} # not ok
i = 0
while i < len(setvar):
print(setvar[i])
i+=1

1.for循环的基本语法

Iterable 可迭代性数据：1.容器类型数据 2.range对象 3.迭代器
for 变量 in Iterable:
code1

#字符串
container = “北京和深圳温差大概20多度”
#列表
container = [1,2,3,4,4,5]
#元组
container = (“孙开洗”,“孙健”,“孙悟空”)
#集合
container = {“陈璐”,“曹静怡”,“王志国”,“邓鹏”,“合力”}
#字典
container = {“cl”:“风流倜傥”,“cjy”:“拳击选手”,“wzg”:“寻花问柳”,“dp”:“帅气,祖国的栋梁”,“hl”:“你是个好人”}

#遍历数据
for i in container:
print(i)

（1）遍历不等长多级容器

container = [1,2,3,4,("嗄","234",{"马春配","李虎凌","刘子涛"})]
for i in container:# 判断当前元素是否是容器,如果是,进行二次遍历,如果不是,直接打印if isinstance(i,tuple):# ("嗄","234",{"马春配","李虎凌","刘子涛"})for j in i:# 判断当前元素是否是集合,如果是,进行三次遍历,如果不是,直接打印if isinstance(j,set):# j = {"马春配","李虎凌","刘子涛"}for k in j :print(k)else:print(j)# 打印数据else:print(i)

（2）遍历不等长多级容器

container = [(“刘玉波”,“历史源”,“张光旭”), (“上朝气”,“于朝志”),(“韩瑞晓”,)]
for i in container:
for j in i:
print(j)

（3）遍历等长的容器

container = [(“马云”,“小马哥”,“马春配”) , [“王健林”,“王思聪”,“王志国”],{“王宝强”,“马蓉”,“宋小宝”}]
for a,b,c in container:
print(a,b,c)

2.变量的解包

a,b,c = “poi”
a,b = (1,2)
a,b = 1,2
a,b,c = [10,11,12]
a,b = {“林明辉”,“家率先”}
a,b = {“lmh”:“林明辉”,“jsx”:“家率先”}
a,b,c = (“马云”,“小马哥”,“马春配”)
print(a,b,c)

3.for…else【详细讲解】

for 临时变量 in 序列:
重复执行的代码
…
else:
循环正常结束后要执行的代码

所谓else指的是循环正常结束后要执行的代码，即如果是break终止循环的情况。else下方缩进的代码将不执行。

4.range对象

range([开始值,] 结束值 [,步长])
取头舍尾,结束值本身获取不到,获取到它之前的那一个数据

#range(一个值) 默认从0开始
for i in range(5): # 0 ~ 4
print(i)

#range(二个值)
for i in range(3,8): # 3 4 5 6 7
print(i)

#range(三个值) 正向的从左到右
for i in range(1,11,3): # 1 4 7 10
print(i)

#range(三个值) 逆向的从右到左
for i in range(10,0,-1): # 10 9 8 7 … 1
print(i)

5.总结

while 一般用于处理复杂的逻辑关系
for 一般用于迭代数据
部分情况下两个循环可以互相转换;

i = 1
while i <= 9:j = 1while j <= i:print("%d*%d=%2d " % (i,j,i*j) ,end="" )j+=1print()    i +=1for i in range(1,10):for j in range(1,i+1):print("%d*%d=%2d " % (i,j,i*j) ,end="" )print()

for循环乘法口诀

方向二：

方向三：

方向四：

6.打印 1 ~ 10 跳过5

i = 1
while i <= 10:    if i == 5:i += 1continueprint(i)i +=1for i in range(1,11):if i == 5:continueprint(i)

7.打印菱形小星星

     *************************
***********
************************************

空格 + 星星 + 换行

总行数:
对于最大行任意个星星n ,总行数: n // 2 + 1
13 -> 7
11 -> 6
9 -> 5
7 -> 4

空格:
对于当前行i , 空格数量 = 总行数 - 当前行
1 => 5
2 => 4
3 => 3
4 => 2
5 => 1
6 => 0

星星:
对于当前行i , 星星数量 = 2 * 当前行 - 1
1 => 1
2 => 3
3 => 5
4 => 7

n = 13
hang = n // 2 + 1
i = 1
while i <= hang:# 打印空格kongge = hang - iprint(" " * kongge, end="")# 打印星星xingxing = 2 * i - 1print("*" * xingxing, end="")# 打印换行print()i += 1i = hang
while i >= 1:# 打印空格kongge = hang - iprint(" " * kongge, end="")# 打印星星xingxing = 2 * i - 1print("*" * xingxing, end="")# 打印换行print()i -= 1

方法二：
#打印菱形

     *************************
***********
************************************

最大行星星数n
n = 2hang - 1
kongge = hang - i
xingxing = 2i -1

n = 13
hang = int((n + 1) / 2)
i = 1
while i <= hang:kongge = hang - iprint(" " * kongge,end='')xingixng = 2*i - 1print(f"*"* xingixng,end='')print()i += 1i = hang
while i >= 1:kongge = hang - iprint(" " * kongge,end='')xingixng = 2*i - 1print(f"*"* xingixng,end='')print()i -= 1

菱形二

   ** ** * *
* * * * 
* * * * * * ** **

行数 n 星星 n

kongge = n - i

hang = 5
i = 1
while i <= hang:kongge = hang - iprint(f" " * kongge,end='')xingxing = iif i == 1:print("*",end='')else:j = 1while j <= 2*i -1:if j % 2 == 1:print("*",end='')else:print(" ",end='')j += 1print()i += 1i = hang
while i >= 1:kongge = hang - iprint(f" " * kongge,end='')xingxing = iif i == 1:print("*",end='')else:j = 1while j <= 2*i -1:if j % 2 == 1:print("*",end='')else:print(" ",end='')j += 1print()i -= 1