如何在Python中使用break跳出多层for循环？

链接：https://www.zhihu.com/question/37076998/answer/935963167


五种方法大合集！下面用一个例子说明：三进制递增计数，从000~222，循环到111退出。
A. flag大法flag = True

for i in range(3):
    for j in range(3):
        for k in range(3):
            print(i, j, k)
            if i == j == k == 1:
                flag = False
                print(break)
                break
        if not flag:
            break
    if not flag:
        break


B. 用循环的else分支for i in range(3):
    for j in range(3):
        for k in range(3):
            print(i, j, k)
            if i == j == k == 1:
                print(break)
                break
        else:
            continue
        break
    else:
        continue
    break

C. 打包进函数def loop():
    for i in range(3):
        for j in range(3):
            for k in range(3):
                print(i, j, k)
                if i == j == k == 1:
                    print(break)
                    return

loop()


D. 抛出异常class Break(Exception):
    pass

try:
    for i in range(3):
        for j in range(3):
            for k in range(3):
                print(i, j, k)
                if i == j == k == 1:
                    raise Break(break)
except Break as e:
    print(e)


E. 笛卡尔积在有篇博文里看到的奇技淫巧 ，看起来很酷的亚子（虽然实用范围有点受限）from itertools import product

for i, j, k in product(range(3), range(3), range(3)):
    print(i, j, k)
    if i == j == k == 1:
        print(break)
        break