如果您只需要在复杂的for循环巢中测试边缘情况，则可以抛出1/0来引发异常。我保证不会告诉任何人。当您希望快速测试深度嵌套的for循环的单个迭代，并且不想跟踪大量break语句或注释掉大量代码时，这种方法非常方便。

是的，您可以将它包装在函数中并使用return，但在某些上下文中，这可能会非常麻烦。

入门级程序员的例子:

for i in first_iter:
    for j in second_iter:
        for k in third_iter:
            print(i_want_to_run_this_once_and_stop_executing(i,j,k))
            1/0
        code_that_takes_a_long_time()
    expensive_code()

在执行大量数据预处理的大型jupyter笔记本脚本中，这尤其方便。

可能像下面这样的小技巧会做，如果不喜欢重构成函数

增加了1个break_level变量来控制while循环条件

break_level = 0
# while break_level < 3: # if we have another level of nested loop here
while break_level < 2:
    #snip: print out current state
    while break_level < 1:
        ok = get_input("Is this ok? (y/n)")
        if ok == "y" or ok == "Y": break_level = 2 # break 2 level
        if ok == "n" or ok == "N": break_level = 1 # break 1 level

下面是一个似乎可以工作的实现:

break_ = False
for i in range(10):
    if break_:
        break
    for j in range(10):
        if j == 3:
            break_ = True
            break
        else:
            print(i, j)

唯一的缺点是你必须在循环之前定义break_。

你可以使用一个标志来打破循环:

if found:
  break

这里，'found'是标志，你最初将它设置为False，然后在循环中使用这段代码。

found = False
for table_height in range(500):
  if found:
    break

下面是三个for循环的完整代码:

found = False
for table_height in range(500):
  if found:
    break
  for cat_height in range(500):
    if found:
      break

    for tort_height in range(500):
      equation1 = table_height + cat_height == tort_height + 170
      equation2 = table_height + tort_height == cat_height + 130
      if equation1 and equation2:
        print('table', table_height, '  cat', cat_height, '  tortoise', tort_height)
        found = True
        break

在这段代码中，如果等式1和等式2为True，它将设置'found'标志为True，并跳出最里面的for循环，它也将跳出其他两个for循环，因为'found'为True。

PEP 3136建议标记中断/继续。Guido拒绝了它，因为“需要这个功能的复杂代码非常罕见”。不过，PEP确实提到了一些变通方法(比如异常技术)，而Guido认为在大多数情况下，使用return进行重构会更简单。

以下是一个非常简短的版本: 创建名为break_out_nested.py的文件

import itertools
import sys

it = sys.modules[__name__] # this allows us to share variables with break_out_nested.py when we import it 


def bol(*args):
    condi = args[-1] # the condition function
    i = args[:-1] # all iterables 
    for p in itertools.product(*i): # itertools.product creates the nested loop
        if condi(): # if the condition is True, we return 
            return
        yield p # if not, we yield the result 

现在你只需要几行就可以打破嵌套的循环(数据来自Rafiq的例子)

from break_out_nested import it, bol # import what we have just created

# you need to create new variables as attributes of it,
# because break_out_nested has only access to these variables
it.i, it.j, it.k = 1, 1, 1
# the break condition
def cond(): return it.i % 3 == 0 and it.j % 3 == 0 and it.k % 3 == 0

# The condition will be checked in each loop 
for it.i, it.j, it.k in bol(range(1, 6, 1), range(1, 11, 2, ), range(1, 21, 4), cond):
    print(it.i, it.j, it.k)

更多的例子:

def cond(): return it.i + it.j + it.k == 777

it.i, it.j, it.k = 0, 0, 0
for it.i, it.j, it.k in bol(range(100), range(1000), range(10000), cond):
    print(it.i, it.j, it.k)




def cond(): return it.i + it.j + it.k >= 100000

it.i, it.j, it.k = 0, 0, 0
# you dont have to use it.i, it.j, it.k as the loop variables, you can
# use anything you want, but you have to update the variables somewhere
for i, j, k in bol(range(100), range(1000), range(10000), cond):
    it.i, it.j, it.k = i * 10, j * 100, k * 100
    print(it.i, it.j, it.k)