给定以下代码(不起作用):

while True:
    # Snip: print out current state
    while True:
        ok = get_input("Is this ok? (y/n)")
        if ok.lower() == "y": break 2 # This doesn't work :(
        if ok.lower() == "n": break

    # Do more processing with menus and stuff

有办法解决这个问题吗?还是我必须先检查一次,跳出输入循环,然后再检查一次,如果用户满意,再跳出外部循环?


当前回答

keeplooping = True
while keeplooping:
    # Do stuff
    while keeplooping:
          # Do some other stuff
          if finisheddoingstuff():
              keeplooping = False

或者类似的东西。

您可以在内部循环中设置一个变量,并在内部循环退出后立即在外部循环中检查它,如果合适的话可以中断。我有点喜欢GOTO方法,前提是你不介意使用愚人节的笑话模块——它不是python的,但它确实有意义。

其他回答

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)

我最近遇到了这个问题,为了避免重复的return语句,因为它会隐藏逻辑错误,我研究了@yak的想法。这在嵌套的for循环中工作得很好,但不是很优雅。另一种方法是在下一次循环之前检查条件:

b = None
for a in range(10):
    if something(a, b): # should never = True if b is None
        break
    for b in range(20):
        pass

这可能并不适用于所有地方,但具有适应性,如果需要,它的优点是允许复制条件,而不是潜在的结果。

要跳出多个嵌套循环,而不需要重构为函数,可以使用带有内置StopIteration异常的“模拟goto语句”:

try:
    for outer in range(100):
        for inner in range(100):
            if break_early():
                raise StopIteration

except StopIteration: pass

请参阅使用goto语句打破嵌套循环的讨论。

keeplooping = True
while keeplooping:
    # Do stuff
    while keeplooping:
          # Do some other stuff
          if finisheddoingstuff():
              keeplooping = False

或者类似的东西。

您可以在内部循环中设置一个变量,并在内部循环退出后立即在外部循环中检查它,如果合适的话可以中断。我有点喜欢GOTO方法,前提是你不介意使用愚人节的笑话模块——它不是python的,但它确实有意义。

和上一台很像,但更紧凑。 (布尔值只是数字)

breaker = False #our mighty loop exiter!
while True:
    while True:
        ok = get_input("Is this ok? (y/n)")
        breaker+= (ok.lower() == "y")
        break

    if breaker: # the interesting part!
        break   # <--- !