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
由于这个问题已经成为进入特定循环的标准问题，我想用Exception给出我的答案。
虽然在多循环构造中不存在名为“循环中断”的标签，但我们可以使用用户定义异常来中断到我们选择的特定循环。考虑下面的例子，让我们在6进制编号系统中打印所有最多4位的数字:
class BreakLoop(Exception):
def __init__(self, counter):
Exception.__init__(self, 'Exception 1')
self.counter = counter
for counter1 in range(6): # Make it 1000
try:
thousand = counter1 * 1000
for counter2 in range(6): # Make it 100
try:
hundred = counter2 * 100
for counter3 in range(6): # Make it 10
try:
ten = counter3 * 10
for counter4 in range(6):
try:
unit = counter4
value = thousand + hundred + ten + unit
if unit == 4 :
raise BreakLoop(4) # Don't break from loop
if ten == 30:
raise BreakLoop(3) # Break into loop 3
if hundred == 500:
raise BreakLoop(2) # Break into loop 2
if thousand == 2000:
raise BreakLoop(1) # Break into loop 1
print('{:04d}'.format(value))
except BreakLoop as bl:
if bl.counter != 4:
raise bl
except BreakLoop as bl:
if bl.counter != 3:
raise bl
except BreakLoop as bl:
if bl.counter != 2:
raise bl
except BreakLoop as bl:
pass
当我们打印输出时，我们永远不会得到任何单位位是4的值。在这种情况下，在同一个循环中引发BreakLoop(4)并捕获时，我们不会中断任何循环。类似地，当十位有3时，我们使用BreakLoop(3)进入第三个循环。当百位是5时，我们使用BreakLoop(2)进入第二个循环，当千位是2时，我们使用BreakLoop(1)进入第一个循环。
简而言之，在内部循环中引发异常(内置或用户定义)，并在循环中从您想恢复控件的位置捕获它。如果想从所有循环中中断，可以在所有循环之外捕获异常。(我没有举例说明)。
其他回答
首先，普通的逻辑是有用的。
如果由于某种原因，终止条件无法确定，例外是一个后备计划。
class GetOutOfLoop( Exception ):
pass
try:
done= False
while not done:
isok= False
while not (done or isok):
ok = get_input("Is this ok? (y/n)")
if ok in ("y", "Y") or ok in ("n", "N") :
done= True # probably better
raise GetOutOfLoop
# other stuff
except GetOutOfLoop:
pass
对于这个特定的示例，可能不需要异常。
另一方面，在字符模式应用程序中，我们经常有“Y”、“N”和“Q”选项。对于"Q"选项，我们要立即退出。这是比较特殊的。
这不是最漂亮的方法，但在我看来，这是最好的方法。
def loop():
while True:
#snip: print out current state
while True:
ok = get_input("Is this ok? (y/n)")
if ok == "y" or ok == "Y": return
if ok == "n" or ok == "N": break
#do more processing with menus and stuff
我很确定你也可以用递归解出一些东西，但我不知道这对你来说是不是一个好选择。
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)
由于这个问题已经成为进入特定循环的标准问题，我想用Exception给出我的答案。
虽然在多循环构造中不存在名为“循环中断”的标签，但我们可以使用用户定义异常来中断到我们选择的特定循环。考虑下面的例子，让我们在6进制编号系统中打印所有最多4位的数字:
class BreakLoop(Exception):
def __init__(self, counter):
Exception.__init__(self, 'Exception 1')
self.counter = counter
for counter1 in range(6): # Make it 1000
try:
thousand = counter1 * 1000
for counter2 in range(6): # Make it 100
try:
hundred = counter2 * 100
for counter3 in range(6): # Make it 10
try:
ten = counter3 * 10
for counter4 in range(6):
try:
unit = counter4
value = thousand + hundred + ten + unit
if unit == 4 :
raise BreakLoop(4) # Don't break from loop
if ten == 30:
raise BreakLoop(3) # Break into loop 3
if hundred == 500:
raise BreakLoop(2) # Break into loop 2
if thousand == 2000:
raise BreakLoop(1) # Break into loop 1
print('{:04d}'.format(value))
except BreakLoop as bl:
if bl.counter != 4:
raise bl
except BreakLoop as bl:
if bl.counter != 3:
raise bl
except BreakLoop as bl:
if bl.counter != 2:
raise bl
except BreakLoop as bl:
pass
当我们打印输出时，我们永远不会得到任何单位位是4的值。在这种情况下，在同一个循环中引发BreakLoop(4)并捕获时，我们不会中断任何循环。类似地，当十位有3时，我们使用BreakLoop(3)进入第三个循环。当百位是5时，我们使用BreakLoop(2)进入第二个循环，当千位是2时，我们使用BreakLoop(1)进入第一个循环。
简而言之，在内部循环中引发异常(内置或用户定义)，并在循环中从您想恢复控件的位置捕获它。如果想从所有循环中中断，可以在所有循环之外捕获异常。(我没有举例说明)。
Break for外层和内部while循环:
while True:
while True:
print('Breaks inner "while" loop')
break # Here
print('Breaks outer "while" loop')
break # Here
或者，用if语句中断外部和内部while循环:
while True:
while True:
if True:
print('Breaks inner "while" loop')
break # Here
print('Breaks outer "while" loop')
break # Here
输出:
Breaks inner "while" loop
Breaks outer "while" loop
Break for outer和inner for循环:
for _ in iter(int, 1):
for _ in iter(int, 1):
print('Breaks inner "for" loop')
break # Here
print('Breaks outer "for" loop')
break # Here
或者，用if语句打破外部和内部for循环:
for _ in iter(int, 1):
for _ in iter(int, 1):
if True:
print('Breaks inner "for" loop')
break # Here
print('Breaks outer "for" loop')
break # Here
输出:
Breaks inner "for" loop
Breaks outer "for" loop
