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

将循环逻辑分解为迭代器,迭代器产生循环变量并在完成时返回——这里有一个简单的迭代器,它将图像按行/列排列,直到我们用完图像或找不到放置它们的位置:

def it(rows, cols, images):
    i = 0
    for r in xrange(rows):
        for c in xrange(cols):
            if i >= len(images):
                return
            yield r, c, images[i]
            i += 1 

for r, c, image in it(rows=4, cols=4, images=['a.jpg', 'b.jpg', 'c.jpg']):
    ... do something with r, c, image ...

这样做的优点是将复杂的循环逻辑和处理分开。

其他回答

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。

将循环逻辑分解为迭代器,迭代器产生循环变量并在完成时返回——这里有一个简单的迭代器,它将图像按行/列排列,直到我们用完图像或找不到放置它们的位置:

def it(rows, cols, images):
    i = 0
    for r in xrange(rows):
        for c in xrange(cols):
            if i >= len(images):
                return
            yield r, c, images[i]
            i += 1 

for r, c, image in it(rows=4, cols=4, images=['a.jpg', 'b.jpg', 'c.jpg']):
    ... do something with r, c, image ...

这样做的优点是将复杂的循环逻辑和处理分开。

# this version breaks up to a certain label

break_label = None
while True:
    # snip: print out current state
    while True:
        ok = get_input("Is this ok? (y/n)")
        if ok == "y" or ok == "Y":
            break_label = "outer"   # specify label to break to
            break
        if ok == "n" or ok == "N":
            break
    if break_label:
        if break_label != "inner":
            break                   # propagate up
        break_label = None          # we have arrived!
if break_label:
    if break_label != "outer":
        break                       # propagate up
    break_label = None              # we have arrived!

#do more processing with menus and stuff

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

使用numpy.ndindex可以简单地将多个循环转换为单个、可破坏的循环

for i in range(n):
  for j in range(n):
    val = x[i, j]
    break # still inside the outer loop!

for i, j in np.ndindex(n, n):
  val = x[i, j]
  break # you left the only loop there was!

您确实需要索引到对象中,而不是显式地遍历值,但至少在简单的情况下,它似乎比大多数答案所建议的要简单大约2-20倍。