[英]Python: Iterator returns None
Here is my code: 这是我的代码:
class Prizes(object):
def __init__(self, purchases, n, d):
self.p = purchases
self.n = n
self.d = d
self.x = 1
def __iter__(self):
return self
def __next__(self):
print(self.x)
if self.x % self.n == 0 and self.p[self.x - 1] % self.d == 0:
self.x = self.x + 1
return self.x - 1
elif self.x > len(self.p):
raise StopIteration
self.x = self.x + 1
def superPrize(purchases, n, d):
return list(Prizes(purchases, n, d))
An example of usage: 用法示例:
superPrize([12, 43, 13, 465, 1, 13], 2, 3)
The output should be: 输出应该是:
[4]
But actual output is: 但实际输出是:
[None, None, None, 4, None, None].
Why does it happen? 为什么会这样?
Your problem is your implementation of __next__
. 你的问题是
__next__
的实现。 When Python calls __next__
, it will always expect a return value . 当Python调用
__next__
, 它总是期望返回值 。 However, in your case, it looks like you may not always have a return value each call. 但是,在您的情况下,看起来您可能并不总是每次调用都有返回值。 Thus, Python uses the default return value of a function -
None
: 因此,Python使用函数的默认返回值 -
None
:
You need some way to keep program control inside of __next__
until you have an actually return value. 你需要一些方法来保持程序控制在
__next__
直到你有一个实际的返回值。 This can be done using a while
-loop: 这可以使用
while
-loop来完成:
def __next__(self):
while True:
if self.x % self.n == 0 and self.p[self.x - 1] % self.d == 0:
self.x = self.x + 1
return self.x - 1
elif self.x > len(self.p):
raise StopIteration
self.x = self.x + 1
Wrap it with while
so that your method doesn't return a value until you found one: 用
while
包装它,这样你的方法就不会返回值,直到找到一个:
def __next__(self):
while True:
if self.x % self.n == 0 and self.p[self.x - 1] % self.d == 0:
self.x = self.x + 1
return self.x - 1
elif self.x > len(self.p):
raise StopIteration
self.x = self.x + 1
Things working with iterators call __next__
expecting it to return a value, but the method returns a value only under a condition, otherwise it reaches the end of the method and it returns None
. 使用迭代器的事情调用
__next__
期望它返回一个值,但该方法仅在条件下返回一个值,否则它到达方法的末尾并返回None
。
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