Question

我有一个过滤列表的生成器功能：

def gen_func(numbers):
    for number in numbers:
        if "?" in number:
            yield number

for item in gen_func(["trunk", "tr?ee", "+lea?f"]):
    print(item)

输出：

tr?ee
lea?f

我尝试编写一个迭代器，期望获得相同的输出，但是我得到的None作为第一项返回：

class GenClass:

    def __init__(self, numbers):
        self.current = 0
        self.numbers = numbers

    def __iter__(self):
        return self

    def __next__(self):
        next_value = self.current   

        if next_value >= len(self.numbers):
            raise StopIteration
        output = self.numbers[self.current]
        self.current += 1

        if "?" in output:
            return output

genclass = GenClass(["trunk", "tr?ee", "+lea?f"])
for num in genclass:
    print(num)

输出：

None
tr?ee
lea?f

如何仅获取tr?ee和lea?f而不是上面的内容？另外，这个迭代器有意义吗？还是我在这里滥用了迭代器？

Answer 1

您需要将__next__的代码更改为：

        next_value = self.current

        if next_value >= len(self.numbers):
            raise StopIteration

        output = self.numbers[self.current]
        while "?" not in output:
            self.current += 1
            output = self.numbers[self.current]

        self.current += 1
        return output

这将返回：

tr?ee
+lea?f

本质上，您遍历数字，并且仅在数字包含?时返回。

Answer 2

您可以使用 yield 代替 return 并仅使用__iter__。

class GenClass:
    def __init__(self, numbers):
        self.numbers = numbers

    def __iter__(self):
        for value  in self.numbers:
            if '?' in value:
                yield value


genclass = GenClass(["trunk", "tr?ee", "+lea?f"])
for num in genclass:
    print(num)

Out:
tr?ee
+lea?f

我上面的解决方案不是 iterator ，而是 iterable ，谢谢juanpa.arrivillaga。我认为最好将 iterable 和 iterator 分开，在这种情况下，您可以分别实现GenClass和 iterator 。因此，如果您要遍历GenClass实例一次，则可以重复它而无需重置计数器（Daniel Mesejo解决方案中的GenClassIterator）。

class GenClass:
    def __init__(self, numbers):
        self.numbers = numbers

    def __iter__(self):
        return GenClassIterator(self.numbers)


class GenClassIterator:
    def __init__(self, numbers):
        self.numbers = numbers
        self.current = 0

    def __next__(self):
        next_value = self.current

        if next_value >= len(self.numbers):
            raise StopIteration

        output = self.numbers[self.current]
        while "?" not in output:
            self.current += 1
            output = self.numbers[self.current]

        self.current += 1
        return output

    def __iter__(self):
        return self

Answer 3

从numbers弹出项目，直到符合条件的项目，然后将其退回。

import collections
class GenClass:

    def __init__(self, numbers):
        self.numbers = collections.deque(numbers)

    def __iter__(self):
        return self

    def __next__(self):
        while True:
            try:
                next_value = self.numbers.popleft()
            except IndexError as e:
                raise StopIteration

            if "?" in next_value:
                return next_value

Answer 4

您可以调用next（self）返回所需的值而无需返回None

class GenClass:

def __init__(self, numbers):
    self.current = 0
    self.numbers = numbers

def __iter__(self):
    return self

def __next__(self):
    next_value = self.current   

    if next_value >= len(self.numbers):
        raise StopIteration
    output = self.numbers[self.current]
    self.current += 1

    if "?" in output:
        return output
    else:
        return next(self)  # Skip value


genclass = GenClass(["trunk", "tr?ee", "+lea?f"])
for num in genclass:
     print(num)

这是更新的版本，在迭代过程中不会存储所有值

class GenClass:

def __init__(self, numbers):
    self.numbers = iter(numbers)

def __iter__(self):
    return self

def __next__(self):
    output = next(self.numbers)
    if "?" in output:
        return output
    else:
        return next(self)

如何制作一个过滤列表的迭代器？

4 个答案: