处理在生成器中抛出的异常 [英] Handle an exception thrown in a generator

问题描述

我有一个生成器和一个功能消耗它：

I've got a generator and a function that consumes it:

def read():
    while something():
        yield something_else()

def process():
    for item in read():
        do stuff

如果生成器抛出异常，我想在消费者函数中处理它，然后继续使用迭代器，直到它耗尽。注意，我不想在生成器中有任何异常处理代码。

If the generator throws an exception, I want to process that in the consumer function and then continue consuming the iterator until it's exhausted. Note that I don't want to have any exception handling code in the generator.

我想到了如下：

reader = read()
while True:
    try:
        item = next(reader)
    except StopIteration:
        break
    except Exception as e:
        log error
        continue
    do_stuff(item)

但这对我看起来相当尴尬。

but this looks rather awkward to me.

推荐答案

当生成器抛出异常时，它将退出。

When a generator throws an exception, it exits. You can't continue consuming the items it generates.

示例：

>>> def f():
...     yield 1
...     raise Exception
...     yield 2
... 
>>> g = f()
>>> next(g)
1
>>> next(g)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in f
Exception
>>> next(g)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
StopIteration

如果您控制生成器代码，可以处理生成器中的异常;如果没有，您应该尽量避免发生异常。

If you control the generator code, you can handle the exception inside the generator; if not, you should try to avoid an exception occurring.

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