从生成器中筛选异常

huangapple
go评论88阅读模式
英文:

Filter on exception from generator

问题

我想要遍历一个生成器/迭代器并将生成的值传递给函数。
这个函数可以返回一个新值，也可以引发错误。
然后我需要返回这个值或者在出现错误时跳过。

我想出了以下的代码：

  1. def yield_or_skip(iter_: Iterable, func: Callable, skip_on_errors: Iterable[Type[Exception]]) -> Iterator:
  2.     skip_on_errors = set(skip_on_errors)
  3.     for item in iter_:
  4.         try:
  5.             yield func(item)
  6.         except Exception as e:
  7.             if type(e) in skip_on_errors:
  8.                 continue
  9.             raise

它似乎有效，但我想知道是否有更好的方法？

注意：它必须是一个迭代器/生成器，我不能将结果收集到列表中。

英文:

I would like to iterate over a generator/iterator and pass the value produced to the function.
The function can either return a new value or raise an error.
I then need to return the value or skip on error.

I came up with the following code:

  1. def yield_or_skip(iter_: Iterable, func: Callable, skip_on_errors: Iterable[Type[Exception]]) -> Iterator:
  2.     skip_on_errors = set(skip_on_errors)
  3.     for item in iter_:
  4.         try:
  5.             yield func(item)
  6.         except Exception as e:
  7.             if type(e) in skip_on_errors:
  8.                 continue
  9.             raise

It seems to work, but I was wondering if there is a better way to do this?

Note: it has to be an iterator/generator, I can't gather results in a list.

答案1

得分: 1

也许是suppress？

  1. from contextlib import suppress
  3. def yield_or_skip(iter_: Iterable, func: Callable, skip_on_errors: Iterable[Type[Exception]]) -> Iterator:
  4.     skipper = suppress(*skip_on_errors)
  5.     for item in iter_:
  6.         with skipper:
  7.             yield func(item)

演示：

  1. for y in yield_or_skip([2, 0, 3], lambda x: 1/x, [ZeroDivisionError]):
  2.     print(y)
  4. print()
  6. for y in yield_or_skip([2,0,3], lambda x: 1/x, [ValueError]):
  7.     print(y)

输出 (在线尝试!):

  1. 0.5
  2. 0.3333333333333333
  4. 0.5
  5. Traceback (most recent call last):
  6.   File "/ATO/code", line 26, in <module>
  7.     for y in yield_or_skip([2,0,3], lambda x: 1/x, [ValueError]):
  8.   File "/ATO/code", line 19, in yield_or_skip
  9.     yield func(item)
  10.   File "/ATO/code", line 26, in <lambda>
  11.     for y in yield_or_skip([2,0,3], lambda x: 1/x, [ValueError]):
  12. ZeroDivisionError: division by zero
英文:

Maybe suppress?

  1. from contextlib import suppress
  3. def yield_or_skip(iter_: Iterable, func: Callable, skip_on_errors: Iterable[Type[Exception]]) -> Iterator:
  4.     skipper = suppress(*skip_on_errors)
  5.     for item in iter_:
  6.         with skipper:
  7.             yield func(item)

Demo:

  1. for y in yield_or_skip([2, 0, 3], lambda x: 1/x, [ZeroDivisionError]):
  2.     print(y)
  4. print()
  6. for y in yield_or_skip([2,0,3], lambda x: 1/x, [ValueError]):
  7.     print(y)

Output (Attempt This Online!):

  1. 0.5
  2. 0.3333333333333333
  4. 0.5
  5. Traceback (most recent call last):
  6.   File "/ATO/code", line 26, in <module>
  7.     for y in yield_or_skip([2,0,3], lambda x: 1/x, [ValueError]):
  8.   File "/ATO/code", line 19, in yield_or_skip
  9.     yield func(item)
  10.   File "/ATO/code", line 26, in <lambda>
  11.     for y in yield_or_skip([2,0,3], lambda x: 1/x, [ValueError]):
  12. ZeroDivisionError: division by zero

答案2

得分: 0

more_itertools.map_except 执行该任务。您可以使用它，或使用它来实现自己的版本（在线尝试！):

  1. from more_itertools import map_except
  3. def yield_or_skip(iter_: Iterable, func: Callable, skip_on_errors: Iterable[Type[Exception]]) -> Iterator:
  4.     return map_except(func, iter_, *skip_on_errors)

或者调整其源代码:

  1.     for item in iterable:
  2.         try:
  3.             yield function(item)
  4.         except exceptions:
  5.             pass
英文:

more_itertools.map_except does that job. You can use it instead, or use it to implement yours (Attempt This Online!):

  1. from more_itertools import map_except
  3. def yield_or_skip(iter_: Iterable, func: Callable, skip_on_errors: Iterable[Type[Exception]]) -> Iterator:
  4.     return map_except(func, iter_, *skip_on_errors)

Or adapt its source code:

  1.     for item in iterable:
  2.         try:
  3.             yield function(item)
  4.         except exceptions:
  5.             pass

huangapple
  • 本文由 发表于 2023年6月2日 07:15:39
  • 转载请务必保留本文链接：https://go.coder-hub.com/76386269.html
  • exception
  • python
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