python是怎么实现过滤 #注释代码的？

Question

今天看python核心编程看到一个问题，题意大概就是过滤一个file-like对象里“#”的注释部分，然后输出其他部分。简单情形下，另写一行的#注释比较好判断，用startwith('#')匹配应该能满足。问题在于那些写在正常业务代码之后的注释，该如何过滤之？举个例子：

if name.find(",") == -1:#Annotations
    pass

请问有没有人了解它是怎么过滤这种注释的？谢谢。

Answer 1

Let’s talk about the idea
If you don’t consider the # in the string, it is very convenient to use re to match

#[^\n]*?\n

is enough
If you consider that the # in the string is slightly more complicated, match:

#[^'"]*?\n

Barely able to cope with most situations

The disadvantage is that for

'a' # 'b'

Such a statement cannot be matched because Python's re does not support balanced groups.

Answer 2

Just use regular matching#It won’t work until the end of the line.