更多pythonic的方式来写这个？

I have this code here:

我在这里有这个代码:

import re
def get_attr(str, attr):
    m = re.search(attr + r'=(\w+)', str)
    return None if not m else m.group(1)

str = 'type=greeting hello=world'

print get_attr(str, 'type')   # greeting    
print get_attr(str, 'hello')  # world
print get_attr(str, 'attr')   # None

Which works, but I am not particularly fond of this line:

哪个有效,但我不是特别喜欢这一行:

return None if not m else m.group(1)

In my opinion this would look cleaner if we could use a ternary operator:

在我看来,如果我们可以使用三元运算符,这将看起来更清晰:

return (m ? m.group(1) : None)

But that of course isn't there. What do you suggest?

但那当然不存在。你有什么建议?

4 个解决方案

#1

Python has a ternary operator. You're using it. It's just in the X if Y else Z form.

Python有一个三元运算符。你正在使用它。它只是在X中,如果是其他Z形式。

That said, I'm prone to writing these things out. Fitting things on one line isn't so great if you sacrifice clarity.

也就是说,我很容易写出这些东西。如果你牺牲清晰度,在一条线上装配东西并不是那么好。

def get_attr(str, attr):
    m = re.search(attr + r'=(\w+)', str)
    if m:
        return m.group(1)

    return None

#2

Another option is to use:

另一种选择是使用:

return m.group(1) if m else m

It's explicit, and you don't have to do any logic puzzles to understand it :)

它是明确的,你不必做任何逻辑谜题来理解它:)

#3

return m and m.group(1)

would be one Pythonic way to do it.

将是一种Pythonic方式来做到这一点。

If m is None (or something else that evaluates "falsely"), it returns m, but if m is "true-ish", then it returns m.group(1).

如果m是None(或其他评估“falsely”的东西),则返回m,但如果m是“true-ish”,则返回m.group(1)。

#4

What you have there is python's conditional operator. IMO it's perfectly pythonic as-is and needs no change. Remember, explicit is better than implicit. What you have now is readable and instantly understandable.

你有什么是python的条件运算符。 IMO它完全是pythonic as-is,不需要改变。请记住,显式优于隐式。你现在拥有的是可读的,可以立即理解的。

#1