[英]Make a subgroup reference (\g<1>) optional in re.sub
How can I make a subgroup reference ( \\g<1>
) optional in re.sub()
? 如何在
re.sub()
创建子组引用( \\g<1>
)可选? For example with: 例如:
import re
regexp = re.compile(r'^http://(lists\.|www\.)?example\.com/')
regexp.sub(
r'https://\g<1>example.com/',
r'http://example.com/helllo-there'
)
I would like \\g<1>
to be replaced with nothing, the optional subgroup isn't matched (and not raise an exception). 我希望
\\g<1>
替换为空,可选子组不匹配(并且不引发异常)。
I know I can use regexp.match(..).groups()
to check which groups are present, but this seems like a lot of work to me (we would need a bunch of replacement patterns, since some examples go up to \\g<6>
). 我知道我可以使用
regexp.match(..).groups()
来检查哪些组存在,但这对我来说似乎很多工作(我们需要一堆替换模式,因为一些例子上升到\\g<6>
)。 It's also not very fast since we need to do a match
and a replace
. 它也不是很快,因为我们需要进行
match
和 replace
。
For example in JavaScript, I can use $1
, if it's not matched it's just ignored: 例如在JavaScript中,我可以使用
$1
,如果它不匹配,它只是被忽略:
'http://example.com/helllo-there'.replace(
RegExp('^http://(lists\.|www\.)?example\.com/'),
'https://$1example.com/')
// Outputs: "https://example.com/helllo-there"
Another option is to provide an explicit empty alternative: 另一种选择是提供一个明确的空替代方案:
regexp = re.compile(r'^http://(lists\.|www\.|)example\.com/')
Also, you can use just \\1
instead of \\g<1>
. 此外,您只能使用
\\1
而不是\\g<1>
。
如果我理解正确,只需做x(y?)z
而不是x(y)?z
I would do like this. 我会这样做的。 Just put the pattern inside a non-capturing group and make it as optional.
只需将模式放在非捕获组中,并将其设置为可选。 Now include that optional non-capturing group inside a capturing group.
现在在捕获组中包含可选的非捕获组。
>>> re.sub(r'^http://((?:lists\.|www\.)?)example\.com/',r'https://\g<1>example.com/', 'http://example.com/helllo-there')
'https://example.com/helllo-there'
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