re.sub replacement text \-escapes woe

[Alexander Schmolck]

Sorry if I missed something obvious, but how do you do this more
intelligently?

def escape(s):
return re.sub(r'([${}\\])', r'\ \1', s).replace('\\ ', '\\')

'as

[Peter Otten]

Alexander Schmolck wrote:

Sorry if I missed something obvious, but how do you do this more
intelligently?

def escape(s):
return re.sub(r'([${}\\])', r'\ \1', s).replace('\\ ', '\\')

'as

Interesting. No direct attack, but another workaround:

re.sub(r'([${}\\])', lambda m: '\\' + m.group(), s)

Peter

[Peter Otten]

Peter Otten wrote:

Alexander Schmolck wrote:

Sorry if I missed something obvious, but how do you do this more
intelligently?

def escape(s):
return re.sub(r'([${}\\])', r'\ \1', s).replace('\\ ', '\\')

'as

Interesting. No direct attack, but another workaround:

re.sub(r'([${}\\])', lambda m: '\\' + m.group(), s)

re.sub(r'([${}\\])', r'\\\1', s)

Either that or I'm slash-blind tonight...

Peter

[David M. Wilson]

Alexander Schmolck wrote:

Sorry if I missed something obvious, but how do you do this more
intelligently?

def escape(s):
return re.sub(r'([${}\\])', r'\ \1', s).replace('\\ ', '\\')

re.escape? :)
David

[Alexander Schmolck]

Peter Otten <__*******@web.de> writes:

Peter Otten wrote:

Alexander Schmolck wrote:

Sorry if I missed something obvious, but how do you do this more
intelligently?

def escape(s):
return re.sub(r'([${}\\])', r'\ \1', s).replace('\\ ', '\\')

'as

Interesting. No direct attack, but another workaround:

re.sub(r'([${}\\])', lambda m: '\\' + m.group(), s)

re.sub(r'([${}\\])', r'\\\1', s)

Either that or I'm slash-blind tonight...

D'oh -- slash-blindness is what happened to me (presumably because I was
unconsciously expecting a raw string when I tested the above in the
interactive console before I posted and thus incorrectly dismissed the
result).

Thanks

'as