Using re.sub with %s

[Thomas Jollans]

On Wednesday 18 August 2010, it occurred to Brandon Harris to exclaim:
> Having trouble using %s with re.sub
> 
> test = '/my/word/whats/wrong'
> re.sub('(/)word(/)', r'\1\%s\2'%'1000', test)
> 
> return is /my/@0/whats/wrong
> 

This has nothing to do with %, of course:

>>> re.sub('(/)word(/)', r'\1\%d\2'%1000, test)
'/my/@0/whats/wrong'
>>> re.sub('(/)word(/)', r'\1\1000\2', test)
'/my/@0/whats/wrong'

let's see if we can get rid of that zero:

>>> re.sub('(/)word(/)', r'\1\100\2', test)
'/my/@/whats/wrong'

so '\100' appears to be getting replaced with '@'. Why?

>>> '\100'
'@'

This is Python's way of escaping characters using octal numbers.

>>> chr(int('100', 8))
'@'

How to avoid this? Well, if you wanted the literal backslash, you'll need to 
escape it properly:

>>> print(re.sub('(/)word(/)', r'\1\\1000\2', test))
/my/\1000/whats/wrong
>>> 

If you didn't want the backslash, then why on earth did you put it there? You 
have to be careful with backslashes, they bite ;-)

Anyway, you can simply do the formatting after the match.

>>> re.sub('(/)word(/)', r'\1%d\2', test) % 1000
'/my/1000/whats/wrong'
>>> 

Or work with match objects to construct the resulting string by hand.

 - Thomas