Using re.sub with %s

2010-08-18 Thread Brandon Harris 

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

however if I cast a value with letters as opposed to numbers

re.sub('(/)word(/)', r'\1\%s\2'%'gosh', test)

return is /my/gosh/whats/wrong


Any help would be good. I've tried passing the value as an int, or 
recasting that value as something else, passing it as a raw string, 
removing the r and just double escaping the groups.


Brandon L. Harris

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Re: Using re.sub with %s

2010-08-18 Thread 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
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Re: Using re.sub with %s

2010-08-18 Thread MRAB 

Thomas Jollans wrote:

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.


You can stop group references which are followed by digits from turning
into octal escapes in the replacement template by using \gn instead:

 print r'\1%s' % '00'
\100
 print r'\g1%s' % '00'
\g100
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