[Tutor] Substring substitution

Bernard Lebel 3dbernard at gmail.com
Fri Sep 9 22:14:11 CEST 2005

Hi Kent,

Once again, thanks a lot. Problem solved now, your suggestions work
like a charm. You were absolutely right about the last group matching.
I modified my matching pattern:

oRe = re.compile( "(\d\d_\d\d\_)(\d\d(\D|$))" )

Instead of

oRe = re.compile( "(\d\d_\d\d\_)(\d\d)" )

I had no idea you could put subgroups in groups, I tried that out of
inspiration and whoo it works. This is fantastic!

So here is the full script, followed by the output:

import os, re

def matchShot( sSubString, oRe ):
	sNewString = oRe.sub( r'\g<1>0\2', sSubString )
	return sNewString

def processPath( sString, bTest ):
	sString (string): the path to process
	bTest (boolean): test the validity of paths at each step of the script
	# Create regular expression object
	oRe = re.compile( "(\d\d_\d\d\_)(\d\d(\D|$))" )
	if bTest == True:
		# Test validity of integral path
		if not os.path.exists( sString ): return None
	# Break-up path
	aString = sString.split( os.sep )
	aNewPath = []
	# Iterate individual components
	for sSubString in aString:
		if bTest == True:
			# Test if this part of the path is working with the current path
			sTempPath = '\\'.join( aNewPath )
			sTempPath += '%s%s' % ( '\\', sSubString )
			if not os.path.exists( sTempPath ): sSubString = matchShot( sSubString, oRe )
		else: sSubString = matchShot( sSubString, oRe )
		aNewPath.append( sSubString )
		if bTest == True:
			# Test again if path is valid with this substring
			sTempPath = '\\'.join( aNewPath )
			if not os.path.exists( sTempPath ): return None
	sNewPath = '\\'.join( aNewPath )
	print sNewPath

processPath( r'C:\temp\MT_03_03_03\allo.txt', False )
processPath( r'C:\temp\MT_03_04_04\mt_03_04_04_anim_v1.scn', False )
processPath( r'C:\temp\MT_03_05_005_anim\mt_03_05_05_anim_v1.scn', False )
processPath( r'C:\temp\MT_03_06_006\mt_03_06_006_anim_v1.scn', False )

# ============================================================


This is exactly what I was after. Thanks a lot!!


On 9/8/05, Kent Johnson <kent37 at tds.net> wrote:
> Bernard Lebel wrote:
> > Ok I think I understand what is going: I'm using a 0 in the
> > replacement argument, between the two groups. If I try with a letter
> > or other types of characters it works fine. So how can use a digit
> > here?
> There is a longer syntax for \1 - \g<1> means the same thing but without the ambiguity of where it ends. So you can use r'\g<1>0\2' as your substitution string.
> >>def matchShot( sSubString ):
> >>
> >>        # Create regular expression object
> >>        oRe = re.compile( "(\d\d_\d\d\_)(\d\d)" )
> >>
> >>        oMatch = oRe.search( sSubString )
> >>        if oMatch != None:
> >>                sNewString = oRe.sub( r'\10\2', sSubString )
> >>                return sNewString
> >>        else:
> >>                return sSubString
> You don't have to do the search, oRe.sub() won't do anything if there is no match. Also if you are doing this a lot you should pull the re.compile() out of the function (so oRe is a module variable), this is an expensive step that only has to be done once.
> You hinted in your original post that you are trying to find strings where the last _\d\d has only two digits. The re you are using will also match something like 'mt_03_04_044_anim' and your matchShot() will change that to 'mt_03_04_0044_anim'. If that is not what you want you have to put some kind of a guard at the end of the re - something that won't match a digit. If you always have the _ at the end it is easy, just use r"(\d\d_\d\d\_)(\d\d_)". If you can't count on the underscore you will have to be more clever.
> Kent
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