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Old 12-20-2010, 06:29 AM   #1
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Question: find and replace with variables?

Can anyone give me a clue about using variables in a calibre recipe?

I want to try to find any digit followed by a space and three other digits, for example '1 000', and replace the space with a non-breaking space.

I thought something like this would work:
Spoiler:

(re.compile(r'([0-9]) ([0-9])([0-9])([0-9])'), lambda match: '\1 \2\3\4'),


This finds the right expressions, but the variables aren't replacing correctly.

What am I doing wrong?

Last edited by veezh; 12-20-2010 at 07:30 AM.
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Old 12-20-2010, 12:08 PM   #2
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use

r'\1 \2\3\4'


Note the r
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Old 12-20-2010, 03:49 PM   #3
veezh
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Quote:
Originally Posted by kovidgoyal View Post
use

r'\1 \2\3\4'


Note the r
Thanks very much for your reply, Kovid.

I've just tried your suggestion, but I'm still getting the same result as before, i.e. that the strings are being found but are being replaced with the literal string I've supplied, not the variables.

As an example, the number '22 100' shows up as '2\1 \2\3\4' (the non-breaking space is correctly being added) in the files generated from the recipe.

Any other hints greatly appreciated!
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Old 12-20-2010, 04:22 PM   #4
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oh right you're using a lambda, sorry, do this

lambda m: m.group(1) + ' ' + m.group(2) + m.group(3) + m.group(4)
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Old 12-21-2010, 05:23 AM   #5
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Thanks a lot, Kovid. It's working!
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