Python regex re.sub : something

Question

I have a program like this :

import re

x='aaaaaaaa;aa;aaa;aaa;aaaaaaaaaa;'
x=re.sub(';','.',x, re.IGNORECASE)

print x

But the output is like this:

aaaaaaaa.aa.aaa;aaa;aaaaaaaaaa;

There are still some ; not replaced by a ., why ?

Using Python 2.6

Using "ignorecase" when replacing ; doesn't seem to make much sense. — georg
– georg, Commented May 13, 2012 at 10:53

jamylak · Accepted Answer · 2012-05-13 10:35:26Z

7

Update - In Python 2.6 you can just do this:

>>> re.sub('(?i);','.',x)
'aaaaaaaa.aa.aaa.aaa.aaaaaaaaaa.'

For Python 2.7+ and 3.0+

Do this instead, the third parameter is actually the count(number of replacements to make) and re.IGNORECASE is simply an integer so it is using that as the count.

>>> re.sub(';','.',x, flags=re.IGNORECASE)
'aaaaaaaa.aa.aaa.aaa.aaaaaaaaaa.'

>>> re.IGNORECASE
2

edited May 13, 2012 at 10:35

answered May 13, 2012 at 10:18

jamylak

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4 Comments

heathen90 Over a year ago

It gives me an error with python 2.6: TypeError: sub() got an unexpected keyword argument 'flags' (but it's work with the 2.7 version thank you ^^)

jamylak Over a year ago

Which version of Python are you using? EDIT: Nvm read your comment

jamylak Over a year ago

You can change your pattern to '(?i);' which does the same thing

mavili Over a year ago

(?i) came very handy as I need to be using python2.6 for the work I'm doing.

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