Python Regex re.sub not working as intended with dictionary lookup function?

Question

I have one Py file working as a lookup like (constants.py):

constants = {'[%IP%]':'0.0.0.0'}

From my main file, I'm trying to replace a part of my string using the lookup.

from constants import constants
import re
path = r"[%IP%]\file.exe"
pattern = r'\[%.*?\%]'
def replace(match):
    return constants.get(match)
print(replace("[%IP%]")) # this bit works fine
print(re.sub(pattern, replace, path)) # this results in the matched pattern just vanishing

When I directly access the 'replace' function in my main file, it correctly outputs the matched result from the constants file dictionary. But when I use the re.sub to substitute the pattern directly using the replace function with the lookup, it results in a string without the matched pattern at all.

So this is essentially the output for this (which isn't correct):

0.0.0.0
\file.exe

The desired output for this is:

0.0.0.0
0.0.0.0\file.exe

Please help me fix this.

trincot · Accepted Answer · 2022-08-03 16:37:16Z

2

The callback you pass to re.sub will be called with a Match object, not with a plain string. So the callback function needs to extract the string from that Match object.

Can be done like this:

def replace(match):
    return constants.get(match[0])

answered Aug 3, 2022 at 16:37

trincot

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3 Comments

Ankan Das Over a year ago

Thank you! It's been a long day at work, I didn't even pass the parameters with the function.

trincot Over a year ago

Not sure what you mean. It is not you that needs to pass the parameters... that is done by the re.sub implementation. The code re.sub(pattern, replace, path)) is fine.

Ankan Das Over a year ago

That's absolutely correct. I just tried out a implementation using lambda and that worked too but I had to pass parameters in that, but I took your solution as the convenient one. Thanks!

Ankan Das · Accepted Answer · 2022-08-03 16:48:13Z

0

The solution that I'll mark as Accepted is definitely one solution. The other way around is to use a lambda function and pass the parameters like:

re.sub(pattern, lambda x: replace(x.group()), path))

answered Aug 3, 2022 at 16:48

Ankan Das

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