Python regex replace alphanumeric with indent

Asked Mar 25 '21 at 15:48

Active Mar 25 '21 at 17:54

Viewed 37 times

I'm looking to leverage regex syntax to create an indent (new line) where certain criteria is met. In this case when there is an alphanumeric value following ), pattern.

Something like this - but it is not quite working.

text_old =

some text some text 
),
some text some text 
), some text some text

which becomes -

text_format = 
some text some text 
),
some text some text 
), 
some text some text

text_format = text_old.lower().replace('), ^[a-zA-Z0-9_.-]*$','), \n')

Do I have the proper syntax here?

EDIT

I understand that replace is not best approach for regex. So I have something like this, which isn't quite working -

re.sub(r'\),\s^\w+$','\), \n',text_old)

EDIT 2

I am very close, now it just cuts off the first letter it finds.

re.sub('\),\s\w','), \n',text_old)

Like this

text_format = 
some text some text 
),
some text some text 
), 
ome text some text

Where I would like it to find the pattern, not replace the first character

edited Mar 25 '21 at 17:54

asked Mar 25 '21 at 15:48

paranormaldist

Not quite: `str.replace` does not support regex based search and replace. It is done with `re.sub`. – Wiktor Stribiżew Mar 25 '21 at 15:49
I see, is there a ```replace``` approach to this question? – paranormaldist Mar 25 '21 at 15:50
1

You want to carry forward the part after the comma, so you will use parentheses to group it, and include \1 in your substitution string. Check the re.sub docs [here](https://docs.python.org/3/library/re.html?highlight=re%20sub#re.sub) – RufusVS Mar 25 '21 at 15:58
something like this? ```re.sub(r'\), ^[a-zA-Z0-9_.-]*$','), \n',text_old)``` doesn't quite work – paranormaldist Mar 25 '21 at 16:15
or something like ```re.sub(r'\),\s^\w+$','\), \n',text_old)``` I would think would work – paranormaldist Mar 25 '21 at 16:39

Python regex replace alphanumeric with indent

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