How to replace partial groups in python regex?

Question

I have a regex

(obligor_id): (\d+);(obligor_id): (\d+):

A sample match like below:

Match 1
Full match  57-95   `obligor_id: 505732;obligor_id: 505732:`
Group 1.    57-67   `obligor_id`
Group 2.    69-75   `505732`
Group 3.    76-86   `obligor_id`
Group 4.    88-94   `505732`

I am trying to partially replace the full match to the following:

obligor_id: 505732;obligor_id: 505732: -> obligor_id: 505732;

Two ways to achieve so,

replace group 3 and 4 with empty string
replace group 1 and 2 with empty string, and then replace group 4 to (\d+);

How can I achieve these 2 in python? I know there is a re.sub function, but I only know how to replace the whole, not partially replace group.

Thanks in advance.

Why (obligor_id) needs to be a group? Do you only replace repeating ids? If so, consider replacing the entire line that matches obligor_id: (\d+);obligor_id: \1 with "obligor_id: " + match.group(1). — 9000
– 9000, Commented Nov 27, 2018 at 22:21
Use re.sub with your pattern and r'\1: \2;' replacement. See the regex demo — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Nov 27, 2018 at 22:22

mrzasa · Accepted Answer · 2018-11-27 22:24:24Z

2

You can change capturing groups and reference them in the substitution string:

s = 'obligor_id: 505732;obligor_id: 505732:' 
re.sub(r'(obligor_id: \d+;)(obligor_id: \d+:)', r'\1', s)
# => 'obligor_id: 505732;

answered Nov 27, 2018 at 22:24

mrzasa

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1 Comment

aksh1618 Over a year ago

For using number variables in replacement strings, named reference syntax can be used: re.sub(pat, r'\g<1>' + f'{my_number}', str)

Pythoner · Accepted Answer · 2018-11-27 22:30:46Z

1

Thanks for answers and advices:

I achieved them as below for future users:

re.sub(regex, r'\1: \2;', str)
re.sub(regex, r'\3: \4;', str)

answered Nov 27, 2018 at 22:30

Pythoner

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