Replace part of string using a regular expression

Question

s = 'scores:start:4.2.1.3.4.3:final:55.34.13.63.44.34'

I'm trying to reset the start scores to 0.0.0.0.0.0 so that it reads

scores:start:0.0.0.0.0.0:final:55.34.13.63.44.34

This works:

re.sub('start\:(\d+\.\d+\.\d+\.\d+\.\d+\.\d+)','0.0.0.0.0.0',s)

But I was looking for something more flexible to use in case the amount of scores change.

EDIT: Actually my example does not work, because start is removed from the string

scores:0.0.0.0.0.0:final:55.34.13.63.44.34

but I would like to keep it:

scores:start:0.0.0.0.0.0:final:55.34.13.63.44.34

You may use: re.sub(r'start:\d+(\.\d+)+', '0.0.0.0.0.0', s) — anubhava
– anubhava, Commented Aug 13, 2021 at 14:43
@HappyPy Really? Regardless of amount of scores, you always want exactly six zeros? — Kelly Bundy
– Kelly Bundy, Commented Aug 13, 2021 at 14:47

anubhava · Accepted Answer · 2021-08-13 14:49:07Z

3

You may use this re.sub with a lambda if you want to replace dot seperated numbers with same number of zeroes:

>>> import re
>>> s = 'scores:start:4.2.1.3.4.3:final:55.34.13.63.44.34'
>>> re.sub(r'start:\d+(?:\.\d+)+', lambda m: re.sub(r'\d+', '0', m.group()), s)
'scores:start:0.0.0.0.0.0:final:55.34.13.63.44.34'

We are using regex as start:\d+(?:\.\d+)+ to match text that starts with start: and followed by digits separated by dot.

In lambda part we replace each 1+ digit with 0 to get same number of zeroes in output as the input.

answered Aug 13, 2021 at 14:49

anubhava

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Comments

Kelly Bundy · Accepted Answer · 2021-08-13 15:27:31Z

2

Could replace all numbers before final with a zero:

re.sub(r'\d+(?=.*final)', '0', s)

Try it online!

Or perhaps more efficient if there were many more scores:

re.sub(r'\d|(final.*)', lambda m: m[1] or '0', s)

edited Aug 13, 2021 at 15:27

answered Aug 13, 2021 at 14:51

Kelly Bundy

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dawg · Accepted Answer · 2021-08-13 15:06:47Z

1

Another lambda in re.sub:

import re 
s = 'scores:start:4.2.1.3.4.3:final:55.34.13.63.44.34'
pat=r'(?<=:start:)([\d.]+)(?=:final:)'

>>> re.sub(pat, lambda m: '.'.join(['0']*len(m.group(1).split('.'))), s)
scores:start:0.0.0.0.0.0:final:55.34.13.63.44.34

answered Aug 13, 2021 at 15:06

dawg

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Comments

The fourth bird · Accepted Answer · 2021-08-13 19:04:30Z

0

A variant using the regex PyPi module:

(?:\bstart:|\G(?!^))\.?\K\d+

The pattern matches

(?: Non capture group
- \bstart: Match start:
- | Or
- \G(?!^) Assert the position at the end of the previous match, not at the start of the string
) Close the non capture group
\.?\K Match an optional dot, clear the current match buffer (forget what is matched so far)
\d+ Match 1 or more digits (to be replaced with a 0)

Regex demo

import regex

s = 'scores:start:4.2.1.3.4.3:final:55.34.13.63.44.34'
pattern = r"(?:\bstart:|\G(?!^))\.?\K\d+"
print(regex.sub(pattern, '0', s))

Output

scores:start:0.0.0.0.0.0:final:55.34.13.63.44.34

edited Aug 13, 2021 at 19:04

answered Aug 13, 2021 at 18:59

The fourth bird

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