REGEX "dynamic" substitution

Question

I have a multi-line string and need to append at the beginning of each row a (varying) character (1 white-space as separator). How can I do that with regex? Is there a way to do that without without too many splittings and merging?

It is assumed that the amount of rows is equal to the amount of characters to be added.

Input

m = """
0 1 2 3 4 5
1 0 1 2 3 4
2 1 1 2 3 4
"""

Desired output (assumed reference string "bip")

b 0 1 2 3 4 5
i 1 0 1 2 3 4
p 2 1 1 2 3 4

Here an example of working code based on \n match

import re

m = """
0 1 2 3 4 5
1 0 1 2 3 4
2 1 1 2 3 4
"""

re.sub(r'\n', '\n{} ', m[:-1]).format(*list('bip'))

but I needed to omit the last character, it would \n. I tried to play around with other pattern such as \A or \^ or ^, flags=re.MULTILINE but with no success.

How can I avoid the string slice m[:-1]?

Is there a more pure regex way to solve the problem?

Answer 1

Here is another approach using 4th parameter count of re.sub that tells how many substitutions are to be made:

As per linked doc:

re.sub(pattern, repl, string, count=0, flags=0)

Code:

import re
 
m = """
0 1 2 3 4 5
1 0 1 2 3 4
2 1 1 2 3 4
"""
repl = 'bip'
 
print (re.sub(r'\n', '\n{} ', m, len(repl)).format(*list(repl)))

# or use this if you want to strip leading and trailing newlines
# print (re.sub(r'\n', '\n{} ', m, len(repl)).format(*list(repl)).strip())

Code Demo

Output:

b 0 1 2 3 4 5
i 1 0 1 2 3 4
p 2 1 1 2 3 4

Answer 2

If you apply .strip() to your m string, you can use

import re
m = """
0 1 2 3 4 5
1 0 1 2 3 4
2 1 1 2 3 4
"""
print(re.sub(r'^', '{} ', m.strip(), flags=re.M).format(*list('bip')))

See the Python demo, the output will be

b 0 1 2 3 4 5
i 1 0 1 2 3 4
p 2 1 1 2 3 4

The ^ regex anchor will match any line start position since the flags argument is set to re.M (equivalent to re.MULTILINE).