get index of leading 0s using regex re python-CodePudding

Using Regex in Python (library re (only)), I want to create a function that gives me the position of all leading 0s in a string.

For example, if the string was: My house has 01 garden and 003 rooms. I would want me the function to return 13, 27 and 28.

I tried for example:

import re
string = "My house has 01 garden and 003 rooms."
pattern = "(0 )[1-9]\d*"

print(re.findall(pattern,string))

Obviously, the output gives me the matches but no position...

CodePudding user response：

You can do the following:

import re

text = "My house has 01 garden and 003 rooms."
pattern = re.compile(r"\b0 ")


def leading_zeros_index(s: str) -> list:
    return [i for m in pattern.finditer(s) for i in range(m.start(), m.end())]


print(leading_zeros_index(text))

output:

[13, 27, 28]

Basically you use .finditer() in order to get the match object, then you create a range() object from match object's .start() and .end().

I used \b0 as the pattern. There is no need to check the other characters come after zeros. \b is word boundary, here means, zeros should be at the start of the words.