Regex: How to match all strings that start with one underscore and print them without the underscore?

Question

My regex so far is (?<=_)[a-zA-Z0-9]+\b

The input text is:

Calculate the _area of the _perfectRectangle object.

The _id and _age variables are both integers.

__invalidVariable _evenMoreInvalidVariable_ _validVariable

The output should be:

area,perfectRectangle

id,age

validVariable

But instead is:

area,perfectRectangle

id,age

invalidVariable,validVariable

How to match strings that start with only 1 underscore?

Answer 1

You can assert an _ to the left that by itself is not preceded by a char:

(?<=_(?<!\S_))[a-zA-Z0-9]+\b

Regex demo | Python demo

import re
 
regex = r"(?<=_(?<!\S_))[a-zA-Z0-9]+\b"
 
s = ("Calculate the _area of the _perfectRectangle object.\n\n"
    "The _id and _age variables are both integers.\n\n"
    "__invalidVariable _evenMoreInvalidVariable_ _validVariable\n\n"
    "_validVariable_test"
    )
 
print(re.findall(regex, s))

Output

['area', 'perfectRectangle', 'id', 'age', 'validVariable']

Or with a capture group:

(?<!\S)_([a-zA-Z0-9]+)\b

Regex demo | Python demo

Answer 2

You can try this: (demo)

\b_{1}([a-zA-Z0-9]+)\b

_{1} : one _

([...]) : get what in this [...]

Code for cheking:

import re
data= '''
Calculate the _area of the _perfectRectangle object.

The _id and _age variables are both integers.

__invalidVariable _evenMoreInvalidVariable_ _validVariable
'''

re.findall(r'\b_{1}([a-zA-Z0-9]+)\b', data)
# ['area', 'perfectRectangle', 'id', 'age', 'validVariable']