Regex to parse percentage from the string [duplicate]

Question

I want to get all the percentage from above string with decimal or without decimal points. I also want to get % sign.

string  = """Bachelor of Engineering in Electronics from  in 2013 with 60%.
Diploma (Electronics & Video Engineering) from  in 2009 with 79.39%.
SSC from  in 2006 with 61.71%."""

I have tried as below

re.findall(r'\d+%|([0-9]\d?)\.\d+%', string)

getting output as

['', '79', '61']

But I want output as below

    ['60%', '79.39%', '61.71%']

Anybody can tell where I am going wrong and regex to get expected output.

Answer 1

You can use the RegEx (\d+(\.\d+)?%)

• \d+ captures 1 or more digits

• (\.\d+)? captures a dot and 1 or more digits 0 or 1 time

• % captures % literally

Demo.