I want to match null or space as optional from the start of the line. The line is as follow:

 Date       Description  Amount
 
 null 12/05/2016 Asian Paints 2,150.65

   13/05/2016 Nerolac GEB 5.86 22,512.65 Cr

 14/05/2016 Hydra 12,412

The regex that I used is :

regex_null = re.compile(r"^(?:null)?\s (\d{2}/\d{2}/\d{4})\s (.*?)\s (\d[\d,]*\.\d{2}\s (?:Cr)?)$", re.M)

And what I got is:

 null 12/05/2016 Asian Paints 2,150.65

     13/05/2016 Nerolac GEB 5.86 22,512.65 Cr

So the null is not optional. It is currently considered compulsory. Can you please help me with this?

CodePudding user response：

You may use this regex with optional groups:

^\s*(?:null)?\s*(\d{2}/\d{2}/\d{4})\s (.*?)\s (\d[\d,]*(?:\.\d{2})?(\s Cr)?)$

RegEx Demo

RegEx Details:

• ^\s*(?:null)?\s*: Match optional null with 0 or more whitespaces on both sides
• (\d{2}/\d{2}/\d{4}): Match date string in capture group #1
• \s : Match 1 whitespaces
• (.*?): Math 0 or more characters in capture group #2
• \s : Match 1 whitespaces
• (\d[\d,]*: Match a digit followed by 0 or more digit/comma characters
• (?:\.\d{2})?: Match optional dot and digits
• (\s Cr)?): Match optional 1 whitespaces followed by Cr
• $: End

CodePudding user response：

You may apply a regex pattern in multiline mode which makes the first, sixth, and seventh values optional in the line.

inp = """ null 12/05/2016 Asian Paints 2,150.65

   13/05/2016 Nerolac GEB 5.86 22,512.65 Cr

 14/05/2016 Hydra 12,412"""

lines = re.findall(r'^\s*(null)?\s*(\d{1,2}/\d{1,2}/\d{4}) (\w (?: \w )*) (\d{1,3}(?:,\d{3})*(?:\.\d )?)?(?: (\d{1,3}(?:,\d{3})*(?:\.\d )?))?(?: (\w ))?', inp, flags=re.M)
print(lines)

This prints:

[('null', '12/05/2016', 'Asian Paints', '2,150.65', '', ''),
 ('', '13/05/2016', 'Nerolac GEB', '5.86', '22,512.65', 'Cr'),
 ('', '14/05/2016', 'Hydra', '12,412', '', '')]