How to use RegEx re.search and group() to get multiple lines of matching data in python

Question

My text is this

text =

  [ BP2572 23,
    BP2345 34,
    BP2457 45,
    BP2866 56 ] 

I want to extract just a portion of the text using search and group() re expression.

What I am hoping for is the following output

[BP2572, BP2345, BP2457, BP2866]
[23, 34, 45, 56]

I tried this code but I didn't get quite what I was expecting

>>re.findall(r"\s+[A-Z]{2}\d{4} \d{2}",text)

['\nBP2572 23', '\nBP2345 34', '\nBP2457 45', '\nBP2866 56']

>>re.search(r"\s+([A-Z]{2}\d{4}) (\d{2})",text).group(1)
>>re.search(r"\s+([A-Z]{2}\d{4}) (\d{2})",text,re.DOTALL).group(1)

'BP2572' #my expected output is [BP2572, BP2345, BP2457, BP2866]

>>re.search(r"\s+([A-Z]{2}\d{4}) (\d{2})",text).group(2)
>>re.search(r"\s+([A-Z]{2}\d{4}) (\d{2})",text,re.DOTALL).group(2)

'23' #my expected output is `[23, 34, 45, 56]`

Here I only got the first match.

How can use re.search and group() to get all the matching results and not just the first match?

Answer 1

You cannot use re.search to find multiple matches. Both re.search and re.match only return the FIRST match.

You need to use re.findall or re.finditer for multiple matches of a pattern.

You can do:

 >>> re.findall(r'^[ \t]*([A-Z]{2}\d{4}[ \t]+\d{2})', text, flags=re.M)
 ['BP2572 23', 'BP2345 34', 'BP2457 45', 'BP2866 56']

Or if you only want the first part:

 >>> re.findall(r'^[ \t]*([A-Z]{2}\d{4})[ \t]+\d{2}', text, flags=re.M)
 ['BP2572', 'BP2345', 'BP2457', 'BP2866']

Or only the second part:

 >>> re.findall(r'^[ \t]*[A-Z]{2}\d{4}[ \t]+(\d{2})', text, flags=re.M)
 ['23', '34', '45', '56']

Or both parts:

 >>> re.findall(r'^[ \t]*([A-Z]{2}\d{4})[ \t]+(\d{2})', text, flags=re.M)
 [('BP2572', '23'), ('BP2345', '34'), ('BP2457', '45'), ('BP2866', '56')]

Or, you can do simple splits:

>>> [line.split()[0] for line in text.splitlines() if line]
['BP2572', 'BP2345', 'BP2457', 'BP2866']
>>> [line.split()[1] for line in text.splitlines() if line]
['23', '34', '45', '56']