Getting captured group in one line

Question

There is a known \"pattern\" to get the captured group value or an empty string if no match:

match = re.search(\'regex\', \'text\')
if match:
    value = mat         


        

Answer 1

One liners, one liners... Why can't you write it on 2 lines?

getattr(re.search('regex', 'text'), 'group', lambda x: '')(1)

Your second solution if fine. Make a function from it if you wish. My solution is for demonstrational purposes and it's in no way pythonic.

Answer 2

Starting Python 3.8, and the introduction of assignment expressions (PEP 572) (:= operator), we can name the regex search expression pattern.search(text) in order to both check if there is a match (as pattern.search(text) returns either None or a re.Match object) and use it to extract the matching group:

# pattern = re.compile(r'key=(\w+)')
match.group(1) if (match := pattern.search('foo=bar,key=value,beer=pub')) else ''
# 'value'
match.group(1) if (match := pattern.search('no match here')) else ''
# ''

Answer 3

You can play with the pattern, using an empty alternative at the end of the string in the capture group:

>>> re.search(r'((?<=key=)\w+|$)', 'foo=bar,key=value').group(1)
'value'
>>> re.search(r'((?<=key=)\w+|$)', 'no match here').group(1)
''

Answer 4

One-line version:

if re.findall(pattern,string): pass

The issue here is that you want to prepare for multiple matches or ensure that your pattern only hits once. Expanded version:

# matches is a list
matches = re.findall(pattern,string)

# condition on the list fails when list is empty
if matches:
    pass

So for your example "extract all alphanumeric characters (and _) from the text after the key= string":

# Returns 
def find_text(text):
    return re.findall("(?<=key=)[a-zA-Z0-9_]*",text)[0]

Answer 5

You can do it as:

value = re.search('regex', 'text').group(1) if re.search('regex', 'text') else ''

Although it's not terribly efficient considering the fact that you run the regex twice.

Or to run it only once as @Kevin suggested:

value = (lambda match: match.group(1) if match else '')(re.search(regex,text))

Answer 6

It's possible to refer to the result of a function call twice in a single one-liner: create a lambda expression and call the function in the arguments.

value = (lambda match: match.group(1) if match else '')(re.search(regex,text))

However, I don't consider this especially readable. Code responsibly - if you're going to write tricky code, leave a descriptive comment!