Python Question

return string with first match Regex

I want to get the first match of a regex.

In this case, I got a list:

text = 'aa33bbb44'

['33', '44']

I could extract the first element of the list:

text = 'aa33bbb44'


But that only works if there is at least one match, otherwise I'll get an error:

text = 'aazzzbbb'

IndexError: list index out of range

In which case I could define a function:

def return_first_match(text):
result = re.findall('\d+',text)[0]
except Exception, IndexError:
result = ''
return result

Is there a way of obtaining that result without defining a new function?

Answer Source

You could embed the '' default in your regex by adding |$:

>>> re.findall('\d+|$', 'aa33bbb44')[0]
>>> re.findall('\d+|$', 'aazzzbbb')[0]
>>> re.findall('\d+|$', '')[0]

Also works with pointed out by others:

>>>'\d+|$', 'aa33bbb44').group()
>>>'\d+|$', 'aazzzbbb').group()
>>>'\d+|$', '').group()
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