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'
    re.findall('\d+',text)

['33', '44']

I could extract the first element of the list:

    text = 'aa33bbb44'
    re.findall('\d+',text)[0]

'33'

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

    text = 'aazzzbbb'
    re.findall('\d+',text)[0]

IndexError: list index out of range

In which case I could define a function:

    def return_first_match(text):
        try:
            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?

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

    >>> re.findall('\d+|$', 'aa33bbb44')[0]
    '33'
    >>> re.findall('\d+|$', 'aazzzbbb')[0]
    ''
    >>> re.findall('\d+|$', '')[0]
    ''

Also works with re.search pointed out by others:

    >>> re.search('\d+|$', 'aa33bbb44').group()
    '33'
    >>> re.search('\d+|$', 'aazzzbbb').group()
    ''
    >>> re.search('\d+|$', '').group()
    ''

From: stackoverflow.com/q/38579725