return string with first match Regex

Solution 1:

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()
''

Solution 2:

If you only need the first match, then use re.search instead of re.findall:

>>> m = re.search('\d+', 'aa33bbb44')
>>> m.group()
'33'
>>> m = re.search('\d+', 'aazzzbbb')
>>> m.group()
Traceback (most recent call last):
  File "<pyshell#281>", line 1, in <module>
    m.group()
AttributeError: 'NoneType' object has no attribute 'group'

Then you can use m as a checking condition as:

>>> m = re.search('\d+', 'aa33bbb44')
>>> if m:
        print('First number found = {}'.format(m.group()))
    else:
        print('Not Found')


First number found = 33

Solution 3:

I'd go with:

r = re.search("\d+", ch)
result = return r.group(0) if r else ""

re.search only looks for the first match in the string anyway, so I think it makes your intent slightly more clear than using findall.

Solution 4:

You shouldn't be using .findall() at all - .search() is what you want. It finds the leftmost match, which is what you want (or returns None if no match exists).

m = re.search(pattern, text)
result = m.group(0) if m else ""

Whether you want to put that in a function is up to you. It's unusual to want to return an empty string if no match is found, which is why nothing like that is built in. It's impossible to get confused about whether .search() on its own finds a match (it returns None if it didn't, or an SRE_Match object if it did).

Solution 5:

You can do:

x = re.findall('\d+', text)
result = x[0] if len(x) > 0 else ''

Note that your question isn't exactly related to regex. Rather, how do you safely find an element from an array, if it has none.