Return string with first match for a regex, handling case where there is no match

Question

I want to get the first match of a regex.

In the following case, I have 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 IndexError:

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?

Encephalomyelitis

For me the accepted answer didn't work. I had to remove the array index access and use len(re.findAll)==0 check instead. – Motionless

Answer 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()
''

Overmatter

Answer 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

Litigable

Answer 3

I'd go with:

r = re.search("\d+", ch)
result = 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.

Tonkin

yes @Tonkin and findall could be critical if the amount of data is very large. waste of resources – Cognizance

Answer 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).

Fivefinger

Answer 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.

Cowshed

Answer 6

Maybe this would perform a bit better in case greater amount of input data does not contain your wanted piece because except has greater cost.

def return_first_match(text):
    result = re.findall('\d+',text)
    result = result[0] if result else ""
    return result

Coffle

Answer 7

just assign the results to a variable then iterate the variable

 text = 'aa33bbb44'
 result=re.findall('\d+',text)
 for item in result:
     print(item)

Winsor

Answer 8

With Assignment expressions (PEP572):

text = 'aa33bbb44'
r = m.group() if (m:=re.search(r'\d+',text)) is not None else ''

Halfon

Answer 9

With re.findall, you can convert the output into an iterator with iter() and call next() on it to get the first result. next() is particularly useful for this task because a default value (e.g. '') can be passed to it; the default is returned if the iterator is empty, i.e. if there are no matches.

next(iter(re.findall('\d+', 'aa33bbb44')), '')      # '33'
next(iter(re.findall('\d+', 'aazzzbbb')), '')       # ''

At this point, next() can used with re.finditer for the job as well.

next(re.finditer('\d+', 'aa33bbb44'), [''])[0]      # '33'
next(re.finditer('\d+', 'aazzzbbb'), [''])[0]       # ''

You can also use the walrus operator with re.search for a one-liner.

m[0] if (m:=re.search('\d+', 'aa33bbb44')) else ''  # '33'
m[0] if (m:=re.search('\d+', 'aazzzbbb')) else ''   # ''

For this specific task, the argument against re.findall is performance and, indeed for large strings, the gap is huge. If there are multiple matches, re.findall is much, much slower than re.search or re.finditer¹. However, if there are no matches, re.search with the walrus and re.finditer are the fastest.².

---

¹ Timings for strings with 1mil characters and 100k matches.

text = 'aabbbccc11'*100_000

%timeit m[0] if (m:=re.search('\d+', text)) else ''
# 1.94 µs ± 192 ns per loop (mean ± std. dev. of 10 runs, 100,000 loops each)
%timeit next(re.finditer('\d+', text), [''])[0]
# 2.38 µs ± 122 ns per loop (mean ± std. dev. of 10 runs, 100,000 loops each)
%timeit next(iter(re.findall('\d+', text)), '')
# 59 ms ± 8.65 ms per loop (mean ± std. dev. of 10 runs, 10 loops each)

%timeit re.search('\d+|$', text)[0]
# 2.32 µs ± 300 ns per loop (mean ± std. dev. of 10 runs, 100,000 loops each)
%timeit re.findall('\d+|$', text)[0]
# 82.7 ms ± 1.64 ms per loop (mean ± std. dev. of 10 runs, 10 loops each)

² Timings for strings with 1mil characters and no matches.

text = 'aabbbcccdd'*100000

%timeit m[0] if (m:=re.search('\d+', text)) else ''
# 26.3 ms ± 662 µs per loop (mean ± std. dev. of 10 runs, 100 loops each)
%timeit next(re.finditer('\d+', text), [''])[0]
# 26 ms ± 195 µs per loop (mean ± std. dev. of 10 runs, 100 loops each)
%timeit next(iter(re.findall('\d+', text)), '')
# 26.2 ms ± 615 µs per loop (mean ± std. dev. of 10 runs, 100 loops each)

%timeit re.search('\d+|$', text)[0]
# 72.9 ms ± 14.1 ms per loop (mean ± std. dev. of 10 runs, 100 loops each)
%timeit re.findall('\d+|$', text)[0]
# 67.8 ms ± 2.38 ms per loop (mean ± std. dev. of 10 runs, 100 loops each)

Hypolimnion