Just use re.search which stops matching once it finds a match.

re.search(r'\d+', headline).group()

or

You must remove the forward slashes present in your regex.

re.findall(r'^\D*(\d+)', headline)

If you were to solve it without regular expressions, you could have used itertools.dropwhile():

>>> from itertools import dropwhile
>>>
>>> ''.join(dropwhile(str.isdigit, "32cl2"))
'cl2'
>>> ''.join(dropwhile(str.isdigit, "4563nh3"))
'nh3'

Or, using re.sub(), replacing one or more digits at the beginning of a string:

>>> import re
>>> re.sub(r"^\d+", "", "32cl2")
'cl2'
>>> re.sub(r"^\d+", "", "4563nh3")
'nh3'

I'd use a regexp:

>>> import re
>>> re.findall(r'\d+', "hello 42 I'm a 32 string 30")
['42', '32', '30']

This would also match 42 from bla42bla. If you only want numbers delimited by word boundaries (space, period, comma), you can use \b:

>>> re.findall(r'\b\d+\b', "he33llo 42 I'm a 32 string 30")
['42', '32', '30']

To end up with a list of numbers instead of a list of strings:

>>> [int(s) for s in re.findall(r'\b\d+\b', "he33llo 42 I'm a 32 string 30")]
[42, 32, 30]

NOTE: this does not work for negative integers

As the comments on that answer note, in Python 3, filter returns a filter generator object, so you must iterate over it and build a new string before you can call int:

>>> s = '3 reviews'
>>> filter(str.isdigit, s)
<filter object at 0x800ad5f98>
>>> int(''.join(filter(str.isdigit, s)))
3

However, as other answers in that same thread point out, this is not necessarily a good way to do the job at all:

>>> s = '3 reviews in 12 hours'
>>> int(''.join(filter(str.isdigit, s)))
312

It might be better to use a regular expression matcher to find the number at the front of the string. You can then decide whether to allow signs (+ and -) and leading white-space:

>>> import re
>>> m = re.match(r'\s*([-+])?\d+', s)
>>> m
<_sre.SRE_Match object; span=(0, 1), match='3'>
>>> m.group()
'3'
>>> int(m.group())
3

Now if your string contains a malformed number, m will be None, and if it contains a sign, the sign is allowed:

>>> m = re.match(r'\s*([-+])?\d+', 'not a number')
>>> print(m)
None
>>> m = re.match(r'\s*([-+])?\d+', '  -42')
>>> m
<_sre.SRE_Match object; span=(0, 5), match='  -42'>
>>> int(m.group())
-42

If you wish to inspect what came after the number, if anything, add more to the regular expression (including some parentheses for grouping) and use m.group(1) to get the matched number. Replace \d+ with \d* to allow an empty number-match, if that's meaningful (but then be mindful of matching a lone - or + sign, if you still allow signs).

I am sure there are plenty of possibilities, here are a few I would consider:

thisLine <- paste(runif(11), collapse = " ")
thisLine
# [1] "0.841216114815325 0.861485596280545 0.973681036382914 0.683699210174382 0.95226536039263 0.368689567316324 0.173984130611643 0.497511914698407 0.870743532432243 0.45606177020818 0.222731305286288"

sub("\\s+.*", "", thisLine)              # assumes no leading space
sub("\\s*(\\S+?)\\s.*", "\\1", thisLine) # handles leading spaces
strsplit(thisLine, " ")[[1]][1]          # more flexible if you want 2nd, 3rd, ...

All give

# [1] "0.841216114815325"

You want to look into regular expressions (regex) and Python's re module.

import re

string = "CORSAIR VENGEANCE RGB 16GB (2X8GB) DDR4 3200MHZ CL 16"

matches = re.findall("(\d+)", string)
print(matches)

Output: ['16', '2', '8', '4', '3200', '16']

The regex pattern \d+ matches 1 or more numerical characters in sequence. By putting it in brackets ( ) it captures each occurrence of that pattern as a group. And the re.findall method returns those groups in a list. Note that they are still strings, not numbers.

You can use regex as below:

import re
str = "Owen White punt for 26 yards, downed at the army 37"
#search using regex
x = re.findall('[0-9]+', str)
print(x[0])