This will be really quick marks for someone...
Here's my string:
Jan 13.BIGGS.04222 ABC DMP 15
I'm looking to match:
Here is what I have so far:
(\w{3} \d{2})\.(\w*)\..*(\d{1,3})$
Through a lot of playing around with http://www.pythonregex.com/ I can get to matching the '5', but not '15'.
What am I doing wrong?
Use .*? to match .* non-greedily:
In [9]: re.search(r'(\w{3} \d{2})\.(\w*)\..*?(\d{1,3})$', text).groups()
Out[9]: ('Jan 13', 'BIGGS', '15')
Without the question mark, .* matches as many characters as possible, including the digit you want to match with \d{1,3}.
Alternatively to what @unutbu has proposed, you can also use word boundary \b - this matches "word border":
(\w{3} \d{2})\.(\w*)\..*\b(\d{1,3})$
From the site you referred:
>>> regex = re.compile("(\w{3} \d{2})\.(\w*)\..*\b(\d{1,3})$")
>>> regex.findall('Jan 13.BIGGS.04222 ABC DMP 15')
[(u'Jan 13', u'BIGGS', u'15')]
.* before numbers are greedy and match as much as it can, leaveing least possible digits to the last block. You either need to make it non-greedy (with ? like unutbu said) or make it do not match digits, replacing . with \D
