I have the string.
st = "12345 hai how r u @3456? Awer12345 7890"
re.findall('([0-9]+)',st)
It should not come like :
['12345', '3456', '12345', '7890']
I should get
['12345','7890']
I should only take the numeric values
and
It should not contain any other chars like alphabets,special chars
No need to use a regular expression:
[i for i in st.split(" ") if i.isdigit()]
Which I think is much more readable than using a regex
Corey's solution is really the right way to go here, but since the question did ask for regex, here is a regex solution that I think is simpler than the others:
re.findall(r'(?<!\S)\d+(?!\S)', st)
And an explanation:
(?<!\S) # Fail if the previous character (if one exists) isn't whitespace
\d+ # Match one or more digits
(?!\S) # Fail if the next character (if one exists) isn't whitespace
Some examples:
>>> re.findall(r'(?<!\S)\d+(?!\S)', '12345 hai how r u @3456? Awer12345 7890')
['12345', '7890']
>>> re.findall(r'(?<!\S)\d+(?!\S)', '12345 hai how r u @3456? Awer12345 7890123ER%345 234 456 789')
['12345', '234', '456', '789']
In [21]: re.findall(r'(?:^|\s)(\d+)(?=$|\s)', st)
Out[21]: ['12345', '7890']
Here,
(?:^|\s) is a non-capture group that matches the start of the string, or a space.(\d+) is a capture group that matches one or more digits.(?=$|\s) is lookahead assertion that matches the end of the string, or a space, without consuming it.
['12345', '234', '456', '789']. Is this not what you expect?use this: (^|\s)[0-9]+(\s|$) pattern. (^|\s) means that your number must be at the start of the string or there must be a whitespace character before the number. And (\s|$) means that there must be a whitespace after number or the number is at the end of the string.
As said Jan Pöschko, 456 won't be found in 123 456. If your "bad" parts (@, Awer) are always prefixes, you can use this (^|\s)[0-9]+ pattern and everything will be OK. It will match all numbers, which have only whitespaces before or are at the start of the string. Hope this helped...
Your expression finds all sequences of digits, regardless of what surrounds them. You need to include a specification of what comes before and after the sequence to get the behavior you want:
re.findall(r"[\D\b](\d+)[\D\b]", st)
will do what you want. In English, it says "match all sequences of one or more digits that are surrounded by a non-digit character.or a word boundary"
