Python regular expression for substring

Question

All I want is to grab the first 3 numeric characters of string:

st = '123_456'
import re
r = re.match('([0-9]{3})', st)
print r.groups()[0]

1. Am I doing the right thing for grabbing first 3 characters?

2. This returns 123 but what if I want to get the first 3 characters regardless of numbers and alphabets or special characters?

When given 12_345 , I want to grab only 12_

Thanks,

Answer 1

If you always need first three characters in a string, then you can use the below:

first_3_charaters = st[:3]

There is no need of regular expression in your case.

Answer 2

If all digits are separated by _ , then you can simply use this regular expression which greedily matches all numeric characters before the first _ .

r = re.match('([0-9]*)_', st)

Actually, the _ in this RE is not necessary，so you can simplify it to (so that any separator is accepted ):

r = re.match('(\d*)', st)

But this solution will give you 1234 if st = '1234_56' . I'm not sure whether it is your intention.

So, if you want at most 3 numeric characters , you can just modify the regular expression to:

r = re.match('(\d{,3})', st)

Answer 3

You are really close, just drop the extra set of parenthesis and use the proper indexing of zero instead of one. Python indexing starts at zero. See below.

This works:

import re
mystring = '123_456'
check = re.search('^[0-9]{3}', mystring)
if check:
    print check.group(0)

the ^ anchors to the beginning of the string which will ensure a match to the first three numeric digits only. If you do not use the carrot the regexp will match any three digits in a row in the string.

Some may suggest \\d but this includes more than 0-9.

As others will surely point out a simple substring operation will do the trick if all the fields start with three numeric digits that you want to extract.

Good luck!