Regex find string then return 10 character after it

[Pedroski55]

You can get the start and end positions of any pattern with re

pattern = re.compile('x:')
# this returns a tuple with the position of the pattern in your string
start_stop = pattern.search(astring).span()

You could do it like this:

import re
astring = 'aaaax:bbbbbbbbbbbbbbby:cccccccccccccccccccz:ddddddddddddddddd'
myvars = ['x:', 'y:', 'z:', 'dog']
length = 10
results = []
for item in myvars:
    pattern = re.compile(item)
    # find where the search string starts
    start = pattern.search(astring)
    # if the pattern is not found search returns None
    if not start == None:
        #get the span tuple
        tup = start.span()
        # change the tuple to get what you want
        newtup = (tup[0] + 2, tup[1] + length)
        # slice the string
        wanted = astring[newtup[0]:newtup[1]]
        results.append(wanted)