# regex_lookarounds.py

# regex_lookarounds.py
import re

# Found many examples that gave faulty results... so I hereby provide my own with... (?!) ... to prove as an accurate explanation

z = 'abcxyzXYZABC'
def x(a):
    print re.findall(a,z)
x( '(?i)abc(?=xyz)'     )   # finds the 1st abc ("abc" which has "ABC" after it)
x( '(?i)abc(?!xyz)'     )   # finds the 2nd abc ("ABC" which does not have "abc" after it)
x( '(?i)(?<=xyz)abc'    )   # finds the 2nd abc ("ABC" which does not have "abc" after it)
x( '(?i)(?<!xyz)abc'    )   # finds the 1st abc ("abc" which has "ABC" after it)

z = 'abc. xyz X. XYZ .A. A.B.C'
print re.sub('(?<= [A-z])\. ',' ',z)