regular expression matching everything except a given regular expression

Question

I am trying to figure out a regular expression which matches any string which doesn't start with mpeg. A generalization of this is matching any string which doesn't start with a given regular expression.

I tried something like as follows:

[^m][^p][^e][^g].*

The problem with this is that it requires at least 4 characters to be present in the string. I was not able to figure out a good way to handle this and a generalized way to handle this in a general purpose manner.

I will be using this in Python.

Thanx in advance.

Answer 1

Try a look-ahead assertion:

(?!mpeg)^.*

Or if you want to use negated classes only:

^(.{0,3}$|[^m]|m([^p]|p([^e]|e([^g])))).*$

Answer 2

^(?!mpeg).*

This uses a negative lookahead to only match a string where the beginning doesn't match mpeg. Essentially, it requires that "the position at the beginning of the string cannot be a position where if we started matching the regex mpeg, we could successfully match" - thus matching anything which doesn't start with mpeg, and not matching anything that does.

However, I'd be curious about the context in which you're using this - there might be other options aside from regex which would be either more efficient or more readable, such as...

if inputstring[:4] != "mpeg":

Answer 3

don't lose your mind with regex.

if len(mystring) >=4 and mystring[:4]=="mpeg":
    print "do something"

or use startswith() with "not" keyword

if len(mystring)>=4 and not mystring.startswith("mpeg")

Answer 4

Your regexp wouldn't match "npeg", I think you would need come up with ^($|[^m]|m($|[^p]|p($|[^e]|e($|[^g])))), which is quite horrible. Another alternative would be ^(.{0,3}$|[^m]|.[^p]|..[^e]|...[^g]) which is only slightly better.

So I think you should really use a look-ahead assertion as suggested by Dav and Gumbo :-)