Question

i am having a python string of format

mystr = "hi.this(is?my*string+"

here i need to get the position of ‘is’ that is surrounded by special characters or non-alphabetic characters (i.e. second ‘is’ in this example). however, using

mystr.find('is')

will return the position if ‘is’ that is associated with ‘this’ which is not desired. how can i find the position of a substring that is surrounded by non-alphabetic characters in a string? using python 2.7

Answer 1

Here the best option is to use a regular expression. Python has the re module for working with regular expressions.

We use a simple search to find the position of the "is":

>>> match = re.search(r"[^a-zA-Z](is)[^a-zA-Z]", mystr)

This returns the first match as a match object. We then simply use MatchObject.start() to get the starting position:

>>> match.start(1)
8

Edit: A good point made, we make "is" a group and match that group to ensure we get the correct position.

As pointed out in the comments, this makes a few presumptions. One is that surrounded means that "is" cannot be at the beginning or end of the string, if that is the case, a different regular expression is needed, as this only matches surrounded strings.

Another is that this counts numbers as the special characters – you stated non-alphabetic, which I take to mean numbers included. If you don’t want numbers to count, then using r"\b(is)\b" is the correct solution.