python string pattern matching

Question

       new_str="@@2@@*##1"
       new_str1="@@3@@*##5##7"

How to split the above string in python

          for val in new_str.split("@@*"):
            logging.debug("=======")
            logging.debug(val[2:]) // will give 
            for st in val.split("@@*"):
                //how to get the values after ##  in new_str and new_str1

Answer 1

I don't understand the question.

Are you trying to split a string by a delimiter? Then use split:

>>> a = "@@2@@*##1"
>>> b = "@@3@@*##5##7"
>>>
>>> a.split("@@*")
['@@2', '##1']
>>> b.split("@@*")
['@@3', '##5##7']

Are you trying to strip extraneous characters from a string? Then use strip:

>>> c = b.split("@@*")[1]
>>> c
'##5##7'
>>> c.strip("#")
'5##7'

Are you trying to remove all the hashes (#) from a string? Then use replace:

>>> c.replace("#","")
'57'

Are you trying to find all the characters after "##"? Then use rsplit with its optional argument to split only once:

>>> a.rsplit("##",1)
['@@2@@*', '1']