how to convert string function to string method?

[Bruno Desthuilliers]

Dr. Phillip M. Feldman a écrit :
> Bruno- You've made some excellent suggestions, and I'm always grateful for
> the opportunity to learn.

Glad to know I've been of any help !-)

>  My revised code appears below.  Philllip
> 
> def strip_pairs(s, open='([{\'"', close=')]}\'"'):
>    """
>    OVERVIEW
> 
>    This function strips matching pairs of characters from the beginning and
>    end of the input string `s`.  If `s` begins with a character in `open`
> and
>    ends with the corresponding character in `close` (see below), both are
>    removed from the string. This process continues until no further matching
>    pairs can be removed.
> 
>    INPUTS
> 
>    `open` and `close`: These arguments, which must be equal-length strings,
>    specify matching start-of-scope and end-of-scope characters.  The same
>    character may appear in both `open` and `close`; single and double quotes
>    conventionally match themselves.  By default, `open` contains a left
>    parenthesis, left square bracket, left curly bracket, single quote, and
>    double quote), and `close` contains the corresponding characters."""
> 
>    if not isinstance(s,(str,unicode)):
>       raise TypeError, '`s` must be a string (str or unicode).'

Might be a bit more helpful (for the programmer using your function) to 
specify what 's' actually is. Also, the recommanded way to raise 
exceptions is to use the 'call' syntax, ie:


if not isinstance(s,(str,unicode)):
    raise TypeError("'s' must be a str or unicode, got '%s'" % type(s))


>    if not isinstance(open,(str,unicode)) or not
> isinstance(close,(str,unicode)):
>       raise TypeError, '`open` and `close` must be strings (str or
> unicode).'

Mmmm.... I still wonder why you wouldn't accept a tuple or list of chars 
here.

>    if len(open) != len(close): raise ValueError, \
>      '\'open\' and \'close\' arguments must be equal-length strings.'
> 
>    while len(s) >= 2:
> 
>       # Check whether first character of `s` is in `open`:
>       i= open.find(s[0])
> 
>       # If `s` does not begin with a character from `open`, there are no
> more
>       # pairs to be stripped:
>       if i == -1: break

wrt/ readability, it might be better to put the break statement on it's 
own line - but you can probably count this one as more of a personnal 
preference than a guideline !-)

>       # If `s` does not begin and end with matching characters, there are no
>       # more pairs to be stripped:
>       if s[-1] != close[i]: break
> 
>       # Strip the first and last character from `s`:
>       s= s[1:-1]
>    return s




Steven (D'Aprano) posted a possibly interesting implementation. Might be 
worth timeit'ing both.