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.
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