can it be shorter?

[MRAB]

kj wrote:
> In <h0e9q8$ni7$1 at reader1.panix.com> kj <no.email at please.post> writes:
> 
>> In <023a8d04$0$20636$c3e8da3 at news.astraweb.com> Steven D'Aprano <steve at REMOVE-THIS-cybersource.com.au> writes:
> 
>>> On Sat, 06 Jun 2009 15:59:37 +0000, kj wrote:
> 
>>>> In <h0e0oi$1es2$1 at adenine.netfront.net> "tsangpo"
>>>> <tsangpo.newsgroup at gmail.com> writes:
>>>>
>>>>> I want to ensure that the url ends with a '/', now I have to do thisa
>>>>> like below.
>>>>> url = url + '' if url[-1] == '/' else '/'
>>>>> Is there a better way?
>>>> It's a pity that in python regexes are an "extra", as it were. Otherwise
>>>> I'd propose:
>>>>
>>>> url = re.sub("/?$", "/", url)
> 
> 
>>> Thank goodness regexs are an "extra" in Python, because it discourages 
>>> noobs from pulling out the 80 pound sledgehammer of the regex engine to 
>>> crack the peanut of a test-and-concatenate:
> 
>> I was just responding to the OP's subject line.  Whatever else one
>> may say about my proposal, it *is* shorter.
> 
>> But thanks for the tip with timeit.  That looks like a good module
>> to know.
> 
> 
> 
> And actually, if speed is the criterion, then one should also avoid endswith:
> 
>>>> from timeit import Timer
>>>> min(Timer("if s[-1] != '/': s += '/'", "s = 'abcd/efgh'").repeat())
> 0.18654584884643555
>>>> min(Timer("if not s.endswith('/'): s += '/'", "s = 'abcd/efgh'").repeat())
> 0.43395113945007324
> 
If there's any chance that the string could be empty (len(s) == 0) then
use:

     if s[-1 : ] != '/'
         s += '/'