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? -- -- http://mail.python.org/mailman/listinfo/python-list
On Jun 6, 8:07 am, "tsangpo" <tsangpo.newsgr... <at> gmail.com> wrote:
> 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?
url+= { '/': '' }.get( url[ -1 ], '/' )
Shorter is always better.
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Aaron Brady wrote:
> Shorter is always better.
>
> url+= { '/': '' }.get( url[ -1 ], '/' )
Why bother with spaces or 3 letter-wide token, check this :o) :
x+={'/':''}.get(x[-1],'/')
Apart from joking, the following proposed solution is by **far** the one
I prefer
> if not url.endswith('/'):
> url += '/'
Maybe not the shorter, but the most concise and clear to me.
Jean-Michel
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Jean-Michel Pichavant wrote:
> Aaron Brady wrote:
>> Shorter is always better.
>> url+= { '/': '' }.get( url[ -1 ], '/' )
>
> Why bother with spaces or 3 letter-wide token, check this :o) :
> x+={'/':''}.get(x[-1],'/')
>
Even shorter:
x+='/'*(x[-1]!='/')
> Apart from joking, the following proposed solution is by **far** the one
> I prefer
>
>> if not url.endswith('/'):
>> url += '/'
>
> Maybe not the shorter, but the most concise and clear to me.
>
Definitely.
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Aaron Brady <castironpi <at> gmail.com> writes:
> url+= { '/': '' }.get( url[ -1 ], '/' )
>
> Shorter is always better.
url = url.rstrip('/') + '/'
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Paul Rubin <http> writes:
> url = url.rstrip('/') + '/'
That's what I use: It has the (nice) side effect of ending the URL with a
*single* slash.
C
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On Jun 7, 6:13 pm, Paul Rubin <http://phr... <at> NOSPAM.invalid> wrote: > Aaron Brady <castiro... <at> gmail.com> writes: > > url+= { '/': '' }.get( url[ -1 ], '/' ) > > > Shorter is always better. > > url = url.rstrip('/') + '/' I was joking. Sheesh. -- -- http://mail.python.org/mailman/listinfo/python-list
On Jun 7, 6:13 pm, Paul Rubin <http://phr... <at> NOSPAM.invalid> wrote:> Aaron Brady <castiro... <at> gmail.com> writes:I was joking. Sheesh.
> > url+= { '/': '' }.get( url[ -1 ], '/' )
>
> > Shorter is always better.
>
> url = url.rstrip('/') + '/'
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Since extra slashes at the end of a URL are ignored, that means I win!
url+='/'
On Jun 6, 8:07 am, "tsangpo" <tsangpo.newsgr... <at> gmail.com> wrote:url+= { '/': '' }.get( url[ -1 ], '/' )
> 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?
Shorter is always better.
-- -- http://mail.python.org/mailman/listinfo/python-list
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)
kynn (lowest-of-the-low python noob)
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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:
>>> from timeit import Timer
>>> min(Timer(
... "if not s.endswith('/'): s += '/'", "s = 'abcd/efgh'").repeat())
0.70030903816223145
>>> min(Timer(
... "sub('/?$', '/', s)", "from re import sub; s = 'abcd/efgh'").repeat())
7.6922709941864014
That's more than ten times slower. Really, a regex is massive overkill
for a task that simple.
--
Steven
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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.
kynn
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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
kynn
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kj wrote:
> ... 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
_but_, try this with s = '', and you are in trouble.
So, try:
if s[-1:] != '/':
s += '/'
To be a trifle more reliable. But, for more normal semantics,
you might prefer either:
if s[-1:] != '/':
s = (s or '.') + '/'
or:
if s and s[-1] != '/':
s += '/'
--Scott David Daniels
Scott.Daniels <at> Acm.Org
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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 += '/'
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"kj" <no.email <at> please.post> 写入消息 news:h0e3p9$85t$1 <at> reader1.panix.com...
> 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)
>
> kynn (lowest-of-the-low python noob)
look nice, but:
>>> re.sub('/?$/', '/', 'aaabbb')
'aaabbb'
has no effect. what a pity.
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tsangpo <tsangpo.newsgroup <at> gmail.com> wrote:
>
> "kj" <no.email <at> please.post> 写入消息 news:h0e3p9$85t$1 <at> reader1.panix.com...
> > 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)
> >
> > kynn (lowest-of-the-low python noob)
>
> look nice, but:
>
> >>> re.sub('/?$/', '/', 'aaabbb')
> 'aaabbb'
>
> has no effect. what a pity.
That is because you typoed what kynn wrote.
>>> re.sub('/?$', '/', 'aaabbb')
'aaabbb/'
>>>
That solution is very perl-ish I'd say, IMHO
if not url.endswith("/"):
url += "/"
is much more pythonic and immediately readable. In fact even someone
who doesn't know python could understand what it does, unlike the
regexp solution which requires a little bit of thought.
--
Nick Craig-Wood <nick <at> craig-wood.com> -- http://www.craig-wood.com/nick
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On Sun, 07 Jun 2009 00:21:45 +0800, tsangpo wrote:
> "kj" <no.email <at> please.post> 写入消息 news:h0e3p9$85t
$1 <at> reader1.panix.com...
>> 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)
>>
>> kynn (lowest-of-the-low python noob)
>
> look nice, but:
>
>>>> re.sub('/?$/', '/', 'aaabbb')
> 'aaabbb'
>
> has no effect. what a pity.
That's because you're doing it wrong. You've added an extra slash: you're
asking the regex engine to match a backslash *after* the end of the
string. Obviously that will never match.
>>> re.sub('/?$/', '/', 'aaabbb') # extra /
'aaabbb'
>>> re.sub('/?$', '/', 'aaabbb') # no extra /
'aaabbb/'
But before you get too excited, see my previous post: the regex solution
is more than ten times slower than test-and-concatenate.
--
Steven
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On Sat, 06 Jun 2009 23:07:58 +0800, tsangpo wrote:
> 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?
if not url.endswith('/'):
url += '/'
--
Steven
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