Re: [Python-Dev] datetime.timedelta total_microseconds
Looks like timedelta has a microseconds property. Would this work for your needs? In [12]: d Out[12]: datetime.timedelta(0, 3, 398407) In [13]: d.microseconds Out[13]: 398407 On Wed, Feb 13, 2019 at 9:08 PM Richard Belleville via Python-Dev < python-dev@python.org> wrote: > In a recent code review, the following snippet was called out as > reinventing the > wheel: > > _MICROSECONDS_PER_SECOND = 100 > > > def _timedelta_to_microseconds(delta): > return int(delta.total_seconds() * _MICROSECONDS_PER_SECOND) > > > The reviewer thought that there must already exist a standard library > function > that fulfills this functionality. After we had both satisfied ourselves > that we > hadn't simply missed something in the documentation, we decided that we had > better raise the issue with a wider audience. > > Does this functionality already exist within the standard library? If not, > would > a datetime.timedelta.total_microseconds function be a reasonable addition? > I > would be happy to submit a patch for such a thing. > > Richard Belleville > ___ > Python-Dev mailing list > Python-Dev@python.org > https://mail.python.org/mailman/listinfo/python-dev > Unsubscribe: > https://mail.python.org/mailman/options/python-dev/tahafut%40gmail.com > ___ Python-Dev mailing list Python-Dev@python.org https://mail.python.org/mailman/listinfo/python-dev Unsubscribe: https://mail.python.org/mailman/options/python-dev/archive%40mail-archive.com
Re: [Python-Dev] datetime.timedelta total_microseconds
Oops. That isn't the TOTAL microseconds, but just the microseconds portion. Sorry for the confusion. On Wed, Feb 13, 2019 at 9:23 PM Henry Chen wrote: > Looks like timedelta has a microseconds property. Would this work for your > needs? > > In [12]: d > Out[12]: datetime.timedelta(0, 3, 398407) > > In [13]: d.microseconds > Out[13]: 398407 > > On Wed, Feb 13, 2019 at 9:08 PM Richard Belleville via Python-Dev < > python-dev@python.org> wrote: > >> In a recent code review, the following snippet was called out as >> reinventing the >> wheel: >> >> _MICROSECONDS_PER_SECOND = 100 >> >> >> def _timedelta_to_microseconds(delta): >> return int(delta.total_seconds() * _MICROSECONDS_PER_SECOND) >> >> >> The reviewer thought that there must already exist a standard library >> function >> that fulfills this functionality. After we had both satisfied ourselves >> that we >> hadn't simply missed something in the documentation, we decided that we >> had >> better raise the issue with a wider audience. >> >> Does this functionality already exist within the standard library? If >> not, would >> a datetime.timedelta.total_microseconds function be a reasonable >> addition? I >> would be happy to submit a patch for such a thing. >> >> Richard Belleville >> ___ >> Python-Dev mailing list >> Python-Dev@python.org >> https://mail.python.org/mailman/listinfo/python-dev >> Unsubscribe: >> https://mail.python.org/mailman/options/python-dev/tahafut%40gmail.com >> > ___ Python-Dev mailing list Python-Dev@python.org https://mail.python.org/mailman/listinfo/python-dev Unsubscribe: https://mail.python.org/mailman/options/python-dev/archive%40mail-archive.com
Re: [Python-Dev] datetime.timedelta total_microseconds
Indeed there is a potential loss of precision:
_timedelta_to_microseconds(timedelta(0, 1, 1)) returns 100
where conversion function is defined according to the initial message in
this thread
On Fri, Feb 15, 2019 at 2:29 PM Paul Ganssle wrote:
> I'm still with Alexander on this. I see functions like total_X as
> basically putting one of the arguments directly in the function name - it
> should be `total_duration(units)`, not `total_units()`, because all of
> those functions do the same thing and only differ in the units they use.
>
> But Alexander's approach of "divide it by the base unit" is *even more
> general* than this, because it allows you to use non-traditional units
> like weeks (timedelta(days=7)) or "two-day periods" or whatever you want.
> If you use this idiom a lot and want a simple "calculate the total"
> function, this should suffice:
>
> def total_duration(td, *args, **kwargs):
> return td / timedelta(*args, **kwargs)
>
> Then you can spell "x.total_microseconds()" as:
>
> total_duration(x, microseconds=1)
>
> Or you can write it like this:
>
> def total_duration(td, units='seconds'):
> return td / timedelta(**{units: 1})
>
> In which case it would be spelled:
>
> total_duration(x, units='microseconds')
>
> I don't see there being any compelling reason to add a bunch of methods
> for a marginal (and I'd say arguable) gain in aesthetics.
> On 2/15/19 4:48 PM, Chris Barker via Python-Dev wrote:
>
> On Fri, Feb 15, 2019 at 11:58 AM Rob Cliffe via Python-Dev <
> python-dev@python.org> wrote:
>
>> A function with "microseconds" in the name IMO misleadingly suggests that
>> it has something closer to microsecond accuracy than a 1-second granularity.
>>
>
> it sure does, but `delta.total_seconds()` is a float, so ms accuracy is
> preserved.
>
> However, if you DO want a "timedelta_to_microseconds" function, it really
> should use the microseconds field in the timedelta object. I haven't
> thought it through, but it makes me nervous to convert to floating point,
> and then back again -- for some large values of timedelta some precision
> may be lost.
>
> Also:
>
> _MICROSECONDS_PER_SECOND = 100
>
>
> really? why in the world would you define a constant for something that
> simple that can never change? (and probably isn't used in more than one
> place anyway
>
> As Alexander pointed out the canonical way to spell this would be:
>
> delta / timedelta(microseconds=1)
>
> but I think that is less than obvious to the usual user, so I think a:
>
> delta.total_microseconds()
>
> would be a reasonable addition.
>
> I know I use .totalseconds() quite a bit, and would not want to have to
> spell it:
>
> delta / timedelta(seconds=1)
>
> (and can't do that in py2 anyway)
>
> -CHB
>
> --
>
> Christopher Barker, Ph.D.
> Oceanographer
>
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Re: [Python-Dev] datetime.timedelta total_microseconds
+1 on the improved docs solution: no new code to maintain and big return on investment in preventing future bugs / confusion :) On Sat, Feb 16, 2019 at 9:40 AM Nick Coghlan wrote: > On Sun, 17 Feb 2019 at 03:20, Paul Ganssle wrote: > > I think if we add such a function, it will essentially be just a slower > version of something that already exists. I suspect the main reason the > "divide the timedelta by the interval" thing isn't a common enough idiom > that people see it all the time is that it's only supported in Python 3. As > more code drops Python 2, I think the "td / interval" idiom will hopefully > become common enough that it will obviate the need for a total_duration > function. > > And personally, the total_seconds() case has always been enough for me. > > > That said, if people feel very strongly that a total_duration function > would be useful, maybe the best thing to do would be for me to add it to > dateutil.utils? In that case it would at least be available in Python 2, so > people who find it more readable and people still writing polyglot code > would be able to use it, without the standard library unnecessarily > providing two ways to do the exact same thing. > > I'm now thinking a slight documentation improvement would have > addressed my own confusion (and I suspect the OPs as well): > > * In the "Supported Operations" section of > https://docs.python.org/3/library/datetime.html#timedelta-objects, > change "Division (3) of t2 by t3." to "Division (3) of overall > duration t2 by interval unit t3." > * In the total_seconds() documentation, add a sentence "For interval > units other than seconds, use the division form directly (e.g. `td / > timedelta(microseconds=1)`)" > > Cheers, > Nick. > > -- > Nick Coghlan | ncogh...@gmail.com | Brisbane, Australia > ___ > Python-Dev mailing list > Python-Dev@python.org > https://mail.python.org/mailman/listinfo/python-dev > Unsubscribe: > https://mail.python.org/mailman/options/python-dev/tahafut%40gmail.com > ___ Python-Dev mailing list Python-Dev@python.org https://mail.python.org/mailman/listinfo/python-dev Unsubscribe: https://mail.python.org/mailman/options/python-dev/archive%40mail-archive.com
Re: [Python-Dev] Compact ordered set
If sets were ordered, then what ought pop() return - first, last, or nevertheless an arbitrary element? I lean toward arbitrary because in existing code, set.pop often implies that which particular element is immaterial. On Wed, Feb 27, 2019 at 2:18 PM Barry Warsaw wrote: > On Feb 27, 2019, at 13:54, Chris Barker via Python-Dev < > python-dev@python.org> wrote: > > > > A mapping and a set type really don't have much to do with each other > other than implementation -- anyone that isn't familiar with python C code, > or hash tables in general, wouldn't likely have any expectation of them > having anything to do with each other. > > I’m just relaying a data point. Some Python folks I’ve worked with do > make the connection between dicts and sets, and have questions about the > ordering guarantees of then (and how they relate). > > -Barry > > ___ > Python-Dev mailing list > Python-Dev@python.org > https://mail.python.org/mailman/listinfo/python-dev > Unsubscribe: > https://mail.python.org/mailman/options/python-dev/tahafut%40gmail.com > ___ Python-Dev mailing list Python-Dev@python.org https://mail.python.org/mailman/listinfo/python-dev Unsubscribe: https://mail.python.org/mailman/options/python-dev/archive%40mail-archive.com
