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time as float since Jan 1, 0001?

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  time as float since Jan 1, 0001? Roy Smith <roy@panix.com> - 2013-02-23 08:29 -0500
    Re: time as float since Jan 1, 0001? Mark Lawrence <breamoreboy@yahoo.co.uk> - 2013-02-23 13:41 +0000
      Re: time as float since Jan 1, 0001? Roy Smith <roy@panix.com> - 2013-02-23 08:49 -0500
        Re: time as float since Jan 1, 0001? Dennis Lee Bieber <wlfraed@ix.netcom.com> - 2013-02-23 14:08 -0500
    Re: time as float since Jan 1, 0001? Chris Angelico <rosuav@gmail.com> - 2013-02-24 00:45 +1100

#39665 — time as float since Jan 1, 0001?

I'm working with matplotlib.plot_date(), which represents time as 
"floats starting at January 1st, year 0001".  Is there any 
straight-forward way to get that out of a datetime?

datetime.toordinal() gives me the number of days since that epoch, but 
as an integer.  I figured it wouldn't be too hard to just do:

t.toordinal() + t.time().total_seconds()

except it turns out that only timedelta supports total_seconds(); time 
doesn't!

I suppose I could do:

t.toordinal() + t.hour / 24.0 \
              + t.minute / 1440.0 \
              + t.second / 86400.0

but that's really ugly.  Is there no cleaner way to do this conversion?

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#39667

On 23/02/2013 13:29, Roy Smith wrote:
> I'm working with matplotlib.plot_date(), which represents time as
> "floats starting at January 1st, year 0001".  Is there any
> straight-forward way to get that out of a datetime?
>
> datetime.toordinal() gives me the number of days since that epoch, but
> as an integer.  I figured it wouldn't be too hard to just do:
>
> t.toordinal() + t.time().total_seconds()
>
> except it turns out that only timedelta supports total_seconds(); time
> doesn't!
>
> I suppose I could do:
>
> t.toordinal() + t.hour / 24.0 \
>                + t.minute / 1440.0 \
>                + t.second / 86400.0
>
> but that's really ugly.  Is there no cleaner way to do this conversion?
>

IIRC you needn't bother, matplotlib will do all the conversions for you. 
  In the highly likely case that I'm wrong this should help 
http://matplotlib.org/api/dates_api.html#module-matplotlib.dates

-- 
Cheers.

Mark Lawrence

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#39669

In article <mailman.2342.1361626870.2939.python-list@python.org>,
 Mark Lawrence <breamoreboy@yahoo.co.uk> wrote:

> On 23/02/2013 13:29, Roy Smith wrote:
> > I'm working with matplotlib.plot_date(), which represents time as
> > "floats starting at January 1st, year 0001".  Is there any
> > straight-forward way to get that out of a datetime?
> >
> > datetime.toordinal() gives me the number of days since that epoch, but
> > as an integer.  I figured it wouldn't be too hard to just do:
> >
> > t.toordinal() + t.time().total_seconds()
> >
> > except it turns out that only timedelta supports total_seconds(); time
> > doesn't!
> >
> > I suppose I could do:
> >
> > t.toordinal() + t.hour / 24.0 \
> >                + t.minute / 1440.0 \
> >                + t.second / 86400.0
> >
> > but that's really ugly.  Is there no cleaner way to do this conversion?
> >
> 
> IIRC you needn't bother, matplotlib will do all the conversions for you. 
>   In the highly likely case that I'm wrong this should help 
> http://matplotlib.org/api/dates_api.html#module-matplotlib.dates

Duh!  I didn't get that far in the docs!  Thanks, that makes life a lot 
easier.

Still, it seems like allowing toordinal() and fromordinal() to handle 
floats would be a useful addition :-)

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#39703

On Sat, 23 Feb 2013 08:49:08 -0500, Roy Smith <roy@panix.com> declaimed
the following in gmane.comp.python.general:

> Still, it seems like allowing toordinal() and fromordinal() to handle 
> floats would be a useful addition :-)

	But "ordinals" aren't floats... <G>

-- 
	Wulfraed                 Dennis Lee Bieber         AF6VN
        wlfraed@ix.netcom.com    HTTP://wlfraed.home.netcom.com/

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#39668

On Sun, Feb 24, 2013 at 12:29 AM, Roy Smith <roy@panix.com> wrote:
> datetime.toordinal() gives me the number of days since that epoch, but
> as an integer.  I figured it wouldn't be too hard to just do:
>
> t.toordinal() + t.time().total_seconds()

What about t.timestamp()? That's since 1970, but you could add (if my
calculations are correct) 62135683200.0 to it.

ChrisA

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