| Re: timegen [message #86363 is a reply to message #86360] |
Wed, 30 October 2013 19:36   |
spluque
Messages: 33
Registered: September 2013
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Member |
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On Wed, 30 Oct 2013 15:16:57 -0700 (PDT),
Phillip Bitzer <bitzerp@uah.edu> wrote:
> This may be pertinent, from the help: Note: If the step size is not an
> integer then the last element may not be equal to the FINAL time. In
> this case, TIMEGEN will return enough elements such that the last
> element is less than or equal to FINAL.
> If you 'caldat' your final time ( end_jd + 1 - (float(step_size) /
> 86400) ), I suspect this is the issue.
> But, I'm not sure why this isn't what you're looking for:
> ts=timegen(start=beg_jd, final=end_jd+1,step_size=step_size,
> units='seconds') caldat, ts, mo, dd, yyyy, hh, mm, ss print, yyyy[-1],
> mo[-1], dd[-1], hh[-1], mm[-1], ss[-1] 2013 10 11 23 59 59.902636
> BTW, your original code contains the line float(step_size) which is
> superfluous - step_size is already a float, yes?
Thanks very much for these pointers. I left the float() call with
step_size by accident here. In the actual code, this is part of a
procedure that needs to be quite general for any step_size, so I'm
coercing it to float. You're quite right about the note in the help
page; that's exactly what's going on. To protect against this, the
following seems to do what I need:
step_d=float(step_time) / 86400
times=timegen(start=beg_jd, $
final=end_jd + 1 - (step_d / 2), $
step_size=step_time, units='seconds')
i.e. adding half a step to 'final' ensures that the last step is always
included.
Thanks,
--
Seb
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