({. + i.@(-~/))&.(todayno :. todate) 2004 1 1,:2006 12 12
solves your problems.
R.E. Boss
> -----Oorspronkelijk bericht-----
> Van: [email protected] [mailto:programming-
> [email protected]] Namens PackRat
> Verzonden: maandag 29 november 2010 3:50
> Aan: Programming forum
> Onderwerp: [Jprogramming] Calendrical "odometer"--how to?
>
> [1] The "London Gold Fix" plot demo I mentioned previously has the
> line:
>
> bgn=. todayno 2004,.(1+i.12),.1
>
> I'd like to change the 2004 to something like (2004+i.3) so that a
> plot could handle multiple years rather than just one. Although that's
> the intuitive approach a beginner like me would take, it doesn't work
> that way. (The problem, of course, is that "stitch" can take a list
> and an atom, but it can't take two lists where each value of the one
> list is repeated as a fill for each value of the other list.) How can
> I adjust that line to get an output like this:
>
> 2004 1 1
> 2004 2 1
> 2004 3 1
> . . .
> 2004 12 1
> 2005 1 1
> 2005 2 1
> 2005 3 1
> . . .
> 2005 12 1
> 2006 1 1
> 2006 2 1
> 2006 3 1
> . . .
> 2006 12 1
>
> [2] Additionally, just as a matter of interest, how would one write a
> more complete "calendrical odometer"? In other words, all three
> elements (year, month, day) cycle through their values. (Leap years
> present problems, though.) The result would look like:
>
> 2004 1 1
> 2004 1 2
> . . .
> 2004 1 31
> 2004 2 1
> 2004 2 2
> . . .
> 2004 12 31
> 2005 1 1
> 2005 1 2
> . . .
> 2006 12 31
>
> (I suppose one could "cheat" by doing the calculation backwards--that
> is, going from a sequence of day numbers to the equivalent yyyy mm dd
> values.)
>
> [3] More challenging (I presume), how would one create such an odometer
> using any starting and ending dates in the format yyyy mm dd? I
> presume this might possibly be a multiline explicit verb (no tacit,
> please). (Again, I suppose one could "cheat", as above.)
>
> I really do appreciate the help you all give--thanks!
>
> Harvey
>
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