[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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