Dear Dan and R.E.Boss:
Thank you for your solutions. I now understand a little better the
all-important "Rank" concept. I agree, one should make the operation return
the right rank, rather than afterward reduce the rank.
My problem came when unboxing a matrix of scalars, in which there were some
empty boxes for missing values (like empty cells from a spreadsheeet range).
I now see that a: (ace) has rank 1. It is not really an empty box, but a
box containing a zero-element vector (I guess that you can't have a really
empty box in J). Hence, if there is an ace in the matrix, the operation >"0
will have the shape (($Matrix), 1). And the cells will unbox as zeros (more
accurately, as 1$0). I solved the problem with ({. & >)"0. This one-step
solution seems more efficient that the two steps >"0 and {."0.
I also just realized that an atom (scalar) can be produced from (i.0) $
scalar, which is just as it ought to be according to J rank theory.
Sometimes primitives seem quirky, but here everything makes sense.
Sincerely,
Leigh
-----Original Message-----
So, now that I've given you the bandaid, I'd like to teach you preventative
medicine. You should (nearly) always be able to predict the ranks of the
results produced by your code.
If you cannot, and you find you "need to remove a trailing axis", it's
better to investigate why that's so, and modify the code that produces the
trailing axis, rather than removing it ex post facto.
If you post a small working example of the code that's causing you trouble,
we'll probably be able to explain what's wrong, and help you fix it.
-Dan
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