I find your post quite convoluted. Here's what I read:
Nick Simicich-2 wrote:
>
> (I figured out an answer for the first question - I don't ask unless I
> search thoroughly, and I finally found it, so I am going to leave it in
> case the answer helps someone else.)
>
> I have an array that can be any of a couple different square sizes. I
> wanted to determine which of the diagonals the coordinates I was
> processing
> was, so I made a list of the coordinates that were on the diagonals,[...]
>
Problem:
Given a square matrix, return diagonal that contains element with
coordinates i j.
Solution:
finddiag =: ,@[ #~ -/@] = ,@:(-/~@i.@#)@[
The solution uses the fact that i-j (i and j are in y) gives the number of
the diagonal,
and then uses the number to extract the corresponding diagonal from x (the
matrix.)
]mat=.i.5 5
0 1 2 3 4
5 6 7 8 9
10 11 12 13 14
15 16 17 18 19
20 21 22 23 24
mat finddiag 3 2
5 11 17 23
(<3 2){mat
17
Note that finddiag does not make coordinates of diagonal elements.
> I have a second question. I wanted to make two x by x matrices, and
> compare one of those matrices to a scalar, building a single matrix by
> extracting from each one. What I wanted to do was to code some variation
> of. I had a matrix where the filled in pieces were ordinary numbers and
> the placeholders were _. I had a second matrix that had candidates, but
> they were not
>
> (a=b) { mat1,.mat2
>
> That does nothing like I want - because it just plain does not work.
> Every
> zero and one selects the whole array, and changing the rank of {, as {"1,
> {"2 etc just does not work.
>
> What I ended up doing was
>
> ((a=b)*mat1)+(-.a=b)*mat2
>
> Now, that worked because I happened to have numeric matrices. If I'd had
> character arrays, I might have had to map them into numeric arrays to get
> this sort of thing to work at all.
>
> Is the math trick the only way to do this?
>
Here I couldn't formulate a problem based on your post.
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