Tracy Harms wrote:
|A simpler boxing of the columns of (a) would have each column result
|in a simple list of the values, like this:
|   ]ca=. <"1 |: a
|+-----+-----+------+------+
||0 4 8|1 5 9|2 6 10|3 7 11|
|+-----+-----+------+------+

Then how do you "flip" the above to get the following, without the
complications you and Dan spoke of?

+-+-+--+--+
|0|1| 2| 3|
|4|5| 6| 7|
|8|9|10|11|
+-+-+--+--+

Essentially, starting with an array of shape 3 4, I'm simply trying to
end up with the boxed rows having a shape of 3 1 and with the boxed
columns having a shape of 1 4.

You additionally wrote: "... it's important to note that each of the
boxes in the boxed-columns is itself a two-axis array" and "A simpler
boxing of the columns ... would have ...":

I was merely using the "Phrases" page as often suggested in this forum,
assuming the formulas given there are the "official", "authorized",
"best practice" way of accomplishing what the comments column indicates.
Maybe that's not the case.  I took the formula as "gospel truth" to use
without having to spend time analyzing and/or proving why it works
(which a user should not have to do with pre-provided formulas).  If the
approach used in formula 45 creates the extra "burden" of arrays within
the boxes that you and Dan mentioned, then perhaps the comments column
ought to indicate that kind of thing.

Thanks for your helpful comments!

Harvey
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