Paul Mansour

@Silas Good to know, thanks.

OK, thanks.

Anyone know how to install a user command on a Mac?

@B.Wilson Ties survive workspace loading, so I wonder if way back in the day of limited workspace size, a constant literal tie number was useful. If you had the tie number in a variable you would lose when loading a different workspace.

Signing off...

Will check it out, thanks.

I assume the under operator will help with transposing twice?

Don't be, I'm a rank beginner, pun intended

Ok. I feel better now.

I think that is my problem. If I want to apply a non-scalar user-defined function to the columns of m, I can't do that directly with rank without transposing first. Is that correct?

Yes, not saying good, trying to apply rank'

Trying to apply user defined function f to column of M

hold, not sure my question is well formed...

What about doing ⍉↑v f¨↓⍉m

This is how I understood it, so the reference to column on APL Cart confused me.

"Effectively, we are taking each scalar from v and pairing up with each row from m."

Please do...

Ah....

What am I not getting?

Yes, but there are only 3 column vectors on the right

I'm sure now I don't understand rank

Is that right, or it swapped for the definition of the corresponding one for rows?

is "Apply f between vector Mv and each column of Nm"

APL Cart says that Mv(f⍤0 1)Nm

@Adám Nice, will check it out.

@Adám Constructing the attribute column for ⎕XML for HTML tables. Pairing up attribute names and values.

Signing off for today, thanks again for your help.

Can you guess the application?

I'm liking e3

Values are all char vecs, not scalars, in my particular case.

 cmpx e1 e2 e3 e4
  ⊂[1 0](d⍴(×/1↓d)/j),[¯.5]k → 2.9E¯4 |    0% ⎕⎕⎕⎕⎕⎕
  ⍉⊂⍤2⊢j,⍤0⍤1⍉k              → 3.8E¯4 |  +32% ⎕⎕⎕⎕⎕⎕⎕⎕
  ⊂[0 3]j,⍤0⍤0 2⊢k           → 2.6E¯4 |   -9% ⎕⎕⎕⎕⎕
  (↑(⊂¨j),¨⊢)¨,⌿⊂¨⊂¨k        → 1.9E¯3 | +575% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕

Ok, now you just caused me another day of study!

often because they don't come up, but often because its easier to cheat and use each

Yes, I really need to study them more. It's remarkable how much coding one can do and avoid them

Ok, noted. I'll take that under advisement.

I'm pretty sure I never would have figured that out!

Cool, thanks so much!

Both do same!

and it seems to do what I want...

I just tried ⊂[0 3] j,⍤0⍤0 2⊢k

Cool! Am actual doing a ⊂[1 0] on my result, so maybe I can get rid of the transpose...

Woah... is that rank twice or an atop? I'm going to have to spend a day studying it!

Funny it actually runs with the extra letter

Yes, sorry

using the rank operator to avoid all of the reshaping of j?

        (d⍴(×/1↓d)/j),[¯.5]k

is there a way to do this:

   d←5 4 3
   k←d⍴⍳×/d
   j←'abcdef'

Given:

In production to would default to ⍺ ;)