(2 3 5 4$i.24)(+"1"_1)(2 4$i.8)
0 2 4 6
4 6 8 10
8 10 12 14
12 14 16 18
16 18 20 22
20 22 24 26
0 2 4 6
4 6 8 10
8 10 12 14
12 14 16 18
16 18 20 22
20 22 24 26
0 2 4 6
4 6 8 10
8 10 12 14
16 18 20 22
20 22 24 26
24 26 28 30
4 6 8 10
8 10 12 14
12 14 16 18
16 18 20 22
20 22 24 26
24 26 28 30
4 6 8 10
8 10 12 14
12 14 16 18
16 18 20 22
20 22 24 26
24 26 28 30
That’s the (almost surely correct) result from a J session.
Here’s an operator R which functions identically to the rank operator in J,
to which I believe the ISO operator is identical:
∇
[0] Z←X(U R N)Y;M;L;R;E
[1] (M P Q)←⌽3⍴⌽N
[2] E←{⊂[⌽(K|(-K←K+1)⌈K⌊⍺-⍺<0)↑⌽⍳K←≢⍴⍵]⍵;K}
[3] Z←⊃⍎(⎕IO+0≠⎕NC'X')⊃'U¨M E Y' '(P E X)U¨Q E Y'
∇
You guys should try every twisted operation you can on R to make sure it works.
I’ll ask the wizards in the J forum what they think.
As a side-note, I’m liking the lambda local vars. They don’t look a bit out of
place.
Hope it helps,
Louis
> On 24 Aug 2016, at 23:16, Xiao-Yong Jin <[email protected]> wrote:
>
> This should work,
>
> (2 3 5 4⍴⍳24)(+⍤1⍤¯1)(2 4⍴⍳8)
> VALUE ERROR
> μ-Z__A_LO_RANK_X7_B[4] →(μ-X7≢¯1)⍴μ-WITH_AXES
> ^
> You could use the axis operator generically
>
> (2 3 5 4⍴⍳24){⍺+[1,(⍴⍴⍺)⌈⍴⍴⍵]⍵}(2 4⍴⍳8)
>
> but the rank operator seems cleaner.
>
> Best,
> Xiao-Yong
>
>
