Higher order arrays in q

Community Support
1

I’m trying to create a simple 3rd order array - like so:
3 3 3 # til 27

I get a 'length. 

I tried with higher dimensions, no luck, same error. Only 2 dimensions allowed. Does the evaluation version limit dimensions of arrays or something? (that is a weird question! )

Do explain, 

Thank you, 

Kumar

2

http://code.kx.com/wiki/Reference/NumberSign

q)0N!3 3#til 5;
(0 1 2;3 4 0;1 2 3)

to create 3-dim matrices

q)0N!2 2#3 3#til 5;
((0 1 2;3 4 0);(1 2 3;0 1 2))
q)0N!2 2#/:3 3#til 5;
((0 1;2 0);(3 4;0 3);(1 2;3 1))

ta, jack

3

This is being added in the upcoming kdb v3.4

You will be able to use 3 3 3#

Terry

4

Ahh, brilliant!
Thank you, Jack and Terry.

Kumar

5

a square matrix of indices in x dimensions:

q)x:3
q)0N!{y#x}/[til(*/)s;1_(0N,')s:x#x];
((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 25 26))

6

or with commute

C:{{x .(z;y)}x}
C[#]/[til(*/)s;1_(0N,')s:x#x]

… i’ve been reading http://nsl.com/k/y.q