CODEWITHSUNDEEP
kdb+
Daniel Krizian
Daniel Krizian

Reputation: 4716

matrix cells of the same row and column

Is there a more efficient/idiomatic way to retrieve cells of a matrix that are on the same row and column as the given cell?

q) f:{except[;y] x[y div n;],x[;y mod n:count first x]}
q) show A:s#til prd s:2 3
0 1 2
3 4 5
q) f[A;4]
3 5 1
q) f[A;2]
0 1 5

Upvotes: 2

Views: 165

Answers (2)

SJT
SJT

Reputation: 1097

g:{
  s:count each 1 first\x; // shape
  rc:s vs y; // y as row-column
  on:rc+/:{x,reverse each x} -1 1,'0; // orthogonal neighbours of rc
  nn:on where all flip[on]within'0,'s-1; // near neighbours: eliminate out of range
  x ./:nn }

q)A:2 3#til prd 2 3
q)g[A;4]
1 3 5
q)g[A;2]
5 1

If A contains only the indices of its raze (raze A) then we need only its shape, and g can return the indexes of the orthogonal neighbours of y.

h:{[s;y]
  rc:s vs y; // y as row-column
  on:rc+/:{x,reverse each x} -1 1,'0; // orthogonal neighbours of rc
  nn:on where all flip[on]within'0,'s-1; // near neighbours: eliminate out of range
  s sv/:nn }

q)h[2 3;4]
1 3 5
q)h[2 3;2]
5 1

Note that this can easily be adapted to diagonal neighbours instead of or as well as orthogonal neighbours; also to vector y.

Key concepts

  • Use sv and vs to encode/decode numbers to any arithmetical base
  • Use of map iterators Each and Each Right to control iteration

Upvotes: 4

terrylynch
terrylynch

Reputation: 13657

I'm not sure if your approach works in the general case? It may only work for your specific setup, e.g.

q)A:3 cut neg[6]?20
q)A
12 13 4
7  9  17

q)f[A;9]
12 7

One alternative approach is to use in to find the columns and rows to include

f2:{except[raze(x where y in'x),f where y in'f:flip x;y]}

q)f2[A;9]
7 17 13

Upvotes: 1

Related Questions