Borrar filtros
Borrar filtros

How to choose values ​​in the second and third column corresponding to the drawn numbers?

2 visualizaciones (últimos 30 días)
%N X Y
AA=[ 1 0 4
2 1 5
21 4 6
81 3 7
92 7 8
73 6 4
65 3 3
36 5 4
16 6 5
6 7 4]
A=AA(:,1);
disp(A);
b=(A(randperm(size(A,1),3),1))
disp(b);
% How to choose values ​​in the second and third column corresponding to the drawn numbers?
for i=1:3 %This solution give me error - Index in position 1 exceeds array bounds (must not exceed 10).
c=b(i,1);
disp(AA(c,2));
disp(AA(c,3));
end

Respuesta aceptada

Arunkumar M
Arunkumar M el 17 de Nov. de 2018
Editada: madhan ravi el 17 de Nov. de 2018
Error occurs because with c you are finding the element which is a part of first column in AA. But this element is not the index. So you have to find the index where it is located and then pull out second and third column values.
for i=1:3
c=b(i,1);
temp1 = find(A == c);
temp2 = temp1(1,1); % in case multiple values are returned in temp1.
disp(AA(temp2,2));
disp(AA(temp2,3));
end
  2 comentarios
hawk5577
hawk5577 el 17 de Nov. de 2018
Error:
Index in position 1 exceeds array bounds.
Error in smiec4 (line 22)
temp2 = temp1(1,1); % in case multiple values are returned in temp1.

Iniciar sesión para comentar.

Más respuestas (1)

Bruno Luong
Bruno Luong el 17 de Nov. de 2018
Why make so complicated? RANDPERM returns the position, store and use it rather than trying to recover it.
AA=[ 1 0 4
2 1 5
21 4 6
81 3 7
92 7 8
73 6 4
65 3 3
36 5 4
16 6 5
6 7 4]
p = randperm(size(A,1),3);
b = AA(p,1)
AA(p,:)

Categorías

Más información sobre Structures en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by