Matrix manipulation using reshape

If A=
1 0 0 0 0 0 0 0 0 0
0 2 0 0 0 0 0 0 0 0
0 0 3 0 0 0 0 0 0 0
0 0 0 4 0 0 0 0 0 0
0 0 0 0 5 0 0 0 0 0
0 0 0 0 0 6 0 0 0 0
0 0 0 0 0 0 7 0 0 0
0 0 0 0 0 0 0 8 0 0
0 0 0 0 0 0 0 0 9 0
0 0 0 0 0 0 0 0 0 10
I got B =
1 2 0 0 0 0 0 0 0 0
1 2 0 0 0 0 0 0 0 0
0 0 3 4 0 0 0 0 0 0
0 0 3 4 0 0 0 0 0 0
0 0 0 0 5 6 0 0 0 0
0 0 0 0 5 6 0 0 0 0
0 0 0 0 0 0 7 8 0 0
0 0 0 0 0 0 7 8 0 0
0 0 0 0 0 0 0 0 9 10
0 0 0 0 0 0 0 0 9 10
using
B = reshape(repmat(max(reshape(A,2,[])),2,1),size(A)).
But if i want to get C =
1 0 3 0 0 0 0 0 0 0
0 2 0 4 0 0 0 0 0 0
1 0 3 0 0 0 0 0 0 0
0 2 0 4 0 0 0 0 0 0
0 0 0 0 5 0 0 8 0 0
0 0 0 0 0 6 0 0 9 0
0 0 0 0 0 0 7 0 0 10
0 0 0 0 5 0 0 8 0 0
0 0 0 0 0 6 0 0 9 0
0 0 0 0 0 0 7 0 0 10
from A how the reshape can be written?

2 comentarios

Roger Stafford
Roger Stafford el 18 de Dic. de 2017
How would you generalize what is desired for an arbitrary size diagonal matrix? It is fairly easy to produce the desired result for your particular size.
Prabha Kumaresan
Prabha Kumaresan el 18 de Dic. de 2017
I need to have random grouping of rows which reults in sharing their values.

Respuestas (1)

Roger Stafford
Roger Stafford el 18 de Dic. de 2017
If the "grouping" is to be randomly determined, do this:
n = size(A,1);
p = mod((0:n-1)+randi(n-1,1,n),n)+1; % p(k) never equals k
B = A;
for k = 1:n
B(p(k),k) = B(k,k);
end
If you have already determined a vector p to be used, then leave out the second line.

1 comentario

The following code executes but I am unable to get the random grouping of users.
N_UE=[10 20 30 40 50];
N_SC=[60 70 80 90 100];
for t= 1:length(N_UE)
for r = 1:length(N_SC)
C=rand(N_UE(t),N_SC(r));
s = size(C,1);
p = mod((0:s-1)+randi(s-1,1,s),s)+1; % p(k) never equals k
B = C;
for k = 1:s
B(p(k),k) = B(k,k);
end
end
end

La pregunta está cerrada.

Preguntada:

el 18 de Dic. de 2017

Cerrada:

el 20 de Ag. de 2021

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