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 commentaires

Roger Stafford
Roger Stafford le 18 Déc 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 le 18 Déc 2017
I need to have random grouping of rows which reults in sharing their values.

Réponses (1)

Roger Stafford
Roger Stafford le 18 Déc 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 commentaire

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

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le 20 Août 2021

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