Matrix manipulation using reshape
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Questa domanda è chiusa. Riaprila per modificarla o per rispondere.
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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 Commenti
Roger Stafford
il 18 Dic 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
il 18 Dic 2017
Risposte (1)
Roger Stafford
il 18 Dic 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 Commento
Prabha Kumaresan
il 26 Dic 2017
Questa domanda è chiusa.
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