How to obtain a matrix from the iterations of a for loop?
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I woud like to obtain a matrix C[84x24] with every row filled, by the moltiplication between each row af a Matrix A[7x24] by each element of vector B[1x12].
I am able to do it only by this loop with "cell command at the end but i'm looking for an easyer way .
A=[ 0.2167 0.2467 0.2520 0.2652 0.2661 0.2907 0.3498 0.4211 0.5348 0.5735 0.5242 0.4361 0.4810 0.5233 0.5365 0.5356 0.5533 0.7629 0.8317 0.8704 0.7770 0.7321 0.6563 0.5427
0.2406 0.2738 0.2797 0.2944 0.2954 0.3227 0.3883 0.4675 0.5936 0.6367 0.5819 0.4841 0.5340 0.5809 0.5956 0.5946 0.6142 0.8469 0.9232 0.9663 0.8626 0.8127 0.7286 0.6024
0.2440 0.2778 0.2837 0.2986 0.2996 0.3274 0.3938 0.4742 0.6021 0.6458 0.5902 0.4910 0.5416 0.5892 0.6041 0.6031 0.6230 0.8591 0.9364 0.9801 0.8749 0.8244 0.7390 0.6111
0.2453 0.2792 0.2851 0.3001 0.3011 0.3290 0.3958 0.4766 0.6052 0.6490 0.5932 0.4935 0.5444 0.5922 0.6072 0.6062 0.6261 0.8634 0.9412 0.9850 0.8794 0.8285 0.7428 0.6142
0.2453 0.2792 0.2851 0.3001 0.3011 0.3290 0.3958 0.4766 0.6052 0.6490 0.5932 0.4935 0.5444 0.5922 0.6072 0.6062 0.6261 0.8634 0.9412 0.9850 0.8794 0.8285 0.7428 0.6142
0.2448 0.2786 0.2846 0.2995 0.3005 0.3284 0.3950 0.4756 0.6040 0.6477 0.5920 0.4925 0.5433 0.5910 0.6060 0.6050 0.6249 0.8617 0.9393 0.9831 0.8776 0.8268 0.7413 0.6129
0.2369 0.2696 0.2754 0.2899 0.2908 0.3178 0.3823 0.4603 0.5845 0.6269 0.5730 0.4767 0.5258 0.5720 0.5865 0.5855 0.6048 0.8340 0.9091 0.9514 0.8494 0.8003 0.7174 0.5932];
B=[0.999; 0.992; 0.939; 0.850; 0.810; 0.803; 0.796; 0.792; 0.821; 0.929; 0.990; 1.000];
for j=1:12;
C=A*B(j);
cell_C{j}=C;
end
D=[cell_C{1,1};cell_C{1,2};cell_C{1,3};cell_C{1,4};cell_C{1,5};cell_C{1,6};cell_C{1,7};cell_C{1,8};cell_C{1,9};cell_C{1,10};cell_C{1,11};cell_C{1,12}];
D
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Réponse acceptée
Voss
le 21 Juin 2022
Modifié(e) : Voss
le 21 Juin 2022
A=[ 0.2167 0.2467 0.2520 0.2652 0.2661 0.2907 0.3498 0.4211 0.5348 0.5735 0.5242 0.4361 0.4810 0.5233 0.5365 0.5356 0.5533 0.7629 0.8317 0.8704 0.7770 0.7321 0.6563 0.5427
0.2406 0.2738 0.2797 0.2944 0.2954 0.3227 0.3883 0.4675 0.5936 0.6367 0.5819 0.4841 0.5340 0.5809 0.5956 0.5946 0.6142 0.8469 0.9232 0.9663 0.8626 0.8127 0.7286 0.6024
0.2440 0.2778 0.2837 0.2986 0.2996 0.3274 0.3938 0.4742 0.6021 0.6458 0.5902 0.4910 0.5416 0.5892 0.6041 0.6031 0.6230 0.8591 0.9364 0.9801 0.8749 0.8244 0.7390 0.6111
0.2453 0.2792 0.2851 0.3001 0.3011 0.3290 0.3958 0.4766 0.6052 0.6490 0.5932 0.4935 0.5444 0.5922 0.6072 0.6062 0.6261 0.8634 0.9412 0.9850 0.8794 0.8285 0.7428 0.6142
0.2453 0.2792 0.2851 0.3001 0.3011 0.3290 0.3958 0.4766 0.6052 0.6490 0.5932 0.4935 0.5444 0.5922 0.6072 0.6062 0.6261 0.8634 0.9412 0.9850 0.8794 0.8285 0.7428 0.6142
0.2448 0.2786 0.2846 0.2995 0.3005 0.3284 0.3950 0.4756 0.6040 0.6477 0.5920 0.4925 0.5433 0.5910 0.6060 0.6050 0.6249 0.8617 0.9393 0.9831 0.8776 0.8268 0.7413 0.6129
0.2369 0.2696 0.2754 0.2899 0.2908 0.3178 0.3823 0.4603 0.5845 0.6269 0.5730 0.4767 0.5258 0.5720 0.5865 0.5855 0.6048 0.8340 0.9091 0.9514 0.8494 0.8003 0.7174 0.5932];
B=[0.999; 0.992; 0.939; 0.850; 0.810; 0.803; 0.796; 0.792; 0.821; 0.929; 0.990; 1.000];
kron method:
D_test = kron(B,A)
Compare to the cell array method:
for j=1:12;
C=A*B(j);
cell_C{j}=C;
end
D=vertcat(cell_C{:});
The result is the same:
isequal(D_test,D)
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