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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采纳的回答
Voss
2022-6-21
编辑:Voss
2022-6-21
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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