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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Voss
2022년 6월 21일
편집: Voss
2022년 6월 21일
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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