How to obtain a matrix from the iterations of a for loop?
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Pietro Fiondella
am 21 Jun. 2022
Kommentiert: Voss
am 21 Jun. 2022
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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Akzeptierte Antwort
Voss
am 21 Jun. 2022
Bearbeitet: Voss
am 21 Jun. 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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