mean and peak for all sample

조회 수: 2 (최근 30일)
Ishaq Alshuaili
Ishaq Alshuaili 2021년 5월 25일
댓글: Ishaq Alshuaili 2021년 5월 26일
this code for one sample :
[M,I] = max(pz3,[],2);
Msg=sprintf('peak is %f and it is locatedat %i index of the array\n ',z(I(1)),I(1));
disp(Msg);
M1 = sum(z.*pz3(1,:))/sum(pz3(1,:)) ;
Msg=sprintf('mean value is %f \n ',M1);
disp(Msg);
how I can generalize it to the entire sample?

답변 (1개)

DGM
DGM 2021년 5월 25일
What is "the entire sample"? Is it the entirety of pz3? Is it each row? Is it z.*pz3?
I'm going to just assume that you want to find the max and mean of a 2D array called A.
A = rand(10)
A = 10×10
0.7105 0.3732 0.6008 0.1505 0.2378 0.3719 0.1127 0.7750 0.9504 0.2775 0.8302 0.0118 0.5578 0.2057 0.1885 0.2546 0.0788 0.2278 0.3845 0.1041 0.0135 0.0674 0.8598 0.5881 0.1280 0.9999 0.2626 0.7047 0.7745 0.5728 0.9912 0.8548 0.5598 0.7705 0.7054 0.4524 0.1961 0.8815 0.0596 0.1660 0.7031 0.8162 0.7211 0.1055 0.6576 0.5292 0.5518 0.2771 0.3455 0.3161 0.4669 0.1692 0.1344 0.3577 0.7394 0.5139 0.6424 0.9441 0.9933 0.8715 0.0840 0.1055 0.7442 0.2946 0.2792 0.2208 0.9380 0.9139 0.4512 0.9012 0.1633 0.1292 0.9782 0.6147 0.6756 0.3635 0.8992 0.0145 0.1988 0.8237 0.9181 0.7343 0.9633 0.9055 0.8752 0.3824 0.1803 0.3478 0.8031 0.7840 0.7233 0.6912 0.0834 0.4913 0.1718 0.1445 0.7990 0.9673 0.3888 0.5010
The mean is simple:
mn = mean(A(:))
mn = 0.5052
If you want the global maximum and the linear index:
[mx idx] = max(A(:))
mx = 0.9999
idx = 53
If you want subscripts instead of a linear index:
[suby subx] = ind2sub(size(A),idx)
suby = 3
subx = 6
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Ishaq Alshuaili
Ishaq Alshuaili 2021년 5월 26일
thank you I'll try this method

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