mean and peak for all sample

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Ishaq Alshuaili
Ishaq Alshuaili 2021 年 5 月 25 日
コメント済み: Ishaq Alshuaili 2021 年 5 月 26 日
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?

回答 (1 件)

DGM
DGM 2021 年 5 月 25 日
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
  1 件のコメント
Ishaq Alshuaili
Ishaq Alshuaili 2021 年 5 月 26 日
thank you I'll try this method

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