mean and peak for all sample

1 Ansicht (letzte 30 Tage)
Ishaq Alshuaili
Ishaq Alshuaili am 25 Mai 2021
Kommentiert: Ishaq Alshuaili am 26 Mai 2021
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?

Antworten (1)

DGM
DGM am 25 Mai 2021
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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