Plotting a %of time vs %of area graph

Dear all,

I strugglIeling with this problem for a while already. I have the matrix A(1000x1000x120) which represents noise levels arranged as (longitude x latitude x time). This matrix has the following representation (the figure considers time=1, which mean A(1000 x 1000 x 1)). Note that the (1000 x 1000) locations are distributed in an uniform grid.

Im interested in what percentage of time and what percentage of area exceeded a specified level, let's say 150 in order to draw the following plot:

With this plot, I can see what percentage of area exceeds a specif level for what percentage of time.

This is the best I get so far:

which is clearly wrong.

This is my code:

threshold = 150; %limite maximo de ruido

percentageofareavector=[];

for i=1:120

tooNoisy = noise(:,:,i) >= threshold;

soma=sum(sum(tooNoisy)); %soma dos valores que excedem o limiar definido

totalnumberofpoints=(size(tooNoisy,1).*size(tooNoisy,2));

percentageofarea=soma*100/totalnumberofpoints;

percentageofareavector=[percentageofareavector percentageofarea];

end

timevector=1:size(noise,3);

percentageoftime=timevector*100/size(noise,3);

figure; area(percentageoftime, percentageofareavector);

xlabel('% of time'); ylabel('% of area');

title(['% of time vs % of area ', num2str(threshold),'dB is exceeded']);

What am I doing wrong?

Thank you in advance.

Answer 1

[KSSV]

I did nothing but removed the loop and tried to make code clearer.

threshold = 150;       %limite maximo de ruido
[m,n,p] = size(noise) ; 
data = reshape(noise,m*n,p) ;   % make 3D a 2D vector 
tooNoisy = data >= threshold;  % get the logical indices obeying 
percentageofarea = sum(tooNoisy)*100/(m*n);
percentageoftime = (1:p)*100/p;

figure; area(percentageoftime, percentageofareavector);  % you need to think about this line use area? or histogram? 
xlabel('% of time'); ylabel('% of area');
title(['% of time vs % of area ', num2str(threshold),'dB is exceeded']);