Matrix integral over layers

Hello! I have a matrix in it the number of steps per time. Is it possible to integrate each column of the matrix layer by layer (layer sizes may differ)
MatrixHelpStep=rand(30,40);
layer=(1:5:30); % for example, I took layer 5 (but it can change)
% trying to create a loop for layers
for i=1:length(layer)-1
Imhs=trapz(MatrixHelpStep(layer(i):layer(i)+1))
end
it gives me only one value

4 comentarios

Walter Roberson
Walter Roberson el 17 de Jul. de 2020
"layer" is usually a term used for the third dimension, but your array is 2D. But in MatrixHelpStep(layer(i):layer(i)+1) you are indexing it with a single index.
As layer is maximum 30 and 30 is the maximum number of rows, we would suspect that
Imhs=trapz(MatrixHelpStep(layer(i):layer(i)+1,:))
but watch out for which dimension you wish to trapz() over.
Lev Mihailov
Lev Mihailov el 17 de Jul. de 2020
Editada: Lev Mihailov el 17 de Jul. de 2020
this is what I want to get, now I am doing it all manually and I want to write a loop
x=MatrixHelpStep(:,1);
x1=trapz(1:5,1);
x2=trapz(5:10,1);
x3=trapz(10:15,1);
%
x=MatrixHelpStep(:,2);
x4=trapz(1:5,2);
x5=trapz(5:10,2);
x6=trapz(10:15,2);
%
x=MatrixHelpStep(:,3);
x7=trapz(1:5,3);
x8=trapz(5:10,3);
x9=trapz(10:15,3);
Walter Roberson
Walter Roberson el 17 de Jul. de 2020
Could you confirm that you want overlap? trapz(x(1:5,1)) uses 5 elements to form the output, but trapz(x(5:10,1)) uses 6 elements to form the output. Are you wanting 1:5, 6:10, 11:15, 16:20 and so on?
Lev Mihailov
Lev Mihailov el 17 de Jul. de 2020
yes, I want such overlaps

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Respuestas (1)

Bjorn Gustavsson
Bjorn Gustavsson el 17 de Jul. de 2020

0 votos

For this I would try cumtrapz - it calculates the cumulative (sp?) trapezoidal integral in the same manner as cumsum calculates the cumulative sum along an array. Doing that would give you the cumulative integrals of your matrix with the same size as your matrix. That way you can easily calculate every following integral between 2 boundaries from that one just by taking the difference between the corresponding boundary-elements. Something like this:
I = peaks(40);
I = I(1:30,:);
ciI = cumtrapz(I);
i10to17ofI = ciI(17,:) - ciI(10,:);
HTH

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el 17 de Jul. de 2020

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