Doubt in the specific line

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Rd
Rd el 24 de Jul. de 2020
Editada: KSSV el 24 de Jul. de 2020
could you please explain the following code?
k = zeros(img_h, img_w, 4);
k(:,:,1) = (fxx./((1 + fx.^2).^(3/2))).*fvr; % hor
k(:,:,2) = (fyy./((1 + fy.^2).^(3/2))).*fvr; % ver
k(:,:,3) = (f11./((1 + f1.^2).^(3/2))).*fvr; % \
k(:,:,4) = (f22./((1 + f2.^2).^(3/2))).*fvr; % /
Wr = 0;
bla = k(:,:,1) > 0;
for y=1:img_h
for x=1:img_w
if(bla(y,x)) %%% I cant understand (bla(y,x))
Wr = Wr + 1;
end

Respuesta aceptada

KSSV
KSSV el 24 de Jul. de 2020
Editada: KSSV el 24 de Jul. de 2020
It is bad to ask such questions......asking to explain a specific line or a code. You have code in hand, play around with that. Print the values, run the code in debug and you can make yourself clear with the code.
bla = k(:,:,1) > 0;
In the above case bla is a logical matrix. It will have 0, and 1's. 0 where the condition >0 is not met and 1's where the condition > 0 is met.
if(bla(y,x))
In the above line a if condition is used, the conditon is met or if is executed if bla(x,y) is 1 or the if contion is not met.

Más respuestas (1)

Walter Roberson
Walter Roberson el 24 de Jul. de 2020
The code
Wr = 0;
bla = k(:,:,1) > 0;
for y=1:img_h
for x=1:img_w
if(bla(y,x)) %%% I cant understand (bla(y,x))
Wr = Wr + 1;
end
end
end
can be entirely replaced with
Wr = nnz(k(:,:,1) > 0);
You are just counting the number of places that k(:,:,1) is true.

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