NEED HELP SOLVING "Array indices must be positive integers or logical values"
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Daniel Hodgson
el 29 de Ag. de 2020
Comentada: Peter O
el 30 de Ag. de 2020
Im looking at a 3x3 grid around specific number within a larger 16x16 and in some cases i land up being outside of my 16x16 grind and therefore getting and error. Want i want to do is say if outside 16x16 grind then =0.
%new attempt (want to say if outside box then =0)
for i=a(1):a(256) %trying to calcule neighbours around each cell containing one
c=[a(i+1),a(i-1),a(i+16),a(i-16),a(i+15),a(i+17),a(-15),a(-17)]
c(isnan(c))=0
A(i)=c; %stores value as vector
end
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Respuesta aceptada
David Hill
el 29 de Ag. de 2020
a=blkdiag(0,a,0);
count=1;
for m=2:17
for n=2:17
A{count}=[a(m+1,n),a(m-1,n),a(m,n+1),a(m,n-1),a(m-1,n+1),a(m+1,n+1),a(m+1,n-1),a(m-1,n-1)];
count=count+1;
end
end
2 comentarios
David Hill
el 30 de Ag. de 2020
If you do not want a cell array, you can combine into a matrix. Count just counts the total in both loops (16*16=256 total rows or cells).
a=blkdiag(0,a,0);
count=1;
for m=2:17
for n=2:17
A(count,:)=[a(m+1,n),a(m-1,n),a(m,n+1),a(m,n-1),a(m-1,n+1),a(m+1,n+1),a(m+1,n-1),a(m-1,n-1)];
count=count+1;
end
end
The above makes a matrix instead of a cell. Each row of the matrix corresponds to a vector. A(1,:) is the vector of the first point, A(2,:) is the vector of the second point, ... You could change the loops if you want to change the order.
a=blkdiag(0,a,0);
count=1;
for n=2:17%run rows second
for m=2:17%run columns first
A(count,:)=[a(m+1,n),a(m-1,n),a(m,n+1),a(m,n-1),a(m-1,n+1),a(m+1,n+1),a(m+1,n-1),a(m-1,n-1)];
count=count+1;
end
end
Más respuestas (1)
Peter O
el 29 de Ag. de 2020
Does it need to be a one-liner? If you can grab the x,y coordinate, you could clamp your index points to the box bounds using
% For a 256 x 256 grid
x_bounds = min(256, max(1, [index_x-16, index_x+16]))
y_bounds = min(256, max(1, [index_y-16, index_y+16]))
%Then get the 16x16 (or smaller) blob:
M = A(x_bounds(1):x_bounds(2),y_bounds(1):y_bounds(2))
You could do the same to clamp the 3x3 if it's near the edge of the image.
2 comentarios
Peter O
el 30 de Ag. de 2020
Sorry for the delay and I see you've found a solution that works, and I think I misinterpreted your question at first because my solution wasn't what you wanted to do. However, to close the loop here:
Assuming you have a grid of 256x256 and you have an (x,y) position indictated by (index_x, index_y), this code will extract the square with a side length of 33 pixels around it and the (index_x,index_y) point at its center (16 pixels to either side). The minmax wrapping ensures that you never exceed the bounds of the grid 1...256. If you access something where the (index_x, index_y) is under 16 pixels from an edge, it clamps to the edge so you don't get an out-of-bounds error. If you were 8 pixels from the left edge and in the middle of the grid vertically, for instance (x,y) is (8, 50), you'd get a 24x33 region returned.
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