Find the perimeter of the shaded region-- the logical "1."

Question

Shilo 2022-11-18

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https://ww2.mathworks.cn/matlabcentral/answers/1854408-find-the-perimeter-of-the-shaded-region-the-logical-1

回答： Image Analyst 2022-11-18

In a binomial array of some random, consecutive shape. I have to create script that will determine the perimeter of each "line."

Example:

random sahpe=

[ 0 1 0 0 0

1 1 1 0 0

0 1 0 0 0

1 1 0 0 0

0 1 0 0 0 ]

ans= 18

I currently have this in a script looking like this:

box=[0 0 0 0 0; 0 0,1,0 0; 0 0,1,0 0; 0 0,0,0 0; 0 0 0 0 0] %in a 3 by 3 with a perimeter of zeros

for row= 2:4 %row number of shape box

for col= 2:4 %col number of shape box

if box(row,col)==1 %if element(1,1)=1 look at row and col of it

row;

col;

end

Sides= box(row, col-1)+ box(row, col+1)+ box(row+1, col)+ box(row-1, col)

eachSides= 4 -Sides

sub= eachSides*(Sides+1)

end

It works for only two conjoined logical arrays. Can anyone help?

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Answer 1

DGM 2022-11-18

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See bwperim() or regionprops() to get the perimeter. If you want to come up with a different way to define the perimeter, that'll be up to you to define.

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Answer 2

Image Analyst 2022-11-18

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在 MATLAB Online 中打开

You can use bwboundaries to get perimeter points. But if it's the length of the perimeter that is wanted regionprops is the way to go

sahpe = [     0    1    0    0    0
    1     1    1    0    0
    0     1    0    0    0
    1     1    0    0    0
    0     1    0    0    0 ]
sahpe = 5×5
     0     1     0     0     0
     1     1     1     0     0
     0     1     0     0     0
     1     1     0     0     0
     0     1     0     0     0
props = regionprops(sahpe, 'Perimeter');
perim = props.Perimeter
perim = 10.2140
% or you could count the number of perimeter pixels.
perim = nnz(bwperim(sahpe))
perim = 7
% Or you could run along the border computing the distance to the last
% pixel.
boundaries = bwboundaries(sahpe);
b = boundaries{1};
x = b(:, 2);
y = b(:, 1);
% Tack on end
x = [x; x(1)];
y = [y; y(1)];
perim = sum(sqrt(diff(x) .^ 2 + diff(y) .^ 2))
perim = 10.4853