# How do I quantify this image based on area and perimeter?

조회 수: 28(최근 30일)
tgohsu 2018년 4월 7일
댓글: John BG 2018년 4월 8일
So I have this image: And I would like to quantify it by area and perimeter. The large object is the particle of interest, everything else is an artifact and can be disregarded. How would I go about doing this?

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### 채택된 답변

John BG 2018년 4월 7일
Hi Tushar
1. acquiring image
h1=figure(1);imshow(A)
ax=gca
2.
even the object of interest has low illumination, so the binarization threshold set low, at 12 only:
A1=A(:,:,1);
% A1=imbinarize(A,10);
A1(A1>10)=255;A1(A1<12)=0;
figure;imshow(A1)
3.-
Getting boundaries of all the objects
[B,L]=bwboundaries(A1);
perim=[];
for k=1:1:size(B,1)
perim=[perim size(B{k,:},1)];
end
4.-
Capturing the object of interest, because it has by far the longest perimeter:
B1_cell=B(find(perim==max(perim)),:);
B1=B1_cell{:};
hold(ax,'all');
plot(ax,B1(:,2),B1(:,1),'r','LineWidth',2) 5.-
the length of the perimeter, in pixels
Obj_perim_length=size(B1,1) % perimeter length in pixels
6.- now for the area
xp=B1(:,2);
yp=B1(:,1);
D=zeros(size(A1));
for k=1:1:numel(xp)
D(yp(k),xp(k))=1;
end
% figure;imshow(D);axd=gca;
D2=imfill(D,'holes')
figure;imshow(D2) obj_area =
5.207962500000000e+04
obj_area/(size(A1,1)*size(A1,2)) =
0.259624444156414
the area is of 52k pixels, about a quarter of the input image rectangle, if one can accurately get a scale of the window, then translation from pixels to mm, cm, microns, whatever .. is straight forward.
thanks in advance for time and attention
John BG
.
.
.
.
What I tried but either time consuming or more complex than the above solution:
1. inpolygon with all points of the initial image against the found perimeter:
% too time consuming, it almost crashes MATLAB twice
[in,on]=inpolygon(yq,xq,B1(:,2),B1(:,1));
in1=find(in>0);
for k=1:1:length(in)
plot(ax,xq(in1(k)),yq(in1(k)),'b*');
end
2. Define an alphashape and then apply command area
% alphasape awkwardly defines the shapes along the perimeter
% leaving almost all inside hollow, therefore the resulting area is almost that of a thin perimeter strip, not of the whole inside
shp1=alphaShape(B1(:,2),B1(:,1))
figure;plot(shp1)
area_obj1=area(shp1)
area_obj1/(size(A1,1)*size(A1,2))
% hold all
% for k = 1:length(B)
% boundary = B{k};
% plot(boundary(:,2), boundary(:,1),'r', 'LineWidth', 2)
% end
too sketchy without filtering first, that it's been already done above
3.
A2=del2(double(A1));figure;imshow(A2)
Laplacian gets the correct perimeter, but along with many other 'false' perimeters, thus requiring filtering before applying del2.

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### 추가 답변(1개)

tgohsu 2018년 4월 8일
편집: tgohsu 2018년 4월 8일
John
Thanks for taking the time to do this.
##### 댓글 수: 1표시숨기기 없음
John BG 2018년 4월 8일
for my friend Tushar, any time, happy to help.

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