Intersecting and non-intersecting box regions

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Elysi Cochin
Elysi Cochin 2021 年 7 月 10 日
コメント済み: Matt J 2021 年 7 月 12 日
Having a set of bounding box values [x y width height] , how can i find the number of bounding box that gets intersected and that do not gets intersected when plotted
From the above example, there are 5 intersecting boxes and 2 non-intersecting boxes
How can i do so with the attached bounding box values

採用された回答

Matt J
Matt J 2021 年 7 月 10 日
編集済み: Matt J 2021 年 7 月 10 日
Using rectint(), you can straightforwardly obtain a binary mask A such that A(i,j)=1 if rectangle bbx(i,:) and bbx(j,:) intersect.
load bbx
A=rectint(bbx,bbx)>0
A = 20×20 logical array
1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
and therefore the number of rectangles that have an intersection another rectangle would be,
N=sum(tril(A,-1),'all')
  6 件のコメント
Elysi Cochin
Elysi Cochin 2021 年 7 月 12 日
編集済み: Elysi Cochin 2021 年 7 月 12 日
load bbx
A = rectint(bbx,bbx)>0
Sir there are total 20 boxes
and non-intersecting box = 13, but there are more zeros, in the matrix A
So Sir, if I use A(i,j), how will i get 13?
Matt J
Matt J 2021 年 7 月 12 日
>> nnz(sum(A,1)==1)
ans =
13

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その他の回答 (2 件)

Simon Chan
Simon Chan 2021 年 7 月 10 日
If viusal inspection is allowed, then the number can be counted by plotting the boxes in a figure:
rawdata=load('bbx.mat');
figure
for k=1:length(rawdata.bbx)
rectangle('Position',rawdata.bbx(k,:))
end
  4 件のコメント
Elysi Cochin
Elysi Cochin 2021 年 7 月 10 日
when i increase the number of rows in bbx, i get error
Matrix dimensions must agree.
Error in demo
B = B+A;
Simon Chan
Simon Chan 2021 年 7 月 10 日
Would you please run the following commands before the above script:
clear;
clc;
If the error happens again, would you please run the following:
size(A) % Check the dimension of Matrix A
size(B) % Check the dimension of Matrix B

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KSSV
KSSV 2021 年 7 月 10 日
Run a loop for each box and find the intersection points. Use this to get the intersection points.
If your output is empty, it means there is no intersection.
  3 件のコメント
KSSV
KSSV 2021 年 7 月 10 日
You have origin, length and width of bounding box. So you can get four vertices of the box. You need to use four coordinates for each bounding box to get the intersection points.
Elysi Cochin
Elysi Cochin 2021 年 7 月 10 日
i didnt get that Sir. Please can you show me an example?

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