# Intersecting and non-intersecting box regions

조회 수: 6 (최근 30일)
Elysi Cochin 2021년 7월 10일
댓글: Matt J 2021년 7월 12일
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

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### 채택된 답변

Matt J 2021년 7월 10일
편집: Matt J 2021년 7월 10일
Using rectint(), you can straightforwardly obtain a binary mask A such that A(i,j)=1 if rectangle bbx(i,:) and bbx(j,:) intersect.
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')
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Elysi Cochin 2021년 7월 12일
편집: Elysi Cochin 2021년 7월 12일
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 2021년 7월 12일
>> nnz(sum(A,1)==1)
ans =
13

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### 추가 답변 (2개)

Simon Chan 2021년 7월 10일
If viusal inspection is allowed, then the number can be counted by plotting the boxes in a figure:
figure
for k=1:length(rawdata.bbx)
rectangle('Position',rawdata.bbx(k,:))
end
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Elysi Cochin 2021년 7월 10일
when i increase the number of rows in bbx, i get error
Matrix dimensions must agree.
Error in demo
B = B+A;
Simon Chan 2021년 7월 10일
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 2021년 7월 10일
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.
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KSSV 2021년 7월 10일
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 2021년 7월 10일
i didnt get that Sir. Please can you show me an example?

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