how to detect empty rows and columns from 3-D matrix and crop them?
6 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Nibras Abo Alzahab
am 31 Dez. 2018
Kommentiert: Nibras Abo Alzahab
am 31 Dez. 2018
I have a 3-D matrix as a video. It contains empty (0) zero values along the tree dimentions.
The matrix is represented as a video. So I need to:
- if the row along the frames is empty (==0), I want to crop it.
- if the column along the frames is empty (==0), I want to crop it
I have an example 3-D dimention but I need a general method.
I have tried to write something but there is something wrong.
Please Help.
%% define the video (3-D matrix)
Frames = [0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0;0 0 0 0 0];
Frames(:,:,2 ) = [0 0 0 0 0;0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0];
Frames(:,:,3 ) = [0 0 0 0 0;0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0];
Frames(:,:,4 ) = [0 0 0 0 0;0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0];
Frames(:,:,5 ) = [0 0 0 0 0;0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0];
Frames(:,:,6 ) = [0 0 0 0 0;0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0];
Frames(:,:,7 ) = [0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0;0 0 0 0 0];
Frames(:,:,8 ) = [0 0 0 0 0;0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0];
Frames(:,:,9 ) = [0 0 0 0 0;0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0];
Frames(:,:,10) = [0 0 0 0 0;0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0];
%% define some varibles
%% n & m are zeros counters
%% x,y,z are lengths along the three dimentions
n=0;
m=0;
x=length(Frames(:,1,1));
y=length(Frames(1,:,1));
z=length(Frames(1,1,:));
%% if the row along the frames is empty (==0),crop it.
%% r for row
% %% c for columns
% %% f for frames
for c=1:y
for r=1:x
for f=1:z
if Frames(r,c,f)==0 %detect
m=m+1;
end
end
if m==25
Frames(:,c,:) = []; %crop
end
end
end
%% %% if the column along the frames is empty (==0),crop it.
%% r for row
% %% c for columns
% %% f for frames
for r=1:x
for c=1:y
for f=1:z
if Frames(r,c,f)==0 %detect
n=n+1;
end
end
if m==25
Frames(r,:,:) = []; %crop
end
end
end
0 Kommentare
Akzeptierte Antwort
Weitere Antworten (2)
Stephan
am 31 Dez. 2018
Bearbeitet: Stephan
am 31 Dez. 2018
Hi,
two lines of code do the job:
Frames = zeros(5,5,10);
Frames(:,:,1 ) = [0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0;0 0 0 0 0];
Frames(:,:,2 ) = [0 0 0 0 0;0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0];
Frames(:,:,3 ) = [0 0 0 0 0;0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0];
Frames(:,:,4 ) = [0 0 0 0 0;0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0];
Frames(:,:,5 ) = [0 0 0 0 0;0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0];
Frames(:,:,6 ) = [0 0 0 0 0;0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0];
Frames(:,:,7 ) = [0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0;0 0 0 0 0];
Frames(:,:,8 ) = [0 0 0 0 0;0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0];
Frames(:,:,9 ) = [0 0 0 0 0;0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0];
Frames(:,:,10) = [0 0 0 0 0;0 0 90 0 0; 0 0 90 0 0; 90 90 90 0 0;0 0 0 0 0];
% crop all empty rows and columns
Frames(:,find(sum(Frames,[1 3])==0),:)=[];
Frames(find(sum(Frames,[2 3])==0),:,:)=[]
disp('Please accept useful answers.')
Best regards
Stephan
2 Kommentare
Luna
am 31 Dez. 2018
Bearbeitet: Luna
am 31 Dez. 2018
Hi,
Try this below:
%% delete zero rows
Frames = reshape(Frames(bsxfun(@times,any(Frames,2),ones(1,size(Frames,2)))>0),[],size(Frames,2),size(Frames,3));
%% delete zero columns
Frames = reshape(Frames(bsxfun(@times,any(Frames,1),ones(size(Frames,1),1))>0),size(Frames,1),[],size(Frames,3));
you will get 3x3x10 matrix
6 Kommentare
Siehe auch
Kategorien
Mehr zu Matrix Indexing finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!