how to detect empty rows and columns from 3-D matrix and crop them?

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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

Respuesta aceptada

Matt J
Matt J el 31 de Dic. de 2018
Editada: Matt J el 31 de Dic. de 2018
tmp=any(Frames,3);
I = any(tmp,2);
J= any(tmp,1);
Frames=Frames(I,J,:);
  3 comentarios

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Más respuestas (2)

Stephan
Stephan el 31 de Dic. de 2018
Editada: Stephan el 31 de Dic. de 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 comentarios
Nibras Abo Alzahab
Nibras Abo Alzahab el 31 de Dic. de 2018
Thank you for your answer.
I got an error while using the
Error using sum
Dimension argument must be a positive integer scalar within indexing range.
Frames(:,find(sum(Frames,[1 3])==0),:)=[];
Stephan
Stephan el 31 de Dic. de 2018
Editada: Stephan el 31 de Dic. de 2018
Hm, works for me. Which Release do you use? however - shortest answer from Matt is a nice one.

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Luna
Luna el 31 de Dic. de 2018
Editada: Luna el 31 de Dic. de 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 comentarios
Stephan
Stephan el 31 de Dic. de 2018
Editada: Stephan el 31 de Dic. de 2018
The questioner will let us know...
Nibras Abo Alzahab
Nibras Abo Alzahab el 31 de Dic. de 2018
Thank you for your intrest and ansewrs.
Luna is right
I souhld get 4x3x10 matrix.
But in your answer: 3x3x10 matrix

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