# How to separate a matrix by zeros?

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Koos Toebes on 5 Jun 2016
Commented: Image Analyst on 18 May 2021
Hi guys. For an assignment, I have to split up a matrix, separated by zeros. For just a vector it worked like this:
A = [1 2 3 0 0 0 0 0 2 3 4 0 0 0 0 0 4 5 6 7 0 0 0 0 1 1 1];
ne0 = find(A~=0); % Nonzero Elements
ix0 = unique([ne0(1) ne0(diff([0 ne0])>1)]); % Segment Start Indices
ix1 = ne0([find(diff([0 ne0])>1)-1 length(ne0)]); % Segment End Indices
for k1 = 1:length(ix0)
section{k1} = A(ix0(k1):ix1(k1));
end
celldisp(section) % Display Results
I found the above script somewhere else here.
But now I have to do more or less the same for a matrix. For example:
2 9
3 5
7 2
0 4
0 2
0 7
1 4
8 5
0 4
2 8
into:
2 9
3 5
7 2
1 4
8 5
2 8
I tried to modify the script above, but that was not successful.
##### 2 CommentsShowHide 1 older comment
Image Analyst on 18 Oct 2018
To concatenate vertically the number of columns of the matrices must match. To concatenate horizontally the number of rows of the matrices must match. Otherwise, how could you stitch them together? If you still have a problem read this link and post in a new question (not here).

Weird Rando on 5 Jun 2016
ne0 = find(A(:,1)~=0)'; % Nonzero Elements (transposed)
ix0 = unique([ne0(1) ne0(diff([0 ne0])>1)]); % Segment Start Indices
ix1 = ne0([find(diff([0 ne0])>1)-1 length(ne0)]); % Segment End Indices
for k1 = 1:length(ix0)
section{k1} = A(ix0(k1):ix1(k1),:); % (Included the column)
end
celldisp(section) % Display Results
Image Analyst on 18 May 2021
@Wirattawut Boonbandansook, I just ran it and it ran fine:
A = [1 2 3 0 0 0 0 0 2 3 4 0 0 0 0 0 4 5 6 7 0 0 0 0 1 1 1];
ne0 = find(A(:,1)~=0)'; % Nonzero Elements (transposed)
ix0 = unique([ne0(1) ne0(diff([0 ne0])>1)]); % Segment Start Indices
ix1 = ne0([find(diff([0 ne0])>1)-1 length(ne0)]); % Segment End Indices
for k1 = 1:length(ix0)
section{k1} = A(ix0(k1):ix1(k1),:); % (Included the column)
end
celldisp(section) % Display Results
section{1} =
Columns 1 through 20
1 2 3 0 0 0 0 0 2 3 4 0 0 0 0 0 4 5 6 7
Columns 21 through 27
0 0 0 0 1 1 1
What did you do to change it?

Image Analyst on 5 Jun 2016
An alternate way, if you have the Image Processing Toolbox is
m = [...
2 9
3 5
7 2
0 4
0 2
0 7
1 4
8 5
0 4
2 8]
% If there is a zero in column 1, make the zero in column 2 also
m(m(:,1)==0, 2) = 0
[labeledMatrix, numberOfRegions] = bwlabel(m) % Identify separate regions
% Extract/crop all the separate sub-arrays from m and save in a cell array
for k = 1 : numberOfRegions
% Get the values
theRegions{k} = m(labeledMatrix(:, 1) == k, :);
end
% Print out all the regions to the command window.
celldisp(theRegions)
Koos Toebes on 5 Jun 2016
Thanks!