Splitting a vector up into unequal sections seperated by zeros

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Bran 2014-9-29

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https://ww2.mathworks.cn/matlabcentral/answers/156650-splitting-a-vector-up-into-unequal-sections-seperated-by-zeros

评论： Adam 2023-7-17

Hi everyone

I was wondering if someone could help me split up a vector into sections seperated by zeros. I have alot of data but if I give you an example with shorter data

if I have 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] and I would like section 1 = [ 1 2 3] section 2 = [ 2 3 4] section 3 = [ 4 5 6 7] section 4 = [ 1 1 1] how would I go about this?

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Star Strider 2014-9-29

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https://ww2.mathworks.cn/matlabcentral/answers/156650-splitting-a-vector-up-into-unequal-sections-seperated-by-zeros#answer_153235

在 MATLAB Online 中打开

This works:

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

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Star Strider 2014-9-30

编辑：Star Strider 2014-9-30

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I haven’t seen the attachment yet. Took a break, then came back to this, adding leading and trailing zeros to ‘A’ as well, since I discovered that those crashed my previous code. They don’t crash this version.

This works, producing ‘pure’ non-zero and zero sections:

A = [0 0 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 0 0 0];
ne0 = find(A~=0);                                   % Nonzero Elements
ix0 = unique([ne0(1) ne0(diff([0 ne0])>1)]);        % Non-Zero Segment Start Indices
eq0 = find(A==0);                                   % Zero Elements
ix1 = unique([eq0(1) eq0(diff([0 eq0])>1)]);        % Zero Segment Start Indices
ixv = sort([ix0 ix1 length(A)]);                    % Consecutive Indices Vector
for k1 = 1:length(ixv)-1
    section{k1} = A(ixv(k1):ixv(k1+1)-1);
end
celldisp(section)

I ended up duplicating the index calculations for the non-zero segments to find the zero segments as well. This was an interesting problem!

EDIT — Just now saw ‘matrix A.mat’. My current code processed it without errors. (I displayed a few sections of it and the original matrix to be sure.) The only changes to my code to run it were to replace the original ‘A’ assignment with:

matA = load('matrix A.mat');
A = matA.ABSXX2;

since it is easier than changing the two references to ‘A’.

Adam 2023-7-17

在 MATLAB Online 中打开

Hello,

Your code helped me a lot, but it doesn't treat well the last section. I propose this code, which, I believe, covers all possible cases without error.

A = [0 0 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 0 0 0];
ne0 = find(A~=0);                                   % Nonzero Elements
if ~isempty(ne0)
    ix0 = unique([ne0(1) ne0(diff([0 ne0])>1)]);    % Non-Zero Segment Start Indices
else
    ix0 = [];
end
eq0 = find(A==0);                                   % Zero Elements
if ~isempty(eq0)
    ix1 = unique([eq0(1) eq0(diff([0 eq0])>1)]);    % Zero Segment Start Indices
else
    ix1 = [];
end
ixv = sort([ix0 ix1]);                              % Consecutive Indices Vector
section = cell(1, length(ixv));
for k1 = 1:length(ixv)
    if k1 ~= length(ixv)
        section{k1} = A(ixv(k1):ixv(k1+1)-1);
    else
        section{k1} = A(ixv(k1):length(A));
    end
end
celldisp(section)