I would like to separate a matrix into different matrices based on a cycle of the first values

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i.e. [1 2 3 1 2 4; 1 1 1 2 2 2]=>[1 2 3; 1 1 1] and [1 2 4; 2 2 2] i can do it with a for loop looking for values that are smaller than the previous, but I feel there must be a faster way
  2 Comments
Adolf Krige
Adolf Krige on 9 Nov 2018
Thank you for the quick reply. I meant that the value in M(1,:) increases up to a point and then drops down to the low value before increasing again, 123 123. The problem that I have is that I need to find where the matrix goes from a high value to a low one, ie the end of the "Cycles"
I also forgot to say that the matrix is 1809x6 or longer, and I will be analyzing several of them, so manually setting the start points is not an option

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Accepted Answer

Bruno Luong
Bruno Luong on 9 Nov 2018
A=[1 2 3 1 2 4; 1 1 1 2 2 2];
lgt = diff(find([true,diff(A(1,:),1,2)<0,true]));
C = mat2cell(A,size(A,1),lgt);
Result:
C{:}
ans =
1 2 3
1 1 1
ans =
1 2 4
2 2 2

More Answers (1)

Luna
Luna on 9 Nov 2018
Edited: Luna on 9 Nov 2018
Hi Adolf,
As far as I understand from your question. The cycle means that if your data is smaller than the previous data according to your first row, you want to divide your matrix into small matrices according to those cycles.
here is a piece of code executes what you want:
A = [1 2 3 1 2 4 1 2 5; 1 1 1 2 2 2 3 3 3];
indexedElementNumber = [1; find((diff(A') < 0))+1];
for i = 1:numel(indexedElementNumber)-1
eval([ 'X_',sprintf('%d',i), ' = A(:,indexedElementNumber(i):indexedElementNumber(i+1)-1)' ]);
end
i = i+1;
eval([ 'X_',sprintf('%d',i), ' = A(:,indexedElementNumber(i):end)' ]);
  3 Comments
Jan
Jan on 9 Nov 2018
Edited: Jan on 9 Nov 2018
This is a very bad idea. See Why and how to avoid eval . Creating variables dynamically has many severe drawbacks. Use an array instead:
A = [1 2 3 1 2 4 1 2 5; 1 1 1 2 2 2 3 3 3];
index = [1, find((diff(A(1, :)) < 0))+1];
X = cell(1, numel(index));
for i = 1:numel(index) - 1
X{i} = A(:, index(i):index(i+1) - 1);
end
X{i} = A(:, index(end):end);
In "i=i+1" outside the loop you rely on the observation, that the loop counter has the maximum value after the loop. But this is not documented as far as I know.
Luna
Luna on 9 Nov 2018
Hi Jan,
Yes, you are absolutely correct. Actually I know why we should avoid eval and I never use it in my codes normally. But I thought he might need the variables in his workspace.
I have tried your code, the max value of the for loop is 2 but we have 3 cycles so yours will get the last cycle which starts with [1 2 5; 3 3 3] and replaces it with the 2nd cycle. So the 3rd element of cell array X is always empty.
I replaced the same code with cell array instead of eval.
A = [1 2 3 1 2 4 1 2 5; 1 1 1 2 2 2 3 3 3];
indexedElementNumber = [1; find((diff(A') < 0))+1];
X = cell(1,numel(indexedElementNumber));
for i = 1:numel(indexedElementNumber)-1
X{i} = A(:,indexedElementNumber(i):indexedElementNumber(i+1)-1);
end
i = i+1;
X{i} = A(:,indexedElementNumber(i):end);

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