Hello, I'm trying to sort the data from vector AA such that if AA(n+1) > AA(n) it'll store the indices into a matrix {data} if theres a break, (AA(N+1) < A(n)) data in AA will occupy a new column
This is the code I got for so far, however the {data} matrix is empty
AA = [ 1 2 3 5 7 10 9 11 13 14 17 19 17 22 25];
data = {};
k =1;
while true
idx = find (AA < AA - 1, 1);
if ~isempty (idx);
data{k} = AA(1:idx-1);
AA = AA(idx:end);
k = k + 1;
else
data {k} = AA(1:end);
break
end
end
The final result should be the matrix {data} contains the following
cell 1x1: [1 2 3 5 7 10]
cell 1x2: [11 13 14 17 19]
cell 1x3: [22 25]
Thanks!

 Akzeptierte Antwort

Guillaume
Guillaume am 26 Jan. 2015

0 Stimmen

Assuming you made a typo and cell{3} should be [17 22 25]:
AA = [ 1 2 3 5 7 10 9 11 13 14 17 19 17 22 25];
data = mat2cell(AA, 1, diff([0 find(diff(AA) < 0) numel(AA)]))
Basically, find the length of the runs, using diff and find and use mat2cell to convert to cell.

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