Indexing to return segments

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Hi. I know this could look like a silly question but can't find a way to do it. I have a matrix (data) that is 14258 x 7, the first six columns are data from a subject and the last column are time markers (0 off and 5 is on).
I used ischange on the last column to have the changing points:
[index,~] = ischange(data(:,7))
The problem is that I should end with 10 segments (of different duration) that include the data of the other 6 columns, but I don't know how to use the index to have it.
Thanks in advance

Accepted Answer

Image Analyst
Image Analyst on 29 Jan 2021
This will take the 10 different regions of 5, and 11 different regions of 0 and put them into cell arrays output5 and output0. I need to use cell arrays because not every run has the same number of rows in it.
The code requires the Image Processing Toolbox.
s = load('data.mat')
data = s.data;
% Get the segments with 5 in the last column.
is5 = data(:, end) == 5;
props = regionprops(is5, 'Area', 'PixelList');
allRunLengths = [props.Area] % Just for our information - not used at all.
% Take these 10 runs of lengths 266 352 347 384 383 454 505 536 549 604
% and put them into a cell array
for k = 1 : numel(props)
rows = props(k).PixelList(:, 2);
output5{k} = data(rows, 1:6);
end
% Get the segments with 0 in the last column.
is0 = data(:, end) == 0;
props = regionprops(is0, 'Area', 'PixelList');
allRunLengths = [props.Area] % Just for our information - not used at all.
% Take these 11 runs of lengths 1526 618 611 627 608 1957 601 603 604 613 1510
% and put them into a cell array
for k = 1 : numel(props)
rows = props(k).PixelList(:, 2);
output0{k} = data(rows, 1:6);
end
  1 Comment
Frank Pernett
Frank Pernett on 30 Jan 2021
Thank you so much!
It ran perfectly

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More Answers (1)

Bob Thompson
Bob Thompson on 29 Jan 2021
I can't actually use ischange in my version of MATLAB, so take what I suggest with a grain of salt.
It looks like the index output is actually a logic array. I recommend using find to identify the actual indexes of the identified changes. I imagine it woud look something like the following, but it is untested:
[index] = find(ischange(data(:,7))==1);
I'm not quite sure how you want to utilize the different sections, but you could capture the different time elements in a cell array kind of like the following:
timedata{1} = data(1:index(1)-1,1:6);
for i = 1:length(index)-1
timedata{i+1} = data(index(i):index(i-1),1:6);
end
timedata{length(index)+1} = data(index(end):end,1:6);
  1 Comment
Frank Pernett
Frank Pernett on 9 Mar 2021
Sorry for the late comment but this solution was also good, with a small change in the loop.

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