# Breaking up a column vector

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Christopher on 10 Oct 2013
Commented: Cedric Wannaz on 10 Oct 2013
Hello everyone,
I have a data set that I am trying to break up based upon peaks in the data. Essentially I finding the peaks and the corresponding locations. From there I am seperationg the data from loation 1 to 2, 2 to 3, etcc. I will illustrate this with my code below.
load=data(:,3) % just a column vector of data
Then what I am trying to do is build indivdual vectors from peak 1 to peak 2, peak 2 to peak 3, etc. I can do it by doing the following:
etc....
However, there are many peaks, so I need a way to automate this procedure. I have tried several times, but cannot get it to work. I know this should be rather simple I would think. Any help would be great.

dpb on 10 Oct 2013
gotta' run, but one way -- use accumarray w/ the inputs from findpeaks to build a cell array of the resulting array elements specified per your above assignment. NB: have to set the 'uniformoutput' property to False to handle the different lengths.
Cedric Wannaz on 10 Oct 2013
So peaks are present in all series, i.e. load(location(1)) is present in peak1 and peak2 ?
Side note: avoid using load as a variable name, because it conflicts with function LOAD.

Jos (10584) on 10 Oct 2013
First of all, you do not want to have a variable named "load" as it clashes with the important function load Second, things become easy when you store the individual vectors in a cell array. Then a simple for-loop suffices:
MyLoad=data(:,3) % just a column vector of data
location = [1 ; location(:) ; numel(MyLoad)] ;
Nloc = numel(location) ;
MyPeak = cell(Nloc-1,1) ;
for k = 1:Nloc-1
end
Easy to debug, easy to maintain, and easy to explain ...

Cedric Wannaz on 10 Oct 2013
arrayfun actually, not accumarray.
Christopher on 10 Oct 2013
Jos, I got it working, but how do I display the results? This is just simply returning a vector with the location points?
Cedric Wannaz on 10 Oct 2013
MyPeak is a cell array of chunks of MyLoad. Note that Jos added the first and the last element positions to the list of locations (not sure of that's what you wanted, but you can easily remove them if not). Assume that your data, MyLoad is made of 100 elements, with two peaks at locations 27 and 64. Then cell 1 of MyPeak
MyPeak{1}
MyPeak{2}
MyPeak{3}
contains MyLoad(64:100). The important point about this solution is that the chuncks of data which are numeric arrays, are store in a cell array. Addressing cell arrays is done using curly brackets when you want to access cells content. Note that you can combine subs for indexing the cell array and subs for indexing the content, so, if you wanted to access for example element 10 of the first chunk, you could do it with
x = MyPeak{1} ;
y = x(10) ;
but also as follows
y = MyPeak{1}(10) ;

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