Take the sum of an array until a limit is reached determined by a vector of the same length

I have a scatter graph made from two arrays of equal length,

How may I take the sum of the intensities per an arbituary track length of 10mm? I know it might have something to do with cumsum, any help would be much apprieciated, cheers.

For any example maybe consider these two smaller arrays:

TL = [1,1,2,2,1,3,3,3,4,5,5];

I = [1,1,1,1,3,3,1,6,6,6,5];

Answer 1

[Turlough Hughes]

For a single result you could write the following to get the sum for an arbitrary track length of 10 mm, and you have the option of replacing 10 with any other track length you wish to find the cumulative sum for.

result = sum(I(TL<10))

You could also use cumsum as well if you sort the arrays according to TL beforehand:

[TL,index] = sort(TL);
I = I(index);
Ics = cumsum(I);

[HenT2020]

Thanks for answering but this doesn't quite solve my problem,

what I'm looking for is to take the sum of intensity every 10mm of the track length

this is the plot that I want, which I have produced by rearranging the matrix, but its only possible if the length of the array is a number that may be easily divided, which is not the case with pratically all my data sets.

This is my method for producing that plot:

m = max(TL); % Find the length of track
dx = 0:10:m; 
I = I(numel(I) - 4793:numel(I)); % remove data points to become divisible
I = reshape(I,[],numel(dx));
sumI = nansum(I);
scatter(dx,sumI)

It is very inefficient and requires removing data points which is not good.

many thanks

[Turlough Hughes]

Ok, I think what you want is to sum up Intensities where TL is between 0 and 10 and where it's between 10 and 20, 20 and 30, and so on. Is that correct?

TL(isnan(I)) = []; I(isnan(I))=[]; %remove nans
m=max(TL);
dTL = 10;   % arbitrary track length
dx = 0:dTL:m;
for c = 1:numel(dx)-1
    sumI(c) = sum(I(TL>dx(c) & TL<=dx(c+1)));
end

But then I would assume you would plot the scatter as follows:

range = dTL/2:dTL:m-dTL/2;
scatter(range,sumI)

Edit: Changed the less than operator in the for loop to less than or equal to.

[HenT2020]

Thank you very much! This solution works perfectly and allows for an easy change in track length. Thanks again for taking the time to help solve this problem

[Turlough Hughes]

Happy to help. I modified to the relational operators slightly, see above edit.