cumsum reset back to zero every 100 times

I have a matrix R = [1x601] size. I want to use cumsum to get an accumulative value but reset cumsum back to zero each 100 positions.

Then I want to plot the result, so there should be 6 times where the plot goes back to zero on the y axis.

Is this possible? Please let me know if you don't understand what I'm trying to achieve, and thanks!! Have a good day, Chris

[Walter Roberson]

It is positions corresponding to R(100:100:end) that should be set to 0? Is it positions corresponding to R(1:100:end) that should be set to 0? Is it positions corresponding to R(101:100:end) that should be set to 0? Your vector is length 601, so if you choose 100:100:end then position corresponding to 600 would be set to 0, leaving position corresponding to 601 to accumulate to its own value: is that what you would like?

[Akira Agata]

The following code works. I would recommend checking your data and/or code again, or upload your data here in .mat file.

% 1x601 sample numeric array
R = rand(1,601);
output = cumsum(R);

Answer 1

[José-Luis]

I interpreted "reset cumsum to zero" as restart cumsum from scratch.

 data = rand(1,601);
 dummy = zeros(100,ceil(size(data,2)./100));
 dummy(1:numel(data)) = data;
 dummy = cumsum(dummy,1);
 plot(data); hold on
 data = dummy(1:numel(data));
 plot(data);

[Jan]

+1: Clean and efficient.

[José-Luis]

Thanks.

Answer 2

[Jan]

s = cumsum(x);
d = [0, s(101:100:numel(x))];
s = s - d(ceil((1:numel(x)) / 100));

Now s(101) is 0, which might match your "reset to zero", but usually there is no zero in a cumulative sum of positive non-zero elements. See Walter's question for clarification.

[Andrei Bobrov]

+1

Answer 3

[Andrei Bobrov]

with "zeros"

m = 100;
n = numel(R);
R1 = R(:);
R1(m+1:m:n) = 0;
d = accumarray(ceil((1:n)'/m),R1);
R1(m+1:m:n) = -d(1:end-1);
out = cumsum(R1);

without "zeros"

m = 100;
n = numel(R);
R1 = R(:);
d = accumarray(ceil((1:n)'/m),R1);
R1(m+1:m:n) = R1(m+1:m:n) - d(1:end-1);
out = cumsum(R1);

[Jan]

I get an error cause by the auto-expanding in

R1(101:100:numel(R)) = R1(101:100:numel(R)) - d(1:end-1);

The right size is a matrix. d(1:end-1).' fixes the problem. But then neither out(100) not out(101) is zero. This is logical for a cumulative sum, but the author asked for "reset to zero".

[Andrei Bobrov]

Thanks Jan for your comment!

I am fixed my answer.

Answer 4

[Walter Roberson]

If you have the signal processing toolbox, use buffer() to put the signal into columns of appropriate length, zero or delete the appropriate row, cumsum()