cumsum reset back to zero every 100 times
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Chris Matthews
am 11 Aug. 2017
Bearbeitet: Andrei Bobrov
am 12 Aug. 2017
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
2 Kommentare
Walter Roberson
am 11 Aug. 2017
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
am 11 Aug. 2017
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);
Akzeptierte Antwort
José-Luis
am 11 Aug. 2017
Bearbeitet: José-Luis
am 11 Aug. 2017
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);
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Weitere Antworten (3)
Jan
am 11 Aug. 2017
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
am 11 Aug. 2017
Bearbeitet: Andrei Bobrov
am 12 Aug. 2017
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);
2 Kommentare
Jan
am 11 Aug. 2017
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
am 12 Aug. 2017
Bearbeitet: Andrei Bobrov
am 12 Aug. 2017
Thanks Jan for your comment!
I am fixed my answer.
Walter Roberson
am 11 Aug. 2017
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()
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