cumsum reset back to zero every 100 times

5 次查看（过去 30 天）
Chris Matthews
Chris Matthews 2017-8-11
编辑： Andrei Bobrov 2017-8-12
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 个评论
Walter Roberson
Walter Roberson 2017-8-11
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
Akira Agata 2017-8-11
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);

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

José-Luis
José-Luis 2017-8-11
编辑：José-Luis 2017-8-11
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);

更多回答（3 个）

Jan
Jan 2017-8-11
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
Andrei Bobrov 2017-8-11
编辑：Andrei Bobrov 2017-8-12
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 个评论
Jan
Jan 2017-8-11
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
Andrei Bobrov 2017-8-12
编辑：Andrei Bobrov 2017-8-12
Thanks Jan for your comment!
I am fixed my answer.

请先登录，再进行评论。

Walter Roberson
Walter Roberson 2017-8-11
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()

类别

在 Help Center 和 File Exchange 中查找有关 Multirate Signal Processing 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by