# How can I create a vector that shows the sum of the 60 values before each value of another vector?

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Philippe Corner on 13 Jan 2017
Commented: Walter Roberson on 13 Jan 2017
If a have this vector --> x=[0:1:1000] how can i create a vector of length 1000 too.. that is: first value 0, second 0+1.. value #59 from 0 to 59 and when the value is greater than 60 it shows the sum of the 60 values before for example the entry 100 should be from 40 to 100 etc..
Walter Roberson on 13 Jan 2017
"60 values before" implies that 60 values are to be added. 40 to 100 is 61 values.

Walter Roberson on 13 Jan 2017
Or you can cumsum() and subtract the value 60-ish earlier in the cumsum
Walter Roberson on 13 Jan 2017
t = cumsum(zeros(1,60),x);
t(61:end) - t(1:end-60)

Star Strider on 13 Jan 2017
I’m not certain a vector is possible, but a matrix approach could be.
Consider the rows of ‘out’:
N = 5; % RowLength (Use: 60)
L = 10; % Vector Length (Use: 1000)
out = hankel(1:N, N:L);
out(2:end,:) = cumsum(out(2:end,:),2);
I used the example with a short vector to illustrate the approach. Experiment to get the result you want.

Image Analyst on 13 Jan 2017
Here's a way to do it in one line of code with conv().
out = conv(x, [zeros(1,59), ones(1,60)], 'same')
To prove it, here's a full demo where I compare that method to a simple intuitive for loop to prove that it's the same.
x=[0:1:1000]
kernel = [zeros(1,59), ones(1,60)]
out = conv(x, kernel, 'same')
% Compare with for loop method:
for k = 1 : length(x)
i2 = k;
i1 = max(1, i2 - 59);
outf(k) = sum(x(i1:i2));
end
diffs = outf - out
maxDiff = max(diffs) % Should be 0