Sum of following elements in array

Is there a faster way to do the following

s = rand(1, 10);
sum_following = zeros(length(s), 1);
for i = 1:length(s)-1
    sum_following(i) = sum(s(i+1:end), 'all');
end

without doing

sum_following = sum(s,"all") - cumsum(s);

?

[Dyuman Joshi]

sum_following = sum(s,"all") - cumsum(s);

I doubt there is a faster method than this.

And as the input is a vector, you don't need to specify 'all' for the sum() call.

[Ivan Bioli]

The point is that doing this is fast but not numerically stable if the last entries of s are close to zero.

[Dyuman Joshi]

Well, the function sum() is instable.

Check out - https://in.mathworks.com/matlabcentral/fileexchange/26800-xsum

[Ivan Bioli]

I am referring to another type of instability. Try the following

s = [1, eps].^2

s = 1×2

1.0000 0.0000

sum_following = zeros(length(s), 1);
for i = 1:length(s)-1
    sum_following(i) = sum(s(i+1:end), 'all');
end
sum_following

sum_following = 2×1

1.0e-31 * 0.4930 0

sum_following_cumsum = sum(s,"all") - cumsum(s);
sum_following_cumsum

sum_following_cumsum = 1×2

0 0

[Dyuman Joshi]

Yes, that shows that sum() is not stable.

Take a look at the FEX submission I linked.

[Matt J]

Technically, this is not "instability". It's just limited precision.

[Dyuman Joshi]

How is that due to limited precision?

[Matt J]

Because

1+eps^2 == 1

ans = logical
   1

[Walter Roberson]

The point is that doing this is fast but not numerically stable if the last entries of s are close to zero.

The loop with

sum(s(i+1:end), 'all');

is not numerically stable either. The trailing suffix [... A B -B -A] is going to come out as -A if abs(A) < eps(B)

Side note: You are using linear indexing. Linear indexing of anything always gives a vector result, and for a vector being added together, 'all' as a parameter does not provide any value.

Linear indexing of row vector: gives a row vector

Linear indexing of column ector: gives a column vector

Linear indexing of non-vectors: gives a result that is the same shape as the indices. Which, for i+1:end would be a row vector.

Answer 1

[Matt J]

sum_following = cumsum([s(2:end),0],"reverse");