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Sum of following elements in array

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Ivan Bioli
Ivan Bioli 2023-12-9
评论： Walter Roberson 2023-12-9
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);
?
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Walter Roberson
Walter Roberson 2023-12-9
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.

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采纳的回答

Matt J
Matt J 2023-12-9
编辑：Matt J 2023-12-9
sum_following = cumsum([s(2:end),0],"reverse");

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