How to avoid that loop

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Alberto Menichetti
评论： Steven Lord 2016-5-3
k(1) = 1;
for i=2:size(k)
k(i) = k(i-1)*v(i)
end
v(i) is a scalar and it's different on every iteration How could I do that without using a loop?
  2 个评论
the cyclist
the cyclist 2016-5-3
编辑：the cyclist 2016-5-3
As written, this loop will never be executed, because size(k) is 1, and
for i = 2:1
<stuff>
end
will have zero iterations.
Some coding mistake? Maybe you meant length(v)?
Alberto Menichetti
Alberto Menichetti 2016-5-3
I'm sorry you're right
k = ones(size(v), 1);

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the cyclist
the cyclist 2016-5-3
If my speculation about what you meant it correct, then
k = cumprod(v)/v(1)
  2 个评论
Alberto Menichetti
Alberto Menichetti 2016-5-3
I think I didn't explain my problem very well:
k(1) = 1
I want
k(i) = k(i-1)*v(i)
for every i>1
Steven Lord
Steven Lord 2016-5-3
No, we understand you. Another approach that doesn't involve division:
v = randperm(8) % Sample data for demonstration purposes
k = cumprod([1 v(2:end)])

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