different answers for implementing summation
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im trying to implement summation in the following 2 ways:
1.
f1=[10 20 30 40 50]
x1=[1 2 3 4 5]
J=0
for i=1:5
J=J+((f1(i)-a*exp(-(x1(i)-mu)^2/sigma))^2)
end
and 2.
f1=[10 20 30 40 50]
x1=[1 2 3 4 5]
J=0
J=@(f,x) ((f-a*exp(-(x-mu)^2/sigma))^2)
for i=1:5
J(f1(i),x1(i))
end
and im getting different final answers for each.
can anyone tell why?
采纳的回答
Guillaume
2015-6-22
Really, the best way of implementing your summation is option 3 which uses vectorised operations:
f1=[10 20 30 40 50]
x1=[1 2 3 4 5]
J = sum((f1 - a*exp(-(x1 - mu).^2 / sigma)) .^ 2)
Your option 2 looks like it wants to use J to store the result as in option 1 (since it has the line J = 0), but then put a function in J on the following line. In the loop you invoke the function but never assign the result to anything. I'm not sure what you expected to happen with that code. If you want to use an anonymous function, you could write your option 2 as:
func = @(f,x) (f-a*exp(-(x-mu)^2/sigma))^2;
J = 0;
for idx = 1 : numel(f1) %don't hardcode bounds, use numel to get the number of elements
J = J + func(f1(idx), x1(idx));
end
But again, vectorised code is better:
func = @(f,x) (f-a*exp(-(x-mu).^2/sigma)).^2; %note the use of .^ instead of ^
J = sum(func(f1, x1))
更多回答（1 个）
Andrei Bobrov
2015-6-22
编辑：Andrei Bobrov
2015-6-22
J = sum(f1-a*exp(-(x1-mu).^2/sigma)).^2)
for 2 variant:
f1=[10 20 30 40 50]
x1=[1 2 3 4 5]
J1=0
J=@(f,x) ((f-a*exp(-(x-mu)^2/sigma))^2)
for ii=1:5
J1 = J1 + J(f1(ii),x1(ii))
end
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