Defining an anonymous function with 3*(N+1) variables
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I need to define a function
f = @(X)...
where X has 3*(N+1) variables in the form: X(:,1), X(:,2), ... X(:,3*(N+1)) . The problem is that the expression for the function is very long. However, there is a pattern which I want to take advantage of.
The function is
Notice that the function can be expressed in terms of summations and products. I want to take advantage of this fact so that I don't have to manually type out the products and sums. Is there any way of doing this using for loops?
4 件のコメント
Cris LaPierre
2021 年 4 月 8 日
Perhaps I'm getting hung up on symantics here, but it appears you have a single variable, X. Is it correct to say that your array X has 3*(N+1) columns?
採用された回答
David Hill
2021 年 4 月 8 日
f=prod(exp(-beta*(diff(x(1:n+1)).^2+diff(x(n+2:2*n+2)).^2+diff(x(2*n+3:3*n+3)).^2)))*sum((x(1:n+1)-x(1:n+1)').^2+(x(n+2:2*n+2)-x(n+2:2*n+2)').^2+(x(2*n+3:3*n+3)-x(2*n+3:3*n+3)).^2,'all');
7 件のコメント
David Hill
2021 年 4 月 9 日
編集済み: David Hill
2021 年 4 月 9 日
Is there a different f for each row? If not, I don't understand your equation. If so, then just do a loop with my equation. How can you not know the number of rows? X must be defined before this equation is processed.
for k=1:size(x,1)
f(k)=prod(exp(-beta*(diff(x(k,1:n+1)).^2+diff(x(k,n+2:2*n+2)).^2+diff(k,x(2*n+3:3*n+3)).^2)))*...
sum((x(k,1:n+1)-x(k,1:n+1)').^2+(x(k,n+2:2*n+2)-x(k,n+2:2*n+2)').^2+...
(x(k,2*n+3:3*n+3)-x(k,2*n+3:3*n+3)).^2,'all');
end
Or,
for k=1:size(x,1)
X=reshape(x(k,:),[],3);
f(k)=prod(exp(-beta*(sum(diff(X,1).^2),2)))*sum((X(:,1)-X(:,1)').^2+...
(X(:,2)-X(:,2)').^2+(X(:,3)-X(:,3)').^2,'all');
end
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