Defining an anonymous function with 3*(N+1) variables

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Anthony Gurunian 2021-4-8

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/796692-defining-an-anonymous-function-with-3-n-1-variables

评论： Anthony Gurunian 2021-4-10

采纳的回答： David Hill

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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?

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Cris LaPierre 2021-4-8

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?

Anthony Gurunian 2021-4-8

Yes that is correct. I'm trying to use the syntax required here: https://www.mathworks.com/matlabcentral/fileexchange/53477-monte-carlo-integration

" f is a vectorized function a la f=@(X)sin(X(:,1))+cos(X(:,2))"

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采纳的回答

David Hill 2021-4-8

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https://ww2.mathworks.cn/matlabcentral/answers/796692-defining-an-anonymous-function-with-3-n-1-variables#answer_670737

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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');

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David Hill 2021-4-9

编辑：David Hill 2021-4-9

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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