Matlab公式优化:径向基函数 [英] Matlab formula optimization: Radial Basis Function

问题描述

• z-双精度矩阵，大小为Nx2；
• x-双精度矩阵，大小为Nx2；

sup = x(i, :);

phi(1, i) = {@(z) exp(-g * sum((z - sup(ones([size(z, 1) 1]),:)) .^ 2, 2))};

这是用于逻辑回归的径向基函数(RBF).这是公式:

我需要您的建议，我可以优化此公式吗?因为它调用了数百万次，并且需要很多时间...

解决方案

似乎在您最近的编辑中，您引入了一些语法错误，但我想我了解您想要做的事情(从第一个版本开始).

而不是使用 REPMAT 或建立索引来重复向量以匹配z的行，请考虑使用高效的 BSXFUN 功能:

rbf(:,i) = exp( -g .* sum(bsxfun(@minus,z,x(i,:)).^2,2) );

上面的内容显然遍历了x的每一行

---

您可以再走一步，并使用 PDIST2 来计算z和x中每对行之间的欧几里得距离:

%# some random data
X = rand(10,2);
Z = rand(10,2);
g = 0.5;

%# one-line solution
rbf = exp(-g .* pdist2(Z,X,'euclidean').^2);

现在矩阵中的每个值:rbf(i,j)对应于z(i,:)和x(j,:)

之间的函数值

---

我为不同的方法计时，这是我使用的代码:

%# some random data
N = 5000;
X = rand(N,2);
Z = rand(N,2);
g = 0.5;

%# PDIST2
tic
rbf1 = exp(-g .* pdist2(Z,X,'euclidean').^2);
toc

%# BSXFUN+loop
tic
rbf2 = zeros(N,N);
for j=1:N
    rbf2(:,j) = exp( -g .* sum(bsxfun(@minus,Z,X(j,:)).^2,2) );
end
toc

%# REPMAT+loop
tic
rbf3 = zeros(N,N);
for j=1:N
    rbf3(:,j) = exp( -g .* sum((Z-repmat(X(j,:),[N 1])).^2,2) );
end
toc

%# check if results are equal
all( abs(rbf1(:)-rbf2(:)) < 1e-15 )
all( abs(rbf2(:)-rbf3(:)) < 1e-15 )

结果:

Elapsed time is 2.108313 seconds.     # PDIST2
Elapsed time is 1.975865 seconds.     # BSXFUN
Elapsed time is 2.706201 seconds.     # REPMAT

• z - matrix of doubles, size Nx2;
• x - matrix of doubles, size Nx2;

sup = x(i, :);

phi(1, i) = {@(z) exp(-g * sum((z - sup(ones([size(z, 1) 1]),:)) .^ 2, 2))};

this is a Radial Basis Function (RBF) for logistic regression. Here is the formula:

I need your advice, can i optimize this formula? coz it calls millions times, and it takes a lot of time...

解决方案

It seems in your recent edits, you introduced some syntax errors, but I think I understood what you were trying to do (from the first version).

Instead of using REPMAT or indexing to repeat the vector x(i,:) to match the rows of z, consider using the efficient BSXFUN function:

rbf(:,i) = exp( -g .* sum(bsxfun(@minus,z,x(i,:)).^2,2) );

The above obviously loops over every row of x

---

You can go one step further, and use the PDIST2 to compute the euclidean distance between every pair of rows in z and x:

%# some random data
X = rand(10,2);
Z = rand(10,2);
g = 0.5;

%# one-line solution
rbf = exp(-g .* pdist2(Z,X,'euclidean').^2);

Now every value in the matrix: rbf(i,j) corresponds to the function value between z(i,:) and x(j,:)

---

EDIT:

I timed the different methods, here is the code I used:

%# some random data
N = 5000;
X = rand(N,2);
Z = rand(N,2);
g = 0.5;

%# PDIST2
tic
rbf1 = exp(-g .* pdist2(Z,X,'euclidean').^2);
toc

%# BSXFUN+loop
tic
rbf2 = zeros(N,N);
for j=1:N
    rbf2(:,j) = exp( -g .* sum(bsxfun(@minus,Z,X(j,:)).^2,2) );
end
toc

%# REPMAT+loop
tic
rbf3 = zeros(N,N);
for j=1:N
    rbf3(:,j) = exp( -g .* sum((Z-repmat(X(j,:),[N 1])).^2,2) );
end
toc

%# check if results are equal
all( abs(rbf1(:)-rbf2(:)) < 1e-15 )
all( abs(rbf2(:)-rbf3(:)) < 1e-15 )

The results:

Elapsed time is 2.108313 seconds.     # PDIST2
Elapsed time is 1.975865 seconds.     # BSXFUN
Elapsed time is 2.706201 seconds.     # REPMAT

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