在matlab中規范化邏輯回歸代碼

Question

我正在嘗試使用正規化的LR，在matlab中使用這個公式很簡單：

成本函數：

J(theta) = 1/m*sum((-y_i)*log(h(x_i)-(1-y_i)*log(1-h(x_i))))+(lambda/2*m)*sum(theta_j)

漸變：

∂J(theta)/∂theta_0 = [(1/m)*(sum((h(x_i)-y_i)*x_j)] if j=0

∂j(theta)/∂theta_n = [(1/m)*(sum((h(x_i)-y_i)*x_j)]+(lambda/m)*(theta_j) if j>1

這不是matlab代碼只是公式。

到目前為止我已經這樣做了：

function [J, grad] = costFunctionReg(theta, X, y, lambda)

J = 0;
grad = zeros(size(theta));

temp_theta = [];

%cost function

%get the regularization term

for jj = 2:length(theta)

    temp_theta(jj) = theta(jj)^2;
end

theta_reg = lambda/(2*m)*sum(temp_theta);

temp_sum =[];

%for the sum in the cost function

for ii =1:m

   temp_sum(ii) = -y(ii)*log(sigmoid(theta'*X(ii,:)'))-(1-y(ii))*log(1-sigmoid(theta'*X(ii,:)'));

end

tempo = sum(temp_sum);

J = (1/m)*tempo+theta_reg;

%regulatization
%theta 0

reg_theta0 = 0;

for jj=1:m
 reg_theta0(jj) = (sigmoid(theta'*X(m,:)') -y(jj))*X(jj,1)
end    

reg_theta0 = (1/m)*sum(reg_theta0)

grad_temp(1) = reg_theta0

%for the rest of thetas

reg_theta  = [];
thetas_sum = 0;

for ii=2:size(theta)
    for kk =1:m
        reg_theta(kk) = (sigmoid(theta'*X(m,:)') - y(kk))*X(kk,ii)
    end
    thetas_sum(ii) = (1/m)*sum(reg_theta)+(lambda/m)*theta(ii)
    reg_theta = []
end

for i=1:size(theta)

    if i == 1
        grad(i) = grad_temp(i)
    else
        grad(i) = thetas_sum(i)
    end
end
end

成本函數給出正確的結果，但我不知道為什么梯度（一步）不是，成本給出J = 0.6931這是正確的，梯度grad = 0.3603 -0.1476 0.0320，這不是，成本開始從2開始，因為參數theta（1）不必正則化，有什么幫助嗎？ 我猜這個代碼有問題，但4天后我看不到它。謝謝

Answer 1

矢量：

function [J, grad] = costFunctionReg(theta, X, y, lambda)

hx = sigmoid(X * theta);
m = length(X);

J = (sum(-y' * log(hx) - (1 - y')*log(1 - hx)) / m) + lambda * sum(theta(2:end).^2) / (2*m);
grad =((hx - y)' * X / m)' + lambda .* theta .* [0; ones(length(theta)-1, 1)] ./ m ;

end

Answer 2

我使用了更多變量，因此您可以清楚地看到常規公式的內容，以及“增加的正則化成本”。 另外，在Matlab / Octave中使用“矢量化”代替循環是一種很好的做法。 通過這樣做，您可以保證更優化的解決方案。

 function [J, grad] = costFunctionReg(theta, X, y, lambda)

    %Hypotheses
    hx = sigmoid(X * theta);

    %%The cost without regularization
    J_partial = (-y' * log(hx) - (1 - y)' * log(1 - hx)) ./ m;


    %%Regularization Cost Added
    J_regularization = (lambda/(2*m)) * sum(theta(2:end).^2);

    %%Cost when we add regularization
    J = J_partial + J_regularization;

    %Grad without regularization
    grad_partial = (1/m) * (X' * (hx -y));

    %%Grad Cost Added
    grad_regularization = (lambda/m) .* theta(2:end);

    grad_regularization = [0; grad_regularization];

    grad = grad_partial + grad_regularization;

Answer 3

最后得到它，在第四次重寫之后，這是正確的代碼：

function [J, grad] = costFunctionReg(theta, X, y, lambda)
J = 0;
grad = zeros(size(theta));

temp_theta = [];

for jj = 2:length(theta)

    temp_theta(jj) = theta(jj)^2;
end

theta_reg = lambda/(2*m)*sum(temp_theta);

temp_sum =[];

for ii =1:m

   temp_sum(ii) = -y(ii)*log(sigmoid(theta'*X(ii,:)'))-(1-y(ii))*log(1-sigmoid(theta'*X(ii,:)'));

end

tempo = sum(temp_sum);

J = (1/m)*tempo+theta_reg;

%regulatization
%theta 0

reg_theta0 = 0;

for i=1:m
    reg_theta0(i) = ((sigmoid(theta'*X(i,:)'))-y(i))*X(i,1)
end

theta_temp(1) = (1/m)*sum(reg_theta0)

grad(1) = theta_temp

sum_thetas = []
thetas_sum = []

for j = 2:size(theta)
    for i = 1:m

        sum_thetas(i) = ((sigmoid(theta'*X(i,:)'))-y(i))*X(i,j)
    end

    thetas_sum(j) = (1/m)*sum(sum_thetas)+((lambda/m)*theta(j))
    sum_thetas = []
end

for z=2:size(theta)
    grad(z) = thetas_sum(z)
end


% =============================================================

end

如果它可以幫助任何人，或者任何人對我如何做得更好有任何意見。 :)

Answer 4

這是一個消除循環的答案

m = length(y); % number of training examples

predictions = sigmoid(X*theta);
reg_term = (lambda/(2*m)) * sum(theta(2:end).^2);
calcErrors = -y.*log(predictions) - (1 -y).*log(1-predictions);
J = (1/m)*sum(calcErrors)+reg_term;

% prepend a 0 column to our reg_term matrix so we can use simple matrix addition
reg_term = [0 (lambda*theta(2:end)/m)'];
grad = sum(X.*(predictions - y)) / m + reg_term;