在Matlab中梯度下降的結果不正確

Question

我在Matlab中學習了這個課程，並且我已經完成了梯度下降實現，但它給出了不正確的結果。

代碼：

for iter = 1:num_iters 

sumTheta1 = 0; 
sumTheta2 = 0; 
for s = 1:m 
    sumTheta1 = theta(1) + theta(2) .* X(s,2) - y(s); 
    sumTheta2 = theta(1) + theta(2) .* X(s,2) - y(s) .* X(s,2); 
end 

theta(1) = theta(1) - alpha .* (1/m) .* sumTheta1; 
theta(2) = theta(2) - alpha .* (1/m) .* sumTheta2; 

J_history(iter) = computeCost(X, y, theta); 

end 

這是重要的組成部分。我認爲公式的實現是正確的，即使它沒有被優化。公式是：

theta1 = theta1 - (alpha)(1/m)(summation_i^m(theta1 + theta2*x(i)-y(i))) 
theta2 = theta2 - (alpha)(1/m)(summation_i^m(theta1 + theta2*x(i)-y(i)))(x(i)) 

那麼問題出在哪裏呢？

編輯：CODE更新

function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters) 

m = length(y); % number of training examples 
J_history = zeros(num_iters, 1); 


for iter = 1:num_iters 

for s = 1:m 

sumTheta1 = ((theta(1) .* X(s,1)) + (theta(2) .* X(s,2))) - (y(s)); 
sumTheta2 = ((theta(1) .* X(s,1)) + (theta(2) .* X(s,2))) - (y(s)) .* X(s,2); 
end 

temp1 = theta(1) - alpha .* (1/m) .* sumTheta1; 
temp2 = theta(2) - alpha .* (1/m) .* sumTheta2; 

theta(1) = temp1; 
theta(2) = temp2; 

J_history(iter) = computeCost(X, y, theta); 

end 

end 

EDIT（2）：固定它，工作碼。

明白了，這是+丹的暗示，做到了這一點，我會接受他的答案，仍然把代碼放在這裏給任何卡住:)，歡呼聲。

function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters) 

m = length(y); % number of training examples 
J_history = zeros(num_iters, 1); 


for iter = 1:num_iters 

sumTheta1 = 0; 
sumTheta2 = 0; 

for s = 1:m 

sumTheta1 = sumTheta1 + ((theta(1) .* X(s,1)) + (theta(2) .* X(s,2))) - (y(s)); 
sumTheta2 = sumTheta2 + (((theta(1) .* X(s,1)) + (theta(2) .* X(s,2))) - (y(s))) .* X(s,2); 
end 

temp1 = theta(1) - alpha .* (1/m) .* sumTheta1; 
temp2 = theta(2) - alpha .* (1/m) .* sumTheta2; 

theta(1) = temp1; 
theta(2) = temp2; 

% Save the cost J in every iteration  
J_history(iter) = computeCost(X, y, theta); 

end 

end 

Answer 1

乍一看我注意到你的sumTheta1實際上不是相加而是更換自己每次迭代。我想你的意思是：

sumTheta1 = sumTheta1 + theta(1) + theta(2) .* X(s,2) - y(s); 

與同爲sumTheta2

但對於未來的參考，你可以更換這（校正）循環：這個量化的公式

sumTheta1 = sum(theta(1) + theta(2)*X(:,2) - y); 
sumTheta2 = sum(theta(1) + theta(2)*X(:,2) - y.*X(:,2)) 

for s = 1:m 
    sumTheta1 = theta(1) + theta(2) .* X(s,2) - y(s); 
    sumTheta2 = theta(1) + theta(2) .* X(s,2) - y(s) .* X(s,2); 
end 

Answer 2

如果我看到這個公式

theta1 = theta1 - (alpha)(1/m)(summation_i^m(theta1 + theta2*x(i)-y(i))) 

我猜的MATLAB相當於將是：

theta1 = theta1 - alpha/m*(theta1 + theta2)*sum(x-y) 

也許你能確定m如下：

m =length(x); 

不過，你的兩個公式讓我懷疑你是否想計算他們順序或同時進行。

在第二種情況下，創建一個臨時變量並在計算中使用它。

myFactor = alpha/m*(theta1_previous + theta2_previous) 

theta1 = theta1_previous - myFactor*sum(x-y) 
theta2 = theta2_previous - myFactor*sum((x-y).*x) 

Answer 3

矢量化版本：

for iter = 1:num_iters 
    theta = theta - (alpha .* X'*(X * theta - y) ./m); 
    J_history(iter) = computeCost(X, y, theta); 
end