5

我在 matlab 中对多个变量进行梯度下降，并且代码没有得到我使用正常 eq 得到的预期 thetas。即：theta = 1.0e+05 * 3.4041 1.1063 -0.0665 我已经实施了。

使用 GDM，我得到的结果是：theta = 1.0e+05 * 2.6618 -2.6718 -0.5954 我不明白为什么会这样，也许有人可以帮助我并告诉我代码中的错误在哪里。

代码：

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

m = length(y); % number of training examples
J_history = zeros(num_iters, 1);
thetas = size(theta,1);
features = size(X,2)

mu = mean(X);
sigma = std(X);
mu_size = size(mu);
sigma_size = size(sigma);

%for all iterations
for iter = 1:num_iters

tempo = [];

result = [];

theta_temp = [];

%for all the thetas    
for t = 1:thetas
    %all the examples
    for examples = 1:m
       tempo(examples) = ((theta' * X(examples, :)') - y(examples)) * X(m,t)
    end

    result(t) = sum(tempo)
    tempo = 0;

end

%theta temp, store the temp 
for c = 1:thetas

    theta_temp(c) = theta(c) - alpha * (1/m) * result(c)
end

%simultaneous update
for j = 1:thetas

    theta(j) = theta_temp(j)

end

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

end

theta
end

谢谢。

编辑：数据。

  X =
    1.0000    0.1300   -0.2237
    1.0000   -0.5042   -0.2237
    1.0000    0.5025   -0.2237
    1.0000   -0.7357   -1.5378
    1.0000    1.2575    1.0904
    1.0000   -0.0197    1.0904
    1.0000   -0.5872   -0.2237
    1.0000   -0.7219   -0.2237
    1.0000   -0.7810   -0.2237
    1.0000   -0.6376   -0.2237
    1.0000   -0.0764    1.0904
    1.0000   -0.0009   -0.2237
    1.0000   -0.1393   -0.2237
    1.0000    3.1173    2.4045
    1.0000   -0.9220   -0.2237
    1.0000    0.3766    1.0904
    1.0000   -0.8565   -1.5378
    1.0000   -0.9622   -0.2237
    1.0000    0.7655    1.0904
    1.0000    1.2965    1.0904
    1.0000   -0.2940   -0.2237
    1.0000   -0.1418   -1.5378
    1.0000   -0.4992   -0.2237
    1.0000   -0.0487    1.0904
    1.0000    2.3774   -0.2237
    1.0000   -1.1334   -0.2237
    1.0000   -0.6829   -0.2237
    1.0000    0.6610   -0.2237
    1.0000    0.2508   -0.2237
    1.0000    0.8007   -0.2237
    1.0000   -0.2034   -1.5378
    1.0000   -1.2592   -2.8519
    1.0000    0.0495    1.0904
    1.0000    1.4299   -0.2237
    1.0000   -0.2387    1.0904
    1.0000   -0.7093   -0.2237
    1.0000   -0.9584   -0.2237
    1.0000    0.1652    1.0904
    1.0000    2.7864    1.0904
    1.0000    0.2030    1.0904
    1.0000   -0.4237   -1.5378
    1.0000    0.2986   -0.2237
    1.0000    0.7126    1.0904
    1.0000   -1.0075   -0.2237
    1.0000   -1.4454   -1.5378
    1.0000   -0.1871    1.0904
    1.0000   -1.0037   -0.2237

y =
      399900
      329900
      369000
      232000
      539900
      299900
      314900
      198999
      212000
      242500
      239999
      347000
      329999
      699900
      259900
      449900
      299900
      199900
      499998
      599000
      252900
      255000
      242900
      259900
      573900
      249900
      464500
      469000
      475000
      299900
      349900
      169900
      314900
      579900
      285900
      249900
      229900
      345000
      549000
      287000
      368500
      329900
      314000
      299000
      179900
      299900
      239500

完整数据集。

4

1 回答 1

8

您计算速度的线是错误的。它应该是

tempo(examples) = ((theta' * X(examples, :)') - y(examples)) * X(examples,t)

还可以尝试在 MATLAB 中使用矩阵运算。你的代码会更快，也更容易理解。例如，您可以将嵌套循环替换为

E = X * theta - y;
for t = 1:thetas
    result(t) = sum(E.*X(:,t));
end

您可以替换随后的两个循环以将 theta 更新为一行

theta = theta - alpha * (1/m) * result';
于 2013-10-21T15:53:10.983 回答