matlab中的多变量梯度下降

时间:2013-10-20 21:04:53

标签: matlab machine-learning gradient-descent

我在matlab中为多个变量做渐变下降,而且代码没有得到我用正常eq得到的预期值。那是: theta =    1.0e + 05 *     3.4041     1.1063    -0.0665 使用Normal eq。我已经实施了。

通过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

感谢。

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。

1 个答案:

答案 0 :(得分:9)

计算速度的行是错误的。它应该是

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

还尝试在MATLAB中使用矩阵运算。您的代码会更快,也更容易理解。例如,您可以用

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

您可以替换后续的两个循环以将theta更新为一行

theta = theta - alpha * (1/m) * result';