Gradient Descent Matlab实现

Question

我在堆栈溢出中经历了很多代码并在同一行上创建了自己的代码.这段代码有些问题我无法理解.我存储值theta1和theta 2以及成本函数用于分析目的.可以从此Openclassroom页面下载x和Y的数据 .它具有.dat文件形式的x和Y数据,您可以在记事本中打开它们.

    %Single Variate Gradient Descent Algorithm%%
    clc
clear all
close all;
% Step 1 Load x series/ Input data and Output data* y series

x=load('D:\Office Docs_Jay\software\ex2x.dat');
y=load('D:\Office Docs_Jay\software\ex2y.dat');
%Plot the input vectors
plot(x,y,'o');
ylabel('Height in meters');
xlabel('Age in years');

% Step 2 Add an extra column of ones in input vector
[m n]=size(x);
X=[ones(m,1) x];%Concatenate the ones column with x;
% Step 3 Create Theta vector
theta=zeros(n+1,1);%theta 0,1
% Create temporary values for storing summation

temp1=0;
temp2=0;
% Define Learning Rate alpha and Max Iterations

alpha=0.07;
max_iterations=1;
      % Step 4 Iterate over loop
      for i=1:1:max_iterations

     %Calculate Hypothesis for all training example
     for k=1:1:m
        h(k)=theta(1,1)+theta(2,1)*X(k,2); %#ok<AGROW>
        temp1=temp1+(h(k)-y(k));
        temp2=temp2+(h(k)-y(k))*X(k,2);
     end
     % Simultaneous Update
      tmp1=theta(1,1)-(alpha*1/(2*m)*temp1);
      tmp2=theta(2,1)-(alpha*(1/(2*m))*temp2);
      theta(1,1)=tmp1;
      theta(2,1)=tmp2;
      theta1_history(i)=theta(2,1); %#ok<AGROW>
      theta0_history(i)=theta(1,1); %#ok<AGROW>
      % Step 5 Calculate cost function
      tmp3=0;
      tmp4=0;
      for p=1:m
        tmp3=tmp3+theta(1,1)+theta(2,1)*X(p,1);
        tmp4=tmp4+theta(1,1)+theta(2,1)*X(p,2);
      end
      J1_theta0(i)=tmp3*(1/(2*m)); %#ok<AGROW>
      J2_theta1(i)=tmp4*(1/(2*m)); %#ok<AGROW>


      end
      theta
      hold on;
      plot(X(:,2),theta(1,1)+theta(2,1)*X);

我正在获得价值

θ为0.0373和0.1900,应为0.0745和0.3800

这个值大约是我期待的两倍.

Answer 1

我一直在尝试用矩阵和向量来实现迭代步骤(即不更新theta的每个参数).这是我想出的(这里只有渐变步骤):

h = X * theta;  # hypothesis
err = h - y;    # error
gradient = alpha * (1 / m) * (X' * err); # update the gradient
theta = theta - gradient;

难以掌握的是前面例子的梯度步骤中的"和"实际上是通过矩阵乘法来执行的X'*err.你也可以把它写成(err'*X)'

Answer 2

我设法创建了一个使用Matlab支持的更多矢量化属性的算法.我的算法与你的算法略有不同,但是你提出的是梯度下降过程.在我执行的执行和验证(使用polyfit函数)之后,我认为在1500次迭代步骤之后,变量theta(0)= 0.0745和theta(1)= 0.3800中预期的openclassroom(练习2)中的值是错误的0.07(我不回应).这就是为什么我用一个图中的数据绘制我的结果,而另一个图中的数据绘制了所需的结果,我发现数据拟合程序有很大差异.

首先看一下代码:

% Machine Learning : Linear Regression

clear all; close all; clc;

%% ======================= Plotting Training Data =======================
fprintf('Plotting Data ...\n')

x = load('ex2x.dat');
y = load('ex2y.dat');

% Plot Data
plot(x,y,'rx');
xlabel('X -> Input') % x-axis label
ylabel('Y -> Output') % y-axis label

%% =================== Initialize Linear regression parameters ===================
 m = length(y); % number of training examples

% initialize fitting parameters - all zeros
theta=zeros(2,1);%theta 0,1

% Some gradient descent settings
iterations = 1500;
Learning_step_a = 0.07; % step parameter

%% =================== Gradient descent ===================

fprintf('Running Gradient Descent ...\n')

%Compute Gradient descent

% Initialize Objective Function History
J_history = zeros(iterations, 1);

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

% run gradient descent    
for iter = 1:iterations

   % In every iteration calculate hypothesis
   hypothesis=theta(1).*x+theta(2);

   % Update theta variables
   temp0=theta(1) - Learning_step_a * (1/m)* sum((hypothesis-y).* x);
   temp1=theta(2) - Learning_step_a * (1/m) *sum(hypothesis-y);

   theta(1)=temp0;
   theta(2)=temp1;

   % Save objective function 
   J_history(iter)=(1/2*m)*sum(( hypothesis-y ).^2);

end

% print theta to screen
fprintf('Theta found by gradient descent: %f %f\n',theta(1),  theta(2));
fprintf('Minimum of objective function is %f \n',J_history(iterations));

% Plot the linear fit
hold on; % keep previous plot visible 
plot(x, theta(1)*x+theta(2), '-')

% Validate with polyfit fnc
poly_theta = polyfit(x,y,1);
plot(x, poly_theta(1)*x+poly_theta(2), 'y--');
legend('Training data', 'Linear regression','Linear regression with polyfit')
hold off 

figure
% Plot Data
plot(x,y,'rx');
xlabel('X -> Input') % x-axis label
ylabel('Y -> Output') % y-axis label

hold on; % keep previous plot visible
% Validate with polyfit fnc
poly_theta = polyfit(x,y,1);
plot(x, poly_theta(1)*x+poly_theta(2), 'y--');

% for theta values that you are saying
theta(1)=0.0745;  theta(2)=0.3800;
plot(x, theta(1)*x+theta(2), 'g--')
legend('Training data', 'Linear regression with polyfit','Your thetas')
hold off 

好的结果如下:

使用由我的算法产生的theta(0)和theta(1),该行适合数据.

以theta(0)和theta(1)作为固定值,结果该行不适合数据.