本题给出的是50个数据样本点，其中x为这50个小朋友到的年龄，年龄为2岁到8岁，年龄可有小数形式呈现。Y为这50个小朋友对应的身高，当然也是小数形式表示的。现在的问题是要根据这50个训练样本，估计出3.5岁和7岁时小孩子的身高。通过画出训练样本点的分布凭直觉可以发现这是一个典型的线性回归问题。

 

matlab函数介绍:

legend:  比如legend('Training data', 'Linear regression')，它表示的是标出图像中各曲线标志所代表的意义,这里图像的第一条曲线（其实是离散的点）表示的是训练样本数据，第二条曲线（其实是一条直线）表示的是回归曲线。

hold on, hold off: hold on指在前一幅图的情况下打开画纸，允许在上面继续画曲线。hold off指关闭前一副画的画纸。

linspace：比如linspace(-3, 3, 100)指的是给出-3到3之间的100个数，均匀的选取，即线性的选取。

logspace: 比如logspace(-2, 2, 15)，指的是在10^(-2)到10^(2)之间选取15个数，这些数按照指数大小来选取，即指数部分是均匀选取的，但是由于都取了10为底的指数，所以最终是服从指数分布选取的。

实验结果：

训练样本散点和回归曲线预测图：

损失函数与参数之间的曲面图:

损失函数的等高线图：

 

采用normal equations方法求解：

%%方法一x = load('ex2x.dat');y = load('ex2y.dat');plot(x,y,'*')xlabel('height')ylabel('age')x = [ones(size(x),1),x];w=inv(x'*x)*x'*yhold onplot(x(:,2),0.0639*x(:,2)+0.7502)%更正后的代码

 

采用gradient descend过程求解：

% Exercise 2 Linear Regression% Data is roughly based on 2000 CDC growth figures% for boys%% x refers to a boy's age% y is a boy's height in meters%clear all; close all; clcx = load('ex2x.dat'); y = load('ex2y.dat');m = length(y); % number of training examples% Plot the training datafigure; % open a new figure windowplot(x, y, 'o');ylabel('Height in meters')xlabel('Age in years')% Gradient descentx = [ones(m, 1) x]; % Add a column of ones to xtheta = zeros(size(x(1,:)))'; % initialize fitting parametersMAX_ITR = 1500; % 迭代次数alpha = 0.07; % 学习率for num_iterations = 1:MAX_ITR    % This is a vectorized version of the     % gradient descent update formula    % It's also fine to use the summation formula from the videos        % Here is the gradient    grad = (1/m).* x' * ((x * theta) - y);        % Here is the actual update    theta = theta - alpha .* grad;        % Sequential update: The wrong way to do gradient descent —— 参数的不同维度要同时更新    % grad1 = (1/m).* x(:,1)' * ((x * theta) - y);    % theta(1) = theta(1) + alpha*grad1;    % grad2 = (1/m).* x(:,2)' * ((x * theta) - y);    % theta(2) = theta(2) + alpha*grad2;end% print theta to screentheta% Plot the linear fithold on; % keep previous plot visibleplot(x(:,2), x*theta, '-')legend('Training data', 'Linear regression')%标出图像中各曲线标志所代表的意义hold off % don't overlay any more plots on this figure，指关掉前面的那幅图% Closed form solution for reference% You will learn about this method in future videosexact_theta = (x' * x)\x' * y% Predict values for age 3.5 and 7predict1 = [1, 3.5] *thetapredict2 = [1, 7] * theta% Calculate J matrix% Grid over which we will calculate Jtheta0_vals = linspace(-3, 3, 100);theta1_vals = linspace(-1, 1, 100);% initialize J_vals to a matrix of 0'sJ_vals = zeros(length(theta0_vals), length(theta1_vals));for i = 1:length(theta0_vals)      for j = 1:length(theta1_vals)      t = [theta0_vals(i); theta1_vals(j)];          J_vals(i,j) = (0.5/m) .* (x * t - y)' * (x * t - y);    endend% Because of the way meshgrids work in the surf command, we need to % transpose J_vals before calling surf, or else the axes will be flippedJ_vals = J_vals';% Surface plotfigure;surf(theta0_vals, theta1_vals, J_vals)xlabel('\theta_0'); ylabel('\theta_1');% Contour plotfigure;% Plot J_vals as 15 contours spaced logarithmically between 0.01 and 100contour(theta0_vals, theta1_vals, J_vals, logspace(-2, 2, 15))%画出等高线xlabel('\theta_0'); ylabel('\theta_1');%类似于转义字符，但是最多只能是到参数0~9