二元線性迴歸

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http://openclassroom.stanford.edu/MainFolder/DocumentPage.php?course=DeepLearning&doc=exercises/ex2/ex2.html

本題給出的是50個資料樣本點，其中x為這50個小朋友到的年齡，年齡為2歲到8歲，年齡可有小數形式呈現。Y為這50個小朋友對應的身高，當然也是小數形式表示的。現在的問題是要根據這50個訓練樣本，估計出3.5歲和7歲時小孩子的身高。通過畫出訓練樣本點的分布憑直覺可以發現這是一個典型的線性迴歸問題。

 

matlab函數介紹:

legend:  比如legend(‘Training data‘, ‘Linear regression‘)，它表示的是標出映像中各曲線標誌所代表的意義,這裡映像的第一條曲線（其實是離散的點）表示的是訓練樣本資料，第二條曲線（其實是一條直線）表示的是迴歸曲線。

hold on, hold off: hold on指在前一幅圖的情況下開啟畫紙，允許在上面繼續畫曲線。hold off指關閉前一副畫的畫紙。

linspace：比如linspace(-3, 3, 100)指的是給出-3到3之間的100個數，均勻的選取，即線性選取。

logspace: 比如logspace(-2, 2, 15)，指的是在10^(-2)到10^(2)之間選取15個數，這些數按照指數大小來選取，即指數部分是均勻選取的，但是由雩都取了10為底的指數，所以最終是服從指數分布選取的。

實驗結果：

訓練樣本散點和迴歸曲線預測圖：

損失函數與參數之間的曲面圖:

損失函數的等高線圖：

 

採用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)%更正後的代碼

 

採用gradient descend過程求解：

% 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; % 迭代次數alpha = 0.07; % 學習率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 —— 參數的不同維度要同時更新    % 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‘)%標出映像中各曲線標誌所代表的意義hold off % don‘t overlay any more plots on this figure，指關掉前面的那幅圖% 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))%畫出等高線xlabel(‘\theta_0‘); ylabel(‘\theta_1‘);%類似於逸出字元，但是最多隻能是到參數0~9

二元線性迴歸