吴恩达 coursera ML 第九课总结+作业答案
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吴恩达 coursera ML 第九课总结+作业答案
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前言
吳恩達的課程堪稱經典,有必要總結一下。
學以致用,以學促用,通過筆記總結,鞏固學習成果,復習新學的概念。
目錄
文章目錄
- 前言
- 目錄
- 正文
- 問題判斷是方差還是誤差
- 正則化以及方差和偏差的關系
- 學習曲線
- 作業答案
- ex5.m
- linearRegCostFunction.m
- learningCurve.m
- validationCurve.m
正文
本節主要介紹在機器學習系統設計的實踐過程中如何進行調節。
在決定采取措施前,首先必須判斷問題的癥結。
機器學習診斷,判斷一項措施可以在多大程度上提升你的性能。
首先評估你的假設。
將你手上的數據集劃分成訓練集和測試集。
通過訓練集選出最優的參數,通過測試集驗證模型的泛化性能。
邏輯回歸時的相應手續。
過擬合問題,模型總是會更加符合訓練集表現出來的樣子。
測試不同的超參數,選擇適宜的模型。
將數據集劃分為,訓練集,交叉驗證集和測試集。
選擇測試集表現最好的模型。
測試集上的表現近似為泛化誤差。
問題判斷是方差還是誤差
欠擬合,剛好,以及過擬合的情況。
誤差函數的表現以及訓練的情況。
根據誤差圖,判斷問題情況的展示。
正則化以及方差和偏差的關系
正則化對于模型擬合情況的影響。
選擇合適的正則化參數。
在對λ\lambdaλ進行測試的時候,選擇方法。
偏差和方差是正則化參數的函數。
學習曲線
高偏差時的學習曲線
高方差時的學習曲線。
作業答案
僅作參考,切勿抄襲。
ex5.m
%% Machine Learning Online Class % Exercise 5 | Regularized Linear Regression and Bias-Variance % % Instructions % ------------ % % This file contains code that helps you get started on the % exercise. You will need to complete the following functions: % % linearRegCostFunction.m % learningCurve.m % validationCurve.m % % For this exercise, you will not need to change any code in this file, % or any other files other than those mentioned above. %%% Initialization clear ; close all; clc%% =========== Part 1: Loading and Visualizing Data ============= % We start the exercise by first loading and visualizing the dataset. % The following code will load the dataset into your environment and plot % the data. %% Load Training Data fprintf('Loading and Visualizing Data ...\n')% Load from ex5data1: % You will have X, y, Xval, yval, Xtest, ytest in your environment load ('ex5data1.mat');% m = Number of examples m = size(X, 1);% Plot training data plot(X, y, 'rx', 'MarkerSize', 10, 'LineWidth', 1.5); xlabel('Change in water level (x)'); ylabel('Water flowing out of the dam (y)');fprintf('Program paused. Press enter to continue.\n'); pause;%% =========== Part 2: Regularized Linear Regression Cost ============= % You should now implement the cost function for regularized linear % regression. %theta = [1 ; 1]; J = linearRegCostFunction([ones(m, 1) X], y, theta, 1);fprintf(['Cost at theta = [1 ; 1]: %f '...'\n(this value should be about 303.993192)\n'], J);fprintf('Program paused. Press enter to continue.\n'); pause;%% =========== Part 3: Regularized Linear Regression Gradient ============= % You should now implement the gradient for regularized linear % regression. %theta = [1 ; 1]; [J, grad] = linearRegCostFunction([ones(m, 1) X], y, theta, 1);fprintf(['Gradient at theta = [1 ; 1]: [%f; %f] '...'\n(this value should be about [-15.303016; 598.250744])\n'], ...grad(1), grad(2));fprintf('Program paused. Press enter to continue.\n'); pause;%% =========== Part 4: Train Linear Regression ============= % Once you have implemented the cost and gradient correctly, the % trainLinearReg function will use your cost function to train % regularized linear regression. % % Write Up Note: The data is non-linear, so this will not give a great % fit. %% Train linear regression with lambda = 0 lambda = 0; [theta] = trainLinearReg([ones(m, 1) X], y, lambda);% Plot fit over the data plot(X, y, 'rx', 'MarkerSize', 10, 'LineWidth', 1.5); xlabel('Change in water level (x)'); ylabel('Water flowing out of the dam (y)'); hold on; plot(X, [ones(m, 1) X]*theta, '--', 'LineWidth', 2) hold off;fprintf('Program paused. Press enter to continue.\n'); pause;%% =========== Part 5: Learning Curve for Linear Regression ============= % Next, you should implement the learningCurve function. % % Write Up Note: Since the model is underfitting the data, we expect to % see a graph with "high bias" -- Figure 3 in ex5.pdf %lambda = 0; [error_train, error_val] = ...learningCurve([ones(m, 1) X], y, ...[ones(size(Xval, 1), 1) Xval], yval, ...lambda);plot(1:m, error_train, 1:m, error_val); title('Learning curve for linear regression') legend('Train', 'Cross Validation') xlabel('Number of training examples') ylabel('Error') axis([0 13 0 150])fprintf('# Training Examples\tTrain Error\tCross Validation Error\n'); for i = 1:mfprintf(' \t%d\t\t%f\t%f\n', i, error_train(i), error_val(i)); endfprintf('Program paused. Press enter to continue.\n'); pause;%% =========== Part 6: Feature Mapping for Polynomial Regression ============= % One solution to this is to use polynomial regression. You should now % complete polyFeatures to map each example into its powers %p = 8;% Map X onto Polynomial Features and Normalize X_poly = polyFeatures(X, p); [X_poly, mu, sigma] = featureNormalize(X_poly); % Normalize X_poly = [ones(m, 1), X_poly]; % Add Ones% Map X_poly_test and normalize (using mu and sigma) X_poly_test = polyFeatures(Xtest, p); X_poly_test = bsxfun(@minus, X_poly_test, mu); X_poly_test = bsxfun(@rdivide, X_poly_test, sigma); X_poly_test = [ones(size(X_poly_test, 1), 1), X_poly_test]; % Add Ones% Map X_poly_val and normalize (using mu and sigma) X_poly_val = polyFeatures(Xval, p); X_poly_val = bsxfun(@minus, X_poly_val, mu); X_poly_val = bsxfun(@rdivide, X_poly_val, sigma); X_poly_val = [ones(size(X_poly_val, 1), 1), X_poly_val]; % Add Onesfprintf('Normalized Training Example 1:\n'); fprintf(' %f \n', X_poly(1, :));fprintf('\nProgram paused. Press enter to continue.\n'); pause;%% =========== Part 7: Learning Curve for Polynomial Regression ============= % Now, you will get to experiment with polynomial regression with multiple % values of lambda. The code below runs polynomial regression with % lambda = 0. You should try running the code with different values of % lambda to see how the fit and learning curve change. %lambda = 0; [theta] = trainLinearReg(X_poly, y, lambda);% Plot training data and fit figure(1); plot(X, y, 'rx', 'MarkerSize', 10, 'LineWidth', 1.5); plotFit(min(X), max(X), mu, sigma, theta, p); xlabel('Change in water level (x)'); ylabel('Water flowing out of the dam (y)'); title (sprintf('Polynomial Regression Fit (lambda = %f)', lambda));figure(2); [error_train, error_val] = ...learningCurve(X_poly, y, X_poly_val, yval, lambda); plot(1:m, error_train, 1:m, error_val);title(sprintf('Polynomial Regression Learning Curve (lambda = %f)', lambda)); xlabel('Number of training examples') ylabel('Error') axis([0 13 0 100]) legend('Train', 'Cross Validation')fprintf('Polynomial Regression (lambda = %f)\n\n', lambda); fprintf('# Training Examples\tTrain Error\tCross Validation Error\n'); for i = 1:mfprintf(' \t%d\t\t%f\t%f\n', i, error_train(i), error_val(i)); endfprintf('Program paused. Press enter to continue.\n'); pause;%% =========== Part 8: Validation for Selecting Lambda ============= % You will now implement validationCurve to test various values of % lambda on a validation set. You will then use this to select the % "best" lambda value. %[lambda_vec, error_train, error_val] = ...validationCurve(X_poly, y, X_poly_val, yval);close all; plot(lambda_vec, error_train, lambda_vec, error_val); legend('Train', 'Cross Validation'); xlabel('lambda'); ylabel('Error');fprintf('lambda\t\tTrain Error\tValidation Error\n'); for i = 1:length(lambda_vec)fprintf(' %f\t%f\t%f\n', ...lambda_vec(i), error_train(i), error_val(i)); end [theta] = trainLinearReg(X, y, lambda_vec(3)); [error_test,~]=linearRegCostFunction(Xtest, ytest, theta, lambda_vec(3)); fprintf('Program paused. Press enter to continue.\n'); pause; fprintf('theta\terror_test\tlambda') fprintf('%f\t%f\t%f',theta,error_test,lambda_vec(3));linearRegCostFunction.m
function [J, grad] = linearRegCostFunction(X, y, theta, lambda) %LINEARREGCOSTFUNCTION Compute cost and gradient for regularized linear %regression with multiple variables % [J, grad] = LINEARREGCOSTFUNCTION(X, y, theta, lambda) computes the % cost of using theta as the parameter for linear regression to fit the % data points in X and y. Returns the cost in J and the gradient in grad% Initialize some useful values m = length(y); % number of training examples% You need to return the following variables correctly J = 0; grad = zeros(size(theta));% ====================== YOUR CODE HERE ====================== % Instructions: Compute the cost and gradient of regularized linear % regression for a particular choice of theta. % % You should set J to the cost and grad to the gradient. %J=J+sum((X*theta-y).^2); J=J+lambda*sum(theta(2:end).^2); J=J/2/m; grad=(X*theta-y)'*X; theta(1)=0; grad=grad+lambda*theta'; grad=grad/m;% =========================================================================grad = grad(:);endlearningCurve.m
function [error_train, error_val] = ...learningCurve(X, y, Xval, yval, lambda) %LEARNINGCURVE Generates the train and cross validation set errors needed %to plot a learning curve % [error_train, error_val] = ... % LEARNINGCURVE(X, y, Xval, yval, lambda) returns the train and % cross validation set errors for a learning curve. In particular, % it returns two vectors of the same length - error_train and % error_val. Then, error_train(i) contains the training error for % i examples (and similarly for error_val(i)). % % In this function, you will compute the train and test errors for % dataset sizes from 1 up to m. In practice, when working with larger % datasets, you might want to do this in larger intervals. %% Number of training examples m = size(X, 1); % You need to return these values correctly error_train = zeros(m, 1); error_val = zeros(m, 1);% ====================== YOUR CODE HERE ====================== % Instructions: Fill in this function to return training errors in % error_train and the cross validation errors in error_val. % i.e., error_train(i) and % error_val(i) should give you the errors % obtained after training on i examples. % % Note: You should evaluate the training error on the first i training % examples (i.e., X(1:i, :) and y(1:i)). % % For the cross-validation error, you should instead evaluate on % the _entire_ cross validation set (Xval and yval). % % Note: If you are using your cost function (linearRegCostFunction) % to compute the training and cross validation error, you should % call the function with the lambda argument set to 0. % Do note that you will still need to use lambda when running % the training to obtain the theta parameters. % % Hint: You can loop over the examples with the following: % % for i = 1:m % % Compute train/cross validation errors using training examples % % X(1:i, :) and y(1:i), storing the result in % % error_train(i) and error_val(i) % .... % % end %% ---------------------- Sample Solution ----------------------for i = 1:m % % Compute train/cross validation errors using training examples % % X(1:i, :) and y(1:i), storing the result in % % error_train(i) and error_val(i) [theta] = trainLinearReg(X(1:i,:), y(1:i), 1); [error_train(i),~]=linearRegCostFunction(X(1:i,:), y(1:i), theta, 0); [error_val(i),~]=linearRegCostFunction(Xval, yval, theta, 0); end% -------------------------------------------------------------% =========================================================================endvalidationCurve.m
function [lambda_vec, error_train, error_val] = ...validationCurve(X, y, Xval, yval) %VALIDATIONCURVE Generate the train and validation errors needed to %plot a validation curve that we can use to select lambda % [lambda_vec, error_train, error_val] = ... % VALIDATIONCURVE(X, y, Xval, yval) returns the train % and validation errors (in error_train, error_val) % for different values of lambda. You are given the training set (X, % y) and validation set (Xval, yval). %% Selected values of lambda (you should not change this) lambda_vec = [0 0.001 0.003 0.01 0.03 0.1 0.3 1 3 10]';% You need to return these variables correctly. error_train = zeros(length(lambda_vec), 1); error_val = zeros(length(lambda_vec), 1);% ====================== YOUR CODE HERE ====================== % Instructions: Fill in this function to return training errors in % error_train and the validation errors in error_val. The % vector lambda_vec contains the different lambda parameters % to use for each calculation of the errors, i.e, % error_train(i), and error_val(i) should give % you the errors obtained after training with % lambda = lambda_vec(i) % % Note: You can loop over lambda_vec with the following: % % for i = 1:length(lambda_vec) % lambda = lambda_vec(i); % % Compute train / val errors when training linear % % regression with regularization parameter lambda % % You should store the result in error_train(i) % % and error_val(i) % .... % % end % %for i = 1:length(lambda_vec) % lambda = lambda_vec(i); % % Compute train / val errors when training linear % % regression with regularization parameter lambda % % You should store the result in error_train(i) % % and error_val(i) % .... % lambda = lambda_vec(i); [theta] = trainLinearReg(X, y, lambda); [error_train(i),~]=linearRegCostFunction(X, y, theta, 0); [error_val(i),~]=linearRegCostFunction(Xval, yval, theta, 0);end % =========================================================================end總結
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