先贴上作业的答案
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=(1/(2*m))*sum((X*theta-y).^2)+(lambda/(2*m))*sum(theta(2:end,:).^2); n=size(X,2); for i=1:n A(i,1)=(1/m)*sum((X*theta-y).*X(:,i)); end theta(1,:)=0; B=(lambda/m)*theta; grad=A+B; % ========================================================================= grad = grad(:); end
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
%利用X(1:i,:),y(1:i),trainLinearReg(),来训练参数theta
theta=trainLinearReg(X(1:i,:),y(1:i), lambda);
%You should evaluate the training error on the first i training examples (i.e., X(1:i, :) and y(1:i)).
%训练误差计算只用X(1:i,:), y(1:i)
[error_train(i),grad]=linearRegCostFunction(X(1:i,:), y(1:i), theta, 0);
%交叉验证用上所有的验证集，即Xval, yval
%For the cross-validation error, you should instead evaluate on the _entire_ cross validation set (Xval and yval).
[error_val(i), grad]=linearRegCostFunction(Xval, yval, theta, 0);
end
% -------------------------------------------------------------
% =========================================================================
end
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);
theta=trainLinearReg(X,y, lambda);
[error_train(i),grad]=linearRegCostFunction(X, y, theta, 0);
[error_val(i), grad]=linearRegCostFunction(Xval, yval, theta, 0);
end
% =========================================================================
end
%% 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" -- slide 8 in ML-advice.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:m
fprintf(' \t%d\t\t%f\t%f\n', i, error_train(i), error_val(i));
end
fprintf('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 Ones
fprintf('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 = 1;
[theta] = trainLinearReg(X_poly, y, lambda);
%[J, grad] = linearRegCostFunction(X_poly_val, ytest, theta, 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:m
fprintf(' \t%d\t\t%f\t%f\n', i, error_train(i), error_val(i));
end
fprintf('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
fprintf('Program paused. Press enter to continue.\n');
pause;
%%---------------------------------------------------------------------------
%%Optional (ungraded) exercise: Computing test set
lambda=3;
theta=trainLinearReg(X_poly,y, lambda);
[error_test,grad]=linearRegCostFunction(X_poly_test, ytest, theta, 0);
fprintf(' %f\n',error_test);
%%-------------------------------------------------------------------------
%%Optional (ungraded) exercise: Plotting learning curves with randomly selected examples
% m=size(X_poly,1);
% X_poly_y=[X_poly,y];
% X_poly_val_y=[X_poly_val,yval];
% lambda=0.01;
% error_train = zeros(m, 1);
% error_val = zeros(m, 1);
% for i=1:m
% error_train_sum=0;
% error_val_sum =0;
% for k=1:20 %50次迭代
% rand_seq=round(rand(1,i)*(m-1))+1;%生成i个随机序列 0~m
% rand_X_poly_y=X_poly_y(rand_seq,:);
% rand_X_poly_val_y=X_poly_val_y(rand_seq,:);
% X=rand_X_poly_y(:,1:end-1);
% y=rand_X_poly_y(:,end);
% Xval=rand_X_poly_val_y(:,1:end-1);
% yval=rand_X_poly_val_y(:,end);
% theta=trainLinearReg(X,y,lambda);
% [error_train_val,grad]=linearRegCostFunction(X, y, theta, 0);
% [error_val_val, grad]=linearRegCostFunction(Xval, yval, theta, 0);
% error_train_sum=error_train_sum+error_train_val;
% error_val_sum=error_val_sum+error_val_val;
% end
% error_train(i)=error_train_sum/20;
% error_val(i)=error_val_sum/20;
% end
% 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');
%%-------------------------------------------------------------------------
一、优化方案
当建模之后发现使用新的数据进行测试的时候，预测结果非常不理想，这个时候，可以尝试下面的方法进行重新建模：
1. 使用更多的训练数据
2. 尝试使用较少的属性
3. 或者尝试更多的属性
4. 尝试使用属性值的多项式形式
5. 增大常数项参数
6. 减小常数项参数
一般情况下，可以将数据集分为：训练集和测试级 7:3的比例。
利用训练集数据去训练出参数，然后用测试集去测试性能。
过拟合：简单的理解就是参数太多，训练集太少，过拟合的结果是训练误差会非常小，因为我们的参数很多，可以很好的拟合几乎所有的训练数据，但是，过拟合情况下，模型的泛化能力就很差，会导致训练误差比较大。
下面的就是过拟合的一个典型图像：
其实：一般情况下，数据集应该分为训练集，交叉验证集，测试集，因为，我们能会假设有好几种可能的模型，然后用数据集分别去训练这几个模型，然后利用交叉验证集去选择一个比较好的模型，最后用测试集去测试选出最优模型的性能。
如果，我们只假设了一个模型，那么就没有选择模型这个过程，那就把数据集分为训练集，测试集就可以了，训练集训练模型，测试集测试模型。
关于训练误差，代价误差，交叉误差的说明。
注：我们在训练模型的过程中，要使用代价函数，我们使用的代价函数必须是上面这个式子，如果，你加入了正则化的话，一定要加入正则項。
但是，我们在模型训练完，计算训练误差，交叉验证误差，测试误差的时候，即使有正则化参数，也不必加进去，只需按照上面的式子计算各种误差即可，很容易理解，就是比较训练结果与实际结果的差异。
现在开始讲述：偏差(Bias)和方差(variance)的概念
解释：偏差就是欠拟合，简单来讲就是参数太少，数据太多，不足以拟合参数，欠拟合情况下，训练误差会比较大，测试误差也会比较大。
方差就是过拟合，简单来说就是参数太多，训练数据太少，参数过分拟合数据，导致泛化能力非常差，过拟合情况下，训练误差比较小，但是因为模型没有泛化能力，所以，测试误差会比较大。
解释： 左图就是参数太少，不足以拟合训练数据的情况-欠拟合。
右图就是参数太多，过分拟合训练数据的情况 -过拟合
中图刚好合适，即比较好的拟合数据，又具有较好的泛化能力。
解释:继续解释上面的那个多项式拟合，d代表拟合的参数的个数，我们关注一下训练误差曲线Jtrain(θ)和交叉验证曲线Jcv(θ),随着
d的增大，我们用更多的多項式，参数来拟合训练数据，刚开始d比较小时，是欠拟合，训练误差会比较大，当d增大时，就会拟合得越来越好，所以，我们看到训练误差曲线是呈下降的趋势。
对于交叉验证曲线，d比较小是，欠拟合状态，模型拟合效果很差，所以，交叉验证误差会比较大，当d增大时，交叉验证误差会先逐渐减少，但是，当d过分大时，就进入了过拟合状态，模型的泛化能力也会比较差，所以，交叉验证误差又会逐渐增大。
所以，交叉验证误差的谷点，最小值就是我们该选择的d值。
接下来是关于学习曲线的绘制，只有绘制学习曲线，能帮助我们选择参数和判断算法现在是处于什么状态，高偏差(欠拟合)，高方差（过拟合），或者两者都有。
解释：正则化参数的大小是来调节过拟合，欠拟合状态的，正则化参数是抑制参数大小的因子，当正则化参数比较大时，模型的参数值会变得比较小，极端情况下，很多参数会变零。就导致了欠拟合状态。
当正则化参数比较小时，其抑制参数大小的能力就几乎等于没有，所以，如果本身模型的参数比较多，那么就容易进入过拟合状态。
左图对应的是large lambda 欠拟合
右图对应的是small lambda 过拟合
解释：此学习曲线，训练误差和交叉验证误差关于正则化参数的学习曲线。
当lambda比较小是，是过拟合状态，所以，训练误差比较小，随着lambda的增大，进入了欠拟合状态，训练误差增长，所以
Jtrain(θ)是上升的曲线。
当lambda比较小时，过拟合，所以泛化能力差，交叉验证误差会比较大，当lambda比较大时，欠拟合，交叉验证误差也会比较大，所以交叉验证误差曲线Jcv(θ)是抛物线型的曲线(理想情况下)。
所以，合适的lambda还是对应Jcv(θ)的谷点。
接下来要讲述的是训练误差关于训练样本规模的学习曲线
先讲结论，对应高偏差(欠拟合)情况下，增加样本的规模对于提高模型的性能其实一点帮助都没有，因为模型本身拟合得不好，增加数据集没用，本来就处于参数过多，样本过少的情况，增加训练样本，只会增大欠拟合的程度。
当处在高方差(过拟合)情况下，增加样本的规模对于提高模型的性能是有帮助的，因为过拟合是处在参数过多，样本过少的情况下，所以，增加样本的数目对于改善过拟合情况是非常有帮助的。
解释：对于high bias情况下，从error-m图中可以看出，随着训练样本m的增大，交叉验证误差很快就达到水平，下降极其缓慢，
所以，增加样本的规模也无济于事。总结，对于high bias的情况，Jtrain(θ)和Jcv(θ)都会比较大，且比较靠近。
解释：对于高方差的情况下，Jtrain(θ)是缓慢增长的，且数值比较小。而Jcv(θ)会比较大，随着样本规模的增大，持续下降，说明一点，在高方差情况下，Jtrain(θ)和Jcv(θ)还是有一段距离(gap)的。所以增大样本的规模，会减少Jcv(θ)。
当我们画出了曲线，诊断出高方差还是高偏差的问题时，那么就要对症下药了。下面提供，对待不同问题该使用的方法：
1、使用更多的训练样本 -解决过拟合问题
2、使用更少的特征 -解决过拟合问题
3、使用更多的特征 -解决欠拟合问题
4、增加使用多项式 -解决欠拟合问题
5、减小正则参数lambda -解决欠拟合问题
6、增大正则参数lambda -解决过拟合问题
接下来就是关于神经网络的学习曲线
解释：神经网络除了可以画出 代价函数-样本规模 代价函数-正则参数 的学习曲线，还可以画出 代价函数-隐层结点数 代价函数-层数的学习曲线。
神经网络也要使用正则参数去调节过拟合问题，如果神经网络的层数过多，隐层结点数过多，会导致过拟合，实际上会倾向于使用比较大型的神经网络，然后用正则参数lambda来调节过拟合。
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