其中 true positive 表示當實際得病,預測得病。False positive 表示實際未得病,預測值得病。
false negative 表示實際得病,預測未得病。True negative 表示實際未得病,預測未得病。
Precision 越高說明預測精度越高,預測得病的得病機率很高,但這樣會致使低 Recall 值,便可能會漏診。
把得病的預測未得病的。最後用一個 Fscore 值來評價預測模型值越高越好。測試
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. % theta_without1 = theta(2:end , :); J = sum((X * theta - y) .^ 2) / ( 2 * m) + sum(lambda * theta_without1 .^ 2 /( 2 * m)) ; theta_without1 = theta; theta_without1(1) = 0; grad = X' * (X * theta - y) / m + lambda * theta_without1 / m; % ========================================================================= 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. % for i = 1:m theta = trainLinearReg(X(1:i , :) , y(1:i) , lambda); error_train(i) = linearRegCostFunction(X(1:i , :) , y(1:i) , theta , 0); error_val(i) = linearRegCostFunction(Xval , yval , theta , 0); end % ------------------------------------------------------------- % ========================================================================= end
function [X_poly] = polyFeatures(X, p) %POLYFEATURES Maps X (1D vector) into the p-th power % [X_poly] = POLYFEATURES(X, p) takes a data matrix X (size m x 1) and % maps each example into its polynomial features where % X_poly(i, :) = [X(i) X(i).^2 X(i).^3 ... X(i).^p]; % % You need to return the following variables correctly. X_poly = zeros(numel(X), p); % ====================== YOUR CODE HERE ====================== % Instructions: Given a vector X, return a matrix X_poly where the p-th % column of X contains the values of X to the p-th power. % % m = numel(X); X1 = X(:); disp(X1); for i = 1:p for j = 1:m X_poly(j,i) = X1(j)^i; end 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) % for i = 1:length(lambda_vec) 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