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Andrew Ng机器学习 二: Logistic Regression

作者:互联网

一:逻辑回归(Logistic Regression)

  背景:假设你是一所大学招生办的领导,你依据学生的成绩,给与他入学的资格。现在有这样一组以前的数据集ex2data1.txt,第一列表示第一次测验的分数,第二列表示第二次测验的分数,第三列1表示允许入学,0表示不允许入学。现在依据这些数据集,设计出一个模型,作为以后的入学标准。

  

  我们通过可视化这些数据集,发现其与某条直线方程有关,而结果又只有两类,故我们接下来使用逻辑回归去拟合该数据集。

  

  1,回归方程的脚本ex2.m:

%% Machine Learning Online Class - Exercise 2: Logistic Regression
%
%  Instructions
%  ------------
% 
%  This file contains code that helps you get started on the logistic
%  regression exercise. You will need to complete the following functions 
%  in this exericse:
%
%     sigmoid.m
%     costFunction.m
%     predict.m
%     costFunctionReg.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

%% Load Data
%  The first two columns contains the exam scores and the third column
%  contains the label.

data = load('ex2data1.txt');
X = data(:, [1, 2]); y = data(:, 3);

%% ==================== Part 1: Plotting ====================
%  We start the exercise by first plotting the data to understand the 
%  the problem we are working with.

fprintf(['Plotting data with + indicating (y = 1) examples and o ' ...
         'indicating (y = 0) examples.\n']);

plotData(X, y);

% Put some labels 
hold on;
% Labels and Legend
xlabel('Exam 1 score')
ylabel('Exam 2 score')

% Specified in plot order
legend('Admitted', 'Not admitted')
hold off;

fprintf('\nProgram paused. Press enter to continue.\n');
pause;


%% ============ Part 2: Compute Cost and Gradient ============
%  In this part of the exercise, you will implement the cost and gradient
%  for logistic regression. You neeed to complete the code in 
%  costFunction.m

%  Setup the data matrix appropriately, and add ones for the intercept term
[m, n] = size(X);

% Add intercept term to x and X_test
X = [ones(m, 1) X];

% Initialize fitting parameters
initial_theta = zeros(n + 1, 1);

% Compute and display initial cost and gradient
[cost, grad] = costFunction(initial_theta, X, y);

fprintf('Cost at initial theta (zeros): %f\n', cost);
fprintf('Expected cost (approx): 0.693\n');
fprintf('Gradient at initial theta (zeros): \n');
fprintf(' %f \n', grad);
fprintf('Expected gradients (approx):\n -0.1000\n -12.0092\n -11.2628\n');

% Compute and display cost and gradient with non-zero theta
test_theta = [-24; 0.2; 0.2];
[cost, grad] = costFunction(test_theta, X, y);

fprintf('\nCost at test theta: %f\n', cost);
fprintf('Expected cost (approx): 0.218\n');
fprintf('Gradient at test theta: \n');
fprintf(' %f \n', grad);
fprintf('Expected gradients (approx):\n 0.043\n 2.566\n 2.647\n');

fprintf('\nProgram paused. Press enter to continue.\n');
pause;


%% ============= Part 3: Optimizing using fminunc  =============
%  In this exercise, you will use a built-in function (fminunc) to find the
%  optimal parameters theta.

%  Set options for fminunc
options = optimset('GradObj', 'on', 'MaxIter', 400);

%  Run fminunc to obtain the optimal theta
%  This function will return theta and the cost 
[theta, cost] = ...
    fminunc(@(t)(costFunction(t, X, y)), initial_theta, options);

% Print theta to screen
fprintf('Cost at theta found by fminunc: %f\n', cost);
fprintf('Expected cost (approx): 0.203\n');
fprintf('theta: \n');
fprintf(' %f \n', theta);
fprintf('Expected theta (approx):\n');
fprintf(' -25.161\n 0.206\n 0.201\n');

% Plot Boundary
plotDecisionBoundary(theta, X, y);

% Put some labels 
hold on;
% Labels and Legend
xlabel('Exam 1 score')
ylabel('Exam 2 score')

% Specified in plot order
legend('Admitted', 'Not admitted')
hold off;

fprintf('\nProgram paused. Press enter to continue.\n');
pause;

%% ============== Part 4: Predict and Accuracies ==============
%  After learning the parameters, you'll like to use it to predict the outcomes
%  on unseen data. In this part, you will use the logistic regression model
%  to predict the probability that a student with score 45 on exam 1 and 
%  score 85 on exam 2 will be admitted.
%
%  Furthermore, you will compute the training and test set accuracies of 
%  our model.
%
%  Your task is to complete the code in predict.m

%  Predict probability for a student with score 45 on exam 1 
%  and score 85 on exam 2 

prob = sigmoid([1 45 85] * theta);
fprintf(['For a student with scores 45 and 85, we predict an admission ' ...
         'probability of %f\n'], prob);
fprintf('Expected value: 0.775 +/- 0.002\n\n');

% Compute accuracy on our training set
p = predict(theta, X);

fprintf('Train Accuracy: %f\n', mean(double(p == y)) * 100);
fprintf('Expected accuracy (approx): 89.0\n');
fprintf('\n');
ex2.m

  

  2,可视化数据plotData.m:

function plotData(X, y)
%PLOTDATA Plots the data points X and y into a new figure 
%   PLOTDATA(x,y) plots the data points with + for the positive examples
%   and o for the negative examples. X is assumed to be a Mx2 matrix.

% Create New Figure
figure;   hold on;


% ====================== YOUR CODE HERE ======================
% Instructions: Plot the positive and negative examples on a
%               2D plot, using the option 'k+' for the positive
%               examples and 'ko' for the negative examples.
%

   pos=find(y==1);
  neg=find(y==0);
  plot(X(pos,1),X(pos,2),'k+','LineWidth',2,'MarkerSize',7);
  plot(X(neg,1),X(neg,2),'ko','MarkerFaceColor','y','MarkerSize',7);

% =========================================================================


hold off;

end
plotData.m

  

  3,逻辑回归的逻辑函数(Sigmoid Function/Logistic Function):

  $h_{\theta}(x)=g(\theta^{T}x)$ :表示在输入为$x$,预测为$y=1$的概率

  $g(z)=\frac{1}{1+e^{-z}}$  

function g = sigmoid(z)
%SIGMOID Compute sigmoid function
%   g = SIGMOID(z) computes the sigmoid of z.

% You need to return the following variables correctly 
g = zeros(size(z));

% ====================== YOUR CODE HERE ======================
% Instructions: Compute the sigmoid of each value of z (z can be a matrix,
%               vector or scalar).

  g=1./(1+exp(-z));
  
% =============================================================

end
sigmoid.m

 

  4,逻辑回归的代价函数:

  $J(\theta)=-\frac{1}{m}\sum_{i=1}^{m}[y^{(i)}log(h_\theta(x^{(i)}))+(1-y^{(i)})log(1-h_{\theta}(x^{(i)}))]$

function [J, grad] = costFunction(theta, X, y)
%COSTFUNCTION Compute cost and gradient for logistic regression
%   J = COSTFUNCTION(theta, X, y) computes the cost of using theta as the
%   parameter for logistic regression and the gradient of the cost
%   w.r.t. to the parameters.

% 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 of a particular choice of theta.
%               You should set J to the cost.
%               Compute the partial derivatives and set grad to the partial
%               derivatives of the cost w.r.t. each parameter in theta
%
% Note: grad should have the same dimensions as theta
%
  
  h=sigmoid(X*theta); %求hθ(x)
  J=-sum(y.*log(h)+(1-y).*log(1-h))/m; %代价函数
  
  grad=(X')*(h-y)./m; %梯度下降,没有学习速率α,之后给我们调用内置函数fminunc使用
  
  
##  h=sigmoid(X*theta);
##J=sum(-y'*log(h)-(1-y)'*log(1-h))/m;
##grad=((h-y)'*X)/m;





% =============================================================

end
costFunction.m

 

  5,带学习速率$\alpha$的梯度下降:

  $\theta_j:=\theta_j-\frac{\alpha}{m }\sum_{i=1}^{m}[(h_\theta(x^{(i)})-y^{(i)})x^{(i)}_j]$

  

  不带学习速率$\alpha$的梯度下降(给之后fminunc作为梯度下降使用):

  $\frac{\partial J(\theta)}{\partial \theta_j}=\frac{1}{m}\sum_{i=1}^{m}[(h_\theta(x^{(i)})-y^{(i)})x^{(i)}_j]$

  使用内置fminunc函数来拟合参数$\theta$,之前我们是使用梯度下降来拟合参数$\theta$的,在这同样也能使用,不过我们这里使用内置fminunc函数来去拟合,它会自动选择学习速率$\alpha$,不需要我们手工选择,我们只需要给定一个迭代次数,一个写好的代价函数,初始化$\theta$,最后它会为我们找到最优的$\theta$,它像可以加强版的梯度下降法。

options = optimset('GradObj', 'on', 'MaxIter', 400);
[theta, cost] = ...
    fminunc(@(t)(costFunction(t, X, y)), initial_theta, options);//自己写好的costFunction函数

   

  6,根据拟合好的参数$\theta$,预测数据,例如我们想预测某学生第一次分数为45,第二次分数为85,该学生能入学的概率为:

prob = sigmoid([1 45 85] * theta); %入学的概率

 

   预测样本X,我们可以看到预测的准确率为89%。

function p = predict(theta, X)
%PREDICT Predict whether the label is 0 or 1 using learned logistic 
%regression parameters theta
%   p = PREDICT(theta, X) computes the predictions for X using a 
%   threshold at 0.5 (i.e., if sigmoid(theta'*x) >= 0.5, predict 1)

m = size(X, 1); % Number of training examples

% You need to return the following variables correctly
p = zeros(m, 1);

% ====================== YOUR CODE HERE ======================
% Instructions: Complete the following code to make predictions using
%               your learned logistic regression parameters. 
%               You should set p to a vector of 0's and 1's
%
  
  %第一种
  for i=1:m
    p(i,1)=sigmoid(X(i,:)*theta)>=0.5; %预测每一个样本的结果,大于0.5为正向类
  end;
  
  %第二种
  %
##  ans=sigmoid(X*theta);
##  for i=1:m
##      if(ans(i,1)>=0.5)
##        p(i,1)=1;
##      else
##        p(i,1)=0;
##  end




% =========================================================================


end
predict.m

 

二:正则化逻辑回归(Regularized logistic regression):

  背景:假如你是某所工厂的管理员,该工厂生产芯片,每片芯片要经过两次测试后,达到标准方可通过,现在有一组以前的数据集ex2data2.txt,第一列为第一次测试的结果,第二列为第二次测试的结果,第三列1表示该芯片合格,0表示不合格。现在要你通过这些数据,拟合出一个模型,这个模型将作为以后判断芯片是否合格的标准。

  

  我们通过可视化这些数据集,发现其与某条复杂的曲线方程有关,而数据集只有两个特征$x_1$和$x_2$,显然是拟合不出曲线,那么我们可以通过原本的两个特征创造出更多的特征,将原本的特征映射为6次幂,这样我们就得到了28维的特征向量。当特征多了的话,很可能会出现过拟合,显然这不是我们想要的(即是它能很好的拟合原训练集,但预测新样本的能力会很低)。

构造更多的特征:

function out = mapFeature(X1, X2)
% MAPFEATURE Feature mapping function to polynomial features
%
%   MAPFEATURE(X1, X2) maps the two input features
%   to quadratic features used in the regularization exercise.
%
%   Returns a new feature array with more features, comprising of 
%   X1, X2, X1.^2, X2.^2, X1*X2, X1*X2.^2, etc..
%
%   Inputs X1, X2 must be the same size
%

degree = 6;
out = ones(size(X1(:,1)));
for i = 1:degree
    for j = 0:i
        out(:, end+1) = (X1.^(i-j)).*(X2.^j);
    end
end

end
mapFeature.m

 

所以这时我们使用正则化(Regularization)来解决过拟合的问题。

  1,正则化回归的脚本ex2.m: 

%% Machine Learning Online Class - Exercise 2: Logistic Regression
%
%  Instructions
%  ------------
%
%  This file contains code that helps you get started on the second part
%  of the exercise which covers regularization with logistic regression.
%
%  You will need to complete the following functions in this exericse:
%
%     sigmoid.m
%     costFunction.m
%     predict.m
%     costFunctionReg.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

%% Load Data
%  The first two columns contains the X values and the third column
%  contains the label (y).

data = load('ex2data2.txt');
X = data(:, [1, 2]); y = data(:, 3);

plotData(X, y);

% Put some labels
hold on;

% Labels and Legend
xlabel('Microchip Test 1')
ylabel('Microchip Test 2')

% Specified in plot order
legend('y = 1', 'y = 0')
hold off;


%% =========== Part 1: Regularized Logistic Regression ============
%  In this part, you are given a dataset with data points that are not
%  linearly separable. However, you would still like to use logistic
%  regression to classify the data points.
%
%  To do so, you introduce more features to use -- in particular, you add
%  polynomial features to our data matrix (similar to polynomial
%  regression).
%

% Add Polynomial Features

% Note that mapFeature also adds a column of ones for us, so the intercept
% term is handled
X = mapFeature(X(:,1), X(:,2)); %c从原来的二维变成了28(27+1截距项)维,m*28

% Initialize fitting parameters
initial_theta = zeros(size(X, 2), 1);

% Set regularization parameter lambda to 1
lambda = 1;

% Compute and display initial cost and gradient for regularized logistic
% regression
[cost, grad] = costFunctionReg(initial_theta, X, y, lambda);

fprintf('Cost at initial theta (zeros): %f\n', cost);
fprintf('Expected cost (approx): 0.693\n');
fprintf('Gradient at initial theta (zeros) - first five values only:\n');
fprintf(' %f \n', grad(1:5));
fprintf('Expected gradients (approx) - first five values only:\n');
fprintf(' 0.0085\n 0.0188\n 0.0001\n 0.0503\n 0.0115\n');

fprintf('\nProgram paused. Press enter to continue.\n');
pause;

% Compute and display cost and gradient
% with all-ones theta and lambda = 10
test_theta = ones(size(X,2),1);
[cost, grad] = costFunctionReg(test_theta, X, y, 10);

fprintf('\nCost at test theta (with lambda = 10): %f\n', cost);
fprintf('Expected cost (approx): 3.16\n');
fprintf('Gradient at test theta - first five values only:\n');
fprintf(' %f \n', grad(1:5));
fprintf('Expected gradients (approx) - first five values only:\n');
fprintf(' 0.3460\n 0.1614\n 0.1948\n 0.2269\n 0.0922\n');

fprintf('\nProgram paused. Press enter to continue.\n');
pause;

%% ============= Part 2: Regularization and Accuracies =============
%  Optional Exercise:
%  In this part, you will get to try different values of lambda and
%  see how regularization affects the decision coundart
%
%  Try the following values of lambda (0, 1, 10, 100).
%
%  How does the decision boundary change when you vary lambda? How does
%  the training set accuracy vary?
%

% Initialize fitting parameters
initial_theta = zeros(size(X, 2), 1);

% Set regularization parameter lambda to 1 (you should vary this)
lambda = 1;

% Set Options
options = optimset('GradObj', 'on', 'MaxIter', 400);

% Optimize
[theta, J, exit_flag] = ...
    fminunc(@(t)(costFunctionReg(t, X, y, lambda)), initial_theta, options);

% Plot Boundary
plotDecisionBoundary(theta, X, y);
hold on;
title(sprintf('lambda = %g', lambda))

% Labels and Legend
xlabel('Microchip Test 1')
ylabel('Microchip Test 2')

legend('y = 1', 'y = 0', 'Decision boundary')
hold off;

% Compute accuracy on our training set
p = predict(theta, X);

fprintf('Train Accuracy: %f\n', mean(double(p == y)) * 100);
fprintf('Expected accuracy (with lambda = 1): 83.1 (approx)\n');
ex2_reg.m

 

  2,正则化逻辑回归代价函数(忽略偏差项$\theta_0$的正则化):

  $J(\theta)=-\frac{1}{m}\sum_{i=1}^{m}[y^{(i)}log(h_\theta(x^{(i)}))+(1-y^{(i)})log(1-h_{\theta}(x^{(i)}))]+\frac{\lambda }{2m}\sum_{j=1}^{n}\theta_j^{2}$

   

  3,梯度下降:

  带学习速率:

    $\theta_0:=\theta_0-\alpha \frac{1}{m }\sum_{i=1}^{m}[(h_\theta(x^{(i)})-y^{(i)})x^{(i)}_0]$   for $j=0$

    $\theta_j:=\theta_j-\alpha (\frac{1}{m }\sum_{i=1}^{m}[(h_\theta(x^{(i)})-y^{(i)})x^{(i)}_j]+\frac{\lambda }{m}\theta_j)$  for $j\geq 1$

  不带学习速率(给之后fminunc作为梯度下降使用):

    $\frac{\partial J(\theta)}{\partial \theta_0}=\frac{1}{m}\sum_{i=1}^{m}[(h_\theta(x^{(i)})-y^{(i)})x^{(i)}_0]$  for $j=0$

    $\frac{\partial J(\theta)}{\partial \theta_j}=(\frac{1}{m}\sum_{i=1}^{m}[(h_\theta(x^{(i)})-y^{(i)})x^{(i)}_j])+\frac{\lambda }{m}\theta_j $ for $j\geq 1$

  

function [J, grad] = costFunctionReg(theta, X, y, lambda)
%COSTFUNCTIONREG Compute cost and gradient for logistic regression with regularization
%   J = COSTFUNCTIONREG(theta, X, y, lambda) computes the cost of using
%   theta as the parameter for regularized logistic regression and the
%   gradient of the cost w.r.t. to the parameters. 

% 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 of a particular choice of theta.
%               You should set J to the cost.
%               Compute the partial derivatives and set grad to the partial
%               derivatives of the cost w.r.t. each parameter in theta


  h=sigmoid(X*theta);
  n=size(X,2);
  J=(-(y')*log(h)-(1-y)'*log(1-h))/m+(lambda/(2*m))*sum(theta([2:n],:).^2); %忽略偏差项θ(0)的影响

  grad(1,1)=((X(:,1)')*(h-y))/m; %梯度下降
  grad([2:n],:)=(X(:,[2:n])')*(h-y)./m+(theta([2:n],:)).*(lambda/m);


##h=sigmoid(X*theta);
##theta(1,1)=0;
##J=sum(-y'*log(h)-(1-y)'*log(1-h))/m+lambda/2/m*sum(power(theta,2));
##grad=((h-y)'*X)/m+lambda/m*theta';
% =============================================================

end
costFunctionReg.m

 

  我们可以选择不同的$\lambda$大小去拟合数据集并可视化,选择一个较优的$lambda$。

 

   4,预测方法跟逻辑回归差不多,只是现在加入要预测第一次分数为45,第二次分数为80时,要先将这两个特征放到mapFeature函数构造。

 

 

 

 

 

 我的标签:做个有情怀的程序员。

标签:sigmoid,Andrew,Ng,cost,fprintf,Logistic,theta,grad,lambda
来源: https://www.cnblogs.com/-jiandong/p/11881503.html