關於演算法這塊，這周看了斯坦福大學Andrew Ng的公開課。還是極力推薦的，每節課10分鐘左右，講的思路清晰，內容豐富，程式設計作業也很值得去做。（在這裡好想吐槽一下國內培訓機構七月演算法的培訓視訊呀，根本看不下去。）

上課形式是這樣滴

程式設計作業提交是這樣滴

Nice work.每次都是一百分也挺有成就感的。每週的程式設計作業會有一份非常詳細的pdf文件解釋

pdf文件是這樣滴

上課也挺搞笑的，很願意聽下去。關鍵是，免費。

每集的內容我也下載下來了，等全部整理完畢會上傳到網盤。有需要請留言。

然後工具就換成了Octave/Matlab，果然還是Matlab用起來比較順手。

下圖分別是用邏輯迴歸對非線性邊界處理的效果，因為非線性，需要將特徵組合成多種多項式，擴大特徵，這樣容易出現過擬合的現象，於是需要正則化。下圖分別是正則化係數取0，1，100的效果。取0相當於不正則化。

具體Matlab程式碼如下：
ex2_reg.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));

% 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');

具體方法實現就不想貼了