吴恩达机器学习代码作业答案ex1
整理方式：
作业原文件：
链接: https://pan.baidu.com/s/1BSiu57cDfiC4grCIx7zalA 提取码: a8sm
可成功运行的答案：
链接: https://pan.baidu.com/s/1S3yvdSMbaKwuukyl3R6t3A 提取码: uyhx
运行结果整理：
machine-learning-ex1-linear regression：
基础部分代码运行结果，图片就不贴了，在作业说明里已经给了运行结果：
Running warmUpExercise ...
5x5 Identity Matrix:
ans =
Diagonal Matrix
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
Program paused. Press enter to continue.
Plotting Data ...
Program paused. Press enter to continue.
Testing the cost function ...
With theta = [0 ; 0]
Cost computed = 32.072734
Expected cost value (approx) 32.07
With theta = [-1 ; 2]
Cost computed = 54.242455
Expected cost value (approx) 54.24
Program paused. Press enter to continue.
Running Gradient Descent ...
Theta found by gradient descent:
-3.630291
1.166362
Expected theta values (approx)
-3.6303
1.1664
For population = 35,000, we predict a profit of 4519.767868
For population = 70,000, we predict a profit of 45342.450129
Program paused. Press enter to continue.
Visualizing J(theta_0, theta_1) ...
Optional Exercises 部分：
这里我修改了ex1_multi.m文件中源代码的迭代次数和theta值，如下：
#
alpha = 0.01;
num_iters = 800;
% Init Theta and Run Gradient Descent
theta = zeros(3, 1);
[theta, J_history] = gradientDescentMulti(X, y, theta, alpha, num_iters);
theta = zeros(3, 1);
[theta, J_history1] = gradientDescentMulti(X, y, theta, 0.03, num_iters);
theta = zeros(3, 1);
[theta, J_history2] = gradientDescentMulti(X, y, theta, 0.005, num_iters);
因此，这里的运行结果Theta是alph=0.005时拟合出来的结果：
# 运行结果
octave:36> ex1_multi
Loading data ...
First 10 examples from the dataset:
x = [2104 3], y = 399900
x = [1600 3], y = 329900
x = [2400 3], y = 369000
x = [1416 2], y = 232000
x = [3000 4], y = 539900
x = [1985 4], y = 299900
x = [1534 3], y = 314900
x = [1427 3], y = 198999
x = [1380 3], y = 212000
x = [1494 3], y = 242500
Program paused. Press enter to continue.
Normalizing Features ...
Running gradient descent ...
Theta computed from gradient descent:
334240.028807
100065.016724
3690.356114
Predicted price of a 1650 sq-ft, 3 br house (using gradient descent):
$289258.577516
Program paused. Press enter to continue.
Solving with normal equations...
Theta computed from the normal equations:
89597.909542
139.210674
-8738.019112
Predicted price of a 1650 sq-ft, 3 br house (using normal equations):
$293081.464335
红蓝黑分别代表theta的值为0.02,0.01.0.005时的结果。
后记：
我运行的第一版代码的结果中gradient descent与normal equations的结果差别较大，当时使用的gradient descent中price计算代码：
% Estimate the price of a 1650 sq-ft, 3 br house
% ====================== YOUR CODE HERE ======================
% Recall that the first column of X is all-ones. Thus, it does
% not need to be normalized.
price = 0; % You should change this
price = [1 1650 3]*theta;
% ============================================================
fprintf(['Predicted price of a 1650 sq-ft, 3 br house ' ...
'(using gradient descent):\n $%f\n'], price);
其实这边在作业要求中有被提醒：
也就是说，因为之前我们拟合theta的值的时候使用的是特征缩放之后的X，所以，在我们有一个新的x需要进行预测时，也要使用特征缩放之后的x值：
% Estimate the price of a 1650 sq-ft, 3 br house
% ====================== YOUR CODE HERE ======================
% Recall that the first column of X is all-ones. Thus, it does
% not need to be normalized.
price = 0; % You should change this
xt=([1650 3]-mu)./sigma; #对x进行特征缩放
price=[1,xt]*theta;
作者：XIAXIAgo
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