吴恩达机器学习Exercise1的部分Octavecommand

在吴恩达视频中的第35节有讲师关于如何提交答案的过程介绍&＃xff0c;以下的代码部分没有提交过&＃xff0c;正确性还有待考量。

主要记录的是生成gradient discent函数的过程&＃xff0c;这次我用Octave生成的gradient descent函数和原数据汇集图的对比图&＃xff1a;

看起来感觉挺对应的

&＃xff08;英语小白&＃xff09;

the first part

Now there are some data and are asked to be visiualized ,and these data are saved in ex1data.txt.The basic steps has been given,so it&＃39;s easy the get the plot.

As the exercise asks to complete plotData.m to draw the plot,here are the code in plotData.m:

function plotData(x, y)
plot(x,y,&＃39;rx&＃39;,&＃39;MarkerSize&＃39;,10);
ylabel(&＃39;Profit in $10,000&＃39;);
xlabel(&＃39;Poputation in 10,000&＃39;);

For the first command,both x and y are vector,saving the dataset of profit and polulation respectively.The second command,the plot function has two propertise,x and y, so the result is a scatter plot with red(&＃39;rx&＃39;) cross(叉号&＃xff09;.The rest two command are just label.

Now the function is finished,and by typing the command in Octave小黑框 and calling this function,the plot can be visualized.

here are the commands:

>>cd E:\Octave\exercise;
>>data &＃61; load(&＃39;ex1data1.txt&＃39;);

The first command is to find the path of dataset,and then downlod them.So now &＃39;data&＃39; is a matrix with two colume.

Define two vector X and y ,x is the first colume of &＃39;data&＃39;,y is the another,here are the commands:

>>X&＃61;data(:,1);
>>y&＃61;data(:,2);

we can see the x vector :

>>X
X &＃61;6.11015.52778.51867.00325.85988.38297.47648.57816.48625.05465.710714.16405.73408.40845.64075.37946.36545.13016.42967.07086.189120.27005.49016.32615.564918.945012.8280

well it&＃39;s clear to see what the vector X saved

now call the function dataPlot,and then we can see the plot

>>cd &＃39;E:\Octave\exercise&＃39;;
>>plotData(X,y);

and here is the plot:

写的啰嗦一点方便以后忘了回看*——*

the second part

The vedio has taught the cost function a lot,the cost function:

you know that h(x) is a linear regression,with two unkonwn parameters θ1,θ2:

Let&＃39;s just hypothesis that we know theta(θ) completely

now define a metrix theta &＃61;{0;0},so we have all the factors to define cost function:x,θ,y,m(m is a real number,meaning the number of training examples,that is the length of y)

well before figure the cost funxtion,it&＃39;s necessary to notice the way to figure the h(x):a transfer mertix &＃39;θ&＃39; times a metrix x,which is also the way where Octave figure the h(x).Thus,as we know now that theta is a two rows one columns mertix,while x is a m rows one column metrix.This means,x need to make some change--it&＃39;s ok to add a colume to x,and all the result will be right,the new x likes this:

    1.0000    6.1101
    1.0000    5.5277
    1.0000    8.5186
    1.0000    7.0032
    1.0000    5.8598
    1.0000    8.3829.........

so now it&＃39;s can time to theta correctly

Following the exercise,now we complete the computeCost.m,here are the code:

function J &＃61; computeCost(X, y, theta)
m &＃61; length(y);
predictions&＃61;X*theta;
sqrErros&＃61;(predictions-y).^2;
J&＃61;sum(sqrErros)/(2*m);

the predictions is the predict result of y,by using the theta that we predict (0,0).

So now it can figure out the J function.

well the last step is to type the command and call the computeCost function in Octave小黑框,here are the commands:

>>m&＃61;length(y);
>>x&＃61;[ones(m,1),data(:,1)];
>>theta&＃61;zeros(2,1);
>>cd &＃39;E:\Octave\exercise&＃39;
>>J&＃61;computeCost(x,y,theta)
J &＃61; 32.073

the second command is to rebuild x

and the result J is super big,it can be changed bu changing the theta.

the third part

now it&＃39;s time to figure the gradient descent,here is the formula:

There are two θ that needed to be figured until getting  the most proper number.

(in the vedio,the teacher has told about the simply way to figure all the θ a time,that is to regard θ as a vetor and multiply metrix to get the answer.However,I didn&＃39;t use this complexed way because there are two θ only .But this is not common in reality&＃64;_&＃64;)

As the formula shows,θj is a real number,and the ultimate result of α...xj would be a real number too,so it&＃39;s better to calculate the sum of &＃39;h(x)-y)xj&＃39; firstly.

the h(x):

As it mentioned before,h(x)&＃61;theta(metrix) times x(metrix),the answer will be a m row one column vector,and then minus y vector.

At this moment,x is still a m row two column metrix(the first row is one[maybe there are some grammer mistakes]).To explain this,x is:

     x1:          x2:

    1.0000    6.1101
    1.0000    5.5277
    1.0000    8.5186
    1.0000    7.0032
    1.0000    5.8598
    1.0000    8.3829.........

So,when it calculates θ1,we use x1,and when it calculates θ2,we use x2:

θ1&＃61;θ1-α* (1/m) * sum {the transfer metrix of (theta*x-y) * x1}

θ2&＃61;θ2-α* (1/m) * sum {the transfer metrix of (theta*x-y) * x2}

α is 0.01,the exercise has given

When I try to complete the gradientDescentMulti.m,I found that I have  no idea to seperate X by using Octave command in the function,I could only get x1 and x2 in Octave 小黑框 .It&＃39;s a sad thing,maybe I can acquire this skill in later learning

So I have to input parameters x,x1 and x2 in function...:

function [theta, J_history] &＃61; gradientDescentMulti(X,x1,x2, y, theta, alpha, num_iters)
m &＃61; length(y); % number of training examples
J_history &＃61; zeros(num_iters, 1);for iter &＃61; 1:num_iterspredictions&＃61;X*theta;vector_1&＃61;(predictions-y).*x1;theta(1)&＃61;theta(1)-alpha/m*sum(vector_1);vector_2&＃61;(predictions-y).*x2;theta(2)&＃61;theta(2)-alpha/m*sum(vector_2);%figure the cost function and same them in J_historypredictions&＃61;X*theta;sqrErros_2&＃61;(predictions-y).^2;J_history(iter)&＃61;sum(sqrErros_2)/(2*m);
end

the meaning of  num_iters is the time of calculate the theta,again and again,until getting the most proper results.

J_history is a num_iters row vector,saving all of the result of cost function(J function)

Now you can define the theta by yourself,I define them with [1;5]

here are the commands in Octave小黑盒:

>>x1&＃61;X(:,1)
>>x2&＃61;X(:,2)
>>theta&＃61;[1;5];
>>cd &＃39;E:\Octave\exercise&＃39;;
>>[theta,J]&＃61;gradientDescentMulti(x,x1,x2,y,theta,alpha,1)
theta &＃61;-2.49851.5015J &＃61; 12.843


the first and second command is to seperate the x metrix

I try one time,and get the J&＃61;12.843.

then I try a few hundreds times,and get all the J&＃61;4.5661

>>[theta,J]&＃61;gradientDescentMulti(x,x1,x2,y,theta,alpha,100)
theta &＃61;-2.90621.0938J &＃61;4.56614.56614.56614.56614.5661

now I suppose the theta gets the good answer

in the end,I make the ultimate plot,commands:

>>a&＃61;[5:4:25]
a &＃61;5 9 13 17 21 25>>b&＃61;theta(1)&＃43;theta(2)*a %use the theta
b &＃61;2.5630 6.9384 11.3137 15.6890 20.0644 24.4397>>plot(a,b)
>>hold on;%keep two lines on one image
>>plotData(X,y);

plot:

&＃xff08;欢迎指错&＃xff09;