我正在尝试用梯度下降实现逻辑回归,
我得到了我的成本函数j_theta的迭代次数,幸运的是,当j_theta与迭代次数相对应时,我的j_theta正在减小。
我使用的数据集如下:
x=
1 20 30
1 40 60
1 70 30
1 50 50
1 50 40
1 60 40
1 30 40
1 40 50
1 10 20
1 30 40
1 70 70
y= 0
1
1
1
0
1
0
0
0
0
1
我设法使用Gradient descent编写的逻辑回归代码是:
%1. The below code would load the data present in your desktop to the octave memory
x=load('stud_marks.dat');
%y=load('ex4y.dat');
y=x(:,3);
x=x(:,1:2);
%2. Now we want to add a column x0 with all the rows as value 1 into the matrix.
%First take the length
[m,n]=size(x);
x=[ones(m,1),x];
X=x;
% Now we limit the x1 and x2 we need to leave or skip the first column x0 because they should stay as 1.
mn = mean(x);
sd = std(x);
x(:,2) = (x(:,2) - mn(2))./ sd(2);
x(:,3) = (x(:,3) - mn(3))./ sd(3);
% We will not use vectorized technique, Because its hard to debug, We shall try using many for loops rather
max_iter=50;
theta = zeros(size(x(1,:)))';
j_theta=zeros(max_iter,1);
for num_iter=1:max_iter
% We calculate the cost Function
j_cost_each=0;
alpha=1;
theta
for i=1:m
z=0;
for j=1:n+1
% theta(j)
z=z+(theta(j)*x(i,j));
z
end
h= 1.0 ./(1.0 + exp(-z));
j_cost_each=j_cost_each + ( (-y(i) * log(h)) - ((1-y(i)) * log(1-h)) );
% j_cost_each
end
j_theta(num_iter)=(1/m) * j_cost_each;
for j=1:n+1
grad(j) = 0;
for i=1:m
z=(x(i,:)*theta);
z
h=1.0 ./ (1.0 + exp(-z));
h
grad(j) += (h-y(i)) * x(i,j);
end
grad(j)=grad(j)/m;
grad(j)
theta(j)=theta(j)- alpha * grad(j);
end
end
figure
plot(0:1999, j_theta(1:2000), 'b', 'LineWidth', 2)
hold off
figure
%3. In this step we will plot the graph for the given input data set just to see how is the distribution of the two class.
pos = find(y == 1); % This will take the postion or array number from y for all the class that has value 1
neg = find(y == 0); % Similarly this will take the position or array number from y for all class that has value 0
% Now we plot the graph column x1 Vs x2 for y=1 and y=0
plot(x(pos, 2), x(pos,3), '+');
hold on
plot(x(neg, 2), x(neg, 3), 'o');
xlabel('x1 marks in subject 1')
ylabel('y1 marks in subject 2')
legend('pass', 'Failed')
plot_x = [min(x(:,2))-2, max(x(:,2))+2]; % This min and max decides the length of the decision graph.
% Calculate the decision boundary line
plot_y = (-1./theta(3)).*(theta(2).*plot_x +theta(1));
plot(plot_x, plot_y)
hold off
%%%%%%% The only difference is In the last plot I used X where as now I use x whose attributes or features are featured scaled %%%%%%%%%%%
如果您查看x1与x2的图形,图形将如下,
运行我的代码后,我创建了一个决策边界。决策线的形状似乎没问题,但有点流离失所。具有决策边界的x1对x2的图形如下:
![在此输入图片说明] [2]
请建议我哪里出错了。
感谢:)
新图::::
![enter image description here][1]
If you see the new graph the coordinated of x axis have changed ..... Thats because I use x(feature scalled) instead of X.
答案 0 :(得分:4)
问题在于成本函数计算和/或梯度计算,您的绘图功能很好。我运行了我为逻辑回归实现的算法的数据集,但是使用了矢量化技术,因为在我看来它更容易调试。 我得到的最终值是
theta = [-76.4242, 0.8214, 0.7948 强> 我还使用了 alpha = 0.3
我绘制了决策边界并且它看起来很好,我建议使用矢量化形式,因为它更容易实现并在我看来调试。
我也认为你的梯度下降实现不太正确。 50次迭代是不够的,最后一次迭代的成本不够好。也许您应该尝试在停止条件下运行它以进行更多迭代。 另请参阅本讲座以了解优化技巧。 https://class.coursera.org/ml-006/lecture/37