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I'm currently taking Andrew Ng's machine learning course and I try implementing the stuff as I learn so as not to forget them, I just finished regularization (chapter 7). I know that theta 0 is updated normally, separate from other parameters, however, I am not sure which of these is the correct implementation.
Implementation 1: in my gradient function, after computing the regularization vector, change theta 0 part to 0 so when it is added to the total, it is as if theta 0 was never regularized.
Implementation 2: store theta in a temp variable: _theta, update it with a reg_step of 0 (so it's as if there's no regularization), store the new theta 0 in a temp variable: t1, then update the original theta value with my desired reg_step and replace theta 0 with t1 (value from non-regularized update).
below is my code for the first implementation, it's not meant to be advanced, I'm just practicing:
I'm using octave which is 1-index, so theta(1) is theta(0)
function ret = gradient(X,Y,theta,reg_step),
H = theta' * X;
dif = H-Y;
mul = dif .* X;
total = sum(mul,2);
m=(size(Y)(1,1));
regular = (reg_step/m)*theta;
regular(1)=0;
ret = (total/m)+regular,
endfunction
Thanks in advance.
A slight tweak to the first implementation worked for me.
First, calculate regularization for every theta. Then go on to perform gradient step and later you can change the first entry of the matrix containing gradients manually to ignore regularization for theta_0.
% Calculate regularization
regularization = (reg_step / m) * theta;
% Gradient Step
gradients = (1 / m) * (X' * (predictions - y)) + regularization;
% Ignore regularization in theta_0
gradients(1) = (1 / m) * (X(:, 1)' * (predictions - y));
I'm trying to understand the Gradient descent algorithm for linear regression.
The question is why we multiply by x(1) at the end of theta1 and don't do that at the end of theta0?
Thanks a lot!
Hypothesis is theta0 + theta1*x. On differentiating with respect to theta0 and theta1 you get 1 and x respectively. So, you get x in update for theta1 but not for theta0. For more details, refer to this document, cs229-notes1.
In short because of partial derivative & Application of the chain rule.
For Theta 0, when you take derivative of the loss function (MSE) with respect to Theta 0 (Or Beta 0 / Intercept ), your derivative is in the form shown the rightmost of eq1.
imagine...
Y = Mx + C
M = Theta 1
C = Theta 0
Loss Function = (Y - (Mx + C))^2
The derivative is in the form of f(x) * f'(x) if that makes sense. f'(x) in Theta 0 is 1 (watch the video to understand the derivate). So
2(Y - (Mx + C)) * derivative of with respect to C of (Y - (Mx + C))
= 2(Y - (Mx + C)) [disregard the 2 in front]
For Theta 1, when you take derivative of the loss function (MSE) with respect to Theta 1 (Or Beta 1 / slope ), your derivative is in the form shown the rightmost of eq1. In this case f'(x) is x, because.....
2(Y - (Mx + C)) * derivative of with respect to M of (Y - (Mx + C))
= 2(Y - (Mx + C)) * (1*x) [because the only term that is left is dx(Mx)]
Here is a video that can help
https://www.youtube.com/watch?v=sDv4f4s2SB8
The loss function for linear regression is given by
J = {(HthetaX(i))-y}^2
And we have gradient = Derivative of Loss.Therefore,
DJ/Dtheta = 2*(HthetaX(i))-y)*X(i). Now for theta0 X(i) ==1 ,hence
DJ/Dtheta for Theta0 = 2*(Htheta*X(i))-y)
I'm doing Andrew Ng's course on Machine Learning and I'm trying to wrap my head around the vectorised implementation of gradient descent for multiple variables which is an optional exercise in the course.
This is the algorithm in question (taken from here):
I just cannot do this in octave using sum though, I'm not sure how to multiply the sum of the hypothesis of x(i) - y(i) by the all variables xj(i). I tried different iterations of the following code but to no avail (either the dimensions are not right or the answer is wrong):
theta = theta - alpha/m * sum(X * theta - y) * X;
The correct answer, however, is entirely non-obvious (to a linear algebra beginner like me anyway, from here):
theta = theta - (alpha/m * (X * theta-y)' * X)';
Is there a rule of thumb for cases where sum is involved that governs transformations like the above?
And if so, is there the opposite version of the above (i.e. going from a sum based solution to a general multiplication one) as I was able to come up with a correct implementation using sum for gradient descent for a single variable (albeit not a very elegant one):
temp0 = theta(1) - (alpha/m * sum(X * theta - y));
temp1 = theta(2) - (alpha/m * sum((X * theta - y)' * X(:, 2)));
theta(1) = temp0;
theta(2) = temp1;
Please note that this only concerns vectorised implementations and although there are several questions on SO as to how this is done, my question is primarily concerned with the implementation of the algorithm in Octave using sum.
The general "rule of the thumb" is as follows, if you encounter something in the form of
SUM_i f(x_i, y_i, ...) g(a_i, b_i, ...)
then you can easily vectorize it (and this is what is done in the above) through
f(x, y, ...)' * g(a, b, ...)
As this is just a typical dot product, which in mathematics (in Euclidean space of finite dimension) looks like
<A, B> = SUM_i A_i B_i = A'B
thus
(X * theta-y)' * X)
is just
<X * theta-y), X> = <H_theta(X) - y, X> = SUM_i (H_theta(X_i) - y_i) X_i
as you can see this works both ways, as this is just a mathematical definition of dot product.
Referring to this part of your question specifically - "I'm not sure how to multiply the sum of the hypothesis of x(i) - y(i) by the all variables xj(i)."
In Octave you can multiply xj(i) to all the predictions using ".", so it can be written as:
m = size(X, 1);
predictions = X * theta;
sqrErrors = (predictions-y).^2;
J = 1 / (2*m) * sum(sqrErrors);
The vector multiplication automatically includes calculating the sum of the products. So you don't have to specify the sum() function. By using the sum() function, you are converting a vector into a scalar and that's bad.
You actually don't want to use summation here, because what you try to calculate are the single values for all thetas, and not the overall cost J. As you do this in one line of code, if you sum it up you end up with a single value (the sum of all thetas).
Summation was correct, though unnecessary, when you computed the values of theta one by one in the previous exercise. This works just the same:
temp0 = theta(1) - (alpha/m * (X * theta - y)' * X(:, 1));
temp1 = theta(2) - (alpha/m * (X * theta - y)' * X(:, 2));
theta(1) = temp0;
theta(2) = temp1;
I'm in the second week of Professor Andrew Ng's Machine Learning course through Coursera. We're working on linear regression and right now I'm dealing with coding the cost function.
The code I've written solves the problem correctly but does not pass the submission process and fails the unit test because I have hard coded the values of theta and not allowed for more than two values for theta.
Here's the code I've got so far
function J = computeCost(X, y, theta)
m = length(y);
J = 0;
for i = 1:m,
h = theta(1) + theta(2) * X(i)
a = h - y(i);
b = a^2;
J = J + b;
end;
J = J * (1 / (2 * m));
end
the unit test is
computeCost( [1 2 3; 1 3 4; 1 4 5; 1 5 6], [7;6;5;4], [0.1;0.2;0.3])
and should produce ans = 7.0175
So I need to add another for loop to iterate over theta, therefore allowing for any number of values for theta, but I'll be damned if I can wrap my head around how/where.
Can anyone suggest a way I can allow for any number of values for theta within this function?
If you need more information to understand what I'm trying to ask, I will try my best to provide it.
You can use vectorize of operations in Octave/Matlab.
Iterate over entire vector - it is really bad idea, if your programm language let you vectorize operations.
R, Octave, Matlab, Python (numpy) allow this operation.
For example, you can get scalar production, if theta = (t0, t1, t2, t3) and X = (x0, x1, x2, x3) in the next way:
theta * X' = (t0, t1, t2, t3) * (x0, x1, x2, x3)' = t0*x0 + t1*x1 + t2*x2 + t3*x3
Result will be scalar.
For example, you can vectorize h in your code in the next way:
H = (theta'*X')';
S = sum((H - y) .^ 2);
J = S / (2*m);
Above answer is perfect but you can also do
H = (X*theta);
S = sum((H - y) .^ 2);
J = S / (2*m);
Rather than computing
(theta' * X')'
and then taking the transpose you can directly calculate
(X * theta)
It works perfectly.
The below line return the required 32.07 cost value while we run computeCost once using θ initialized to zeros:
J = (1/(2*m)) * (sum(((X * theta) - y).^2));
and is similar to the original formulas that is given below.
It can be also done in a line-
m- # training sets
J=(1/(2*m)) * ((((X * theta) - y).^2)'* ones(m,1));
J = sum(((X*theta)-y).^2)/(2*m);
ans = 32.073
Above answer is perfect,I thought the problem deeply for a day and still unfamiliar with Octave,so,Just study together!
If you want to use only matrix, so:
temp = (X * theta - y); % h(x) - y
J = ((temp')*temp)/(2 * m);
clear temp;
This would work just fine for you -
J = sum((X*theta - y).^2)*(1/(2*m))
This directly follows from the Cost Function Equation
Python code for the same :
def computeCost(X, y, theta):
m = y.size # number of training examples
J = 0
H = (X.dot(theta))
S = sum((H - y)**2);
J = S / (2*m);
return J
function J = computeCost(X, y, theta)
m = length(y);
J = 0;
% Hypothesis h(x)
h = X * theta;
% Error function (h(x) - y) ^ 2
squaredError = (h-y).^2;
% Cost function
J = sum(squaredError)/(2*m);
end
I think we needed to use iteration for much general solution for cost rather one iteration, also the result shows in the PDF 32.07 may not be correct answer that grader is looking for reason being its a one case out of many training data.
I think it should loop through like this
for i in 1:iteration
theta = theta - alpha*(1/m)(theta'*x-y)*x
j = (1/(2*m))(theta'*x-y)^2
I implemented a gradient descent algorithm to minimize a cost function in order to gain a hypothesis for determining whether an image has a good quality. I did that in Octave. The idea is somehow based on the algorithm from the machine learning class by Andrew Ng
Therefore I have 880 values "y" that contains values from 0.5 to ~12. And I have 880 values from 50 to 300 in "X" that should predict the image's quality.
Sadly the algorithm seems to fail, after some iterations the value for theta is so small, that theta0 and theta1 become "NaN". And my linear regression curve has strange values...
here is the code for the gradient descent algorithm:
(theta = zeros(2, 1);, alpha= 0.01, iterations=1500)
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
m = length(y); % number of training examples
J_history = zeros(num_iters, 1);
for iter = 1:num_iters
tmp_j1=0;
for i=1:m,
tmp_j1 = tmp_j1+ ((theta (1,1) + theta (2,1)*X(i,2)) - y(i));
end
tmp_j2=0;
for i=1:m,
tmp_j2 = tmp_j2+ (((theta (1,1) + theta (2,1)*X(i,2)) - y(i)) *X(i,2));
end
tmp1= theta(1,1) - (alpha * ((1/m) * tmp_j1))
tmp2= theta(2,1) - (alpha * ((1/m) * tmp_j2))
theta(1,1)=tmp1
theta(2,1)=tmp2
% ============================================================
% Save the cost J in every iteration
J_history(iter) = computeCost(X, y, theta);
end
end
And here is the computation for the costfunction:
function J = computeCost(X, y, theta) %
m = length(y); % number of training examples
J = 0;
tmp=0;
for i=1:m,
tmp = tmp+ (theta (1,1) + theta (2,1)*X(i,2) - y(i))^2; %differenzberechnung
end
J= (1/(2*m)) * tmp
end
If you are wondering how the seemingly complex looking for loop can be vectorized and cramped into a single one line expression, then please read on. The vectorized form is:
theta = theta - (alpha/m) * (X' * (X * theta - y))
Given below is a detailed explanation for how we arrive at this vectorized expression using gradient descent algorithm:
This is the gradient descent algorithm to fine tune the value of θ:
Assume that the following values of X, y and θ are given:
m = number of training examples
n = number of features + 1
Here
m = 5 (training examples)
n = 4 (features+1)
X = m x n matrix
y = m x 1 vector matrix
θ = n x 1 vector matrix
xi is the ith training example
xj is the jth feature in a given training example
Further,
h(x) = ([X] * [θ]) (m x 1 matrix of predicted values for our training set)
h(x)-y = ([X] * [θ] - [y]) (m x 1 matrix of Errors in our predictions)
whole objective of machine learning is to minimize Errors in predictions. Based on the above corollary, our Errors matrix is m x 1 vector matrix as follows:
To calculate new value of θj, we have to get a summation of all errors (m rows) multiplied by jth feature value of the training set X. That is, take all the values in E, individually multiply them with jth feature of the corresponding training example, and add them all together. This will help us in getting the new (and hopefully better) value of θj. Repeat this process for all j or the number of features. In matrix form, this can be written as:
This can be simplified as:
[E]' x [X] will give us a row vector matrix, since E' is 1 x m matrix and X is m x n matrix. But we are interested in getting a column matrix, hence we transpose the resultant matrix.
More succinctly, it can be written as:
Since (A * B)' = (B' * A'), and A'' = A, we can also write the above as
This is the original expression we started out with:
theta = theta - (alpha/m) * (X' * (X * theta - y))
i vectorized the theta thing...
may could help somebody
theta = theta - (alpha/m * (X * theta-y)' * X)';
I think that your computeCost function is wrong.
I attended NG's class last year and I have the following implementation (vectorized):
m = length(y);
J = 0;
predictions = X * theta;
sqrErrors = (predictions-y).^2;
J = 1/(2*m) * sum(sqrErrors);
The rest of the implementation seems fine to me, although you could also vectorize them.
theta_1 = theta(1) - alpha * (1/m) * sum((X*theta-y).*X(:,1));
theta_2 = theta(2) - alpha * (1/m) * sum((X*theta-y).*X(:,2));
Afterwards you are setting the temporary thetas (here called theta_1 and theta_2) correctly back to the "real" theta.
Generally it is more useful to vectorize instead of loops, it is less annoying to read and to debug.
If you are OK with using a least-squares cost function, then you could try using the normal equation instead of gradient descent. It's much simpler -- only one line -- and computationally faster.
Here is the normal equation:
http://mathworld.wolfram.com/NormalEquation.html
And in octave form:
theta = (pinv(X' * X )) * X' * y
Here is a tutorial that explains how to use the normal equation: http://www.lauradhamilton.com/tutorial-linear-regression-with-octave
While not scalable like a vectorized version, a loop-based computation of a gradient descent should generate the same results. In the example above, the most probably case of the gradient descent failing to compute the correct theta is the value of alpha.
With a verified set of cost and gradient descent functions and a set of data similar with the one described in the question, theta ends up with NaN values just after a few iterations if alpha = 0.01. However, when set as alpha = 0.000001, the gradient descent works as expected, even after 100 iterations.
Using only vectors here is the compact implementation of LR with Gradient Descent in Mathematica:
Theta = {0, 0}
alpha = 0.0001;
iteration = 1500;
Jhist = Table[0, {i, iteration}];
Table[
Theta = Theta -
alpha * Dot[Transpose[X], (Dot[X, Theta] - Y)]/m;
Jhist[[k]] =
Total[ (Dot[X, Theta] - Y[[All]])^2]/(2*m); Theta, {k, iteration}]
Note: Of course one assumes that X is a n * 2 matrix, with X[[,1]] containing only 1s'
This should work:-
theta(1,1) = theta(1,1) - (alpha*(1/m))*((X*theta - y)'* X(:,1) );
theta(2,1) = theta(2,1) - (alpha*(1/m))*((X*theta - y)'* X(:,2) );
its cleaner this way, and vectorized also
predictions = X * theta;
errorsVector = predictions - y;
theta = theta - (alpha/m) * (X' * errorsVector);
If you remember the first Pdf file for Gradient Descent form machine Learning course, you would take care of learning rate. Here is the note from the mentioned pdf.
Implementation Note: If your learning rate is too large, J(theta) can di-
verge and blow up', resulting in values which are too large for computer
calculations. In these situations, Octave/MATLAB will tend to return
NaNs. NaN stands fornot a number' and is often caused by unde ned
operations that involve - infinity and +infinity.