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Re: how to vectorize in the octave


From: Herier chen
Subject: Re: how to vectorize in the octave
Date: Mon, 12 Nov 2012 14:52:55 -0500

yep, thanks for  your help. after considering your advice and I have just gotten rid of the for loop and the i and take replace of the " * " with the " .* " .

grad = zeros(size(theta));
grad = (1.0/m) .* X' * (h-y);

it works now!!
Thank for you help. and hoping you have a nice day


On Mon, Nov 12, 2012 at 2:45 PM, Ed Meyer <address@hidden> wrote:


On Mon, Nov 12, 2012 at 11:33 AM, Herier chen <address@hidden> wrote:
yes, it is from professor Andrew Ng  the on-line coursesa class about machine learning. 

I am doing the ex3 now. I started the class a little late.


On Mon, Nov 12, 2012 at 2:27 PM, Ed Meyer <address@hidden> wrote:


On Mon, Nov 12, 2012 at 9:39 AM, Herier chen <address@hidden> wrote:

Hi, all, I am new for the octave, and I want to know how to vectorize the following code, I have tried lots of time ,but failed. would anybody do me a favor to help. thanks.

grad = zeros(size(theta));

 m1 = size(grad); 

h = sigmoid(X*theta); 

for i=1:m1,

 grad(i) = (1.0/m)X(:,i)'(h-y); 

end

and another question about the octave.

what is the difference between "/" and "./" . and how to find these operators definition in the octave help document. I have not find these all kinds of operators in octave.

thanks a lot/


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which exercise is this? (I assume it is from Dr. Ng's ML class)

--
Ed Meyer



You can almost always figure out the correct form by considering the dimensions
of the matrices;  in this case X is m by n, theta is (n,1), h = sigmoid(X*theta) is (m,1)
so to vectorize the gradient calculation you want something like

grad = X' * (h - y)/m

(n,1) = (n,m) * (m,1)

--
Ed Meyer



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