machine learning octave code gradient descent question
$begingroup$
I'm taking Coursera Machine learning course. so who take this courses will able to help this problem.
this is the octave code to find the delta for gradient descent.
theta = theta - alpha / m * ((X * theta - y)'* X)';//this is the answerkey provided
First question)
the way i know to solve the gradient descent theta(0) and theta(1) should have different approach to get value as follow
theta(0) = theta(0) - alpha / m * ((X * theta(0) - y)')'; //my answer key
theta(1) = theta(1) - alpha / m * ((X * theta(1) - y)')'; //my answer key
but i'm not sure why the answer key only show
theta = theta - alpha / m * ((X * theta - y)'* X)';
this equation.
Second question) what is the ' ' doing in octave code?
theta = theta - alpha / m * ((X * theta - y)'* X)';
'* X)' // what ' ' thing do in here
machine-learning octave
asked May 25 '16 at 10:34
james Milerjames Miler
159412
$endgroup$
add a comment |
$begingroup$
I'm taking Coursera Machine learning course. so who take this courses will able to help this problem.
this is the octave code to find the delta for gradient descent.
theta = theta - alpha / m * ((X * theta - y)'* X)';//this is the answerkey provided
First question)
the way i know to solve the gradient descent theta(0) and theta(1) should have different approach to get value as follow
theta(0) = theta(0) - alpha / m * ((X * theta(0) - y)')'; //my answer key
theta(1) = theta(1) - alpha / m * ((X * theta(1) - y)')'; //my answer key
but i'm not sure why the answer key only show
theta = theta - alpha / m * ((X * theta - y)'* X)';
this equation.
Second question) what is the ' ' doing in octave code?
theta = theta - alpha / m * ((X * theta - y)'* X)';
'* X)' // what ' ' thing do in here
machine-learning octave
asked May 25 '16 at 10:34
james Milerjames Miler
159412
$endgroup$
$begingroup$
in octave/matlab, $x = u+v*w$ can be : $x,u,w$ : 3 vectors, $v$ : a matrix with $v*w$ the multiplication of a matrix with a vector. the main idea of matlab is that the basic datatypes instead of being integers and floating point numbers, are arrays / matrices of numbers.
$endgroup$
– reuns
May 25 '16 at 10:40
$begingroup$
In Octave, $X'$ corresponds to the transpose of the matrix (or the vector) $X$.
$endgroup$
– zuggg
May 25 '16 at 11:36
$begingroup$
oh ok so X' means transpose of X. is there someone who knows gradient descent ? I do not understand why they used transpose to find theta here
$endgroup$
– james Miler
May 26 '16 at 0:28
add a comment |
$begingroup$
I'm taking Coursera Machine learning course. so who take this courses will able to help this problem.
this is the octave code to find the delta for gradient descent.
theta = theta - alpha / m * ((X * theta - y)'* X)';//this is the answerkey provided
First question)
the way i know to solve the gradient descent theta(0) and theta(1) should have different approach to get value as follow
theta(0) = theta(0) - alpha / m * ((X * theta(0) - y)')'; //my answer key
theta(1) = theta(1) - alpha / m * ((X * theta(1) - y)')'; //my answer key
but i'm not sure why the answer key only show
theta = theta - alpha / m * ((X * theta - y)'* X)';
this equation.
Second question) what is the ' ' doing in octave code?
theta = theta - alpha / m * ((X * theta - y)'* X)';
'* X)' // what ' ' thing do in here
machine-learning octave
asked May 25 '16 at 10:34
james Milerjames Miler
159412
$endgroup$
I'm taking Coursera Machine learning course. so who take this courses will able to help this problem.
this is the octave code to find the delta for gradient descent.
theta = theta - alpha / m * ((X * theta - y)'* X)';//this is the answerkey provided
First question)
the way i know to solve the gradient descent theta(0) and theta(1) should have different approach to get value as follow
theta(0) = theta(0) - alpha / m * ((X * theta(0) - y)')'; //my answer key
theta(1) = theta(1) - alpha / m * ((X * theta(1) - y)')'; //my answer key
but i'm not sure why the answer key only show
theta = theta - alpha / m * ((X * theta - y)'* X)';
this equation.
Second question) what is the ' ' doing in octave code?
theta = theta - alpha / m * ((X * theta - y)'* X)';
'* X)' // what ' ' thing do in here
machine-learning octave
machine-learning octave
asked May 25 '16 at 10:34
james Milerjames Miler
159412
asked May 25 '16 at 10:34
james Milerjames Miler
159412
asked May 25 '16 at 10:34
james Milerjames Miler
159412
asked May 25 '16 at 10:34
james Milerjames Miler
159412
asked May 25 '16 at 10:34
james Milerjames Miler
159412
159412
$begingroup$
in octave/matlab, $x = u+v*w$ can be : $x,u,w$ : 3 vectors, $v$ : a matrix with $v*w$ the multiplication of a matrix with a vector. the main idea of matlab is that the basic datatypes instead of being integers and floating point numbers, are arrays / matrices of numbers.
$endgroup$
– reuns
May 25 '16 at 10:40
$begingroup$
In Octave, $X'$ corresponds to the transpose of the matrix (or the vector) $X$.
$endgroup$
– zuggg
May 25 '16 at 11:36
$begingroup$
oh ok so X' means transpose of X. is there someone who knows gradient descent ? I do not understand why they used transpose to find theta here
$endgroup$
– james Miler
May 26 '16 at 0:28
add a comment |
$begingroup$
in octave/matlab, $x = u+v*w$ can be : $x,u,w$ : 3 vectors, $v$ : a matrix with $v*w$ the multiplication of a matrix with a vector. the main idea of matlab is that the basic datatypes instead of being integers and floating point numbers, are arrays / matrices of numbers.
$endgroup$
– reuns
May 25 '16 at 10:40
$begingroup$
In Octave, $X'$ corresponds to the transpose of the matrix (or the vector) $X$.
$endgroup$
– zuggg
May 25 '16 at 11:36
$begingroup$
oh ok so X' means transpose of X. is there someone who knows gradient descent ? I do not understand why they used transpose to find theta here
$endgroup$
– james Miler
May 26 '16 at 0:28
$begingroup$
in octave/matlab, $x = u+v*w$ can be : $x,u,w$ : 3 vectors, $v$ : a matrix with $v*w$ the multiplication of a matrix with a vector. the main idea of matlab is that the basic datatypes instead of being integers and floating point numbers, are arrays / matrices of numbers.
$endgroup$
– reuns
May 25 '16 at 10:40
$begingroup$
in octave/matlab, $x = u+v*w$ can be : $x,u,w$ : 3 vectors, $v$ : a matrix with $v*w$ the multiplication of a matrix with a vector. the main idea of matlab is that the basic datatypes instead of being integers and floating point numbers, are arrays / matrices of numbers.
$endgroup$
– reuns
May 25 '16 at 10:40
$begingroup$
In Octave, $X'$ corresponds to the transpose of the matrix (or the vector) $X$.
$endgroup$
– zuggg
May 25 '16 at 11:36
$begingroup$
In Octave, $X'$ corresponds to the transpose of the matrix (or the vector) $X$.
$endgroup$
– zuggg
May 25 '16 at 11:36
$begingroup$
oh ok so X' means transpose of X. is there someone who knows gradient descent ? I do not understand why they used transpose to find theta here
$endgroup$
– james Miler
May 26 '16 at 0:28
$begingroup$
oh ok so X' means transpose of X. is there someone who knows gradient descent ? I do not understand why they used transpose to find theta here
$endgroup$
– james Miler
May 26 '16 at 0:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Transpose here is used for matching the columns of the X with rows of theta.
Ex:
size of
X=97x2;
y=97x1;
theta=2x1;
first calc is X * theta. The size of the resulting matrix will be 97x1. Then, the sub of two same size matrices. Now, we have to multiply X with the matrix obtained from the previous step. But, the sizes are different
(97x1) * (97x2)
Thus transposing the first matrix makes multiplication possible.
This results in a new matrix of size 1x2 (row vector). But, theta is of size 2x1 (column vector). Hence the final transpose.
answered Mar 3 '17 at 20:28
Bharani K DharanBharani K Dharan
1
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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oldest
votes
$begingroup$
Transpose here is used for matching the columns of the X with rows of theta.
Ex:
size of
X=97x2;
y=97x1;
theta=2x1;
first calc is X * theta. The size of the resulting matrix will be 97x1. Then, the sub of two same size matrices. Now, we have to multiply X with the matrix obtained from the previous step. But, the sizes are different
(97x1) * (97x2)
Thus transposing the first matrix makes multiplication possible.
This results in a new matrix of size 1x2 (row vector). But, theta is of size 2x1 (column vector). Hence the final transpose.
answered Mar 3 '17 at 20:28
Bharani K DharanBharani K Dharan
1
$endgroup$
add a comment |
$begingroup$
Transpose here is used for matching the columns of the X with rows of theta.
Ex:
size of
X=97x2;
y=97x1;
theta=2x1;
first calc is X * theta. The size of the resulting matrix will be 97x1. Then, the sub of two same size matrices. Now, we have to multiply X with the matrix obtained from the previous step. But, the sizes are different
(97x1) * (97x2)
Thus transposing the first matrix makes multiplication possible.
This results in a new matrix of size 1x2 (row vector). But, theta is of size 2x1 (column vector). Hence the final transpose.
answered Mar 3 '17 at 20:28
Bharani K DharanBharani K Dharan
1
$endgroup$
add a comment |
$begingroup$
Transpose here is used for matching the columns of the X with rows of theta.
Ex:
size of
X=97x2;
y=97x1;
theta=2x1;
first calc is X * theta. The size of the resulting matrix will be 97x1. Then, the sub of two same size matrices. Now, we have to multiply X with the matrix obtained from the previous step. But, the sizes are different
(97x1) * (97x2)
Thus transposing the first matrix makes multiplication possible.
This results in a new matrix of size 1x2 (row vector). But, theta is of size 2x1 (column vector). Hence the final transpose.
answered Mar 3 '17 at 20:28
Bharani K DharanBharani K Dharan
1
$endgroup$
Transpose here is used for matching the columns of the X with rows of theta.
Ex:
size of
X=97x2;
y=97x1;
theta=2x1;
first calc is X * theta. The size of the resulting matrix will be 97x1. Then, the sub of two same size matrices. Now, we have to multiply X with the matrix obtained from the previous step. But, the sizes are different
(97x1) * (97x2)
Thus transposing the first matrix makes multiplication possible.
This results in a new matrix of size 1x2 (row vector). But, theta is of size 2x1 (column vector). Hence the final transpose.
answered Mar 3 '17 at 20:28
Bharani K DharanBharani K Dharan
1
answered Mar 3 '17 at 20:28
Bharani K DharanBharani K Dharan
1
answered Mar 3 '17 at 20:28
Bharani K DharanBharani K Dharan
1
answered Mar 3 '17 at 20:28
Bharani K DharanBharani K Dharan
1
1
add a comment |
add a comment |
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$begingroup$
in octave/matlab, $x = u+v*w$ can be : $x,u,w$ : 3 vectors, $v$ : a matrix with $v*w$ the multiplication of a matrix with a vector. the main idea of matlab is that the basic datatypes instead of being integers and floating point numbers, are arrays / matrices of numbers.
$endgroup$
– reuns
May 25 '16 at 10:40
$begingroup$
In Octave, $X'$ corresponds to the transpose of the matrix (or the vector) $X$.
$endgroup$
– zuggg
May 25 '16 at 11:36
$begingroup$
oh ok so X' means transpose of X. is there someone who knows gradient descent ? I do not understand why they used transpose to find theta here
$endgroup$
– james Miler
May 26 '16 at 0:28