I have a provided standardize function for a machine learning course that wasn't well documented and I'm still new to MATLAB so I'm just trying to break down the function. Any explanation of the syntax or the general idea of standardizing would greatly help. We use this function to standardize a set of training data provided in a large matrix. A break down of most of the lines of the code snippet would help me greatly. Thank you so much.
function [X, mean_X, std_X] = standardize(varargin)
switch nargin
case 1
mean_X = mean(varargin{1});
std_X = std(varargin{1});
X = varargin{1} - repmat(mean_X, [size(varargin{1}, 1) 1]);
for i = 1:size(X, 2)
X(:, i) = X(:, i) / std(X(:, i));
end
case 3
mean_X = varargin{2};
std_X = varargin{3};
X = varargin{1} - repmat(mean_X, [size(varargin{1}, 1) 1]);
for i = 1:size(X, 2)
X(:, i) = X(:, i) / std_X(:, i);
end
end
This code accepts a data matrix of size M x N, where M is the dimensionality of one data sample from this matrix and N is the total number of samples. Therefore, one column of this matrix is one data sample. Data samples are all stacked horizontally and are columns.
Now, the true purpose of this code is to take all of the columns of your matrix and standardize / normalize the data so that each data sample exhibits zero mean and unit variance. This means that after this transform, if you found the mean value of any column in this matrix, it would be 0 and the variance would be 1. This is a very standard method for normalizing values in statistical analysis, machine learning, and computer vision.
This actually comes from the z-score in statistical analysis. Specifically, the equation for normalization is:
Given a set of data points, we subtract the value in question by the mean of these data points, then divide by the respective standard deviation. How you'd call this code is the following. Given this matrix, which we will call X, there are two ways you can call this code:
Method #1: [X, mean_X, std_X] = standardize(X);
Method #2: [X, mean_X, std_X] = standardize(X, mu, sigma);
The first method automatically infers the mean of each column of X and the standard deviation of each column of X. mean_X and std_X will both return 1 x N vectors that give you the mean and standard deviation of each column in the matrix X. The second method allows you to manually specify a mean (mu) and standard deviation (sigma) for each column of X. This is possibly for use in debugging, but you would specify both mu and sigma as 1 x N vectors in this case. What is returned for mean_X and std_X is identical to mu and sigma.
The code is a bit poorly written IMHO, because you can certainly achieve this vectorized, but the gist of the code is that it finds the mean of every column of the matrix X if we are are using Method #1, duplicates this vector so that it becomes a M x N matrix, then we subtract this matrix with X. This will subtract each column by its respective mean. We also compute the standard deviation of each column before the mean subtraction.
Once we do that, we then normalize our X by dividing each column by its respective standard deviation. BTW, doing std_X(:, i) is superfluous as std_X is already a 1 x N vector. std_X(:, i) means to grab all of the rows at the ith column. If we already have a 1 x N vector, this can simply be replaced with std_X(i) - a bit overkill for my taste.
Method #2 performs the same thing as Method #1, but we provide our own mean and standard deviation for each column of X.
For the sake of documentation, this is how I would have commented the code:
function [X, mean_X, std_X] = standardize(varargin)
switch nargin %// Check how many input variables we have input into the function
case 1 %// If only one variable - this is the input matrix
mean_X = mean(varargin{1}); %// Find mean of each column
std_X = std(varargin{1}); %// Find standard deviation of each column
%// Take each column of X and subtract by its corresponding mean
%// Take mean_X and duplicate M times vertically
X = varargin{1} - repmat(mean_X, [size(varargin{1}, 1) 1]);
%// Next, for each column, normalize by its respective standard deviation
for i = 1:size(X, 2)
X(:, i) = X(:, i) / std(X(:, i));
end
case 3 %// If we provide three inputs
mean_X = varargin{2}; %// Second input is a mean vector
std_X = varargin{3}; %// Third input is a standard deviation vector
%// Apply the code as seen in the first case
X = varargin{1} - repmat(mean_X, [size(varargin{1}, 1) 1]);
for i = 1:size(X, 2)
X(:, i) = X(:, i) / std_X(:, i);
end
end
If I can suggest another way to write this code, I would use the mighty and powerful bsxfun function. This avoids having to do any duplication of elements and we can do this under the hood. I would rewrite this function so that it looks like this:
function [X, mean_X, std_X] = standardize(varargin)
switch nargin
case 1
mean_X = mean(varargin{1}); %// Find mean of each column
std_X = std(varargin{1}); %// Find std. dev. of each column
X = bsxfun(#minus, varargin{1}, mean_X); %// Subtract each column by its respective mean
X = bsxfun(#rdivide, X, std_X); %// Take each column and divide by its respective std dev.
case 3
mean_X = varargin{2};
std_X = varargin{3};
%// Same code as above
X = bsxfun(#minus, varargin{1}, mean_X);
X = bsxfun(#rdivide, X, std_X);
end
I would argue that the new code above is much faster than using for and repmat. In fact, it is known that bsxfun is faster than the former approach - especially for larger matrices.
Related
Is there a matlab command for generating a random n by n matrix, with elements taken in the interval [0,1], with x% of the entries on the off-diagonal to be 0. Then, additionally setting the element in the diagonal to be the sum of every element in its respective column? In order to create a diagonally dominant dense/sparse matrix? This may be easy enough to write a code for but I was wondering if there was already a built in function with this capability.
EDIT:
I am new to Matlab/programming so this was an easier said than done. I'm having trouble making the matrix with the percentage ignoring the diagonal. It's a n x n matrix, so there are $n^2$ entries, with n of them on the diagonal, I want the percentage of zeros to be taken from $n^2 - n$ elements, i.e. all the off-diagonal elements. I cannot implement this correctly. I do not know how to initialize my M (see below) to correspond correctly.
% Enter percentage as a decimal
function [M] = DiagDomSparse(n,x)
M = rand(n);
disp("Original matrix");
disp(M);
x = sum(M);
for i=1:n
for j=1:n
if(i == j)
M(i,j) = x(i);
end
end
end
disp(M);
Here is one approach that you could use. I'm sure you will get some other answers now with a more clever approach, but I like to keep things simple and understandable.
What I'm doing below is creating the data to be put in the off-diagonal elements first. I create an empty matrix and copy this data into the off-diagonal elements using linear indexing. Now I can compute the sum of columns and write those into the diagonal elements using linear indexing again. Because the matrix was initialized to zero, the diagonal elements are still zero when I compute the sum of columns, so they don't interfere.
n = 5;
x = 0.3; % fraction of zeros in off-diagonal
k = round(n*(n-1)*x); % number of zeros in off-diagonal
data = randn(n*(n-1)-k,1); % random numbers, pick your distribution here!
data = [data;zeros(k,1)]; % the k zeros
data = data(randperm(length(data))); % shuffle
diag_index = 1:n+1:n*n; % linear index to all diagonal elements
offd_index = setdiff(1:n*n,diag_index); % linear index to all other elements
M = zeros(n,n);
M(offd_index) = data; % set off-diagonal elements to data
M(diag_index) = sum(M,1); % set diagonal elements to sum of columns
To refer to the diagonal you want eye(n,'logical'). Here is a solution:
n=5;
M = rand(n);
disp("Original matrix");
disp(M);
x = sum(M);
for i=1:n
for j=1:n
if(i == j)
M(i,j) = x(i);
end
end
end
disp('loop solution:')
disp(M);
M(eye(n,'logical'))=x;
disp('eye solution:')
disp(M);
I want to solve the following system of equations shown in the image below,
The matrix system
where the component of the matrix A is complex numbers with the angle (theta) runs from 0 to 2*pi which has m divisions, and n = 9. The known value z = x + iy. Suppose the x and y of matrix z is
z =
0 1.0148
0.1736 0.9848
0.3420 0.9397
0.5047 0.8742
0.6748 0.8042
0.8419 0.7065
0.9919 0.5727
1.1049 0.4022
1.1757 0.2073
1.1999 0
1.1757 -0.2073
1.1049 -0.4022
0.9919 -0.5727
0.8419 -0.7065
0.6748 -0.8042
0.5047 -0.8742
0.3420 -0.9397
0.1736 -0.9848
0 -1.0148
How do you solve them iteratively? Notice that the value of the first component of the desired constants must equal 1. I am working with Matlab.
You can apply simple multilinear regression for complex valued data.
Step 1. Get the matrix ready for linear regression
Your linear system
written without matrices, becomes
that rearranged yelds
If you rewrite it with matrices you get
Step 2. Apply multiple linear regression
Let the system above be
where
Now you can apply linear regression, that returns the best fit for α when
where
is the conjugate transpose.
In MATLAB
Y = Z - A(:,1); % Calculate Y subtracting the first col of A from Z
R = A(:,:); R(:,1) = []; % Calculate R as an exact copy of A, just without first column
Rs = ctranspose(R); % Calculate R-star (conjugate transpose of R)
alpha = (Rs*R)^(-1)*Rs*Y; % Finally apply multiple linear regression
alpha = cat(1, 1, alpha); % Add alpha1 back, whose value is 1
or, if you prefer built-ins, have a look at regress function:
Y = Z - A(:,1); % Calculate Y subtracting the first col of A from Z
R = A(:,:); R(:,1) = []; % Calculate R as an exact copy of A, just without first column
alpha = regress(Y, R); % Finally apply multiple linear regression
alpha = cat(1, 1, alpha); % Add alpha1 back, whose value is 1
I have a provided standardize function for a machine learning course that wasn't well documented and I'm still new to MATLAB so I'm just trying to break down the function. Any explanation of the syntax or the general idea of standardizing would greatly help. We use this function to standardize a set of training data provided in a large matrix. A break down of most of the lines of the code snippet would help me greatly. Thank you so much.
function [X, mean_X, std_X] = standardize(varargin)
switch nargin
case 1
mean_X = mean(varargin{1});
std_X = std(varargin{1});
X = varargin{1} - repmat(mean_X, [size(varargin{1}, 1) 1]);
for i = 1:size(X, 2)
X(:, i) = X(:, i) / std(X(:, i));
end
case 3
mean_X = varargin{2};
std_X = varargin{3};
X = varargin{1} - repmat(mean_X, [size(varargin{1}, 1) 1]);
for i = 1:size(X, 2)
X(:, i) = X(:, i) / std_X(:, i);
end
end
This code accepts a data matrix of size M x N, where M is the dimensionality of one data sample from this matrix and N is the total number of samples. Therefore, one column of this matrix is one data sample. Data samples are all stacked horizontally and are columns.
Now, the true purpose of this code is to take all of the columns of your matrix and standardize / normalize the data so that each data sample exhibits zero mean and unit variance. This means that after this transform, if you found the mean value of any column in this matrix, it would be 0 and the variance would be 1. This is a very standard method for normalizing values in statistical analysis, machine learning, and computer vision.
This actually comes from the z-score in statistical analysis. Specifically, the equation for normalization is:
Given a set of data points, we subtract the value in question by the mean of these data points, then divide by the respective standard deviation. How you'd call this code is the following. Given this matrix, which we will call X, there are two ways you can call this code:
Method #1: [X, mean_X, std_X] = standardize(X);
Method #2: [X, mean_X, std_X] = standardize(X, mu, sigma);
The first method automatically infers the mean of each column of X and the standard deviation of each column of X. mean_X and std_X will both return 1 x N vectors that give you the mean and standard deviation of each column in the matrix X. The second method allows you to manually specify a mean (mu) and standard deviation (sigma) for each column of X. This is possibly for use in debugging, but you would specify both mu and sigma as 1 x N vectors in this case. What is returned for mean_X and std_X is identical to mu and sigma.
The code is a bit poorly written IMHO, because you can certainly achieve this vectorized, but the gist of the code is that it finds the mean of every column of the matrix X if we are are using Method #1, duplicates this vector so that it becomes a M x N matrix, then we subtract this matrix with X. This will subtract each column by its respective mean. We also compute the standard deviation of each column before the mean subtraction.
Once we do that, we then normalize our X by dividing each column by its respective standard deviation. BTW, doing std_X(:, i) is superfluous as std_X is already a 1 x N vector. std_X(:, i) means to grab all of the rows at the ith column. If we already have a 1 x N vector, this can simply be replaced with std_X(i) - a bit overkill for my taste.
Method #2 performs the same thing as Method #1, but we provide our own mean and standard deviation for each column of X.
For the sake of documentation, this is how I would have commented the code:
function [X, mean_X, std_X] = standardize(varargin)
switch nargin %// Check how many input variables we have input into the function
case 1 %// If only one variable - this is the input matrix
mean_X = mean(varargin{1}); %// Find mean of each column
std_X = std(varargin{1}); %// Find standard deviation of each column
%// Take each column of X and subtract by its corresponding mean
%// Take mean_X and duplicate M times vertically
X = varargin{1} - repmat(mean_X, [size(varargin{1}, 1) 1]);
%// Next, for each column, normalize by its respective standard deviation
for i = 1:size(X, 2)
X(:, i) = X(:, i) / std(X(:, i));
end
case 3 %// If we provide three inputs
mean_X = varargin{2}; %// Second input is a mean vector
std_X = varargin{3}; %// Third input is a standard deviation vector
%// Apply the code as seen in the first case
X = varargin{1} - repmat(mean_X, [size(varargin{1}, 1) 1]);
for i = 1:size(X, 2)
X(:, i) = X(:, i) / std_X(:, i);
end
end
If I can suggest another way to write this code, I would use the mighty and powerful bsxfun function. This avoids having to do any duplication of elements and we can do this under the hood. I would rewrite this function so that it looks like this:
function [X, mean_X, std_X] = standardize(varargin)
switch nargin
case 1
mean_X = mean(varargin{1}); %// Find mean of each column
std_X = std(varargin{1}); %// Find std. dev. of each column
X = bsxfun(#minus, varargin{1}, mean_X); %// Subtract each column by its respective mean
X = bsxfun(#rdivide, X, std_X); %// Take each column and divide by its respective std dev.
case 3
mean_X = varargin{2};
std_X = varargin{3};
%// Same code as above
X = bsxfun(#minus, varargin{1}, mean_X);
X = bsxfun(#rdivide, X, std_X);
end
I would argue that the new code above is much faster than using for and repmat. In fact, it is known that bsxfun is faster than the former approach - especially for larger matrices.
We are given a D-dimensional tensor, represented as a vector of size n^D.
The vector represents a D-dimensional distribution of a random variable X \in {0,1,..,n}^d. That is the (i_1,i_2,...,i_d) entry in the tensor represents the probability of X_1 = i_1, X_2 = i_2, ... X_d = i_d.
I need to compute, for each dimension d, and value i\in [n] the marginal distribution P(X_d = i).
i.e., this means that the answer of P(X_d = i) is the sum of n^(D-1) entries of the vector.
For example, if D=2 and n=4, we have a vector x of size (16,1) and the probability of the first dimension being equal to 1 is
P(X_1 = 1) = x(1) + x(2) + x(3) + x(4)
The probability of the second dimension being equal to 3 is '
P(X_2 = 3) = x(3) + x(7) + x(11) + x(15)
I'm writing Matlab code that needs to compute these marginal distributions, but I'm not familiar enough with Matlab to do it in a simple way (it is doable using some ugly recursion, but there has to be a better option).
To calculate P(X_k=z) for a D-dimensional matrix you can use
xD = reshape(x, n*ones(1,D));
B = permute(xD, [k setdiff(1:D, k)]);
P = sum(B(z,:));
It first makes it a D-dimensional matrix. It brings the dimension of interest k to the beginning and then chooses the z-th element and sums over elements corresponding to that.
Mohsen Nosratinia's answer would be my first option. As an alternative, it can be done without reshaping or permuting dimensions, which can result in faster code:
k = 2; %// chosen dimension
z = 3; %// chosen value (along d-th dimension)
result = sum(x(mod(floor((0:end-1)/n^(k-1)), n)==z-1));
I have a matrix 2NxN.
And I want get some parametrs by this matrix. For example it:
How, I can do it?
You may want to break your 12x6 matrix, into two 6x6 matrix; let's say: Z and Zb (last one for z bar). Odd rows are Z and evens are Zb.
Considering M to be the combined matrices:
Z = M(1:2:end,:)
Zb = M(2:2:end,:)
read about the colon(:) operator and end to see what 1:2:end means.
Hope it helps.
From what I understand here are the first three:
% Random Matrix
% Needs to be defined before the functions since the functions look for
% the m variable
m = rand(12,6);
% Function 1
p = #(i,j) sign(m(i,j)+m(i+1,j)) * max(abs(m(i,j)),abs(m(i+1,j)));
p(2,2)
% Function 2 - Avg of row
pavg = #(i) mean(m(i,:));
pavg(2)
% Function 3
c = #(i,j) abs(m(i,j)+m(i+1,j)) / (abs(m(i,j)) + abs(m(i+1,j)));
c(2,2)