I have a feature vector of size [4096 x 180], where 180 is the number of samples and 4096 is the feature vector length of each sample.
I want to reduce the dimensionality of the data using PCA.
I tried using the built in pca function of MATLAB [V U]=pca(X) and reconstructed the data by X_rec= U(:, 1:n)*V(:, 1:n)', n being the dimension I chose. This returns a matrix of 4096 x 180.
Now I have 3 questions:
How to obtain the reduced dimension?
When I put n as 200, it gave an error as matrix dimension increased, which gave me the assumption that we cannot reduce dimension lesser than the sample size. Is this true?
How to find the right number of reduced dimensions?
I have to use the reduced dimension feature set for further classification.
If anyone can provide a detailed step by step explanation of the pca code for this I would be grateful. I have looked at many places but my confusion still persists.
You may want to refer to Matlab example to analyse city data.
Here is some simplified code:
load cities;
[~, pca_scores, ~, ~, var_explained] = pca(ratings);
Here, pca_scores are the pca components with respective variances of each component in var_explained. You do not need to do any explicit multiplication after running pca. Matlab will give you the components directly.
In your case, consider that data X is a 4096-by-180 matrix, i.e. you have 4096 samples and 180 features. Your goal is to reduce dimensionality such that you have p features, where p < 180. In Matlab, you can simply run the following,
p = 100;
[~, pca_scores, ~, ~, var_explained] = pca(X, 'NumComponents', p);
pca_scores will be a 4096-by-p matrix and var_explained will be a vector of length p.
To answer your questions:
How to obtain the reduced dimension? I above example, pca_scores is your reduced dimension data.
When I put n as 200, it gave an error as matrix dimension increased, which gave me the assumption that we cannot reduce dimension lesser than the sample size. Is this true? You can't use 200, since the reduced dimensions have to be less than 180.
How to find the right number of reduced dimensions? You can make this decision by checking the var_explained vector. Typically you want to retain about 99% variance of the features. You can read more about this here.
Related
I have a matrix like M = K x N ,where k is 49152 and is the dimension of the problem and N is 52 and is the number of observations.
I have tried to use [U,S,V]=SVD(M) but doing this I get less memory space.
I found another code which uses [U,S,V]=SVD(COV(M)) and it works well. My questions are what is the meaning of using the COV(M) command inside the SVD and what is the meaning of the resultant [U,S,V]?
Finding the SVD of the covariance matrix is a method to perform Principal Components Analysis or PCA for short. I won't get into the mathematical details here, but PCA performs what is known as dimensionality reduction. If you like a more formal treatise on the subject, you can read up on my post about it here: What does selecting the largest eigenvalues and eigenvectors in the covariance matrix mean in data analysis?. However, simply put dimensionality reduction projects your data stored in the matrix M onto a lower dimensional surface with the least amount of projection error. In this matrix, we are assuming that each column is a feature or a dimension and each row is a data point. I suspect the reason why you are getting more memory occupied by applying the SVD on the actual data matrix M itself rather than the covariance matrix is because you have a significant amount of data points with a small amount of features. The covariance matrix finds the covariance between pairs of features. If M is a m x n matrix where m is the total number of data points and n is the total number of features, doing cov(M) would actually give you a n x n matrix, so you are applying SVD on a small amount of memory in comparison to M.
As for the meaning of U, S and V, for dimensionality reduction specifically, the columns of V are what are known as the principal components. The ordering of V is in such a way where the first column is the first axis of your data that describes the greatest amount of variability possible. As you start going to the second columns up to the nth column, you start to introduce more axes in your data and the variability starts to decrease. Eventually when you hit the nth column, you are essentially describing your data in its entirety without reducing any dimensions. The diagonal values of S denote what is called the variance explained which respect the same ordering as V. As you progress through the singular values, they tell you how much of the variability in your data is described by each corresponding principal component.
To perform the dimensionality reduction, you can either take U and multiply by S or take your data that is mean subtracted and multiply by V. In other words, supposing X is the matrix M where each column has its mean computed and the is subtracted from each column of M, the following relationship holds:
US = XV
To actually perform the final dimensionality reduction, you take either US or XV and retain the first k columns where k is the total amount of dimensions you want to retain. The value of k depends on your application, but many people choose k to be the total number of principal components that explains a certain percentage of your variability in your data.
For more information about the link between SVD and PCA, please see this post on Cross Validated: https://stats.stackexchange.com/q/134282/86678
Instead of [U, S, V] = svd(M), which tries to build a matrix U that is 49152 by 49152 (= 18 GB 😱!), do svd(M, 'econ'). That returns the “economy-class” SVD, where U will be 52 by 52, S is 52 by 52, and V is also 52 by 52.
cov(M) will remove each dimension’s mean and evaluate the inner product, giving you a 52 by 52 covariance matrix. You can implement your own version of cov, called mycov, as
function [C] = mycov(M)
M = bsxfun(#minus, M, mean(M, 1)); % subtract each dimension’s mean over all observations
C = M' * M / size(M, 1);
(You can verify this works by looking at mycov(randn(49152, 52)), which should be close to eye(52), since each element of that array is IID-Gaussian.)
There’s a lot of magical linear algebraic properties and relationships between the SVD and EVD (i.e., singular value vs eigenvalue decompositions): because the covariance matrix cov(M) is a Hermitian matrix, it’s left- and right-singular vectors are the same, and in fact also cov(M)’s eigenvectors. Furthermore, cov(M)’s singular values are also its eigenvalues: so svd(cov(M)) is just an expensive way to get eig(cov(M)) 😂, up to ±1 and reordering.
As #rayryeng explains at length, usually people look at svd(M, 'econ') because they want eig(cov(M)) without needing to evaluate cov(M), because you never want to compute cov(M): it’s numerically unstable. I recently wrote an answer that showed, in Python, how to compute eig(cov(M)) using svd(M2, 'econ'), where M2 is the 0-mean version of M, used in the practical application of color-to-grayscale mapping, which might help you get more context.
Suppose there is a matrix B, where its size is a 500*1000 double(Here, 500 represents the number of observations and 1000 represents the number of features).
sigma is the covariance matrix of B, and D is a diagonal matrix whose diagonal elements are the eigenvalues of sigma. Assume A is the eigenvectors of the covariance matrix sigma.
I have the following questions:
I need to select the first k = 800 eigenvectors corresponding to the eigenvalues with the largest magnitude to rank the selected features. The final matrix named Aq. How can I do this in MATLAB?
What is the meaning of these selected eigenvectors?
It seems the size of the final matrix Aq is 1000*800 double once I calculate Aq. The time points/observation information of 500 has disappeared. For the final matrix Aq, what does the value 1000 in matrix Aq represent now? Also, what does the value 800 in matrix Aq represent now?
I'm assuming you determined the eigenvectors from the eig function. What I would recommend to you in the future is to use the eigs function. This not only computes the eigenvalues and eigenvectors for you, but it will compute the k largest eigenvalues with their associated eigenvectors for you. This may save computational overhead where you don't have to compute all of the eigenvalues and associated eigenvectors of your matrix as you only want a subset. You simply supply the covariance matrix of your data to eigs and it returns the k largest eigenvalues and eigenvectors for you.
Now, back to your problem, what you are describing is ultimately Principal Component Analysis. The mechanics behind this would be to compute the covariance matrix of your data and find the eigenvalues and eigenvectors of the computed result. It has been known that doing it this way is not recommended due to numerical instability with computing the eigenvalues and eigenvectors for large matrices. The most canonical way to do this now is via Singular Value Decomposition. Concretely, the columns of the V matrix give you the eigenvectors of the covariance matrix, or the principal components, and the associated eigenvalues are the square root of the singular values produced in the diagonals of the matrix S.
See this informative post on Cross Validated as to why this is preferred:
https://stats.stackexchange.com/questions/79043/why-pca-of-data-by-means-of-svd-of-the-data
I'll throw in another link as well that talks about the theory behind why the Singular Value Decomposition is used in Principal Component Analysis:
https://stats.stackexchange.com/questions/134282/relationship-between-svd-and-pca-how-to-use-svd-to-perform-pca
Now let's answer your question one at a time.
Question #1
MATLAB generates the eigenvalues and the corresponding ordering of the eigenvectors in such a way where they are unsorted. If you wish to select out the largest k eigenvalues and associated eigenvectors given the output of eig (800 in your example), you'll need to sort the eigenvalues in descending order, then rearrange the columns of the eigenvector matrix produced from eig then select out the first k values.
I should also note that using eigs will not guarantee sorted order, so you will have to explicitly sort these too when it comes down to it.
In MATLAB, doing what we described above would look something like this:
sigma = cov(B);
[A,D] = eig(sigma);
vals = diag(D);
[~,ind] = sort(abs(vals), 'descend');
Asort = A(:,ind);
It's a good thing to note that you do the sorting on the absolute value of the eigenvalues because scaled eigenvalues are also eigenvalues themselves. These scales also include negatives. This means that if we had a component whose eigenvalue was, say -10000, this is a very good indication that this component has some significant meaning to your data, and if we sorted purely on the numbers themselves, this gets placed near the lower ranks.
The first line of code finds the covariance matrix of B, even though you said it's already stored in sigma, but let's make this reproducible. Next, we find the eigenvalues of your covariance matrix and the associated eigenvectors. Take note that each column of the eigenvector matrix A represents one eigenvector. Specifically, the ith column / eigenvector of A corresponds to the ith eigenvalue seen in D.
However, the eigenvalues are in a diagonal matrix, so we extract out the diagonals with the diag command, sort them and figure out their ordering, then rearrange A to respect this ordering. I use the second output of sort because it tells you the position of where each value in the unsorted result would appear in the sorted result. This is the ordering we need to rearrange the columns of the eigenvector matrix A. It's imperative that you choose 'descend' as the flag so that the largest eigenvalue and associated eigenvector appear first, just like we talked about before.
You can then pluck out the first k largest vectors and values via:
k = 800;
Aq = Asort(:,1:k);
Question #2
It's a well known fact that the eigenvectors of the covariance matrix are equal to the principal components. Concretely, the first principal component (i.e. the largest eigenvector and associated largest eigenvalue) gives you the direction of the maximum variability in your data. Each principal component after that gives you variability of a decreasing nature. It's also good to note that each principal component is orthogonal to each other.
Here's a good example from Wikipedia for two dimensional data:
I pulled the above image from the Wikipedia article on Principal Component Analysis, which I linked you to above. This is a scatter plot of samples that are distributed according to a bivariate Gaussian distribution centred at (1,3) with a standard deviation of 3 in roughly the (0.878, 0.478) direction and of 1 in the orthogonal direction. The component with a standard deviation of 3 is the first principal component while the one that is orthogonal is the second component. The vectors shown are the eigenvectors of the covariance matrix scaled by the square root of the corresponding eigenvalue, and shifted so their tails are at the mean.
Now let's get back to your question. The reason why we take a look at the k largest eigenvalues is a way of performing dimensionality reduction. Essentially, you would be performing a data compression where you would take your higher dimensional data and project them onto a lower dimensional space. The more principal components you include in your projection, the more it will resemble the original data. It actually begins to taper off at a certain point, but the first few principal components allow you to faithfully reconstruct your data for the most part.
A great visual example of performing PCA (or SVD rather) and data reconstruction is found by this great Quora post I stumbled upon in the past.
http://qr.ae/RAEU8a
Question #3
You would use this matrix to reproject your higher dimensional data onto a lower dimensional space. The number of rows being 1000 is still there, which means that there were originally 1000 features in your dataset. The 800 is what the reduced dimensionality of your data would be. Consider this matrix as a transformation from the original dimensionality of a feature (1000) down to its reduced dimensionality (800).
You would then use this matrix in conjunction with reconstructing what the original data was. Concretely, this would give you an approximation of what the original data looked like with the least amount of error. In this case, you don't need to use all of the principal components (i.e. just the k largest vectors) and you can create an approximation of your data with less information than what you had before.
How you reconstruct your data is very simple. Let's talk about the forward and reverse operations first with the full data. The forward operation is to take your original data and reproject it but instead of the lower dimensionality, we will use all of the components. You first need to have your original data but mean subtracted:
Bm = bsxfun(#minus, B, mean(B,1));
Bm will produce a matrix where each feature of every sample is mean subtracted. bsxfun allows the subtraction of two matrices in unequal dimension provided that you can broadcast the dimensions so that they can both match up. Specifically, what will happen in this case is that the mean of each column / feature of B will be computed and a temporary replicated matrix will be produced that is as large as B. When you subtract your original data with this replicated matrix, the effect will subtract every data point with their respective feature means, thus decentralizing your data so that the mean of each feature is 0.
Once you do this, the operation to project is simply:
Bproject = Bm*Asort;
The above operation is quite simple. What you are doing is expressing each sample's feature as a linear combination of principal components. For example, given the first sample or first row of the decentralized data, the first sample's feature in the projected domain is a dot product of the row vector that pertains to the entire sample and the first principal component which is a column vector.. The first sample's second feature in the projected domain is a weighted sum of the entire sample and the second component. You would repeat this for all samples and all principal components. In effect, you are reprojecting the data so that it is with respect to the principal components - which are orthogonal basis vectors that transform your data from one representation to another.
A better description of what I just talked about can be found here. Look at Amro's answer:
Matlab Principal Component Analysis (eigenvalues order)
Now to go backwards, you simply do the inverse operation, but a special property with the eigenvector matrix is that if you transpose this, you get the inverse. To get the original data back, you undo the operation above and add the means back to the problem:
out = bsxfun(#plus, Bproject*Asort.', mean(B, 1));
You want to get the original data back, so you're solving for Bm with respect to the previous operation that I did. However, the inverse of Asort is just the transpose here. What's happening after you perform this operation is that you are getting the original data back, but the data is still decentralized. To get the original data back, you must add the means of each feature back into the data matrix to get the final result. That's why we're using another bsxfun call here so that you can do this for each sample's feature values.
You should be able to go back and forth from the original domain and projected domain with the above two lines of code. Now where the dimensionality reduction (or the approximation of the original data) comes into play is the reverse operation. What you need to do first is project the data onto the bases of the principal components (i.e. the forward operation), but now to go back to the original domain where we are trying to reconstruct the data with a reduced number of principal components, you simply replace Asort in the above code with Aq and also reduce the amount of features you're using in Bproject. Concretely:
out = bsxfun(#plus, Bproject(:,1:k)*Aq.', mean(B, 1));
Doing Bproject(:,1:k) selects out the k features in the projected domain of your data, corresponding to the k largest eigenvectors. Interestingly, if you just want the representation of the data with regards to a reduced dimensionality, you can just use Bproject(:,1:k) and that'll be enough. However, if you want to go forward and compute an approximation of the original data, we need to compute the reverse step. The above code is simply what we had before with the full dimensionality of your data, but we use Aq as well as selecting out the k features in Bproject. This will give you the original data that is represented by the k largest eigenvectors / eigenvalues in your matrix.
If you'd like to see an awesome example, I'll mimic the Quora post that I linked to you but using another image. Consider doing this with a grayscale image where each row is a "sample" and each column is a feature. Let's take the cameraman image that's part of the image processing toolbox:
im = imread('camerman.tif');
imshow(im); %// Using the image processing toolbox
We get this image:
This is a 256 x 256 image, which means that we have 256 data points and each point has 256 features. What I'm going to do is convert the image to double for precision in computing the covariance matrix. Now what I'm going to do is repeat the above code, but incrementally increasing k at each go from 3, 11, 15, 25, 45, 65 and 125. Therefore, for each k, we are introducing more principal components and we should slowly start to get a reconstruction of our data.
Here's some runnable code that illustrates my point:
%%%%%%%// Pre-processing stage
clear all;
close all;
%// Read in image - make sure we cast to double
B = double(imread('cameraman.tif'));
%// Calculate covariance matrix
sigma = cov(B);
%// Find eigenvalues and eigenvectors of the covariance matrix
[A,D] = eig(sigma);
vals = diag(D);
%// Sort their eigenvalues
[~,ind] = sort(abs(vals), 'descend');
%// Rearrange eigenvectors
Asort = A(:,ind);
%// Find mean subtracted data
Bm = bsxfun(#minus, B, mean(B,1));
%// Reproject data onto principal components
Bproject = Bm*Asort;
%%%%%%%// Begin reconstruction logic
figure;
counter = 1;
for k = [3 11 15 25 45 65 125 155]
%// Extract out highest k eigenvectors
Aq = Asort(:,1:k);
%// Project back onto original domain
out = bsxfun(#plus, Bproject(:,1:k)*Aq.', mean(B, 1));
%// Place projection onto right slot and show the image
subplot(4, 2, counter);
counter = counter + 1;
imshow(out,[]);
title(['k = ' num2str(k)]);
end
As you can see, the majority of the code is the same from what we have seen. What's different is that I loop over all values of k, project back onto the original space (i.e. computing the approximation) with the k highest eigenvectors, then show the image.
We get this nice figure:
As you can see, starting with k=3 doesn't really do us any favours... we can see some general structure, but it wouldn't hurt to add more in. As we start increasing the number of components, we start to get a clearer picture of what the original data looks like. At k=25, we actually can see what the cameraman looks like perfectly, and we don't need components 26 and beyond to see what's happening. This is what I was talking about with regards to data compression where you don't need to work on all of the principal components to get a clear picture of what's going on.
I'd like to end this note by referring you to Chris Taylor's wonderful exposition on the topic of Principal Components Analysis, with code, graphs and a great explanation to boot! This is where I got started on PCA, but the Quora post is what solidified my knowledge.
Matlab - PCA analysis and reconstruction of multi dimensional data
My data matrix is X which is 4999*37152. Then I use this command in Matlab:
[coeff, score, latent, tsquared1, explained1] = pca(X);
The output: coeff is 37152*4998, score is 4999*4998, latent is 4998*1. According to http://www.mathworks.com/help/stats/pca.html, the coeff should be p*p. So what is wrong with my code ?
As Matlab documentation says, "Rows of X correspond to observations and columns correspond to variables". So you are feeding in a matrix with only 4999 observations for 37152 observations. Geometrically, you have 4999 points in a 37152-dimensional space. These points are contained in a 4998-dimensional affine subspace, so Matlab gets you 4998 directions there (each expressed as a vector with 37152 components).
For more, see the Statistics site:
Why are there only n-1 principal components for n data points if the number of dimensions is larger than n?
PCA when the dimensionality is greater than the number of samples
The MATLAB documentation is written under assumption that you have at least as many observations as variables, which is how people normally use PCA.
Of course, it's possible that your data actually has 37152 observations for 4999 variables, in which case you need to transpose X.
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MATLAB is running out of memory but it should not be
I want to perform PCA analysis on a huge data set of points. To be more specific, I have size(dataPoints) = [329150 132] where 328150 is the number of data points and 132 are the number of features.
I want to extract the eigenvectors and their corresponding eigenvalues so that I can perform PCA reconstruction.
However, when I am using the princomp function (i.e. [eigenVectors projectedData eigenValues] = princomp(dataPoints); I obtain the following error :
>> [eigenVectors projectedData eigenValues] = princomp(pointsData);
Error using svd
Out of memory. Type HELP MEMORY for your options.
Error in princomp (line 86)
[U,sigma,coeff] = svd(x0,econFlag); % put in 1/sqrt(n-1) later
However, if I am using a smaller data set, I have no problem.
How can I perform PCA on my whole dataset in Matlab? Have someone encountered this problem?
Edit:
I have modified the princomp function and tried to use svds instead of svd, but however, I am obtaining pretty much the same error. I have dropped the error bellow :
Error using horzcat
Out of memory. Type HELP MEMORY for your options.
Error in svds (line 65)
B = [sparse(m,m) A; A' sparse(n,n)];
Error in princomp (line 86)
[U,sigma,coeff] = svds(x0,econFlag); % put in 1/sqrt(n-1) later
Solution based on Eigen Decomposition
You can first compute PCA on X'X as #david said. Specifically, see the script below:
sz = [329150 132];
X = rand(sz);
[V D] = eig(X.' * X);
Actually, V holds the right singular vectors, and it holds the principal vectors if you put your data vectors in rows. The eigenvalues, D, are the variances among each direction. The singular vectors, which are the standard deviations, are computed as the square root of the variances:
S = sqrt(D);
Then, the left singular vectors, U, are computed using the formula X = USV'. Note that U refers to the principal components if your data vectors are in columns.
U = X*V*S^(-1);
Let us reconstruct the original data matrix and see the L2 reconstruction error:
X2 = U*S*V';
L2ReconstructionError = norm(X(:)-X2(:))
It is almost zero:
L2ReconstructionError =
6.5143e-012
If your data vectors are in columns and you want to convert your data into eigenspace coefficients, you should do U.'*X.
This code snippet takes around 3 seconds in my moderate 64-bit desktop.
Solution based on Randomized PCA
Alternatively, you can use a faster approximate method which is based on randomized PCA. Please see my answer in Cross Validated. You can directly compute fsvd and get U and V instead of using eig.
You may employ randomized PCA if the data size is too big. But, I think the previous way is sufficient for the size you gave.
My guess is that you have a huge data set. You don't need all of the svd coefficients. In this case, use svds instead of svd :
Taken directly from Matlab help:
s = svds(A,k) computes the k largest singular values and associated singular vectors of matrix A.
From your question, I understand that you don't call svd directly. But you might as well take a look at princomp (It is editable!) and alter the line that calls it.
You probably needed to calculate an n by n matrix in your computation somehow that is to say:
329150 * 329150 * 8btyes ~ 866GB`
of space which explains why you're getting a memory error. There seems to be an efficient way to calculate pca using princomp(X, 'econ') which I suggest you give it a try.
More on this in stackoverflow and mathworks..
Manually compute X'X (132x132) and svd on it. Or find NIPALS script.
I try to calculate the correlation matrix of a set of histogram vectors. But the result is a truncated version of what I (think) I want. I have 200 histograms by 32 bins each. The result from
correlation_matrix = corrcoef(set_of_histograms)
is a 32 by 32 matrix.
I want to use this to calculate how my original histograms match up. (this by later using eigs and other stuff).
But which correlation method is right for this? I have tried "corrcoef" but there are "corr" and "cov" as well. Can't understand their differences by reading matlab help...
correlation_matrix = corrcoef(set_of_histograms')
(Note the ')
1) corrcoef treats every column as an observation, and calculates the correlations between each pair. I'm assuming your histograms matrix is 200x32; hence, in your case, every row is an observation. If you transpose your histograms matrix before running corrcoef, you should get the 200x200 result you're looking for:
[rho, p] = corrcoef( set_of_histograms' );
(' transposes the matrix)
2) cov returns the covariance matrix, not the correlation; while the covariance matrix is used in calculating the correlation, it is not the measure you're looking for.
3) As for corr and corrcoef, they have a few implementation differences between them. As long as you are only interested in Pearson's correlation, they are identical for your purposes. corr also has an option to calculate Spearman's or Kendall's correlations, which corrcoef does not have.