How to apply PCA correctly?
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Hello
I'm currently struggling with PCA and Matlab. Let's say we have a data matrix X and a response y (classification task). X consists of 12 rows and 4 columns. The rows are the data points, the columns are the predictors (features).
Now, I can do PCA with the following command:
[coeff, score] = pca(X);
As I understood from the matlab documentation, coeff contains the loadings and score contains the principal components in the columns. That mean first column of score contains the first principal component (associated with the highest variance) and the first column of coeff contains the loadings for the first principal component.
Is this correct?
But if this is correct, why is then X * coeff not equal to score?
1 commentaire
Sepp @Sepp
your doubt can be clarified by this tutorial (eventhough in another program context) .. specially after 5' in https://www.youtube.com/watch?v=eJ08Gdl5LH0
the cliclist
fabulous and generous explanation
Réponse acceptée
the cyclist
le 12 Déc 2015
Modifié(e) : the cyclist
le 25 Juin 2024
==============================================================================
EDIT: I recommend looking at my answer to this other question for a more detailed discussion of topics mentioned here.
==============================================================================
Maybe this script will help.
rng 'default'
M = 7; % Number of observations
N = 5; % Number of variables observed
X = rand(M,N);
% De-mean
X = bsxfun(@minus,X,mean(X));
% Do the PCA
[coeff,score,latent] = pca(X);
% Calculate eigenvalues and eigenvectors of the covariance matrix
covarianceMatrix = cov(X);
[V,D] = eig(covarianceMatrix);
% "coeff" are the principal component vectors.
% These are the eigenvectors of the covariance matrix.
% Compare the columns of coeff and V.
% (Note that the columns are not necessarily in the same *order*,
% and they might be slightly different from each other
% due to floating-point error.)
coeff
V
% Multiply the original data by the principal component vectors
% to get the projections of the original data on the
% principal component vector space. This is also the output "score".
% Compare ...
dataInPrincipalComponentSpace = X*coeff
score
% The columns of X*coeff are orthogonal to each other. This is shown with ...
corrcoef(dataInPrincipalComponentSpace)
% The variances of these vectors are the eigenvalues of the covariance matrix, and are also the output "latent". Compare
% these three outputs
var(dataInPrincipalComponentSpace)'
latent
sort(diag(D),'descend')
17 commentaires
Sepp
le 12 Déc 2015
the cyclist
le 12 Déc 2015
I think it is difficult to make the language crystal clear. I would trust MATLAB's documentation language more than mine. My script was intended to just illustrate what was going on mathematically.
Yes, I think one would use score as you suggest, although it will only be a dimensional reduction if you use less than the full N.
You can judge how components to keep by seeing how much of the variance is explained by each additional component. (The output explained tells you that.)
Image Analyst
le 12 Déc 2015
The first PC or the one that explains most variation is not necessarily that one that you want. I know this from very recent first hand experience. I wanted to find the "blueness" of an image that had a gamut not exactly aligned with the axes of the gamut is HSV color space. So I took the PCA of the HSV image. The first PC was along the main axis of the gamut and explained most of the variation. But it mostly corresponded to the lightness-to-darkness range of the image and nothing related to how pastel or deep the blue was. The PC2 seemed to correlate well with the saturation ("purity" of the blue color) and this is what I wanted. The third PC was virtually flat noise - nothing of interest.
So you should look at your PCs and figure out which one contains the most relevant information.
Sepp
le 12 Déc 2015
Geetanjali Geetanjali
le 28 Mar 2018
Modifié(e) : Geetanjali Geetanjali
le 28 Mar 2018
in the above code where you load the data & what is m & n ????
Ashesh Shah
le 20 Mar 2019
Nice explaination of PCA using programming. However, It couldn't be understood that why de-mean is taken and what is purpose of bsxfun function? Is this function is inbuilt?
the cyclist
le 22 Mar 2019
Yes, bsxfun is a built-in function. It applies the element-wise operation, implicitly expanding either array, if necessary. With more modern versions of MATLAB, implicit expansion will happen automatically, so one could actually replace that line with
X = X - mean(X);
Jaime de la Mota
le 24 Juil 2019
This is very interesting, but a question comes to my mind. Coeffs are the eigenvectors and scores are the projection of the data in the principal component space.
Are these then equivalent to the eigenfunctions and random variables of the Karhunen-Loève expansion?
the cyclist
le 24 Juil 2019
That's a math question, not a MATLAB question. :-)
I don't really know, but this abstract -- I did not access or read the paper itself -- suggests that KL and PCA are not strictly equivalent.
Xiying Deng
le 14 Avr 2020
Hi. What does these lines do? Thank you!
% "coeff" are the principal component vectors. These are the eigenvectors of the covariance matrix. Compare ...
coeff
V
the cyclist
le 14 Avr 2020
The first line (in green) is a comment that describes what these lines mean.
The second line
coeff
displays the value of the variable coeff to the screen. coeff is one of the outputs of MATLAB's pca function. It is a numeric array in which each column is a principal component vector.
The third line
V
displays the value of the variable V to the screen. V is one of the outputs of MATLAB's eig function. I applied eig to the covariance matrix of the data, to calculate its eigenvectors. My code shows that coeff and V are equal to each other.
V_sal
le 18 Avr 2020
Thanks, you explanation is really good, I'm working with EEG data , and I cant make 'V' and 'coeff' be the same, don't really understand why, do you have any idea?
the cyclist
le 18 Avr 2020
coeff and V are not necessarily identical.
Their columns will be the same (within floating-point error), but those columns will not necessarily appear in the same order.
I've edited the comments in the code above to reflect that.
If that does not fix the issue, maybe you could upload your data and I could take a look.
Keinan Poradosu
le 31 Mar 2022
Hi, thanks for the explanation, however for some reason when I run made up data coeff*X doesnt equal score. Any insights? here's the code:
%%
X=[80,90,30;
90,90,70;
95,85,50;
92,92,20];
[coeff,score,latent,~,explained] = pca(X);
Pca_space_Dat=X*coeff;
%%
What I get is:
Pca_space_Dat =
31.6028 61.1270 103.2703
72.2528 67.1491 106.6326
53.0821 74.7708 101.6937
22.5825 73.5387 106.8180
score =
-13.2773 -8.0194 -1.3334
27.3728 -1.9973 2.0290
8.2020 5.6244 -2.9099
-22.2975 4.3923 2.2144
%%
Thanks in advance
You skipped the step where the means are subtracted:
%%
X=[80,90,30;
90,90,70;
95,85,50;
92,92,20];
% De-mean
X = bsxfun(@minus,X,mean(X)); % <------ YOU MISSED THIS STEP
[coeff,score,latent,~,explained] = pca(X);
Pca_space_Dat=X*coeff
score
The reason for this step is mentioned in the comments above. Also, in more recent versions of MATLAB, you can do
X = X - mean(X);
rather than
X = bsxfun(@minus,X,mean(X));
evelyn
le 21 Juin 2024
Déplacé(e) : the cyclist
le 25 Juin 2024
appreciate the explanation.
Has anyone try when data is complex-value number.
When I generate complex-value number data, the result is equite different when using 'pca function' directly and using eigenvectors of covariance matrix. Could anyone help me! code is shown below.
M = 13; % Number of observations
N = 4; % Number of variables observed
PCA_component_Num = 4;
data_matrix = rand(M,N) + 1i * randn(M,N);
X = data_matrix;
% De-mean
centredX = bsxfun(@minus,X,mean(X))
%% PCA using Matlab function directly
[coeff,score,latent] = pca(centredX);
% latent is eigenvalues of the covariance matrix of X.
% "coeff" are the principal component vectors.
dataInPrincipalComponentSpace = centredX * coeff;
% Multiply the original data by the principal component vectors
% to get the projections of the original data on the
% principal component vector space. This is also the output "score".
% Compare ...
dataInPrincipalComponentSpace
score
% The columns of X*coeff are orthogonal to each other. This is shown with ...
corrcoef(dataInPrincipalComponentSpace)
%% eigenvector
C = cov(centredX);
[W, Lambda] = eig(C);
% Compare the columns of coeff and V.
% (Note that the columns are not necessarily in the same *order*,
% and they might be *lightly different from each other
coeff
W
% The variances of these vectors are the eigenvalues of the covariance matrix, and are also the output "latent". Compare
% these three outputs
var(dataInPrincipalComponentSpace)'
latent
sort(diag(Lambda),'descend')
ev = (diag(Lambda))'; % 提取特征值
ev = ev(:, end:-1:1); % eig计算出的特征值是升序的，这里手动倒序（W同理）
W = W(:, end:-1:1);
sum(W.*W, 1) % 可以验证每个特征向量各元素的平方和均为1
Wr = W(:, 1:4); % 提取前两个主成分的特征向量
% compare Tr with dataInPrincipalComponentSpace
% both PCA component
Tr = centredX * Wr % 新坐标空间的数据点
%% SVD
[U, S, V] = svd(centredX);
PCA_result = centredX* V(:,1:PCA_component_Num)
'PCA_result', 'dataInPrincipalComponentSpace' and 'Tr' are different from each other.(they are same with real-value data)
Besides, I see someone using 'centredX* V(:,1:PCA_component_Num)' while someone using 'U(:,rk)*S(rk,:)*V'' to get principle component. Which one is correct?
the cyclist
le 25 Juin 2024
Short answer
I don't know.
Longer answer
I've never done PCA on complex numbers, and I've never even heard of someone doing PCA on complex numbers. (I didn't find anything on a quick web search, and ChatGPT seems to hallucinate on some of the questions I've asked.)
At first, I thought the issue with your version might be due to the fact that you did not use all the principal components in some of the computation. I also thought there might be severe numerical instability because of cancelation in the random input matrix. But, I don't think either of those explanations are correct. (I've tried other input matrices, and get the same issues.)
I definitely don't understand why one would get different results from SVD, eig(cov matrix), and PCA.
I think this is worth crafting a smaller example with non-random input, and posting a brand-new question.
Plus de réponses (2)
Yaser Khojah
le 17 Avr 2019
2 votes
Dear the cyclist, thanks for showing this example. I have a question regarding to the order of the COEFF since they are different than the V. Is there anyway to see which order of these columns? In another word, what are the variables of each column?
8 commentaires
the cyclist
le 17 Avr 2019
Modifié(e) : the cyclist
le 17 Avr 2019
Quoting from the first section of the documentation for the pca function.
"Each column of coeff contains coefficients for one principal component, and the columns are in descending order of component variance."
You can see that
var(dataInPrincipalComponentSpace)
has values in descending order.
Yaser Khojah
le 17 Avr 2019
i understand that but I do not see how the PC is related to the column of the original data (X). How can I know which variables from the original data has the strength impact?
Nyssa Capman
le 5 Jan 2020
Modifié(e) : Nyssa Capman
le 11 Mar 2020
I believe each row of coeff corresponds to the variables, in the order they were input as.
So, the first column has the coefficients for the 1st* PC, for each variable. The second column has the coefficints for the 2nd PC, for each variable, and so on.
This post is now several months old, and not really the original question, however I was also confused by this when getting started so I wanted to add this in case someone else is confused in the future and finds this post.
*[edited typo from '2nd' to '1st']
Image Analyst
le 5 Jan 2020
"So, the first column has the coefficients for the 2nd PC, for each variable. " ??? Huh? And this is supposed to reduce confusion?
Alex
le 31 Mar 2020
Hello,
I have some doubts on pca.
I have 2 variables with n observations each, and the coeff matrix is the following:
0.9999 -0.00944
0.0094 0.9999
As I understood, the first column represents the coefficient of the first principal component, 0.9999 is for the first variable in the initial matrix and 0.0094 for the second one.
But why the linear combination of coeff*variable does not give the same result as the first column of score?
Thank you
the cyclist
le 31 Mar 2020
As you can see in my code above it is
X * coeff
that should equal score, not
coeff * X
(where X is the de-meaned input to pca).
Yuan Luo
le 8 Nov 2020
why X need to be de-meaned? since pca by defualt will center the data.
the cyclist
le 26 Déc 2020
Sorry it took me a while to see this question.
If you do
[coeff,score] = pca(X);
it is true that pca() will internally de-mean the data. So, score is derived from de-meaned data.
But it does not mean that X itself [outside of pca()] has been de-meaned. So, if you are trying to re-create what happens inside pca(), you need to manually de-mean X first.
Greg Heath
le 13 Déc 2015
0 votes
Hope this helps.
Thank you for formally accepting my answer
Greg
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