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I came across an article where the authors did a Principal Component Analysis on gene expression data, and found out the genes that are most correlated to the 1st principal component, and they used that gene list for further analysis. Can somebody tell me how to find out entities (genes in this case) that are most correlated to the 1st principal component?

Here's the link to the original free article, and this is how they've calculated it:

Results of the gene set enrichment analysis of the genes most correlated to the 1st principal component. The correlations of the genes with the 1st principal component were transformed to SDs from the mean, then genes with values > 1.5 (positive correlation) or < -1.5 (negative correlation) were selected.

I can make a toy example of 6 samples and 20 gene matrix and do the PCA the following way but how to proceed next:

rm(list=ls())
set.seed(12345)
my.mat <- matrix(rnorm(120,0,0.5),nrow=6,byrow=TRUE)
rownames(my.mat) <- paste("s",1:6,sep="")
colnames(my.mat) <- paste("g",1:20,sep="")
head(my.mat)

#Ensure that input data is Z-transformed 
pca.object <- prcomp(my.mat,center=TRUE,scale.=TRUE)
summary(pca.object)
par(mfrow=c(1,2))
plot(pca.object)
biplot(pca.object)

#The Rotation
pca.object$rotation
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2 Answers 2

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Summary: if the original variables were standardized, then you should simply look at the first principal axis (rotation in the R terminology) and select variables with highest absolute values.


Consider dataset $\mathbf{X}$ with centered variables in columns and $N$ data points in rows. Performing PCA of this dataset amounts to singular value decomposition $\mathbf{X} = \mathbf{U} \mathbf{S} \mathbf{V}^\top$. Here columns of $\mathbf{V}$ are principal axes, $\mathbf{S}$ is a diagonal matrix with singular values, and columns of $\mathbf{U}$ are principal components scaled to unit norm. Standardized PCs are given by $\sqrt{N-1}\mathbf{U}$. PCs themselves (also known as "scores") are given by columns of $\mathbf{US}$.

Note that covariance matrix is given by $\frac{1}{N-1}\mathbf{X}^\top \mathbf{X} = \mathbf{V}\frac{\mathbf{S}^2}{{N-1}}\mathbf{V}^\top=\mathbf{VEV}^\top$, so principal axes $\mathbf{V}$ are eigenvectors of the covariance matrix and $\mathbf E=\frac{\mathbf S^2}{N-1}$ are its eigenvalues.

We can now compute cross-covariance matrix between original variables and standardized PCs: $$\frac{1}{N-1}\mathbf{X}^\top(\sqrt{N-1}\mathbf{U}) = \frac{1}{\sqrt{N-1}}\mathbf{V}\mathbf{S}\mathbf{U}^\top\mathbf{U} = \mathbf{V}\frac{\mathbf{S}}{\sqrt{N-1}}=\mathbf{V}\sqrt{\mathbf E}=\mathbf{L}.$$ This matrix is called loadings matrix: it is given by the eigenvectors of the covariance matrix scaled by the square roots of the respective eigenvalues.

Cross-correlation matrix between original variables and PCs is given by the same expression divided by the standard deviations of the original variables (by definition of correlation). If the original variables were standardized prior to performing PCA (i.e. PCA was performed on the correlation matrix) they are all equal to $1$. In this last case the cross-correlation matrix is again given simply by $\mathbf{L}$.

You are only interested in the top correlations with the first PC, which means that you should look at the first column of $\mathbf{L}$ and select variables with highest absolute values. But notice that the first column of $\mathbf{L}$ is equal to the first column of $\mathbf{V}$, up to a scaling factor (given by the square root of the first eigenvalue of the covariance matrix). So equivalently, you can look at the first column of $\mathbf{V}$ (i.e. the first principal axis, or rotation in the R terminology) and select variables with highest absolute values.

But again, this is only true if the original variables were standardized.

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  • $\begingroup$ 1. I'm bit confused. My data set is only centered since all data points are in same scale and I want to preserve most info - which I think, will be lost if I standardize the data. Now I want to get the contribution of variables to PC1. Should I look at the first column of L or V, Are they both equivalent in centered case as well? I might be wrong here. Please feel free to correct me. [Continued] $\endgroup$
    – AshlinJP
    Commented Jun 30, 2020 at 17:12
  • $\begingroup$ 2. I was reading through @ttnphs answer - stats.stackexchange.com/a/143949/254880 - and was confused with the last part,"Eigenvector value squared has the meaning of the contribution of a variable into a pr. component; if it is high (close to 1) the component is well defined by that variable alone.". Can you comment on this as well? Thanks $\endgroup$
    – AshlinJP
    Commented Jun 30, 2020 at 17:13
  • $\begingroup$ If the original variables were not standardized prior to PCA, can one still retrieve their precise influence on a specific PC (if not precise, then approximated)? $\endgroup$
    – Sam
    Commented Feb 13 at 15:21
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I got some more clarification on the problem after reading this link on correlations between the principal components and the original variables

It appears to me it is same as finding those variables which contribute mostly to the principal component. So after scaling/z-transform , variable values that are 1.5σ away from mean(σ equals one and mean zero in this case) were considered most important contributors. so the following lines of code will do the job I guess:

rot1.scaled <- scale(pca.object$rotation[,1])
names( which(rot1.scaled[,1] > 1.5 | rot1.scaled[,1] < -1.5) )

In another way of selecting, if I choose the Top 'N' contributing genes, the code would be:

topN <- 5
load.rot <- pca.object$rotation
names(load.rot[,1][order(abs(load.rot[,1]),decreasing=TRUE)][1:topN])

Please correct me if I'm wrong.

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    $\begingroup$ That is correct, but only if the original variables were standardized prior to PCA. pca.object$rotation contains eigenvectors of the covariance matrix (i.e. principal axes), presumably normalized to unit length. If the first eigenvector is $\mathbf{v}$, then covariance between original variables and standardized principal components are proportional to the elements of this vector (coefficient of proportionality is given by the standard deviation $s$ of this PC). But this is equal to correlation only if the variables are standardized. $\endgroup$
    – amoeba
    Commented Sep 11, 2014 at 11:38
  • $\begingroup$ Thanks, it will be great if you can post as a separate answer, still better if you can provide some links, notes, equations etc. $\endgroup$
    – The August
    Commented Sep 11, 2014 at 15:08

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