First of all, you really do want to center the data. If not, the geometric interpretation of PCA shows that the first principal component will be close to the vector of means and all subsequent PCs will be orthogonal to it, which will prevent them from approximating any PCs that happen to be close to that first vector. We can hope that most of the later PCs will be approximately correct, but the value of that is questionable when it's likely the first several PCs--the most important ones--will be quite wrong.
So, what to do? PCA proceeds by means of a singular value decomposition of the matrix $X$. The essential information will be contained in $X X'$, which in this case is a $10000$ by $10000$ matrix: that may be manageable. Its computation involves about 50 million calculations of dot products of one column with the next.
Consider any two columns, then, $Y$ and $Z$ (each one of them is a $500000$-vector; let this dimension be $n$). Let their means be $m_Y$ and $m_Z$, respectively. What you want to compute is, writing $\mathbf{1}$ for the $n$-vector of $1$'s,
$$(Y - m_Y\mathbf{1}) \cdot (Z - m_Z\mathbf{1}) = Y\cdot Z - m_Z\mathbf{1}\cdot Y - m_Y\mathbf{1}.Z + m_Z m_Y \mathbf{1}\cdot \mathbf{1}\\ = Y\cdot Z -n (m_Ym_Z),$$
because $m_Y = \mathbf{1}\cdot Y / n$ and $m_Z = \mathbf{1}\cdot Z/n$.
This allows you to use sparse matrix techniques to compute $X X'$, whose entries provide the values of $Y\cdot Z$, and then adjust its coefficients based on the $10000$ column means. The adjustment shouldn't hurt, because it seems unlikely $X X'$ will be very sparse.
Example
The following R
code demonstrates this approach. It uses a stub, get.col
, which in practice might read one column of $X$ at a time from an external data source, thereby reducing the amount of RAM required (at some cost in computation speed, of course). It computes PCA in two ways: via SVD applied to the preceding construction and directly using prcomp
. It then compares the output of the two approaches. The computation time is about 50 seconds for 100 columns and scales approximately quadratically: be prepared to wait when performing SVD on a 10K by 10K matrix!
m <- 500000 # Will be 500,000
n <- 100 # will be 10,000
library("Matrix")
x <- as(matrix(pmax(0,rnorm(m*n, mean=-2)), nrow=m), "sparseMatrix")
#
# Compute centered version of x'x by having at most two columns
# of x in memory at any time.
#
get.col <- function(i) x[,i] # Emulates reading a column
system.time({
xt.x <- matrix(numeric(), n, n)
x.means <- rep(numeric(), n)
for (i in 1:n) {
i.col <- get.col(i)
x.means[i] <- mean(i.col)
xt.x[i,i] <- sum(i.col * i.col)
if (i < n) {
for (j in (i+1):n) {
j.col <- get.col(j)
xt.x[i,j] <- xt.x[j,i] <- sum(j.col * i.col)
}
}
}
xt.x <- (xt.x - m * outer(x.means, x.means, `*`)) / (m-1)
svd.0 <- svd(xt.x / m)
}
)
system.time(pca <- prcomp(x, center=TRUE))
#
# Checks: all should be essentially zero.
#
max(abs(pca$center - x.means))
max(abs(xt.x - cov(x)))
max(abs(abs(svd.0$v / pca$rotation) - 1)) # (This is an unstable calculation.)
Best Answer
I believe what you are getting at in your question concerns data truncation using a smaller number of principal components (PC). For such operations, I think the function
prcomp
is more illustrative in that it is easier to visualize the matrix multiplication used in reconstruction.First, give a synthetic dataset,
Xt
, you perform the PCA (typically you would center samples in order to describe PC's relating to a covariance matrix:In the results or
prcomp
, you can see the PC's (res$x
), the eigenvalues (res$sdev
) giving information on the magnitude of each PC, and the loadings (res$rotation
).By squaring the eigenvalues, you get the variance explained by each PC:
Finally, you can create a truncated version of your data by using only the leading (important) PCs:
You can see that the result is a slightly smoother data matrix, with small scale features filtered out:
And here is a very basic approach that you can do outside of the prcomp function:
Now, deciding which PCs to retain is a separate question - one that I was interested in a while back. Hope that helps.