/edit: Further follow up now you can use irlba::prcomp_irlba

/edit: following up on my own post. irlba now has "center" and "scale" arguments, which let you use it to calculate principle components, e.g:

pc <- M %*% irlba(M, nv=5, nu=0, center=colMeans(M), right_only=TRUE)$v

I have a large, sparse Matrix of features I would like to use in a machine learning algorithm:

rows <- 500000
cols <- 10000
i <- unlist(lapply(1:rows, function(i) rep(i, sample(1:5,1))))
j <- sample(1:cols, length(i), replace=TRUE)
M <- sparseMatrix(i, j)

Because this matrix has many columns, I would like to reduce its dimensionality to something more manageable. I can use the excellent irlba package to perform SVD and return the first n principal components (5 shown here; I'll probably use 100 or 500 on my actual dataset):

pc <- irlba(M, nu=5)$u

However, I've read that prior to performing PCA, one should center the matrix (subtract the column mean from each column). This is very difficult to do on my dataset, and furthermore would destroy the sparsity of the matrix.

How "bad" is it to perform SVD on the un-scaled data, and feed it straight into a machine learning algorithm? Are there any efficient ways I could scale this data, while preserving the sparsity of the matrix?

/edit: A brought to my attention by B_miner, the "PCs" should really be:

pc <- M %*% irlba(M, nv=5, nu=0)$v 

Also, I think whuber's answer should be pretty easy to implement, via the crossprod function, which is extremely fast on sparse matrices:

system.time(M_Mt <- crossprod(M)) # 0.463 seconds
system.time(means <- colMeans(M)) #0.003 seconds

Now I'm not quite sure what to do to the means vector before subtracting from M_Mt, but will post as soon as I figure it out.

/edit3: Here's modified version of whuber's code, using sparse matrix operations for each step of the process. If you can store the entire sparse matrix in memory, it works very quickly:

m <- 500000
n <- 100
i <- unlist(lapply(1:m, function(i) rep(i, sample(25:50,1))))
j <- sample(1:n, length(i), replace=TRUE)
x <- sparseMatrix(i, j, x=runif(length(i)))

n_comp <- 50
  xt.x <- crossprod(x)
  x.means <- colMeans(x)
  xt.x <- (xt.x - m * tcrossprod(x.means)) / (m-1)
  svd.0 <- irlba(xt.x, nu=0, nv=n_comp, tol=1e-10)
#user  system elapsed 
#0.148   0.030   2.923 

system.time(pca <- prcomp(x, center=TRUE))
#user  system elapsed 
#32.178   2.702  12.322

max(abs(pca$center - x.means))
max(abs(xt.x - cov(as.matrix(x))))
max(abs(abs(svd.0$v / pca$rotation[,1:n_comp]) - 1))

If you set the number of columns to 10,000 and the number of principal components to 25, the irlba-based PCA takes about 17 minutes to calculate 50 approximate principal components and consumes about 6GB of RAM, which isn't too bad.

  • $\begingroup$ Zach, curious if you ever solved this. $\endgroup$
    – B_Miner
    Jul 17, 2013 at 17:45
  • $\begingroup$ @B_Miner: Basically, I've been doing SVD without bothering to center or scale first, because I've never found a good way to do this without converting my sparse matrix to a dense matrix. The original matrix %*% the svd's V component gives the "principle components." Sometimes, I get better results if I "fold in" the eigen values, e.g v %*% diag(d), where d is the vector of eigenvalues from the SVD. $\endgroup$
    – Zach
    Jul 17, 2013 at 17:49
  • $\begingroup$ Do you treat v %*% diag(d) by itself or still multiplied by the original matrix X (i.e. X%*% v %*% diag(d)). It seems above you are using the u matrix as the principal component scores? $\endgroup$
    – B_Miner
    Jul 17, 2013 at 18:32
  • $\begingroup$ I use X %*% v %*% diag(d, ncol=length(d)). The v matrix in the svd is equivalent to the "rotation" element of a prcomp object, and X %*% v or X %*% v %*% diag(d, ncol=length(d)) represents the x element of a prcomp object. Take a look a stats:::prcomp.default. $\endgroup$
    – Zach
    Jul 17, 2013 at 18:41
  • $\begingroup$ Yes, X %*% v is the x element from prcomp. It looks like when you use the u matrix as in your question, you are actually using X %*% v %*% diag(1/d). $\endgroup$
    – B_Miner
    Jul 17, 2013 at 18:49

1 Answer 1


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.


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
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
  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.)
  • $\begingroup$ Thank you for the detailed answer. One of the advantages of irlba is that you can specify nu to limit the algorithm to the first n principle components, which greatly increases it's efficacy and (I think) bypasses the calculation of the XX' matrix. $\endgroup$
    – Zach
    Aug 27, 2012 at 20:50
  • 1
    $\begingroup$ But what do you want to work with? A sparse $10000$ by $500000$ matrix with $5\times 10^9$ coefficients that does not represent the problem you need to solve, or a $10000$ by $10000$ with $10^8$ coefficients that does represent the problem you want to solve? irlba can be applied to the latter to obtain just the first few principal components, anyway, so you get the best of both worlds. $\endgroup$
    – whuber
    Aug 27, 2012 at 20:54
  • $\begingroup$ I suppose the latter. =). So I need to calculate the dot product for each pair of columns in my sparse matrix, subtract the colMeans of the sparse matrix from the dot product matrix, then run irlba on the result? $\endgroup$
    – Zach
    Aug 27, 2012 at 21:11
  • $\begingroup$ Almost: notice you're subtracting products of column means, not the column means themselves. Your formulation of the algorithm otherwise is excellent, because although abstractly you're computing $X X'$, you don't really want R to create $X'$ in order to do the matrix multiplication. Instead, if RAM is really limited, you can perform the column dot products in batches by reading in subsets of the columns at a time. It would be wise to experiment with much smaller matrices at first :-). $\endgroup$
    – whuber
    Aug 27, 2012 at 21:20
  • 5
    $\begingroup$ I added code to illustrate. $\endgroup$
    – whuber
    Jul 17, 2013 at 19:56

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