An algorithm for partial SVD (or PCA) of a Filebacked Big Matrix through the eigen decomposition of the covariance between variables (primal) or observations (dual). Use this algorithm only if there is one dimension that is much smaller than the other. Otherwise use big_randomSVD.

big_SVD(
X,
fun.scaling = big_scale(center = FALSE, scale = FALSE),
ind.row = rows_along(X),
ind.col = cols_along(X),
k = 10,
block.size = block_size(nrow(X))
)

## Arguments

X

An object of class FBM.

fun.scaling

A function with parameters X, ind.row and ind.col, and that returns a data.frame with $center and $scale for the columns corresponding to ind.col, to scale each of their elements such as followed: $$\frac{X_{i,j} - center_j}{scale_j}.$$ Default doesn't use any scaling. You can also provide your own center and scale by using as_scaling_fun().

ind.row

An optional vector of the row indices that are used. If not specified, all rows are used. Don't use negative indices.

ind.col

An optional vector of the column indices that are used. If not specified, all columns are used. Don't use negative indices.

k

Number of singular vectors/values to compute. Default is 10. This algorithm should be used to compute only a few singular vectors/values. If more is needed, have a look at https://stackoverflow.com/a/46380540/6103040.

block.size

Maximum number of columns read at once. Default uses block_size.

## Value

A named list (an S3 class "big_SVD") of

• d, the singular values,

• u, the left singular vectors,

• v, the right singular vectors,

• center, the centering vector,

• scale, the scaling vector.

Note that to obtain the Principal Components, you must use predict on the result. See examples.

## Details

To get $$X = U \cdot D \cdot V^T$$,

• if the number of observations is small, this function computes $$K_(2) = X \cdot X^T \approx U \cdot D^2 \cdot U^T$$ and then $$V = X^T \cdot U \cdot D^{-1}$$,

• if the number of variable is small, this function computes $$K_(1) = X^T \cdot X \approx V \cdot D^2 \cdot V^T$$ and then $$U = X \cdot V \cdot D^{-1}$$,

• if both dimensions are large, use big_randomSVD instead.

## Matrix parallelization

Large matrix computations are made block-wise and won't be parallelized in order to not have to reduce the size of these blocks. Instead, you may use Microsoft R Open or OpenBLAS in order to accelerate these block matrix computations. You can also control the number of cores used with bigparallelr::set_blas_ncores().

## Examples

set.seed(1)

X <- big_attachExtdata()
n <- nrow(X)

# Using only half of the data
ind <- sort(sample(n, n/2))

test <- big_SVD(X, fun.scaling = big_scale(), ind.row = ind)
str(test)
#> List of 5
#>  $d : num [1:10] 178.5 114.5 91 87.1 86.3 ... #>$ u     : num [1:258, 1:10] -0.1092 -0.0928 -0.0806 -0.0796 -0.1028 ...
#>  $v : num [1:4542, 1:10] 0.00607 0.00739 0.02921 -0.01283 0.01473 ... #>$ center: num [1:4542] 1.34 1.63 1.51 1.64 1.09 ...
#>  $scale : num [1:4542] 0.665 0.551 0.631 0.55 0.708 ... #> - attr(*, "class")= chr "big_SVD" plot(test$u)

pca <- prcomp(X[ind, ], center = TRUE, scale. = TRUE)

# same scaling
all.equal(test$center, pca$center)
#> [1] TRUE
all.equal(test$scale, pca$scale)
#> [1] TRUE

# scores and loadings are the same or opposite
# except for last eigenvalue which is equal to 0
# due to centering of columns
scores <- test$u %*% diag(test$d)
class(test)
#> [1] "big_SVD"
scores2 <- predict(test) # use this function to predict scores
all.equal(scores, scores2)
#> [1] TRUE
dim(scores)
#> [1] 258  10
dim(pca$x) #> [1] 258 258 tail(pca$sdev)
#> [1] 3.023287e+00 3.008386e+00 2.990514e+00 2.984375e+00 2.965688e+00
#> [6] 1.130391e-14
plot(scores2, pca$x[, 1:ncol(scores2)]) plot(test$v[1:100, ], pca$rotation[1:100, 1:ncol(scores2)]) # projecting on new data X2 <- sweep(sweep(X[-ind, ], 2, test$center, '-'), 2, test$scale, '/') scores.test <- X2 %*% test$v
ind2 <- setdiff(rows_along(X), ind)
scores.test2 <- predict(test, X, ind.row = ind2) # use this
all.equal(scores.test, scores.test2)
#> [1] TRUE
scores.test3 <- predict(pca, X[-ind, ])
plot(scores.test2, scores.test3[, 1:ncol(scores.test2)])