CODEWITHSUNDEEP
rperformancematrix
Dnaiel
Dnaiel

Reputation: 7832

Speed-up inverse of weighted least squares covariance matrix in R

I need to speed up the calculation of the inverse of a WLS covariance matrix in R, where the matrix, wls.cov.matrix, is given by (full example below): 

n = 10000
X = matrix(c(rnorm(n,1,2), sample(c(1,-1), n, replace = TRUE), rnorm(n,2,0.5)), nrow = 1000, ncol = 3)
Q = diag(rnorm(n, 1.5, 0.3))
wls.cov.matrix = solve(t(X)%*%diag(1/diag(Q))%*%X)

Is it possible to speed up this calculation?

MORE INFO VERY RELEVANT TO THE FINAL GOAL: This is still little information, let me explain more my goal and will be clearer if there are ways to speed up my code.
I run 1,0000s of times wls.cov.matrix so I need it to be much faster.

However, each time I run it I use the same X, the only matrix that changes is Q, which is a diagonal matrix.

If X was a square matrix, of same dim as Q, I could just precompute X^-1 and (X^T)^(-1),

X.inv = solve(X)
X.inv.trans = solve(t(X))

and then for each iteration run:

Q.inv = diag(1/diag(Q))
wls.cov.matrix = X.inv%*%Q.inv%*%X.inv.trans

But my X is not square, so is there any other trick?

Upvotes: 1

Views: 637

Answers (2)

Andrey Shabalin
Andrey Shabalin

Reputation: 4614

The main time-consuming part here is t(X)%*%diag(1/diag(Q))%*%X, not the calculation of its inverse.

A nice trick is to calculate it as 

crossprod(X / sqrt(diag(Q)));

Confirmation:

all.equal( (t(X) %*% diag(1/diag(Q)) %*% X) , crossprod(X / sqrt(diag(Q))) );

[1] TRUE

To compare the timing run:

Qdiag = diag(Q);
system.time({(t(X) %*% diag(1/Qdiag) %*% X)})
system.time({crossprod(X / sqrt(Qdiag))})

Upvotes: 2

Hong Ooi
Hong Ooi

Reputation: 57686

Well, Q is a diagonal matrix, so its inverse is just given by the inverses of the diagonal entries. You can thus do

X = matrix(c(rnorm(n,1,2), sample(c(1,-1), n, replace = TRUE), rnorm(n,2,0.5)), nrow = 1000, ncol = 3)
Qinv = diag(1/rnorm(n, 1.5, 0.3))
wls.cov.matrix = solve(t(X)%*%Qinv%*%X)

And in fact, this speeds things up by about a factor of 20.

Upvotes: 2

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