How to solve a least squares (underdetermined system) quickly?

问题

I have a program in R that is computing a large amount of least squares solutions (>10,000: typically 100,000+) and, after profiling, these are the current bottlenecks for the program. I have a matrix A with column vectors that correspond to spanning vectors and a solution b. I am attempting to solve for the least-squares solution x of Ax=b. The matrices are typically 4xj in size - many of them are not square (j < 4) and so general solutions to under-determined systems are what I am looking for.

The main question: What is the fastest way to solve an under-determined system in R? I have many solutions that utilize the Normal Equation but am looking for a routine in R that is faster than any of the methods below.

For example: Solve the system for x given by Ax = b given the following constraints:

The system is not necessary determined [usually under-determined] (ncol (A) <= length(b) always holds). Thus solve(A,b) does not work because solve requires a square matrix.
You can assume that t(A) %*% A (equivalent to crossprod(A)) is non-singular - it is checked earlier in the program
You can use any package freely available in R
The solution need not be pretty - it just needs to be fast
An upper-bound on size of A is reasonably 10x10 and zero elements occur infrequently - A is usually pretty dense

Two random matrices for testing...

A = matrix(runif(12), nrow = 4)
b = matrix(runif(4), nrow = 4)

All of the functions below have been profiled. They are reproduced here:

f1 = function(A,b)
{
  solve(t(A) %*% A, t(A) %*% b)
}
f2 = function(A,b)
{
  solve(crossprod(A), crossprod(A, b))
}
f3 = function(A,b)
{
  ginv(crossprod(A)) %*% crossprod(A,b) # From the `MASS` package
}
f4 = function(A,b)
{
  matrix.inverse(crossprod(A)) %*% crossprod(A,b) # From the `matrixcalc` package
}
f5 = function(A,b)
{
  qr.solve(crossprod(A), crossprod(A,b))
}
f6 = function(A,b)
{
  svd.inverse(crossprod(A)) %*% crossprod(A,b)
}
f7 = function(A,b)
{
  qr.solve(A,b)
}
f8 = function(A,b)
{
  Solve(A,b) # From the `limSolve` package
}

After testing, f2 is the current winner. I have also tested linear model methods - they were ridiculously slow given all of the other information they produce. The code was profiled using the following:

library(ggplot2)
library(microbenchmark)

all.equal(
  f1(A,b),
  f2(A,b),
  f3(A,b),
  f4(A,b),
  f5(A,b),
  f6(A,b),
  f7(A,b),
  f8(A,b),
          )

compare = microbenchmark(
  f1(A,b),
  f2(A,b),
  f3(A,b),
  f4(A,b),
  f5(A,b),
  f6(A,b),
  f7(A,b),
  f8(A,b),
  times = 1000)

autoplot(compare)

回答1:

How about Rcpp?

library(Rcpp)
cppFunction(depends='RcppArmadillo', code='
            arma::mat fRcpp (arma::mat A, arma::mat b) {
            arma::mat betahat ;
            betahat = (A.t() * A ).i() * A.t() * b ;
            return(betahat) ;
            }                                
            ')

all.equal(f1(A, b), f2(A, b), fRcpp(A, b))
#[1] TRUE
microbenchmark(f1(A, b), f2(A, b), fRcpp(A, b))
#Unit: microseconds
#        expr    min     lq     mean  median      uq     max neval
#    f1(A, b) 55.110 57.136 67.42110 59.5680 63.0120 160.873   100
#    f2(A, b) 34.444 37.685 43.86145 39.7120 41.9405 117.920   100
# fRcpp(A, b)  3.242  4.457  7.67109  8.1045  8.9150  39.307   100

来源：https://stackoverflow.com/questions/27674866/how-to-solve-a-least-squares-underdetermined-system-quickly

标签

matrix

linear-algebra

least-squares