\documentclass{article} \usepackage{myVignette} \usepackage[authoryear,round]{natbib} \bibliographystyle{plainnat} %%\VignetteIndexEntry{Comparisons of Least Squares calculation speeds} %%\VignetteDepends{Matrix} \begin{document} \SweaveOpts{engine=R,eps=FALSE,pdf=TRUE,width=5,height=3,strip.white=TRUE} \setkeys{Gin}{width=\textwidth} \title{Comparing Least Squares Calculations} \author{Douglas Bates\\R Development Core Team\\\email{Douglas.Bates@R-project.org}} \date{\today} \maketitle \begin{abstract} Many statistics methods require one or more least squares problems to be solved. There are several ways to perform this calculation, using objects from the base R system and using objects in the classes defined in the \code{Matrix} package. We compare the speed of some of these methods on a very small example and on a example for which the model matrix is large and sparse. \end{abstract} <>= options(width=75) @ \section{Linear least squares calculations} \label{sec:LeastSquares} Many statistical techniques require least squares solutions $$\label{eq:LeastSquares} \widehat{\bm{\beta}}= \arg\min_{\bm{\beta}}\left\|\bm{y}-\bX\bm{\beta}\right\|^2$$ where $\bX$ is an $n\times p$ model matrix ($p\leq n$), $\bm{y}$ is $n$-dimensional and $\bm{\beta}$ is $p$ dimensional. Most statistics texts state that the solution to (\ref{eq:LeastSquares}) is $$\label{eq:XPX} \widehat{\bm{\beta}}=\left(\bX\trans\bX\right)^{-1}\bX\trans\bm{y}$$ when $\bX$ has full column rank (i.e. the columns of $\bX$ are linearly independent) and all too frequently it is calculated in exactly this way. \subsection{A small example} \label{sec:smallLSQ} As an example, let's create a model matrix, \code{mm}, and corresponding response vector, \code{y}, for a simple linear regression model using the \code{Formaldehyde} data. <>= data(Formaldehyde) str(Formaldehyde) print(mm <- cbind(1, Formaldehyde$carb)) print(y <- Formaldehyde$optden) @ Using \code{t} to evaluate the transpose, \code{solve} to take an inverse, and the \code{\%*\%} operator for matrix multiplication, we can translate \ref{eq:XPX} into the \Slang{} as <>= solve(t(mm) %*% mm) %*% t(mm) %*% y @ On modern computers this calculation is performed so quickly that it cannot be timed accurately in \RR{} <>= system.time(solve(t(mm) %*% mm) %*% t(mm) %*% y) @ and it provides essentially the same results as the standard \code{lm.fit} function that is called by \code{lm}. <>= dput(c(solve(t(mm) %*% mm) %*% t(mm) %*% y)) dput(lm.fit(mm, y)$coefficients) @ \subsection{A large example} \label{sec:largeLSQ} For a large, ill-conditioned least squares problem, such as that described in \citet{koen:ng:2003}, the literal translation of (\ref{eq:XPX}) does not perform well. <>= library(Matrix) data(mm, package = "Matrix") data(y, package = "Matrix") mm = as(mm, "matrix") dim(mm) naive.sol = solve(t(mm) %*% mm) %*% t(mm) %*% y system.time(solve(t(mm) %*% mm) %*% t(mm) %*% y) system.time(solve(t(mm) %*% mm) %*% t(mm) %*% y) system.time(solve(t(mm) %*% mm) %*% t(mm) %*% y) @ (Here and in what follows we will repeat timings four times to obtain timings that are close to the steady-state values.) Because the calculation of a cross-product'' matrix, such as$\bX\trans\bX$or$\bX\trans\bm{y}$, is a common operation in statistics, the \code{crossprod} function has been provided to do this efficiently. In the single argument form \code{crossprod(mm)} calculates$\bX\trans\bX$, taking advantage of the symmetry of the product. That is, instead of calculating the$712^2=506944$elements of$\bX\trans\bX$separately, it only calculates the$(712\cdot 713)/2=253828$elements in the upper triangle and replicates them in the lower triangle. Furthermore, there is no need to calculate the inverse of a matrix explicitly when solving a linear system of equations. When the two argument form of the \code{solve} function is used the linear system $$\label{eq:LSQsol} \left(\bX\trans\bX\right) \widehat{\bm{\beta}} = \bX\trans\by$$ is solved directly. Combining these optimizations we obtain <>= cpod.sol = solve(crossprod(mm), crossprod(mm,y)) system.time(solve(crossprod(mm), crossprod(mm,y))) system.time(solve(crossprod(mm), crossprod(mm,y))) system.time(solve(crossprod(mm), crossprod(mm,y))) all.equal(naive.sol, cpod.sol) @ On this computer (2.0 GHz Pentium-4, 1 GB Memory, Goto's BLAS) the crossprod form of the calculation is about four times as fast as the naive calculation. In fact, the entire crossprod solution is faster than simply calculating$\bX\trans\bX\$ the naive way. <>= system.time(t(mm) %*% mm) system.time(t(mm) %*% mm) system.time(t(mm) %*% mm) system.time(t(mm) %*% mm) @ \subsection{Least squares calculations with Matrix classes} \label{sec:MatrixLSQ} The \code{crossprod} function applied to a single matrix takes advantage of symmetry when calculating the product but does not retain the information that the product is symmetric (and positive semidefinite). As a result the solution of (\ref{eq:LSQsol}) is performed using general linear system solver based on an LU decomposition when it would be faster, and more stable numerically, to use a Cholesky decomposition. The Cholesky decomposition could be used but it is rather awkward <>= ch = chol(crossprod(mm)) system.time(chol(crossprod(mm))) system.time(chol(crossprod(mm))) system.time(chol(crossprod(mm))) chol.sol = backsolve(ch, forwardsolve(ch, crossprod(mm, y), upper = TRUE, trans = TRUE)) system.time(backsolve(ch, forwardsolve(ch, crossprod(mm, y), upper = TRUE, trans = TRUE))) system.time(backsolve(ch, forwardsolve(ch, crossprod(mm, y), upper = TRUE, trans = TRUE))) system.time(backsolve(ch, forwardsolve(ch, crossprod(mm, y), upper = TRUE, trans = TRUE))) all.equal(chol.sol, naive.sol) @ The \code{Matrix} package uses the S4 class system \citep{R:Chambers:1998} to retain information on the structure of matrices from the intermediate calculations. A general matrix in dense storage, created by the \code{Matrix} function, has class \code{"geMatrix"} but its cross-product has class \code{"poMatrix"}. The \code{solve} methods for the \code{"poMatrix"} class use the Cholesky decomposition. <>= data(mm, package = "Matrix") mm = as(mm, "geMatrix") class(crossprod(mm)) Mat.sol = solve(crossprod(mm), crossprod(mm, y)) system.time(solve(crossprod(mm), crossprod(mm, y))) system.time(solve(crossprod(mm), crossprod(mm, y))) system.time(solve(crossprod(mm), crossprod(mm, y))) all.equal(naive.sol, as(Mat.sol, "matrix")) @ Furthermore, any method that calculates a decomposition or factorization stores the resulting factorization with the original object so that it can be reused without recalculation. <>= xpx = crossprod(mm) xpy = crossprod(mm, y) system.time(solve(xpx, xpy)) system.time(solve(xpx, xpy)) system.time(solve(xpx, xpy)) @ The model matrix \code{mm} is sparse; that is, most of the elements of \code{mm} are zero. The \code{Matrix} package incorporates special methods for sparse matrices, which produce the fastest results of all. <>= data(mm, package = "Matrix") class(mm) sparse.sol = solve(crossprod(mm), crossprod(mm, y)) system.time(solve(crossprod(mm), crossprod(mm, y))) system.time(solve(crossprod(mm), crossprod(mm, y))) system.time(solve(crossprod(mm), crossprod(mm, y))) all.equal(naive.sol, as(sparse.sol, "matrix")) @ As with other classes in the \code{Matrix} package, the \code{sscMatrix} retains any factorization that has been calculated although, in this case, the decomposition is so fast that it is difficult to determine the difference in the solution times. <>= xpx = crossprod(mm) xpy = crossprod(mm, y) system.time(solve(xpx, xpy)) system.time(solve(xpx, xpy)) system.time(solve(xpx, xpy)) @ \bibliography{Matrix} \end{document}