eigen-2.1.7: Eigen C++ library (linear algebra: matrices, sparse matrices, vectors, numerical solvers).

Safe HaskellNone
LanguageHaskell98

Data.Eigen.LA

Contents

Description

The problem: You have a system of equations, that you have written as a single matrix equation

Ax = b

Where A and b are matrices (b could be a vector, as a special case). You want to find a solution x.

The solution: You can choose between various decompositions, depending on what your matrix A looks like, and depending on whether you favor speed or accuracy. However, let's start with an example that works in all cases, and is a good compromise:

import Data.Eigen.Matrix
import Data.Eigen.LA

main = do
    let
        a :: MatrixXd
        a = fromList [[1,2,3], [4,5,6], [7,8,10]]
        b = fromList [[3],[3],[4]]
        x = solve ColPivHouseholderQR a b
    putStrLn "Here is the matrix A:" >> print a
    putStrLn "Here is the vector b:" >> print b
    putStrLn "The solution is:" >> print x

produces the following output

Here is the matrix A:
Matrix 3x3
1.0 2.0 3.0
4.0 5.0 6.0
7.0 8.0 10.0

Here is the vector b:
Matrix 3x1
3.0
3.0
4.0

The solution is:
Matrix 3x1
-2.0000000000000004
1.0000000000000018
0.9999999999999989

Checking if a solution really exists: Only you know what error margin you want to allow for a solution to be considered valid.

You can compute relative error using norm (ax - b) / norm b formula or use relativeError function which provides the same calculation implemented slightly more efficient.

Synopsis

Basic linear solving

data Decomposition Source #

Decomposition           Requirements on the matrix          Speed   Accuracy  Rank  Kernel  Image

PartialPivLU            Invertible                          ++      +         -     -       -
FullPivLU               None                                -       +++       +     +       +
HouseholderQR           None                                ++      +         -     -       -
ColPivHouseholderQR     None                                +       ++        +     -       -
FullPivHouseholderQR    None                                -       +++       +     -       -
LLT                     Positive definite                   +++     +         -     -       -
LDLT                    Positive or negative semidefinite   +++     ++        -     -       -
JacobiSVD               None                                -       +++       +     -       -

The best way to do least squares solving for square matrices is with a SVD decomposition (JacobiSVD)

Constructors

PartialPivLU

LU decomposition of a matrix with partial pivoting.

FullPivLU

LU decomposition of a matrix with complete pivoting.

HouseholderQR

Householder QR decomposition of a matrix.

ColPivHouseholderQR

Householder rank-revealing QR decomposition of a matrix with column-pivoting.

FullPivHouseholderQR

Householder rank-revealing QR decomposition of a matrix with full pivoting.

LLT

Standard Cholesky decomposition (LL^T) of a matrix.

LDLT

Robust Cholesky decomposition of a matrix with pivoting.

JacobiSVD

Two-sided Jacobi SVD decomposition of a rectangular matrix.

solve :: Elem a b => Decomposition -> Matrix a b -> Matrix a b -> Matrix a b Source #

x = solve d a b
finds a solution x of ax = b equation using decomposition d

relativeError :: Elem a b => Matrix a b -> Matrix a b -> Matrix a b -> a Source #

e = relativeError x a b
computes norm (ax - b) / norm b where norm is L2 norm

Rank-revealing decompositions

Certain decompositions are rank-revealing, i.e. are able to compute the rank of a matrix. These are typically also the decompositions that behave best in the face of a non-full-rank matrix (which in the square case means a singular matrix).

import Data.Eigen.Matrix
import Data.Eigen.LA

main = do
    let a = fromList [[1,2,5],[2,1,4],[3,0,3]] :: MatrixXd
    putStrLn "Here is the matrix A:" >> print a
    putStrLn "The rank of A is:" >> print (rank FullPivLU a)
    putStrLn "Here is a matrix whose columns form a basis of the null-space of A:" >> print (kernel FullPivLU a)
    putStrLn "Here is a matrix whose columns form a basis of the column-space of A:" >> print (image FullPivLU a)

produces the following output

Here is the matrix A:
Matrix 3x3
1.0 2.0 5.0
2.0 1.0 4.0
3.0 0.0 3.0

The rank of A is:
2
Here is a matrix whose columns form a basis of the null-space of A:
Matrix 3x1
0.5000000000000001
1.0
-0.5

Here is a matrix whose columns form a basis of the column-space of A:
Matrix 3x2
5.0 1.0
4.0 2.0
3.0 3.0

rank :: Elem a b => Decomposition -> Matrix a b -> Int Source #

The rank of the matrix

kernel :: Elem a b => Decomposition -> Matrix a b -> Matrix a b Source #

Return matrix whose columns form a basis of the null-space of A

image :: Elem a b => Decomposition -> Matrix a b -> Matrix a b Source #

Return a matrix whose columns form a basis of the column-space of A

Multiple linear regression

A linear regression model that contains more than one predictor variable.

linearRegression :: [[Double]] -> ([Double], Double) Source #

(coeffs, error) = linearRegression points
computes multiple linear regression y = a1 x1 + a2 x2 + ... + an xn + b using ColPivHouseholderQR decomposition
  • point format is [y, x1..xn]
  • coeffs format is [b, a1..an]
  • error is calculated using relativeError
import Data.Eigen.LA
main = print $ linearRegression [
    [-4.32, 3.02, 6.89],
    [-3.79, 2.01, 5.39],
    [-4.01, 2.41, 6.01],
    [-3.86, 2.09, 5.55],
    [-4.10, 2.58, 6.32]]

produces the following output

([-2.3466569233817127,-0.2534897541434826,-0.1749653335680988],1.8905965120153139e-3)