Multiple regression: Difference between revisions
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128.813 -143.162 61.960</lang> |
128.813 -143.162 61.960</lang> |
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Instead of "/", the slash operator, one can also write : |
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<lang Matlab> b = y * X' * inv(X * X') </lang> |
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or |
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<lang Matlab> b = y * pinv(X) </lang> |
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=={{header|PicoLisp}}== |
=={{header|PicoLisp}}== |
Revision as of 00:55, 11 January 2012
You are encouraged to solve this task according to the task description, using any language you may know.
Given a set of data vectors in the following format:
Compute the vector using ordinary least squares regression using the following equation:
You can assume y is given to you as a vector (a one-dimensional array), and X is given to you as a two-dimensional array (i.e. matrix).
Ada
Extension of Reduced row echelon form#Ada:
matrices.ads: <lang Ada>generic
type Element_Type is private; Zero : Element_Type; One : Element_Type; with function "+" (Left, Right : Element_Type) return Element_Type is <>; with function "-" (Left, Right : Element_Type) return Element_Type is <>; with function "*" (Left, Right : Element_Type) return Element_Type is <>; with function "/" (Left, Right : Element_Type) return Element_Type is <>;
package Matrices is
type Vector is array (Positive range <>) of Element_Type; type Matrix is array (Positive range <>, Positive range <>) of Element_Type;
function "*" (Left, Right : Matrix) return Matrix; function Invert (Source : Matrix) return Matrix; function Reduced_Row_Echelon_Form (Source : Matrix) return Matrix; function Regression_Coefficients (Source : Vector; Regressors : Matrix) return Vector; function To_Column_Vector (Source : Matrix; Row : Positive := 1) return Vector; function To_Matrix (Source : Vector; Column_Vector : Boolean := True) return Matrix; function To_Row_Vector (Source : Matrix; Column : Positive := 1) return Vector; function Transpose (Source : Matrix) return Matrix;
Size_Mismatch : exception; Not_Square_Matrix : exception; Not_Invertible : exception;
end Matrices;</lang>
matrices.adb: <lang Ada>package body Matrices is
function "*" (Left, Right : Matrix) return Matrix is Result : Matrix (Left'Range (1), Right'Range (2)) := (others => (others => Zero)); begin if Left'Length (2) /= Right'Length (1) then raise Size_Mismatch; end if; for I in Result'Range (1) loop for K in Result'Range (2) loop for J in Left'Range (2) loop Result (I, K) := Result (I, K) + Left (I, J) * Right (J, K); end loop; end loop; end loop; return Result; end "*";
function Invert (Source : Matrix) return Matrix is Expanded : Matrix (Source'Range (1), Source'First (2) .. Source'Last (2) * 2); Result : Matrix (Source'Range (1), Source'Range (2)); begin -- Matrix has to be square. if Source'Length (1) /= Source'Length (2) then raise Not_Square_Matrix; end if; -- Copy Source into Expanded matrix and attach identity matrix to right for Row in Source'Range (1) loop for Col in Source'Range (2) loop Expanded (Row, Col) := Source (Row, Col); Expanded (Row, Source'Last (2) + Col) := Zero; end loop; Expanded (Row, Source'Last (2) + Row) := One; end loop; Expanded := Reduced_Row_Echelon_Form (Source => Expanded); -- Copy right side to Result (= inverted Source) for Row in Result'Range (1) loop for Col in Result'Range (2) loop Result (Row, Col) := Expanded (Row, Source'Last (2) + Col); end loop; end loop; return Result; end Invert;
function Reduced_Row_Echelon_Form (Source : Matrix) return Matrix is procedure Divide_Row (From : in out Matrix; Row : Positive; Divisor : Element_Type) is begin for Col in From'Range (2) loop From (Row, Col) := From (Row, Col) / Divisor; end loop; end Divide_Row;
procedure Subtract_Rows (From : in out Matrix; Subtrahend, Minuend : Positive; Factor : Element_Type) is begin for Col in From'Range (2) loop From (Minuend, Col) := From (Minuend, Col) - From (Subtrahend, Col) * Factor; end loop; end Subtract_Rows;
procedure Swap_Rows (From : in out Matrix; First, Second : Positive) is Temporary : Element_Type; begin for Col in From'Range (2) loop Temporary := From (First, Col); From (First, Col) := From (Second, Col); From (Second, Col) := Temporary; end loop; end Swap_Rows;
Result : Matrix := Source; Lead : Positive := Result'First (2); I : Positive; begin Rows : for Row in Result'Range (1) loop exit Rows when Lead > Result'Last (2); I := Row; while Result (I, Lead) = Zero loop I := I + 1; if I = Result'Last (1) then I := Row; Lead := Lead + 1; exit Rows when Lead = Result'Last (2); end if; end loop; if I /= Row then Swap_Rows (From => Result, First => I, Second => Row); end if; Divide_Row (From => Result, Row => Row, Divisor => Result (Row, Lead)); for Other_Row in Result'Range (1) loop if Other_Row /= Row then Subtract_Rows (From => Result, Subtrahend => Row, Minuend => Other_Row, Factor => Result (Other_Row, Lead)); end if; end loop; Lead := Lead + 1; end loop Rows; return Result; end Reduced_Row_Echelon_Form;
function Regression_Coefficients (Source : Vector; Regressors : Matrix) return Vector is Result : Matrix (Regressors'Range (2), 1 .. 1); begin if Source'Length /= Regressors'Length (1) then raise Size_Mismatch; end if; declare Regressors_T : constant Matrix := Transpose (Regressors); begin Result := Invert (Regressors_T * Regressors) * Regressors_T * To_Matrix (Source); end; return To_Row_Vector (Source => Result); end Regression_Coefficients;
function To_Column_Vector (Source : Matrix; Row : Positive := 1) return Vector is Result : Vector (Source'Range (2)); begin for Column in Result'Range loop Result (Column) := Source (Row, Column); end loop; return Result; end To_Column_Vector;
function To_Matrix (Source : Vector; Column_Vector : Boolean := True) return Matrix is Result : Matrix (1 .. 1, Source'Range); begin for Column in Source'Range loop Result (1, Column) := Source (Column); end loop; if Column_Vector then return Transpose (Result); else return Result; end if; end To_Matrix;
function To_Row_Vector (Source : Matrix; Column : Positive := 1) return Vector is Result : Vector (Source'Range (1)); begin for Row in Result'Range loop Result (Row) := Source (Row, Column); end loop; return Result; end To_Row_Vector;
function Transpose (Source : Matrix) return Matrix is Result : Matrix (Source'Range (2), Source'Range (1)); begin for Row in Result'Range (1) loop for Column in Result'Range (2) loop Result (Row, Column) := Source (Column, Row); end loop; end loop; return Result; end Transpose;
end Matrices;</lang>
Example multiple_regression.adb: <lang Ada>with Ada.Text_IO; with Matrices; procedure Multiple_Regression is
package Float_Matrices is new Matrices ( Element_Type => Float, Zero => 0.0, One => 1.0); subtype Vector is Float_Matrices.Vector; subtype Matrix is Float_Matrices.Matrix; use type Matrix;
procedure Output_Matrix (X : Matrix) is begin for Row in X'Range (1) loop for Col in X'Range (2) loop Ada.Text_IO.Put (Float'Image (X (Row, Col)) & ' '); end loop; Ada.Text_IO.New_Line; end loop; end Output_Matrix;
-- example from Ruby solution V : constant Vector := (1.0, 2.0, 3.0, 4.0, 5.0); M : constant Matrix := ((1 => 2.0), (1 => 1.0), (1 => 3.0), (1 => 4.0), (1 => 5.0)); C : constant Vector := Float_Matrices.Regression_Coefficients (Source => V, Regressors => M); -- Wikipedia example Weight : constant Vector (1 .. 15) := (52.21, 53.12, 54.48, 55.84, 57.20, 58.57, 59.93, 61.29, 63.11, 64.47, 66.28, 68.10, 69.92, 72.19, 74.46); Height : Vector (1 .. 15) := (1.47, 1.50, 1.52, 1.55, 1.57, 1.60, 1.63, 1.65, 1.68, 1.70, 1.73, 1.75, 1.78, 1.80, 1.83); Height_Matrix : Matrix (1 .. 15, 1 .. 3);
begin
Ada.Text_IO.Put_Line ("Example from Ruby solution:"); Ada.Text_IO.Put_Line ("V:"); Output_Matrix (Float_Matrices.To_Matrix (V)); Ada.Text_IO.Put_Line ("M:"); Output_Matrix (M); Ada.Text_IO.Put_Line ("C:"); Output_Matrix (Float_Matrices.To_Matrix (C)); Ada.Text_IO.New_Line; Ada.Text_IO.Put_Line ("Example from Wikipedia:"); for I in Height'Range loop Height_Matrix (I, 1) := 1.0; Height_Matrix (I, 2) := Height (I); Height_Matrix (I, 3) := Height (I) ** 2; end loop; Ada.Text_IO.Put_Line ("Matrix:"); Output_Matrix (Height_Matrix); declare Coefficients : constant Vector := Float_Matrices.Regression_Coefficients (Source => Weight, Regressors => Height_Matrix); begin Ada.Text_IO.Put_Line ("Coefficients:"); Output_Matrix (Float_Matrices.To_Matrix (Coefficients)); end;
end Multiple_Regression;</lang>
Output:
Example from Ruby solution: V: 1.00000E+00 2.00000E+00 3.00000E+00 4.00000E+00 5.00000E+00 M: 2.00000E+00 1.00000E+00 3.00000E+00 4.00000E+00 5.00000E+00 C: 9.81818E-01 Example from Wikipedia: Matrix: 1.00000E+00 1.47000E+00 2.16090E+00 1.00000E+00 1.50000E+00 2.25000E+00 1.00000E+00 1.52000E+00 2.31040E+00 1.00000E+00 1.55000E+00 2.40250E+00 1.00000E+00 1.57000E+00 2.46490E+00 1.00000E+00 1.60000E+00 2.56000E+00 1.00000E+00 1.63000E+00 2.65690E+00 1.00000E+00 1.65000E+00 2.72250E+00 1.00000E+00 1.68000E+00 2.82240E+00 1.00000E+00 1.70000E+00 2.89000E+00 1.00000E+00 1.73000E+00 2.99290E+00 1.00000E+00 1.75000E+00 3.06250E+00 1.00000E+00 1.78000E+00 3.16840E+00 1.00000E+00 1.80000E+00 3.24000E+00 1.00000E+00 1.83000E+00 3.34890E+00 Coefficients: 1.35403E+02 -1.51161E+02 6.43514E+01
C
Using GNU gsl and c99, with the WP data<lang C>#include <stdio.h>
- include <gsl/gsl_matrix.h>
- include <gsl/gsl_math.h>
- include <gsl/gsl_multifit.h>
double w[] = { 52.21, 53.12, 54.48, 55.84, 57.20, 58.57, 59.93, 61.29, 63.11, 64.47, 66.28, 68.10, 69.92, 72.19, 74.46 }; double h[] = { 1.47, 1.50, 1.52, 1.55, 1.57, 1.60, 1.63, 1.65, 1.68, 1.70, 1.73, 1.75, 1.78, 1.80, 1.83 };
int main() { int n = sizeof(h)/sizeof(double); gsl_matrix *X = gsl_matrix_calloc(n, 3); gsl_vector *Y = gsl_vector_alloc(n); gsl_vector *beta = gsl_vector_alloc(3);
for (int i = 0; i < n; i++) { gsl_vector_set(Y, i, w[i]);
gsl_matrix_set(X, i, 0, 1); gsl_matrix_set(X, i, 1, h[i]); gsl_matrix_set(X, i, 2, h[i] * h[i]); }
double chisq; gsl_matrix *cov = gsl_matrix_alloc(3, 3); gsl_multifit_linear_workspace * wspc = gsl_multifit_linear_alloc(n, 3); gsl_multifit_linear(X, Y, beta, cov, &chisq, wspc);
printf("Beta:"); for (int i = 0; i < 3; i++) printf(" %g", gsl_vector_get(beta, i)); printf("\n");
gsl_matrix_free(X); gsl_matrix_free(cov); gsl_vector_free(Y); gsl_vector_free(beta); gsl_multifit_linear_free(wspc);
}</lang>
Common Lisp
Uses the routine (chol A) from Cholesky decomposition, (mmul A B) from Matrix multiplication, (mtp A) from Matrix transposition.
<lang lisp>
- Solve a linear system AX=B where A is symmetric and positive definite, so it can be Cholesky decomposed.
(defun linsys (A B)
(let* ((n (car (array-dimensions A))) (m (cadr (array-dimensions B))) (y (make-array n :element-type 'long-float :initial-element 0.0L0)) (X (make-array `(,n ,m) :element-type 'long-float :initial-element 0.0L0)) (L (chol A))) ; A=LL'
(loop for col from 0 to (- m 1) do ;; Forward substitution: y = L\B (loop for k from 0 to (- n 1) do (setf (aref y k) (/ (- (aref B k col) (loop for j from 0 to (- k 1) sum (* (aref L k j) (aref y j)))) (aref L k k))))
;; Back substitution. x=L'\y (loop for k from (- n 1) downto 0 do (setf (aref X k col) (/ (- (aref y k) (loop for j from (+ k 1) to (- n 1) sum (* (aref L j k) (aref X j col)))) (aref L k k))))) X))
- Solve a linear least squares problem. Ax=b, with A being mxn, with m>n.
- Solves the linear system A'Ax=A'b.
(defun lsqr (A b)
(linsys (mmul (mtp A) A) (mmul (mtp A) b)))
</lang>
To show an example of multiple regression, (polyfit x y n) from Polynomial regression, which itself uses (linsys A B) and (lsqr A b), will be used to fit a second degree order polynomial to data.
<lang lisp>(let ((x (make-array '(1 11) :initial-contents '((0 1 2 3 4 5 6 7 8 9 10))))
(y (make-array '(1 11) :initial-contents '((1 6 17 34 57 86 121 162 209 262 321))))) (polyfit x y 2))
- 2A((0.9999999999999759d0) (2.000000000000005d0) (3.0d0))</lang>
Go
The example on WP happens to be a polynomial regression example, and so code from the Polynomial regression task can be reused here. The only difference here is that givens x and y are computed in a separate function as a task prerequisite. <lang go>package main
import (
"fmt" "gomatrix.googlecode.com/hg/matrix"
)
func givens() (x, y *matrix.DenseMatrix) {
height := []float64{1.47, 1.50, 1.52, 1.55, 1.57, 1.60, 1.63, 1.65, 1.68, 1.70, 1.73, 1.75, 1.78, 1.80, 1.83} weight := []float64{52.21, 53.12, 54.48, 55.84, 57.20, 58.57, 59.93, 61.29, 63.11, 64.47, 66.28, 68.10, 69.92, 72.19, 74.46} m := len(height) n := 3 y = matrix.MakeDenseMatrix(weight, m, 1) x = matrix.Zeros(m, n) for i := 0; i < m; i++ { ip := float64(1) for j := 0; j < n; j++ { x.Set(i, j, ip) ip *= height[i] } } return
}
func main() {
x, y := givens() n := x.Cols() q, r := x.QR() qty, err := q.Transpose().Times(y) if err != nil { fmt.Println(err) return } c := make([]float64, n) for i := n - 1; i >= 0; i-- { c[i] = qty.Get(i, 0) for j := i + 1; j < n; j++ { c[i] -= c[j] * r.Get(i, j) } c[i] /= r.Get(i, i) } fmt.Println(c)
}</lang> Output:
[128.8128035784373 -143.16202286476116 61.960325442472865]
Haskell
Using package hmatrix from HackageDB <lang haskell>import Numeric.LinearAlgebra import Numeric.LinearAlgebra.LAPACK
m :: Matrix Double m = (3><3)
[7.589183,1.703609,-4.477162, -4.597851,9.434889,-6.543450, 0.4588202,-6.115153,1.331191]
v :: Matrix Double v = (3><1)
[1.745005,-4.448092,-4.160842]</lang>
Using lapack::dgels <lang haskell>*Main> linearSolveLSR m v (3><1)
[ 0.9335611922087276 , 1.101323491272865 , 1.6117769115824 ]</lang>
Or <lang haskell>*Main> inv m `multiply` v (3><1)
[ 0.9335611922087278 , 1.101323491272865 , 1.6117769115824006 ]</lang>
J
<lang j> NB. Wikipedia data
x=: 1.47 1.50 1.52 1.55 1.57 1.60 1.63 1.65 1.68 1.70 1.73 1.75 1.78 1.80 1.83 y=: 52.21 53.12 54.48 55.84 57.20 58.57 59.93 61.29 63.11 64.47 66.28 68.10 69.92 72.19 74.46
y %. x ^/ i.3 NB. calculate coefficients b1, b2 and b3 for 2nd degree polynomial
128.813 _143.162 61.9603</lang>
Breaking it down: <lang j> X=: x ^/ i.3 NB. form Design matrix
X=: (x^0) ,. (x^1) ,. (x^2) NB. equivalent of previous line 4{.X NB. show first 4 rows of X
1 1.47 2.1609 1 1.5 2.25 1 1.52 2.3104 1 1.55 2.4025
NB. Where y is a set of observations and X is the design matrix NB. y %. X does matrix division and gives the regression coefficients y %. X
128.813 _143.162 61.9603</lang>
In other words beta=: y %. X is the equivalent of:
To confirm: <lang j> mp=: +/ .* NB. matrix product
NB. %.X is matrix inverse of X NB. |:X is transpose of X (%.(|:X) mp X) mp (|:X) mp y
128.814 _143.163 61.9606</lang>
LAPACK routines are also available via the Addon math/lapack.
JavaScript
for the print()
and Array.map()
functions.
Extends the Matrix class from Matrix Transpose#JavaScript, Matrix multiplication#JavaScript, Reduced row echelon form#JavaScript. Uses the IdentityMatrix from Matrix exponentiation operator#JavaScript <lang javascript>// modifies the matrix "in place" Matrix.prototype.inverse = function() {
if (this.height != this.width) { throw "can't invert a non-square matrix"; }
var I = new IdentityMatrix(this.height); for (var i = 0; i < this.height; i++) this.mtx[i] = this.mtx[i].concat(I.mtx[i]) this.width *= 2;
this.toReducedRowEchelonForm();
for (var i = 0; i < this.height; i++) this.mtx[i].splice(0, this.height); this.width /= 2;
return this;
}
function ColumnVector(ary) {
return new Matrix(ary.map(function(v) {return [v]}))
} ColumnVector.prototype = Matrix.prototype
Matrix.prototype.regression_coefficients = function(x) {
var x_t = x.transpose(); return x_t.mult(x).inverse().mult(x_t).mult(this);
}
// the Ruby example var y = new ColumnVector([1,2,3,4,5]); var x = new ColumnVector([2,1,3,4,5]); print(y.regression_coefficients(x)); print();
// the Tcl example y = new ColumnVector([
52.21, 53.12, 54.48, 55.84, 57.20, 58.57, 59.93, 61.29, 63.11, 64.47, 66.28, 68.10, 69.92, 72.19, 74.46
]); x = new Matrix(
[1.47,1.50,1.52,1.55,1.57,1.60,1.63,1.65,1.68,1.70,1.73,1.75,1.78,1.80,1.83].map( function(v) {return [Math.pow(v,0), Math.pow(v,1), Math.pow(v,2)]} )
); print(y.regression_coefficients(x));</lang> output
0.9818181818181818 128.8128035798277 -143.1620228653037 61.960325442985436
Mathematica
<lang Mathematica>A={{1.47,52.21},{1.5,53.12},{1.52,54.48},{1.55,55.84},{1.57,57.2},{1.6,58.57},{1.63,59.93},{1.65,61.29},{1.68,63.11}, {1.7,64.47},{1.73,66.28},{1.75,68.1},{1.78,69.92},{1.8,72.19},{1.83,74.46}} Fit[A, {1,z,z^2}, z] ->128.813 -143.162*z + 61.9603*z^2</lang>
MATLAB
The slash and backslash operator can be used for solving this problem. Here some random data is generated.
<lang Matlab> n=100; k=10;
y = randn (1,n); % generate random vector y X = randn (k,n); % generate random matrix X b = y / X b = 0.1457109 -0.0777564 -0.0712427 -0.0166193 0.0292955 -0.0079111 0.2265894 -0.0561589 -0.1752146 -0.2577663 </lang>
In its transposed form yt = Xt * bt, the backslash operator can be used.
<lang Matlab> yt = y'; Xt = X';
bt = Xt \ yt bt = 0.1457109 -0.0777564 -0.0712427 -0.0166193 0.0292955 -0.0079111 0.2265894 -0.0561589 -0.1752146 -0.2577663</lang>
Here is the example for estimating the polynomial fit
<lang Matlab> x = [1.47 1.50 1.52 1.55 1.57 1.60 1.63 1.65 1.68 1.70 1.73 1.75 1.78 1.80 1.83]
y = [52.21 53.12 54.48 55.84 57.20 58.57 59.93 61.29 63.11 64.47 66.28 68.10 69.92 72.19 74.46] X = [x.^0;x.^1;x.^2]; b = y/X
128.813 -143.162 61.960</lang>
Instead of "/", the slash operator, one can also write : <lang Matlab> b = y * X' * inv(X * X') </lang> or <lang Matlab> b = y * pinv(X) </lang>
PicoLisp
<lang PicoLisp>(scl 20)
- Matrix transposition
(de matTrans (Mat)
(apply mapcar Mat list) )
- Matrix multiplication
(de matMul (Mat1 Mat2)
(mapcar '((Row) (apply mapcar Mat2 '(@ (sum */ Row (rest) (1.0 .))) ) ) Mat1 ) )
- Matrix identity
(de matIdent (N)
(let L (need N (1.0) 0) (mapcar '(() (copy (rot L))) L) ) )
- Reduced row echelon form
(de reducedRowEchelonForm (Mat)
(let (Lead 1 Cols (length (car Mat))) (for (X Mat X (cdr X)) (NIL (loop (T (seek '((R) (n0 (get R 1 Lead))) X) @ ) (T (> (inc 'Lead) Cols)) ) ) (xchg @ X) (let D (get X 1 Lead) (map '((R) (set R (*/ (car R) 1.0 D))) (car X) ) ) (for Y Mat (unless (== Y (car X)) (let N (- (get Y Lead)) (map '((Dst Src) (inc Dst (*/ N (car Src) 1.0)) ) Y (car X) ) ) ) ) (T (> (inc 'Lead) Cols)) ) ) Mat )</lang>
<lang PicoLisp>(de matInverse (Mat)
(let N (length Mat) (unless (= N (length (car Mat))) (quit "can't invert a non-square matrix") ) (mapc conc Mat (matIdent N)) (mapcar '((L) (tail N L)) (reducedRowEchelonForm Mat)) ) )
(de columnVector (Ary)
(mapcar cons Ary) )
(de regressionCoefficients (Mat X)
(let Xt (matTrans X) (matMul (matMul (matInverse (matMul Xt X)) Xt) Mat) ) )
(setq
Y (columnVector (1.0 2.0 3.0 4.0 5.0)) X (columnVector (2.0 1.0 3.0 4.0 5.0)) )
(round (caar (regressionCoefficients Y X)) 17)</lang> Output:
-> "0.98181818181818182"
Python
Using
The following
session gives:
<lang python>In [7]: x = [1.47, 1.50, 1.52, 1.55, 1.57, 1.60, 1.63, 1.65, 1.68, 1.70, 1.73, 1.75, 1.78, 1.80, 1.83]
In [8]: y = [52.21, 53.12, 54.48, 55.84, 57.20, 58.57, 59.93, 61.29, 63.11, 64.47, 66.28, 68.10, 69.92, 72.19, 74.46]
In [9]: polyfit(x, y, 2) Out[9]: array([ 61.96032544, -143.16202287, 128.81280358])</lang>
R
R provides the lm() function for linear regression.
<lang R>## Wikipedia Data x <- c(1.47, 1.50, 1.52, 1.55, 1.57, 1.60, 1.63, 1.65, 1.68, 1.70, 1.73, 1.75, 1.78, 1.80, 1.83) } y <- c(52.21, 53.12, 54.48, 55.84, 57.20, 58.57, 59.93, 61.29, 63.11, 64.47, 66.28, 68.10, 69.92, 72.19, 74.46)
lm( y ~ x + I(x^2))</lang> Producing output,
Call: lm(formula = y ~ x + I(x^2)) Coefficients: (Intercept) x I(x^2) 128.81 -143.16 61.96
A simple implementation of multiple regression in native R is useful to illustrate R's model description and linear algebra capabilities.
<lang R>simpleMultipleReg <- function(formula) {
## parse and evaluate the model formula mf <- model.frame(formula)
## create design matrix X <- model.matrix(attr(mf, "terms"), mf)
## create dependent variable Y <- model.response(mf)
## solve solve(t(X) %*% X) %*% t(X) %*% Y
}
simpleMultipleReg(y ~ x + I(x^2))</lang>
This produces the same coefficients as lm()
[,1] (Intercept) 128.81280 x -143.16202 I(x^2) 61.96033
A more efficient way to solve , than
the method above, is to solve the linear system directly and use the crossprod function.
<lang R>solve( crossprod(X), crossprod(X, Y))</lang>
Ruby
Using the standard library Matrix class:
<lang ruby>require 'matrix'
def regression_coefficients y, x
y = Matrix.column_vector y.map { |i| i.to_f } x = Matrix.columns x.map { |xi| xi.map { |i| i.to_f }}
(x.t * x).inverse * x.t * y
end</lang>
Testing: <lang ruby>puts regression_coefficients([1, 2, 3, 4, 5], [ [2, 1, 3, 4, 5] ])</lang> Output:
Matrix[[0.981818181818182]]
Tcl
<lang tcl>package require math::linearalgebra namespace eval multipleRegression {
namespace export regressionCoefficients namespace import ::math::linearalgebra::*
# Matrix inversion is defined in terms of Gaussian elimination # Note that we assume (correctly) that we have a square matrix proc invert {matrix} {
solveGauss $matrix [mkIdentity [lindex [shape $matrix] 0]]
} # Implement the Ordinary Least Squares method proc regressionCoefficients {y x} {
matmul [matmul [invert [matmul $x [transpose $x]]] $x] $y
}
} namespace import multipleRegression::regressionCoefficients</lang> Using an example from the Wikipedia page on the correlation of height and weight: <lang tcl># Simple helper just for this example proc map {n exp list} {
upvar 1 $n v set r {}; foreach v $list {lappend r [uplevel 1 $exp]}; return $r
}
- Data from wikipedia
set x {
1.47 1.50 1.52 1.55 1.57 1.60 1.63 1.65 1.68 1.70 1.73 1.75 1.78 1.80 1.83
} set y {
52.21 53.12 54.48 55.84 57.20 58.57 59.93 61.29 63.11 64.47 66.28 68.10 69.92 72.19 74.46
}
- Wikipedia states that fitting up to the square of x[i] is worth it
puts [regressionCoefficients $y [map n {map v {expr {$v**$n}} $x} {0 1 2}]]</lang> Produces this output (a 3-vector of coefficients):
128.81280358170625 -143.16202286630732 61.96032544293041
Ursala
This exact problem is solved by the DGELSD function from the Lapack library [1], which is callable in Ursala like this. <lang Ursala>regression_coefficients = lapack..dgelsd</lang> test program: <lang Ursala>x =
<
<7.589183e+00,1.703609e+00,-4.477162e+00>, <-4.597851e+00,9.434889e+00,-6.543450e+00>, <4.588202e-01,-6.115153e+00,1.331191e+00>>
y = <1.745005e+00,-4.448092e+00,-4.160842e+00>
- cast %eL
example = regression_coefficients(x,y)</lang> The matrix x needn't be square, and has one row for each data point. The length of y must equal the number of rows in x, and the number of coefficients returned will be the number of columns in x. It would be more typical in practice to initialize x by evaluating a set of basis functions chosen to model some empirical data, but the regression solver is indifferent to the model.
output:
<9.335612e-01,1.101323e+00,1.611777e+00>
A similar method can be used for regression with complex numbers by substituting zgelsd for dgelsd, above.