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This means it might contain formatting issues, incorrect code, conceptual problems, or other severe issues.
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{{task|Probability and statistics}}
;Task: Given a set of data vectors in the following format:
<math>y = \{ y_1, y_2, ..., y_n \}\,</math>
<math>X_i = \{ x_{i1}, x_{i2}, ..., x_{in} \}, i \in 1..k\,</math>
Compute the vector using [[wp:Ordinary least squares|ordinary least squares]] regression using the following equation:
<math>y_j = \Sigma_i \beta_i \cdot x_{ij} , j \in 1..n</math>
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:
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;
matrices.adb:
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;
Example multiple_regression.adb:
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;
{{out}}
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
BBC BASIC
{{works with|BBC BASIC for Windows}}
*FLOAT 64
INSTALL @lib$+"ARRAYLIB"
DIM y(14), x(2,14), c(2)
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() = 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
FOR row% = DIM(x(),1) TO 0 STEP -1
FOR col% = 0 TO DIM(x(),2)
x(row%,col%) = x(0,col%) ^ row%
NEXT
NEXT row%
PROCmultipleregression(y(), x(), c())
FOR i% = 0 TO DIM(c(),1) : PRINT c(i%) " "; : NEXT
PRINT
END
DEF PROCmultipleregression(y(), x(), c())
LOCAL m(), t()
DIM m(DIM(x(),1), DIM(x(),1)), t(DIM(x(),2),DIM(x(),1))
PROC_transpose(x(), t())
m() = x().t()
PROC_invert(m())
t() = t().m()
c() = y().t()
ENDPROC
{{out}}
128.812804 -143.162023 61.9603254
C
Using GNU gsl and c99, with the WP data
#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);
}
C#
{{libheader|Math.Net}}
using System;
using MathNet.Numerics.LinearRegression;
using MathNet.Numerics.LinearAlgebra;
using MathNet.Numerics.LinearAlgebra.Double;
class Program
{
static void Main(string[] args)
{
var col = DenseVector.OfArray(new double[] { 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 });
var X = DenseMatrix.OfColumns(new Vector<double>[] { col.PointwisePower(0), col, col.PointwisePower(2) });
var y = DenseVector.OfArray(new double[] { 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 });
var β = MultipleRegression.QR(X, y);
Console.WriteLine(β);
}
}
{{out}}
DenseVector 3-Double
128.813
-143.162
61.9603
Common Lisp
Uses the routine (chol A) from [[Cholesky decomposition]], (mmul A B) from [[Matrix multiplication]], (mtp A) from [[Matrix transposition]].
;; 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)))
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.
(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))
Emacs Lisp
Multiple regression analysis by Emacs Lisp and built-in Emacs Calc.
{{out}}
```txt
"35.2014388489 X1 - 3.95683453237 X2 - 42.7410071942"
ERRE
PROGRAM MULTIPLE_REGRESSION
!$DOUBLE
CONST N=14,M=2,Q=3 ! number of points and M.R. polynom degree
DIM X[N],Y[N] ! data points
DIM S[N],T[N] ! linear system coefficient
DIM A[M,Q] ! sistem to be solved
BEGIN
DATA(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)
DATA(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)
FOR I%=0 TO N DO
READ(X[I%])
END FOR
FOR I%=0 TO N DO
READ(Y[I%])
END FOR
FOR K%=0 TO 2*M DO
S[K%]=0 T[K%]=0
FOR I%=0 TO N DO
S[K%]=S[K%]+X[I%]^K%
IF K%<=M THEN T[K%]=T[K%]+Y[I%]*X[I%]^K% END IF
END FOR
END FOR
! build linear system
FOR ROW%=0 TO M DO
FOR COL%=0 TO M DO
A[ROW%,COL%]=S[ROW%+COL%]
END FOR
A[ROW%,COL%]=T[ROW%]
END FOR
PRINT("LINEAR SYSTEM COEFFICENTS") PRINT
FOR I%=0 TO M DO
FOR J%=0 TO M+1 DO
WRITE(" ######.#";A[I%,J%];)
END FOR
PRINT
END FOR
PRINT
FOR J%=0 TO M DO
FOR I%=J% TO M DO
EXIT IF A[I%,J%]<>0
END FOR
IF I%=M+1 THEN
PRINT("SINGULAR MATRIX !")
!$STOP
END IF
FOR K%=0 TO M+1 DO
SWAP(A[J%,K%],A[I%,K%])
END FOR
Y=1/A[J%,J%]
FOR K%=0 TO M+1 DO
A[J%,K%]=Y*A[J%,K%]
END FOR
FOR I%=0 TO M DO
IF I%<>J% THEN
Y=-A[I%,J%]
FOR K%=0 TO M+1 DO
A[I%,K%]=A[I%,K%]+Y*A[J%,K%]
END FOR
END IF
END FOR
END FOR
PRINT
PRINT("SOLUTIONS") PRINT
FOR I%=0 TO M DO
PRINT("c";I%;"=";)
WRITE("#####.#######";A[I%,M+1])
END FOR
END PROGRAM
{{out}}
LINEAR SYSTEM COEFFICENTS
15.0 24.8 41.1 931.2
24.8 41.1 68.4 1548.2
41.1 68.4 114.3 2585.5
SOLUTIONS
c 0 = 128.8128036
c 1 = -143.1620229
c 2 = 61.9603254
Fortran
{{libheader|SLATEC}} [http://netlib.org/slatec/ Available at the Netlib]
*-----------------------------------------------------------------------
* MR - multiple regression using the SLATEC library routine DHFTI
*
* Finds the nearest approximation to BETA in the system of linear equations:
*
* X(j,i) . BETA(i) = Y(j)
* where
* 1 ... j ... N
* 1 ... i ... K
* and
* K .LE. N
*
* INPUT ARRAYS ARE DESTROYED!
*
*___Name___________Type_______________In/Out____Description_____________
* X(N,K) Double precision In Predictors
* Y(N) Double precision Both On input: N Observations
* On output: K beta weights
* N Integer In Number of observations
* K Integer In Number of predictor variables
* DWORK(N+2*K) Double precision Neither Workspace
* IWORK(K) Integer Neither Workspace
*-----------------------------------------------------------------------
SUBROUTINE MR (X, Y, N, K, DWORK, IWORK)
IMPLICIT NONE
INTEGER K, N, IWORK
DOUBLE PRECISION X, Y, DWORK
DIMENSION X(N,K), Y(N), DWORK(N+2*K), IWORK(K)
* local variables
INTEGER I, J
DOUBLE PRECISION TAU, TOT
* maximum of all column sums of magnitudes
TAU = 0.
DO J = 1, K
TOT = 0.
DO I = 1, N
TOT = TOT + ABS(X(I,J))
END DO
IF (TOT > TAU) TAU = TOT
END DO
TAU = TAU * EPSILON(TAU) ! tolerance argument
* call function
CALL DHFTI (X, N, N, K, Y, N, 1, TAU,
$ J, DWORK(1), DWORK(N+1), DWORK(N+K+1), IWORK)
IF (J < K) PRINT *, 'mr: solution is rank deficient!'
RETURN
END ! of MR
*-----------------------------------------------------------------------
PROGRAM t_mr ! polynomial regression example
IMPLICIT NONE
INTEGER N, K
PARAMETER (N=15, K=3)
INTEGER IWORK(K), I, J
DOUBLE PRECISION XIN(N), X(N,K), Y(N), DWORK(N+2*K)
DATA XIN / 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 /
DATA 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 /
* make coefficient matrix
DO J = 1, K
DO I = 1, N
X(I,J) = XIN(I) **(J-1)
END DO
END DO
* solve
CALL MR (X, Y, N, K, DWORK, IWORK)
* print result
10 FORMAT ('beta: ', $)
20 FORMAT (F12.4, $)
30 FORMAT ()
PRINT 10
DO J = 1, K
PRINT 20, Y(J)
END DO
PRINT 30
STOP 'program complete'
END
{{out}}
beta: 128.8126 -143.1618 61.9603
STOP program complete
Go
The [http://en.wikipedia.org/wiki/Ordinary_least_squares#Example_with_real_data 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.
Library gonum/matrix
package main
import (
"fmt"
"github.com/gonum/matrix/mat64"
)
func givens() (x, y *mat64.Dense) {
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}
degree := 2
x = Vandermonde(height, degree)
y = mat64.NewDense(len(weight), 1, weight)
return
}
func Vandermonde(a []float64, degree int) *mat64.Dense {
x := mat64.NewDense(len(a), degree+1, nil)
for i := range a {
for j, p := 0, 1.; j <= degree; j, p = j+1, p*a[i] {
x.Set(i, j, p)
}
}
return x
}
func main() {
x, y := givens()
fmt.Printf("%.4f\n", mat64.Formatted(mat64.QR(x).Solve(y)))
}
{{out}}
⎡ 128.8128⎤
⎢-143.1620⎥
⎣ 61.9603⎦
Library go.matrix
package main
import (
"fmt"
"github.com/skelterjohn/go.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)
}
{{out}}
[128.8128035784373 -143.16202286476116 61.960325442472865]
Haskell
Using package [http://hackage.haskell.org/package/hmatrix hmatrix] from HackageDB
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]
Using lapack::dgels
*Main> linearSolveLSR m v
(3><1)
[ 0.9335611922087276
, 1.101323491272865
, 1.6117769115824 ]
Or
*Main> inv m `multiply` v
(3><1)
[ 0.9335611922087278
, 1.101323491272865
, 1.6117769115824006 ]
Hy
(import
[numpy [ones column-stack]]
[numpy.random [randn]]
[numpy.linalg [lstsq]])
(setv n 1000)
(setv x1 (randn n))
(setv x2 (randn n))
(setv y (+ 3 (* 1 x1) (* -2 x2) (* .25 x1 x2) (randn n)))
(print (first (lstsq
(column-stack (, (ones n) x1 x2 (* x1 x2)))
y)))
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
Breaking it down:
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
In other words beta=: y %. X is the equivalent of:
To confirm:
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
xpy=: mp~ |: NB. Or factoring out "X prime y" (monadically "X prime X")
X (%.@:xpy@[ mp xpy) y
128.814 _143.163 61.9606
LAPACK routines are also available via the Addon math/lapack.
load 'math/lapack'
load 'math/lapack/gels'
gels_jlapack_ X;y
128.813 _143.162 61.9603
Julia
{{trans|MATLAB}}
As in Matlab, the backslash or slash operator (depending on the matrix ordering) can be used for solving this problem, for example:
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 = X \ y
{{out}}
3-element Array{Float64,1}:
128.813
-143.162
61.9603
JavaScript
{{works with|SpiderMonkey}} for the print() and ''Array''.map() functions.
{{trans|Ruby}}
Extends the Matrix class from [[Matrix Transpose#JavaScript]], [[Matrix multiplication#JavaScript]], [[Reduced row echelon form#JavaScript]]. Uses the IdentityMatrix from [[Matrix exponentiation operator#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));
{{out}}
0.9818181818181818
128.8128035798277
-143.1620228653037
61.960325442985436
Kotlin
As neither the JDK nor the Kotlin Standard Library has matrix operations built-in, we re-use functions written for various other tasks.
// Version 1.2.31
typealias Vector = DoubleArray
typealias Matrix = Array<Vector>
operator fun Matrix.times(other: Matrix): Matrix {
val rows1 = this.size
val cols1 = this[0].size
val rows2 = other.size
val cols2 = other[0].size
require(cols1 == rows2)
val result = Matrix(rows1) { Vector(cols2) }
for (i in 0 until rows1) {
for (j in 0 until cols2) {
for (k in 0 until rows2) {
result[i][j] += this[i][k] * other[k][j]
}
}
}
return result
}
fun Matrix.transpose(): Matrix {
val rows = this.size
val cols = this[0].size
val trans = Matrix(cols) { Vector(rows) }
for (i in 0 until cols) {
for (j in 0 until rows) trans[i][j] = this[j][i]
}
return trans
}
fun Matrix.inverse(): Matrix {
val len = this.size
require(this.all { it.size == len }) { "Not a square matrix" }
val aug = Array(len) { DoubleArray(2 * len) }
for (i in 0 until len) {
for (j in 0 until len) aug[i][j] = this[i][j]
// augment by identity matrix to right
aug[i][i + len] = 1.0
}
aug.toReducedRowEchelonForm()
val inv = Array(len) { DoubleArray(len) }
// remove identity matrix to left
for (i in 0 until len) {
for (j in len until 2 * len) inv[i][j - len] = aug[i][j]
}
return inv
}
fun Matrix.toReducedRowEchelonForm() {
var lead = 0
val rowCount = this.size
val colCount = this[0].size
for (r in 0 until rowCount) {
if (colCount <= lead) return
var i = r
while (this[i][lead] == 0.0) {
i++
if (rowCount == i) {
i = r
lead++
if (colCount == lead) return
}
}
val temp = this[i]
this[i] = this[r]
this[r] = temp
if (this[r][lead] != 0.0) {
val div = this[r][lead]
for (j in 0 until colCount) this[r][j] /= div
}
for (k in 0 until rowCount) {
if (k != r) {
val mult = this[k][lead]
for (j in 0 until colCount) this[k][j] -= this[r][j] * mult
}
}
lead++
}
}
fun printVector(v: Vector) {
println(v.asList())
println()
}
fun multipleRegression(y: Vector, x: Matrix): Vector {
val cy = (arrayOf(y)).transpose() // convert 'y' to column vector
val cx = x.transpose() // convert 'x' to column vector array
return ((x * cx).inverse() * x * cy).transpose()[0]
}
fun main(args: Array<String>) {
var y = doubleArrayOf(1.0, 2.0, 3.0, 4.0, 5.0)
var x = arrayOf(doubleArrayOf(2.0, 1.0, 3.0, 4.0, 5.0))
var v = multipleRegression(y, x)
printVector(v)
y = doubleArrayOf(3.0, 4.0, 5.0)
x = arrayOf(
doubleArrayOf(1.0, 2.0, 1.0),
doubleArrayOf(1.0, 1.0, 2.0)
)
v = multipleRegression(y, x)
printVector(v)
y = doubleArrayOf(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)
val a = doubleArrayOf(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)
x = arrayOf(DoubleArray(a.size) { 1.0 }, a, a.map { it * it }.toDoubleArray())
v = multipleRegression(y, x)
printVector(v)
}
{{out}}
[0.9818181818181818]
[0.9999999999999996, 2.000000000000001]
[128.8128035798277, -143.1620228653037, 61.960325442985436]
Mathematica
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.PseudoInverse[X]
->{128.813, -143.162, 61.9603}
MATLAB
The slash and backslash operator can be used for solving this problem. Here some random data is generated.
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
In its transposed form yt = Xt * bt, the backslash operator can be used.
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
Here is the example for estimating the polynomial fit
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
Instead of "/", the slash operator, one can also write :
b = y * X' * inv(X * X')
or
b = y * pinv(X)
PARI/GP
pseudoinv(M)=my(sz=matsize(M),T=conj(M))~;if(sz[1]<sz[2],T/(M*T),(T*M)^-1*T)
addhelp(pseudoinv, "pseudoinv(M): Moore pseudoinverse of the matrix M.");
y*pseudoinv(X)
Perl 6
We're going to solve the example on the Wikipedia article using [https://github.com/grondilu/clifford Clifford], a [https://en.wikipedia.org/wiki/Geometric_algebra geometric algebra] module. Optimization for large vector space does not quite work yet, so it's going to take (a lof of) time and a fair amount of memory, but it should work.
Let's create four vectors containing our input data:
Then what we're looking for are three scalars , and such that:
To get for instance we can first make the and terms disappear:
Noting , we then get:
'''Note:''' a number of the formulae above are invisible to the majority of browsers, including Chrome, IE/Edge, Safari and Opera. They may (subject to the installation of necessary fronts) be visible to Firefox.
use Clifford;
my @height = <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>;
my @weight = <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>;
my $w = [+] @weight Z* @e;
my $h0 = [+] @e[^@weight];
my $h1 = [+] @height Z* @e;
my $h2 = [+] (@height X** 2) Z* @e;
my $I = $h0∧$h1∧$h2;
my $I2 = ($I·$I.reversion).Real;
say "α = ", ($w∧$h1∧$h2)·$I.reversion/$I2;
say "β = ", ($w∧$h2∧$h0)·$I.reversion/$I2;
say "γ = ", ($w∧$h0∧$h1)·$I.reversion/$I2;
{{out}}
α = 128.81280357844
β = -143.1620228648
γ = 61.960325442
This computation took over an hour with the april 2016 version of rakudo on MoarVM, running in a VirtualBox linux system guest hosted by a windows laptop with a i7 intel processor.
Phix
{{trans|ERRE}}
constant N = 15, M=3
sequence 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},
s = repeat(0,N),
t = repeat(0,N),
a = repeat(repeat(0,M+1),M)
for k=1 to 2*M do
for i=1 to N do
s[k] += power(x[i],k-1)
if k<=M then t[k] += y[i]*power(x[i],k-1) end if
end for
end for
-- build linear system
for row=1 to M do
for col=1 to M do
a[row,col] = s[row+col-1]
end for
a[row,M+1] = t[row]
end for
puts(1,"Linear system coefficents:\n")
pp(a,{pp_Nest,1,pp_IntFmt,"%7.1f",pp_FltFmt,"%7.1f"})
for j=1 to M do
integer i = j
while a[i,j]=0 do i += 1 end while
if i=M+1 then
?"SINGULAR MATRIX !"
?9/0
end if
for k=1 to M+1 do
{a[j,k],a[i,k]} = {a[i,k],a[j,k]}
end for
atom Y = 1/a[j,j]
a[j] = sq_mul(a[j],Y)
for i=1 to M do
if i<>j then
Y=-a[i,j]
for k=1 to M+1 do
a[i,k] += Y*a[j,k]
end for
end if
end for
end for
puts(1,"Solutions:\n")
?columnize(a,M+1)[1]
{{out}}
Linear system coefficents:
{{ 15.0, 24.8, 41.1, 931.2},
{ 24.8, 41.1, 68.4, 1548.2},
{ 41.1, 68.4, 114.3, 2585.5}}
Solutions:
{128.8128036,-143.1620229,61.96032544}
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 )
{{trans|JavaScript}}
(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)
{{out}}
-> "0.98181818181818182"
Python
{{libheader|NumPy}} '''Method with matrix operations'''
import numpy as np
height = [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 = [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 = np.mat(height**np.arange(3)[:, None])
y = np.mat(weight)
print(y * X.T * (X*X.T).I)
{{out}}
[[ 128.81280359 -143.16202288 61.96032545]]
'''Using numpy lstsq function'''
import numpy as np
height = [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 = [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 = np.array(height)[:, None]**range(3)
y = weight
print(np.linalg.lstsq(X, y)[0])
{{out}}
[ 128.81280358 -143.16202286 61.96032544]
R
R provides the lm() function for linear regression.
## 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))
{{out}}
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.
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))
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:
solve( crossprod(X), crossprod(X, Y))
Racket
#lang racket
(require math)
(define T matrix-transpose)
(define (fit X y)
(matrix-solve (matrix* (T X) X) (matrix* (T X) y)))
Test:
(fit (matrix [[1 2]
[2 5]
[3 7]
[4 9]])
(matrix [[1]
[2]
[3]
[9]]))
{{out}}
(array #[#[9 1/3] #[-3 1/3]])
Ruby
Using the standard library Matrix class:
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
Testing 2-dimension:
puts regression_coefficients([1, 2, 3, 4, 5], [ [2, 1, 3, 4, 5] ])
{{out}}
Matrix[[0.981818181818182]]
Testing 3-dimension: Points(x,y,z): [1,1,3], [2,1,4] and [1,2,5]
puts regression_coefficients([3,4,5], [ [1,2,1], [1,1,2] ])
{{out}}
Matrix[[0.9999999999999996], [2.0]]
Stata
First, build a random dataset:
clear
set seed 17760704
set obs 200
forv i=1/4 {
gen x`i'=rnormal()
}
gen y=1.5+0.8*x1-0.7*x2+1.1*x3-1.7*x4+rnormal()
Now, use the '''[https://www.stata.com/help.cgi?regress regress]''' command:
reg y x*
'''Output'''
The command shows the coefficients along with a bunch of useful information, such as R2, F statistic, standard errors of the coefficients...
Source | SS df MS Number of obs = 200
-------------+---------------------------------- F(4, 195) = 355.15
Model | 1343.81757 4 335.954392 Prob > F = 0.0000
Residual | 184.458622 195 .945941649 R-squared = 0.8793
-------------+---------------------------------- Adj R-squared = 0.8768
Total | 1528.27619 199 7.67977985 Root MSE = .9726
------------------------------------------------------------------------------
y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
x1 | .7525247 .0689559 10.91 0.000 .6165295 .8885198
x2 | -.7036303 .0697456 -10.09 0.000 -.8411828 -.5660778
x3 | 1.157477 .072189 16.03 0.000 1.015106 1.299849
x4 | -1.718201 .0621758 -27.63 0.000 -1.840824 -1.595577
_cons | 1.399131 .0697862 20.05 0.000 1.261499 1.536764
------------------------------------------------------------------------------
The regress command also sets a number of '''[https://www.stata.com/help.cgi?ereturn ereturn]''' values, which can be used by subsequent commands. The coefficients and their standard errors also have a [https://www.stata.com/help.cgi?_variables special syntax]:
. di _b[x1]
.75252466
. di _b[_cons]
1.3991314
. di _se[x1]
.06895593
. di _se[_cons]
.06978623
See '''[https://www.stata.com/help.cgi?estat estat]''', '''[https://www.stata.com/help.cgi?predict predict]''', '''[https://www.stata.com/help.cgi?estimates estimates]''', '''[https://www.stata.com/help.cgi?margins margins]''' for examples of commands that can be used after a regression.
Here we compute [[wp:Akaike information criterion|Akaike's AIC]], the covariance matrix of the estimates, the predicted values and residuals:
. estat ic
Akaike's information criterion and Bayesian information criterion
-----------------------------------------------------------------------------
Model | Obs ll(null) ll(model) df AIC BIC
-------------+---------------------------------------------------------------
. | 200 -487.1455 -275.6985 5 561.397 577.8886
-----------------------------------------------------------------------------
Note: N=Obs used in calculating BIC; see [R] BIC note.
. estat vce
Covariance matrix of coefficients of regress model
e(V) | x1 x2 x3 x4 _cons
-------------+------------------------------------------------------------
x1 | .00475492
x2 | -.00040258 .00486445
x3 | -.00042516 .00017355 .00521125
x4 | -.00011915 -.0002568 .00054646 .00386583
_cons | .00030777 -.00031109 -.00023794 .00058926 .00487012
. predict yhat, xb
. predict r, r
Tcl
{{tcllib|math::linearalgebra}}
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
Using an example from the Wikipedia page on the correlation of height and weight:
# 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}]]
{{out}} (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 [http://www.netlib.org/lapack/lug/node27.html], which is callable in Ursala like this:
regression_coefficients = lapack..dgelsd
test program:
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)
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.
{{out}}
<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.
zkl
Using the GNU Scientific Library:
var [const] GSL=Import("zklGSL"); // libGSL (GNU Scientific Library)
height:=GSL.VectorFromData(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:=GSL.VectorFromData(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);
v:=GSL.polyFit(height,weight,2);
v.format().println();
GSL.Helpers.polyString(v).println();
GSL.Helpers.polyEval(v,height).format().println();
{{out}}
128.81,-143.16,61.96
128.813 - 143.162x + 61.9603x^2
52.25,53.48,54.36,55.77,56.77,58.37,60.08,61.28,63.18,64.50,66.58,68.03,70.30,71.87,74.33
Or, using Lists: {{trans|Common Lisp}}
// Solve a linear system AX=B where A is symmetric and positive definite, so it can be Cholesky decomposed.
fcn linsys(A,B){
n,m:=A.len(),B[1].len(); // A.rows,B.cols
y:=n.pump(List.createLong(n).write,0.0); // writable vector of n zeros
X:=make_array(n,m,0.0);
L:=cholesky(A); // A=LL'
foreach col in (m){
foreach k in (n){ // Forward substitution: y = L\B
y[k]=( B[k][col] - k.reduce('wrap(s,j){ s + L[k][j]*y[j] },0.0) )
/L[k][k];
}
foreach k in ([n-1..0,-1]){ // Back substitution. x=L'\y
X[k][col]=
( y[k] - (k+1).reduce(n-k-1,'wrap(s,j){ s + L[j][k]*X[j][col] },0.0) )
/L[k][k];
}
}
X
}
fcn cholesky(mat){ // Cholesky decomposition task
rows:=mat.len();
r:=(0).pump(rows,List().write, (0).pump(rows,List,0.0).copy); // matrix of zeros
foreach i,j in (rows,i+1){
s:=(0).reduce(j,'wrap(s,k){ s + r[i][k]*r[j][k] },0.0);
r[i][j]=( if(i==j)(mat[i][i] - s).sqrt()
else 1.0/r[j][j]*(mat[i][j] - s) );
}
r
}
// Solve a linear least squares problem. Ax=b, with A being mxn, with m>n.
// Solves the linear system A'Ax=A'b.
fcn lsqr(A,b){
at:=transpose(A);
linsys(matMult(at,A), matMult(at,b));
}
// Least square fit of a polynomial of order n the x-y-curve.
fcn polyfit(x,y,n){
n+=1;
m:=x[0].len(); // columns
A:=make_array(m,n,0.0);
foreach i,j in (m,n){ A[i][j]=x[0][i].pow(j); }
lsqr(A, transpose(y));
}
fcn make_array(n,m,v){ (m).pump(List.createLong(m).write,v)*n }
fcn matMult(a,b){
n,m,p:=a[0].len(),a.len(),b[0].len();
ans:=make_array(m,p,0.0);
foreach i,j,k in (m,p,n){ ans[i][j]+=a[i][k]*b[k][j]; }
ans
}
fcn transpose(M){
if(M.len()==1) M[0].pump(List,List.create); // 1 row --> n columns
else M[0].zip(M.xplode(1));
}
height:=T(T(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:=T(T(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));
polyfit(height,weight,2).flatten().println();
{{out}}
L(128.813,-143.162,61.9603)