Multiple regression: Difference between revisions
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# + 2.92065133466547 x4 + 1.76076442997642 c</lang> |
# + 2.92065133466547 x4 + 1.76076442997642 c</lang> |
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=={{header|Mathematica}}== |
=={{header|Mathematica}}/{{header|Wolfram Language}}== |
||
<lang 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}; |
<lang 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}; |
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}; |
X = {x^0, x^1, x^2}; |
||
b = y.PseudoInverse[X] |
b = y.PseudoInverse[X]</lang> |
||
{{out}} |
|||
<pre>{128.813, -143.162, 61.9603}</pre> |
|||
=={{header|MATLAB}}== |
=={{header|MATLAB}}== |
||
Revision as of 20:04, 30 July 2021
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
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
BBC BASIC
<lang bbcbasic> *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</lang>
- Output:
128.812804 -143.162023 61.9603254
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>
C++
<lang cpp>#include <array>
- include <iostream>
void require(bool condition, const std::string &message) {
if (condition) {
return;
}
throw std::runtime_error(message);
}
template<typename T, size_t N> std::ostream &operator<<(std::ostream &os, const std::array<T, N> &a) {
auto it = a.cbegin(); auto end = a.cend();
os << '[';
if (it != end) {
os << *it;
it = std::next(it);
}
while (it != end) {
os << ", " << *it;
it = std::next(it);
}
return os << ']';
}
template <size_t RC, size_t CC> class Matrix {
std::array<std::array<double, CC>, RC> data;
public:
Matrix() : data{} {
// empty
}
Matrix(std::initializer_list<std::initializer_list<double>> values) {
size_t rp = 0;
for (auto row : values) {
size_t cp = 0;
for (auto col : row) {
data[rp][cp] = col;
cp++;
}
rp++;
}
}
double get(size_t row, size_t col) const {
return data[row][col];
}
void set(size_t row, size_t col, double value) {
data[row][col] = value;
}
std::array<double, CC> get(size_t row) {
return data[row];
}
void set(size_t row, const std::array<double, CC> &values) {
std::copy(values.begin(), values.end(), data[row].begin());
}
template <size_t D>
Matrix<RC, D> operator*(const Matrix<CC, D> &rhs) const {
Matrix<RC, D> result;
for (size_t i = 0; i < RC; i++) {
for (size_t j = 0; j < D; j++) {
for (size_t k = 0; k < CC; k++) {
double prod = get(i, k) * rhs.get(k, j);
result.set(i, j, result.get(i, j) + prod);
}
}
}
return result;
}
Matrix<CC, RC> transpose() const {
Matrix<CC, RC> trans;
for (size_t i = 0; i < RC; i++) {
for (size_t j = 0; j < CC; j++) {
trans.set(j, i, data[i][j]);
}
}
return trans;
}
void toReducedRowEchelonForm() {
size_t lead = 0;
for (size_t r = 0; r < RC; r++) {
if (CC <= lead) {
return;
}
auto i = r;
while (get(i, lead) == 0.0) {
i++;
if (RC == i) {
i = r;
lead++;
if (CC == lead) {
return;
}
}
}
auto temp = get(i);
set(i, get(r));
set(r, temp);
if (get(r, lead) != 0.0) {
auto div = get(r, lead);
for (size_t j = 0; j < CC; j++) {
set(r, j, get(r, j) / div);
}
}
for (size_t k = 0; k < RC; k++) {
if (k != r) {
auto mult = get(k, lead);
for (size_t j = 0; j < CC; j++) {
auto prod = get(r, j) * mult;
set(k, j, get(k, j) - prod);
}
}
}
lead++;
}
}
Matrix<RC, RC> inverse() {
require(RC == CC, "Not a square matrix");
Matrix<RC, 2 * RC> aug;
for (size_t i = 0; i < RC; i++) {
for (size_t j = 0; j < RC; j++) {
aug.set(i, j, get(i, j));
}
// augment identify matrix to right
aug.set(i, i + RC, 1.0);
}
aug.toReducedRowEchelonForm();
// remove identity matrix to left
Matrix<RC, RC> inv;
for (size_t i = 0; i < RC; i++) {
for (size_t j = RC; j < 2 * RC; j++) {
inv.set(i, j - RC, aug.get(i, j));
}
}
return inv;
}
template <size_t RC, size_t CC> friend std::ostream &operator<<(std::ostream &, const Matrix<RC, CC> &);
};
template <size_t RC, size_t CC> std::ostream &operator<<(std::ostream &os, const Matrix<RC, CC> &m) {
for (size_t i = 0; i < RC; i++) {
os << '[';
for (size_t j = 0; j < CC; j++) {
if (j > 0) {
os << ", ";
}
os << m.get(i, j);
}
os << "]\n";
}
return os;
}
template <size_t RC, size_t CC> std::array<double, RC> multiple_regression(const std::array<double, CC> &y, const Matrix<RC, CC> &x) {
Matrix<1, CC> tm; tm.set(0, y);
auto cy = tm.transpose(); auto cx = x.transpose(); return ((x * cx).inverse() * x * cy).transpose().get(0);
}
void case1() {
std::array<double, 5> y{ 1.0, 2.0, 3.0, 4.0, 5.0 };
Matrix<1, 5> x{ {2.0, 1.0, 3.0, 4.0, 5.0} };
auto v = multiple_regression(y, x);
std::cout << v << '\n';
}
void case2() {
std::array<double, 3> y{ 3.0, 4.0, 5.0 };
Matrix<2, 3> x{
{1.0, 2.0, 1.0},
{1.0, 1.0, 2.0}
};
auto v = multiple_regression(y, x);
std::cout << v << '\n';
}
void case3() {
std::array<double, 15> 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 };
std::array<double, 15> a{ 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 };
Matrix<3, 15> x;
for (size_t i = 0; i < 15; i++) {
x.set(0, i, 1.0);
}
x.set(1, a);
for (size_t i = 0; i < 15; i++) {
x.set(2, i, a[i] * a[i]);
}
auto v = multiple_regression(y, x); std::cout << v << '\n';
}
int main() {
case1(); case2(); case3();
return 0;
}</lang>
- Output:
[0.981818] [1, 2] [128.813, -143.162, 61.9603]
C#
<lang csharp>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);
}
}</lang>
- Output:
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.
<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>
D
<lang d>import std.algorithm; import std.array; import std.exception; import std.range; import std.stdio;
public class Matrix {
private double[][] data; private size_t rowCount; private size_t colCount;
public this(size_t size)
in(size > 0, "Must have at least one element")
{
this(size, size);
}
public this(size_t rows, size_t cols)
in(rows > 0, "Must have at least one row")
in(cols > 0, "Must have at least one column")
{
rowCount = rows;
colCount = cols;
data = uninitializedArray!(double[][])(rows, cols);
foreach (ref row; data) {
row[] = 0.0;
}
}
public this(const double[][] source) {
enforce(source.length > 0, "Must have at least one row");
rowCount = source.length;
enforce(source[0].length > 0, "Must have at least one column");
colCount = source[0].length;
data = uninitializedArray!(double[][])(rowCount, colCount);
foreach (i; 0 .. rowCount) {
enforce(source[i].length == colCount, "All rows must have equal columns");
data[i] = source[i].dup;
}
}
public auto opIndex(size_t r, size_t c) const {
return data[r][c];
}
public auto opIndex(size_t r) const {
return data[r];
}
public auto opBinary(string op)(const Matrix rhs) const {
static if (op == "*") {
auto rc1 = rowCount;
auto cc1 = colCount;
auto rc2 = rhs.rowCount;
auto cc2 = rhs.colCount;
enforce(cc1 == rc2, "Cannot multiply if the first columns does not equal the second rows");
auto result = new Matrix(rc1, cc2);
foreach (i; 0 .. rc1) {
foreach (j; 0 .. cc2) {
foreach (k; 0 .. rc2) {
result[i, j] += this[i, k] * rhs[k, j];
}
}
}
return result;
} else {
assert(false, "Not implemented");
}
}
public void opIndexAssign(double value, size_t r, size_t c) {
data[r][c] = value;
}
public void opIndexAssign(const double[] value, size_t r) {
enforce(colCount == value.length, "Slice size must match column size");
data[r] = value.dup;
}
public void opIndexOpAssign(string op)(double value, size_t r, size_t c) {
mixin("data[r][c] " ~ op ~ "= value;");
}
public auto transpose() const {
auto rc = rowCount;
auto cc = colCount;
auto t = new Matrix(cc, rc);
foreach (i; 0 .. cc) {
foreach (j; 0 .. rc) {
t[i, j] = this[j, i];
}
}
return t;
}
public void toReducedRowEchelonForm() {
auto lead = 0;
auto rc = rowCount;
auto cc = colCount;
foreach (r; 0 .. rc) {
if (cc <= lead) {
return;
}
auto i = r;
while (this[i, lead] == 0.0) {
i++;
if (rc == i) {
i = r;
lead++;
if (cc == lead) {
return;
}
}
}
auto temp = this[i];
this[i] = this[r];
this[r] = temp;
if (this[r, lead] != 0.0) {
auto div = this[r, lead];
foreach (j; 0 .. cc) {
this[r, j] = this[r, j] / div;
}
}
foreach (k; 0 .. rc) {
if (k != r) {
auto mult = this[k, lead];
foreach (j; 0 .. cc) {
this[k, j] -= this[r, j] * mult;
}
}
}
lead++;
}
}
public auto inverse() const {
enforce(rowCount == colCount, "Not a square matrix");
auto len = rowCount;
auto aug = new Matrix(len, 2 * len);
foreach (i; 0 .. len) {
foreach (j; 0 .. len) {
aug[i, j] = this[i, j];
}
// augment identity matrix to right
aug[i, i + len] = 1.0;
}
aug.toReducedRowEchelonForm;
auto inv = new Matrix(len);
// remove identify matrix to left
foreach (i; 0 .. len) {
foreach (j; len .. 2 * len) {
inv[i, j - len] = aug[i, j];
}
}
return inv;
}
void toString(scope void delegate(const(char)[]) sink) const {
import std.format;
auto fmt = FormatSpec!char("%s");
put(sink, "[");
foreach (i; 0 .. rowCount) {
if (i > 0) {
put(sink, " [");
} else {
put(sink, "[");
}
formatValue(sink, this[i, 0], fmt);
foreach (j; 1 .. colCount) {
put(sink, ", ");
formatValue(sink, this[i, j], fmt);
}
if (i + 1 < rowCount) {
put(sink, "]\n");
} else {
put(sink, "]");
}
}
put(sink, "]");
}
}
auto multipleRegression(double[] y, Matrix x) {
auto tm = new Matrix([y]); auto cy = tm.transpose; auto cx = x.transpose; return ((x * cx).inverse * x * cy).transpose[0].dup;
}
void main() {
auto y = [1.0, 2.0, 3.0, 4.0, 5.0]; auto x = new Matrix(2.0, 1.0, 3.0, 4.0, 5.0); auto v = multipleRegression(y, x); v.writeln;
y = [3.0, 4.0, 5.0];
x = new Matrix([
[1.0, 2.0, 1.0],
[1.0, 1.0, 2.0]
]);
v = multipleRegression(y, x);
v.writeln;
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];
auto a = [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 = new Matrix([
repeat(1.0, a.length).array,
a,
a.map!"a * a".array
]);
v = multipleRegression(y, x);
v.writeln;
}</lang>
- Output:
[0.981818] [1, 2] [128.813, -143.162, 61.9603]
Emacs Lisp
Multiple regression analysis by Emacs Lisp and built-in Emacs Calc.
<lang emacs-lisp> (setq X1 '[0 1 2 3 4 5 6 7 8 9 10]) (setq X2 '[0 1 1 3 3 7 6 7 3 9 8]) (setq Y '[1 6 17 34 57 86 121 162 209 262 321]) (calc-eval
(format "fit(a*X1+b*X2+c,[X1,X2],[a,b,c],[%s %s %s])" X1 X2 Y))
</lang>
- Output:
"35.2014388489 X1 - 3.95683453237 X2 - 42.7410071942"
ERRE
<lang 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</lang>
- Output:
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
<lang Fortran>*-----------------------------------------------------------------------
- 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
</lang>
- Output:
beta: 128.8126 -143.1618 61.9603 STOP program complete
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.
Library gonum/matrix
<lang go>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)))
}</lang>
- Output:
⎡ 128.8128⎤ ⎢-143.1620⎥ ⎣ 61.9603⎦
Library go.matrix
<lang go>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)
}</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>
Hy
<lang lisp>(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)))</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
xpy=: mp~ |: NB. Or factoring out "X prime y" (monadically "X prime X") X (%.@:xpy@[ mp xpy) y
128.814 _143.163 61.9606 </lang>
LAPACK routines are also available via the Addon math/lapack. <lang j> load 'math/lapack'
load 'math/lapack/gels' gels_jlapack_ X;y
128.813 _143.162 61.9603</lang>
Java
<lang java>import java.util.Arrays; import java.util.Objects;
public class MultipleRegression {
public static void require(boolean condition, String message) {
if (condition) {
return;
}
throw new IllegalArgumentException(message);
}
public static class Matrix {
private final double[][] data;
private final int rowCount;
private final int colCount;
public Matrix(int rows, int cols) {
require(rows > 0, "Need at least one row");
this.rowCount = rows;
require(cols > 0, "Need at least one column");
this.colCount = cols;
this.data = new double[rows][cols];
for (double[] row : this.data) {
Arrays.fill(row, 0.0);
}
}
public Matrix(double[][] source) {
require(source.length > 0, "Need at least one row");
this.rowCount = source.length;
require(source[0].length > 0, "Need at least one column");
this.colCount = source[0].length;
this.data = new double[this.rowCount][this.colCount];
for (int i = 0; i < this.rowCount; i++) {
set(i, source[i]);
}
}
public double[] get(int row) {
Objects.checkIndex(row, this.rowCount);
return this.data[row];
}
public void set(int row, double[] data) {
Objects.checkIndex(row, this.rowCount);
require(data.length == this.colCount, "The column in the row must match the number of columns in the matrix");
System.arraycopy(data, 0, this.data[row], 0, this.colCount);
}
public double get(int row, int col) {
Objects.checkIndex(row, this.rowCount);
Objects.checkIndex(col, this.colCount);
return this.data[row][col];
}
public void set(int row, int col, double value) {
Objects.checkIndex(row, this.rowCount);
Objects.checkIndex(col, this.colCount);
this.data[row][col] = value;
}
@SuppressWarnings("UnnecessaryLocalVariable")
public Matrix times(Matrix that) {
var rc1 = this.rowCount;
var cc1 = this.colCount;
var rc2 = that.rowCount;
var cc2 = that.colCount;
require(cc1 == rc2, "Cannot multiply if the first columns does not equal the second rows");
var result = new Matrix(rc1, cc2);
for (int i = 0; i < rc1; i++) {
for (int j = 0; j < cc2; j++) {
for (int k = 0; k < rc2; k++) {
var prod = get(i, k) * that.get(k, j);
result.set(i, j, result.get(i, j) + prod);
}
}
}
return result;
}
public Matrix transpose() {
var rc = this.rowCount;
var cc = this.colCount;
var trans = new Matrix(cc, rc);
for (int i = 0; i < cc; i++) {
for (int j = 0; j < rc; j++) {
trans.set(i, j, get(j, i));
}
}
return trans;
}
public void toReducedRowEchelonForm() {
int lead = 0;
var rc = this.rowCount;
var cc = this.colCount;
for (int r = 0; r < rc; r++) {
if (cc <= lead) {
return;
}
var i = r;
while (get(i, lead) == 0.0) {
i++;
if (rc == i) {
i = r;
lead++;
if (cc == lead) {
return;
}
}
}
var temp = get(i);
set(i, get(r));
set(r, temp);
if (get(r, lead) != 0.0) {
var div = get(r, lead);
for (int j = 0; j < cc; j++) {
set(r, j, get(r, j) / div);
}
}
for (int k = 0; k < rc; k++) {
if (k != r) {
var mult = get(k, lead);
for (int j = 0; j < cc; j++) {
var prod = get(r, j) * mult;
set(k, j, get(k, j) - prod);
}
}
}
lead++;
}
}
public Matrix inverse() {
require(this.rowCount == this.colCount, "Not a square matrix");
var len = this.rowCount;
var aug = new Matrix(len, 2 * len);
for (int i = 0; i < len; i++) {
for (int j = 0; j < len; j++) {
aug.set(i, j, get(i, j));
}
// augment identity matrix to right
aug.set(i, i + len, 1.0);
}
aug.toReducedRowEchelonForm();
var inv = new Matrix(len, len);
// remove identity matrix to left
for (int i = 0; i < len; i++) {
for (int j = len; j < 2 * len; j++) {
inv.set(i, j - len, aug.get(i, j));
}
}
return inv;
}
}
public static double[] multipleRegression(double[] y, Matrix x) {
var tm = new Matrix(new double[][]{y});
var cy = tm.transpose();
var cx = x.transpose();
return x.times(cx).inverse().times(x).times(cy).transpose().get(0);
}
public static void printVector(double[] v) {
System.out.println(Arrays.toString(v));
System.out.println();
}
public static double[] repeat(int size, double value) {
var a = new double[size];
Arrays.fill(a, value);
return a;
}
public static void main(String[] args) {
double[] y = new double[]{1.0, 2.0, 3.0, 4.0, 5.0};
var x = new Matrix(new double[][]Template:2.0, 1.0, 3.0, 4.0, 5.0);
var v = multipleRegression(y, x);
printVector(v);
y = new double[]{3.0, 4.0, 5.0};
x = new Matrix(new double[][]{
{1.0, 2.0, 1.0},
{1.0, 1.0, 2.0}
});
v = multipleRegression(y, x);
printVector(v);
y = 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 a = 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};
x = new Matrix(new double[][]{
repeat(a.length, 1.0),
a,
Arrays.stream(a).map(it -> it * it).toArray()
});
v = multipleRegression(y, x);
printVector(v);
}
}</lang>
- Output:
[0.9818181818181818] [0.9999999999999996, 2.000000000000001] [128.8128035798277, -143.1620228653037, 61.960325442985436]
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
Julia
As in Matlab, the backslash or slash operator (depending on the matrix ordering) can be used for solving this problem, for example:
<lang julia>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</lang>
- Output:
3-element Array{Float64,1}:
128.813
-143.162
61.9603
Kotlin
As neither the JDK nor the Kotlin Standard Library has matrix operations built-in, we re-use functions written for various other tasks. <lang scala>// 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)
}</lang>
- Output:
[0.9818181818181818] [0.9999999999999996, 2.000000000000001] [128.8128035798277, -143.1620228653037, 61.960325442985436]
Maple
First build a random dataset.
<lang maple>n:=200: X:=<ArrayTools[RandomArray](n,4,distribution=normal)|Vector(n,1,datatype=float[8])>: Y:=X.<4.2,-2.8,-1.4,3.1,1.75>+convert(ArrayTools[RandomArray](n,1,distribution=normal),Vector):</lang>
Now the linear regression, with either the LinearAlgebra package, or the Statistics package.
<lang maple>LinearAlgebra[LeastSquares](X,Y)^+;
- [4.33701132468683, -2.78654498997457, -1.41840666085642, 2.92065133466547, 1.76076442997642]
Statistics[LinearFit]([x1,x2,x3,x4,c],X,Y,[x1,x2,x3,x4,c],summarize=true)
- Summary:
- ----------------
- Model: 4.3370113*x1-2.7865450*x2-1.4184067*x3+2.9206513*x4+1.7607644*c
- ----------------
- Coefficients:
- Estimate Std. Error t-value P(>|t|)
- Parameter 1 4.3370 0.0691 62.7409 0.0000
- Parameter 2 -2.7865 0.0661 -42.1637 0.0000
- Parameter 3 -1.4184 0.0699 -20.2937 0.0000
- Parameter 4 2.9207 0.0687 42.5380 0.0000
- Parameter 5 1.7608 0.0701 25.1210 0.0000
- ----------------
- R-squared: 0.9767, Adjusted R-squared: 0.9761
- 4.33701132468683 x1 - 2.78654498997457 x2 - 1.41840666085642 x3
- + 2.92065133466547 x4 + 1.76076442997642 c</lang>
Mathematica/Wolfram Language
<lang 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]</lang>
- Output:
{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.
<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>
Nim
<lang Nim># Using Wikipedia data sample.
import math import arraymancer, sequtils
var
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].toTensor()
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].toTensor()
- Create Vandermonde matrix.
var a = stack(height.ones_like, height, height *. height, axis = 1)
echo toSeq(least_squares_solver(a, weight).solution.items)</lang>
- Output:
@[128.8128035784397, -143.1620228647656, 61.96032544247468]
PARI/GP
<lang parigp>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)</lang>
Perl
<lang perl>use strict; use warnings; use Statistics::Regression;
my @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); my @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);
my @model = ('const', 'X', 'X**2'); my $reg = Statistics::Regression->new( , [@model] ); $reg->include( $y[$_], [ 1.0, $x[$_], $x[$_]**2 ]) for 0..@y-1; my @coeff = $reg->theta();
printf "%-6s %8.3f\n", $model[$_], $coeff[$_] for 0..@model-1;</lang>
- Output:
const 128.813 X -143.162 X**2 61.960
Phix
<lang Phix>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]</lang>
- Output:
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
<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
Method with matrix operations <lang python>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)</lang>
- Output:
[[ 128.81280359 -143.16202288 61.96032545]]
Using numpy lstsq function <lang python>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])</lang>
- Output:
[ 128.81280358 -143.16202286 61.96032544]
R
R provides the lm function for linear regression.
<lang R>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>
- 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>
A numerically more stable way is to use the QR decomposition of the design matrix:
<lang R>qr.coef(qr(X), y)</lang>
Racket
<lang racket>
- lang racket
(require math) (define T matrix-transpose)
(define (fit X y)
(matrix-solve (matrix* (T X) X) (matrix* (T X) y)))
</lang> Test: <lang racket> (fit (matrix [[1 2]
[2 5]
[3 7]
[4 9]])
(matrix [[1]
[2]
[3]
[9]]))
- Output:
(array #[#[9 1/3] #[-3 1/3]]) </lang>
Raku
(formerly Perl 6) We're going to solve the example on the Wikipedia article using Clifford, a 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.
<lang perl6>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;</lang>
- Output:
α = 128.81280357844 β = -143.1620228648 γ = 61.960325442
April 2016: This computation took over an hour with MoarVM, running in a VirtualBox linux system guest hosted by a windows laptop with a i7 intel processor.
March 2020: With various improvements to the language, 6.d releases of Raku now run the code 2x to 3x as fast, depending on the hardware used, but even so it is generous to describe it as 'slow'.
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 2-dimension: <lang ruby>puts regression_coefficients([1, 2, 3, 4, 5], [ [2, 1, 3, 4, 5] ])</lang>
- Output:
Matrix[[0.981818181818182]]
Testing 3-dimension: Points(x,y,z): [1,1,3], [2,1,4] and [1,2,5] <lang ruby>puts regression_coefficients([3,4,5], [ [1,2,1], [1,1,2] ])</lang>
- Output:
Matrix[[0.9999999999999996], [2.0]]
SPSS
First, build a random dataset:
<lang>set rng=mc seed=17760704. new file. input program.
vector x(4).
loop #i=1 to 200.
loop #j=1 to 4.
compute x(#j)=rv.normal(0,1).
end loop.
end case.
end loop.
end file.
end input program. compute y=1.5+0.8*x1-0.7*x2+1.1*x3-1.7*x4+rv.normal(0,1). execute.</lang>
Now use the regression command:
<lang>regression /dependent=y /method=enter x1 x2 x3 x4.</lang>
- Output:
<lang>Regression Notes |--------------------------------------------------------------------|---------------------------------------------------------------------------| |Output Created |21-MAR-2020 23:17:33 | |--------------------------------------------------------------------|---------------------------------------------------------------------------| |Comments | | |----------------------|---------------------------------------------|---------------------------------------------------------------------------| |Input |Filter |<none> | | |---------------------------------------------|---------------------------------------------------------------------------| | |Weight |<none> | | |---------------------------------------------|---------------------------------------------------------------------------| | |Split File |<none> | | |---------------------------------------------|---------------------------------------------------------------------------| | |N of Rows in Working Data File |200 | |----------------------|---------------------------------------------|---------------------------------------------------------------------------| |Missing Value Handling|Definition of Missing |User-defined missing values are treated as missing. | | |---------------------------------------------|---------------------------------------------------------------------------| | |Cases Used |Statistics are based on cases with no missing values for any variable used.| |--------------------------------------------------------------------|---------------------------------------------------------------------------| |Syntax |regression /dependent=y /method=enter x1 x2 x3 x4. | |----------------------|---------------------------------------------|---------------------------------------------------------------------------| |Resources |Processor Time |00:00:00,00 | | |---------------------------------------------|---------------------------------------------------------------------------| | |Elapsed Time |00:00:00,00 | | |---------------------------------------------|---------------------------------------------------------------------------| | |Memory Required |4080 bytes | | |---------------------------------------------|---------------------------------------------------------------------------| | |Additional Memory Required for Residual Plots|0 bytes | |------------------------------------------------------------------------------------------------------------------------------------------------|
Variables Entered/Removeda |-----|-----------------|-----------------|------| |Model|Variables Entered|Variables Removed|Method| |-----|-----------------|-----------------|------| |1 |x4, x3, x2, x1b |. |Enter | |------------------------------------------------|
a Dependent Variable: y b All requested variables entered.
Model Summary |-----|-----|--------|-----------------|--------------------------| |Model|R |R Square|Adjusted R Square|Std. Error of the Estimate| |-----|-----|--------|-----------------|--------------------------| |1 |,929a|,863 |,860 |,94928 | |-----------------------------------------------------------------|
a Predictors: (Constant), x4, x3, x2, x1
ANOVAa |----------------|--------------|---|-----------|-------|-----| |Model |Sum of Squares|df |Mean Square|F |Sig. | |-----|----------|--------------|---|-----------|-------|-----| |1 |Regression|1106,659 |4 |276,665 |307,021|,000b| | |----------|--------------|---|-----------|-------|-----| | |Residual |175,720 |195|,901 | | | | |----------|--------------|---|-----------|-------|-----| | |Total |1282,379 |199| | | | |-------------------------------------------------------------|
a Dependent Variable: y b Predictors: (Constant), x4, x3, x2, x1
Coefficientsa |----------------|--------------------------------------|-------------------------|-------|----| |Model |Unstandardized Coefficients |Standardized Coefficients|t |Sig.| | |---------------------------|----------|-------------------------| | | | |B |Std. Error|Beta | | | |-----|----------|---------------------------|----------|-------------------------|-------|----| |1 |(Constant)|1,550 |,067 | |23,003 |,000| | |----------|---------------------------|----------|-------------------------|-------|----| | |x1 |,831 |,062 |,360 |13,457 |,000| | |----------|---------------------------|----------|-------------------------|-------|----| | |x2 |-,604 |,075 |-,215 |-8,051 |,000| | |----------|---------------------------|----------|-------------------------|-------|----| | |x3 |1,098 |,065 |,451 |16,989 |,000| | |----------|---------------------------|----------|-------------------------|-------|----| | |x4 |-1,770 |,073 |-,656 |-24,306|,000| |----------------------------------------------------------------------------------------------|
a Dependent Variable: y</lang>
Stata
First, build a random dataset:
<lang stata>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()</lang>
Now, use the regress command:
<lang stata>reg y x*</lang>
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 ereturn values, which can be used by subsequent commands. The coefficients and their standard errors also have a special syntax:
<lang stata>. di _b[x1] .75252466
. di _b[_cons] 1.3991314
. di _se[x1] .06895593
. di _se[_cons] .06978623</lang>
See regress postestimation for a list of commands that can be used after a regression.
Here we compute Akaike's AIC, the covariance matrix of the estimates, the predicted values and residuals:
<lang stata>. 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</lang>
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>
- Output:
(a 3-vector of coefficients)
128.81280358170625 -143.16202286630732 61.96032544293041
TI-83 BASIC
<lang ti83b>{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}→L₁ {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}→L₂ QuadReg L₁,L₂ </lang>
- Output:
y=ax²+bx+c a=61.96032544 b=–143.1620229 c=128.8128036
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.
Visual Basic .NET
<lang vbnet>Module Module1
Sub Swap(Of T)(ByRef x As T, ByRef y As T)
Dim temp = x
x = y
y = temp
End Sub
Sub Require(condition As Boolean, message As String)
If condition Then
Return
End If
Throw New ArgumentException(message)
End Sub
Class Matrix
Private data As Double(,)
Private rowCount As Integer
Private colCount As Integer
Public Sub New(rows As Integer, cols As Integer)
Require(rows > 0, "Need at least one row")
rowCount = rows
Require(cols > 0, "Need at least one column")
colCount = cols
data = New Double(rows - 1, cols - 1) {}
End Sub
Public Sub New(source As Double(,))
Dim rows = source.GetLength(0)
Require(rows > 0, "Need at least one row")
rowCount = rows
Dim cols = source.GetLength(1)
Require(cols > 0, "Need at least one column")
colCount = cols
data = New Double(rows - 1, cols - 1) {}
For i = 1 To rows
For j = 1 To cols
data(i - 1, j - 1) = source(i - 1, j - 1)
Next
Next
End Sub
Default Public Property Index(i As Integer, j As Integer) As Double
Get
Return data(i, j)
End Get
Set(value As Double)
data(i, j) = value
End Set
End Property
Public Property Slice(i As Integer) As Double()
Get
Dim m(colCount - 1) As Double
For j = 1 To colCount
m(j - 1) = Index(i, j - 1)
Next
Return m
End Get
Set(value As Double())
Require(colCount = value.Length, "Slice must match the number of columns")
For j = 1 To colCount
Index(i, j - 1) = value(j - 1)
Next
End Set
End Property
Public Shared Operator *(m1 As Matrix, m2 As Matrix) As Matrix
Dim rc1 = m1.rowCount
Dim cc1 = m1.colCount
Dim rc2 = m2.rowCount
Dim cc2 = m2.colCount
Require(cc1 = rc2, "Cannot multiply if the first columns does not equal the second rows")
Dim result As New Matrix(rc1, cc2)
For i = 1 To rc1
For j = 1 To cc2
For k = 1 To rc2
result(i - 1, j - 1) += m1(i - 1, k - 1) * m2(k - 1, j - 1)
Next
Next
Next
Return result
End Operator
Public Function Transpose() As Matrix
Dim rc = rowCount
Dim cc = colCount
Dim trans As New Matrix(cc, rc)
For i = 1 To cc
For j = 1 To rc
trans(i - 1, j - 1) = Index(j - 1, i - 1)
Next
Next
Return trans
End Function
Public Sub ToReducedRowEchelonForm()
Dim lead = 0
Dim rc = rowCount
Dim cc = colCount
For r = 1 To rc
If cc <= lead Then
Return
End If
Dim i = r
While Index(i - 1, lead) = 0.0
i += 1
If rc = i Then
i = r
lead += 1
If cc = lead Then
Return
End If
End If
End While
Dim temp = Slice(i - 1)
Slice(i - 1) = Slice(r - 1)
Slice(r - 1) = temp
If Index(r - 1, lead) <> 0.0 Then
Dim div = Index(r - 1, lead)
For j = 1 To cc
Index(r - 1, j - 1) /= div
Next
End If
For k = 1 To rc
If k <> r Then
Dim mult = Index(k - 1, lead)
For j = 1 To cc
Index(k - 1, j - 1) -= Index(r - 1, j - 1) * mult
Next
End If
Next
lead += 1
Next
End Sub
Public Function Inverse() As Matrix
Require(rowCount = colCount, "Not a square matrix")
Dim len = rowCount
Dim aug As New Matrix(len, 2 * len)
For i = 1 To len
For j = 1 To len
aug(i - 1, j - 1) = Index(i - 1, j - 1)
Next
REM augment identity matrix to right
aug(i - 1, i + len - 1) = 1.0
Next
aug.ToReducedRowEchelonForm()
Dim inv As New Matrix(len, len)
For i = 1 To len
For j = len + 1 To 2 * len
inv(i - 1, j - len - 1) = aug(i - 1, j - 1)
Next
Next
Return inv
End Function
End Class
Function ConvertArray(source As Double()) As Double(,)
Dim dest(0, source.Length - 1) As Double
For i = 1 To source.Length
dest(0, i - 1) = source(i - 1)
Next
Return dest
End Function
Function MultipleRegression(y As Double(), x As Matrix) As Double()
Dim tm As New Matrix(ConvertArray(y))
Dim cy = tm.Transpose
Dim cx = x.Transpose
Return ((x * cx).Inverse * x * cy).Transpose.Slice(0)
End Function
Sub Print(v As Double())
Dim it = v.GetEnumerator()
Console.Write("[")
If it.MoveNext() Then
Console.Write(it.Current)
End If
While it.MoveNext
Console.Write(", ")
Console.Write(it.Current)
End While
Console.Write("]")
End Sub
Sub Main()
Dim y() = {1.0, 2.0, 3.0, 4.0, 5.0}
Dim x As New Matrix(Template:2.0, 1.0, 3.0, 4.0, 5.0)
Dim v = MultipleRegression(y, x)
Print(v)
Console.WriteLine()
y = {3.0, 4.0, 5.0}
x = New Matrix({
{1.0, 2.0, 1.0},
{1.0, 1.0, 2.0}
})
v = MultipleRegression(y, x)
Print(v)
Console.WriteLine()
y = {52.21, 53.12, 54.48, 55.84, 57.2, 58.57, 59.93, 61.29, 63.11, 64.47, 66.28, 68.1, 69.92, 72.19, 74.46}
Dim a = {1.47, 1.5, 1.52, 1.55, 1.57, 1.6, 1.63, 1.65, 1.68, 1.7, 1.73, 1.75, 1.78, 1.8, 1.83}
Dim xs(2, a.Length - 1) As Double
For i = 1 To a.Length
xs(0, i - 1) = 1.0
xs(1, i - 1) = a(i - 1)
xs(2, i - 1) = a(i - 1) * a(i - 1)
Next
x = New Matrix(xs)
v = MultipleRegression(y, x)
Print(v)
Console.WriteLine()
End Sub
End Module</lang>
- Output:
[0.981818181818182] [1, 2] [128.812803579828, -143.162022865304, 61.9603254429854]
Wren
<lang ecmascript>import "/matrix" for Matrix
var multipleRegression = Fn.new { |y, x|
var cy = y.transpose var cx = x.transpose return ((x * cx).inverse * x * cy).transpose[0]
}
var ys = [
Matrix.new([ [1, 2, 3, 4, 5] ]),
Matrix.new([ [3, 4, 5] ]),
Matrix.new([ [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 a = [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 xs = [
Matrix.new([ [2, 1, 3, 4, 5] ]),
Matrix.new([ [1, 2, 1], [1, 1, 2] ]),
Matrix.new([ List.filled(a.count, 1), a, a.map { |e| e * e }.toList ])
]
for (i in 0...ys.count) {
var v = multipleRegression.call(ys[i], xs[i]) System.print(v) System.print()
}</lang>
- Output:
[0.98181818181818] [1, 2] [128.81280357983, -143.1620228653, 61.960325442985]
zkl
Using the GNU Scientific Library: <lang zkl>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();</lang>
- Output:
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:
<lang zkl>// 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));
}</lang> <lang zkl>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();</lang>
- Output:
L(128.813,-143.162,61.9603)
- Programming Tasks
- Probability and statistics
- Ada
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