牛顿法线性规划octave实现_octave线性规划-优快云博客

本文链接：https://blog.youkuaiyun.com/mingspy/article/details/8881197

本文介绍了一个使用Newton梯度算法实现的线性回归函数newton_regression。该函数接收训练样本矩阵x和目标值向量y作为输入，返回最佳拟合参数θ。通过迭代更新θ来最小化预测误差。

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#newton_regression.m
## Copyright (C) 2013 xiuleili
## 
## This program is free software; you can redistribute it and/or modify
## it under the terms of the GNU General Public License as published by
## the Free Software Foundation; either version 3 of the License, or
## (at your option) any later version.
## 
## This program is distributed in the hope that it will be useful,
## but WITHOUT ANY WARRANTY; without even the implied warranty of
## MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
## GNU General Public License for more details.
## 
## You should have received a copy of the GNU General Public License
## along with Octave; see the file COPYING.  If not, see
## <http://www.gnu.org/licenses/>.

## -*- texinfo -*-
## @deftypefn {Function File} {} [ theta ] = newton_regression ( x, y)
## Return the parameter of linear founction where y = theta[2:n+1]*x + theta(1).
##	where n is the row of matrix x.
## It use newton gradient algorithm obviously.
## For example:
##
## @example
## @group
## x=[1 4;2 5;5 1; 4 2] y = [ 19 26 19 20]
## newton_regression(x, y)
##   @result{} [0.0060406   2.9990063   3.9990063]
## @end group
## @end example
## @seealso{stichastic_gradient}
## @end deftypefn
## Author: xiuleili <xiuleili@XIULEILI>
## Created: 2013-05-03

function [ theta ] = newton_regression (x,y)
[n,m]=size(x);
[my,ny]=size(y);
if(ny ~= n | my!= 1)
	error("Error: x should be a matrix with(n,m) and y must be (1,n), where n is the count of training samples.");
end;

X = [ones(n,1) x]';
theta = rand(1, m+1);
learning_rate = 0.1;
errors = 1;
threshold=0.000001;
times = 0;

% thetaj = thetaj + sum[(y-h(x))*xj]/sum[xj^2]
while errors > threshold
	%sum((y-h(x))*xj)
	sum_of_error = (y-theta*X)*X';
	%sum(xj^2)
	sum_of_xj2 = (diag(X*X'))';
	theta += learning_rate * sum_of_error./sum_of_xj2;
	errors = sum((y-theta*X).^2)/2;
	times ++;
	printf("[%d] errors = %f", times, errors);
	disp(theta);
	if(times > 100000)
		break;
	end
end
endfunction