Help with linear regression function

11 次查看(过去 30 天)
Hello, community.
Can someone help me with the following assignment.
Given a set of approximate x and y coordinates of points in a plane, determine the best fitting line in the least square sense. Using the standard formula of a line ax + b = y, compute a and b. That is, write a function that takes two row verctors of the same length called x and y as input arguments (containing x and y coordinates of points) and returns two scalars, a and b specifying the line, as output arguments.
I can't use polyfit
Thank you in advance!

采纳的回答

Image Analyst
Image Analyst 2020-10-1
Homework hint: Use the backslash operator. See the FAQ:
They do it there.

更多回答(2 个)

Erandi T. Sandarenu
This was done without using backslash operator. But it works!
function [a b] = lin_reg(x,y)
X = mean(x);
Y = mean(y);
a = sum((x-X).*(y-Y))./sum((x-X).^2);
b = Y - a*X;
end

Erandi T. Sandarenu
This was done by using the backslash operator.
function [a b] = lin_reg(x,y)
matrix = [x; ones(1,length(x))]';
x = matrix \ y';
a = x(1);
b = x(2);
end
  2 个评论
Mohaddeseh Mohammadi
Hello, Could you please explain the method you used? I can not understand what is matrix = [x; ones(1,length(x))]'; for.
Image Analyst
Image Analyst 2022-10-4
It's easy just to try something and see. Make x a row vector of 4 elements and see what it gives:
x = [1,2,3,4]
x = 1×4
1 2 3 4
matrix = [x; ones(1,length(x))]'
matrix = 4×2
1 1 2 1 3 1 4 1
So it takes a row vector and puts a row of ones below it
m = [x; ones(1,length(x))] % Append row of 1s below our x row vector.
m = 2×4
1 2 3 4 1 1 1 1
matrix = m' % Transpose it.
matrix = 4×2
1 1 2 1 3 1 4 1
and then transposes it, with the apostrophe operator, to make the 1s be in the right column instead of the bottom row.

请先登录,再进行评论。

类别

Help CenterFile Exchange 中查找有关 Descriptive Statistics 的更多信息

产品


版本

R2020b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by