Least Squares for Solving Linear Regression Equations

Preface

Write this little thing because he was driven mad by his math homework (Big Brother Assassination List)
The principle is least squares.

\[\hat=\frac{\sum\limits^n_x_iy_i-n\overline\overline}{\sum\limits^n_x_i^2-n\overline^2} \]

\[\hat=\overline-\hat\overline \]

Usage method

The number of input data groups in the first row.
The second line enters the value of \(x\) separated by spaces.
The third line enters the corresponding value of \(y\), separated by spaces.
The number of data groups cannot exceed 10000.
Note: Small errors in actual calculations may result from different digits \(\hat\) reserved for \hat\).Six decimal places are reserved by default.
UPD: Increase the sum of squares of residuals \(sum\limits_The calculation of ^n\hat^2\) and related index \(R^2\).

\[\sum\limits_^n\hat^2=\sum\limits_^n(y_i-\hat_i)^2 \]

\[R^2=1-\frac{\sum\limits_^n(y_i-\hat_i)^2}{\sum\limits^n_(y_i-\overline)^2} \]

UPD: Add system("pause"); for easy viewing of results.

sample input

5
15.0 25.8 30.0 36.6 44.4
39.4 42.9 42.9 43.1 49.2

sample output

b=0.291046
a=34.663848
e^2=8.426560
R^2=0.832073

Code

#include <bits/stdc++.h> using namespace std; const int maxn=1e4+10; int n; double x[maxn],y[maxn]; double a,b,R2; double cal(double k){ return b*k+a; } int main(){ scanf("%d",&n); double avex=0,avey=0; for(int i=1;i<=n;i++){ scanf("%lf",&x[i]); avex+=x[i]; } for(int i=1;i<=n;i++){ scanf("%lf",&y[i]); avey+=y[i]; } avex/=n;avey/=n; double sum1=0,sum2=0; for(int i=1;i<=n;i++){ sum1+=x[i]*y[i]; sum2+=x[i]*x[i]; } b=(sum1-n*avex*avey)/(sum2-n*avex*avex); a=avey-b*avex; sum1=0,sum2=0; for(int i=1;i<=n;i++){ sum1+=(y[i]-cal(x[i]))*(y[i]-cal(x[i])); sum2+=(y[i]-avey)*(y[i]-avey); } R2=1-sum1/sum2; printf("b=%lf\na=%lf\ne^2=%lf\nR^2=%lf\n",b,a,sum1,R2); system("pause"); return 0; }

Linear regression equation transformed from function model

UPD: Added calculation of linear regression equation transformed from quadratic and exponential function models:

\[\hat=c_1x^2+c_2 \]

\[\hat=c_3e^ \]

Quadratic function model

#include <bits/stdc++.h> using namespace std; const int maxn=1e4+10; int n; double x[maxn],y[maxn]; double a,b,R2; double cal(double k){ return b*k+a; } int main(){ scanf("%d",&n); double avex=0,avey=0; for(int i=1;i<=n;i++){ scanf("%lf",&x[i]); x[i]*=x[i];//Actually, there is just one more sentence... avex+=x[i]; } for(int i=1;i<=n;i++){ scanf("%lf",&y[i]); avey+=y[i]; } avex/=n;avey/=n; double sum1=0,sum2=0; for(int i=1;i<=n;i++){ sum1+=x[i]*y[i]; sum2+=x[i]*x[i]; } b=(sum1-n*avex*avey)/(sum2-n*avex*avex); a=avey-b*avex; sum1=0,sum2=0; for(int i=1;i<=n;i++){ sum1+=(y[i]-cal(x[i]))*(y[i]-cal(x[i])); sum2+=(y[i]-avey)*(y[i]-avey); } R2=1-sum1/sum2; printf("b=%lf\na=%lf\ne^2=%lf\nR^2=%lf\n",b,a,sum1,R2); system("pause"); return 0; }

Exponential function model

For \(z=ln\y\), the output of this code corresponds to \(z=\hatx+\hat\) and \(hat\) respectively.The corresponding regression equation is (\hat=e^{\hatx+hat}\)
Note: Different values of this code \(e\) may also result in minor errors, which is 2.718281.Daily calculations usually take 2.7, which may lead to errors in \([-0.01,0.01]\).

#include <bits/stdc++.h> using namespace std; const int maxn=1e4+10; int n; double x[maxn],y[maxn],z[maxn]; double a,b,R2; double cal(double k){ return pow(2.718281,b*k+a); } int main(){ scanf("%d",&n); double avex=0,avez=0; for(int i=1;i<=n;i++){ scanf("%lf",&x[i]); avex+=x[i]; } for(int i=1;i<=n;i++){ scanf("%lf",&y[i]); z[i]=log(y[i]); avez+=z[i]; } avex/=n;avez/=n; double sum1=0,sum2=0; for(int i=1;i<=n;i++){ sum1+=x[i]*z[i]; sum2+=x[i]*x[i]; } b=(sum1-n*avex*avez)/(sum2-n*avex*avex); a=avez-b*avex; sum1=0,sum2=0; for(int i=1;i<=n;i++){ sum1+=(y[i]-cal(x[i]))*(y[i]-cal(x[i])); sum2+=(y[i]-avez)*(y[i]-avez); } R2=1-sum1/sum2; printf("b=%lf\na=%lf\ne^2=%lf\nR^2=%lf",b,a,sum1,R2); system("pause"); return 0; }

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