利用R语言进行线性/非线性回归拟合实例（1）

利用R语言进行线性/非线性回归拟合实例（1）
1、 生成一组数据

vector<float>xxvec;

vector<float>yyvec;

ofstreamfout("data2.txt");

for (int i =1;i<200;i++)

{

float x =i*0.8;

float randdnum= rand()%10 * 10;

floatrandomflag = (rand()%10)%2==0?(1):(-1);

float y = 3 *x*x + 2*x + 5 + randomflag*randdnum;

fout<<x<<" "<<y<<endl;

xxvec.push_back(x);

yyvec.push_back(y);

}

fout.close();

将生成的数据存为txt文件，命名为“data1”

2、线性拟合

#-------------------------------------------------------------#载入数据
> fire <- read.table('D:/data.txt',header = TRUE)
#-------------------------------------------------------------#回归分析
> plot(fire$y ~ fire$x)



> fire.reg <- lm(fire$y ~ fire$x,data = fire) #回归拟合
> data1.reg
Call:
lm(formula = data1$y ~ data1$x, data = data1)
Coefficients:
(Intercept) data1$x
6.202 3.003

>summary(data1.reg) #回归分析表
Call:
lm(formula = data1$y ~ data1$x, data = data1)
Residuals:
Min 1Q Median 3Q Max
-93.345 -42.929 -1.948 46.717 88.793

Coefficients:
Estimate Std. Error tvalue Pr(>|t|)
(Intercept) 6.202084 3.352055 1.85 0.0646 .
data1$x 3.002826 0.007259 413.66 <2e-16 ***
标红数字即为线性回归系数，由于生成数据时加了一个随机数，所以拟合出来的直线为：
y=3.002826 x+6.202084
---
Signif. codes: 0 ‘***’ 0.001 ‘**’0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 52.93 on 997 degrees of freedom
Multiple R-squared: 0.9942, Adjusted R-squared: 0.9942
F-statistic: 1.711e+05 on 1 and 997 DF, p-value: < 2.2e-16

>anova(data1.reg) #方差分析表

Analysis of Variance Table

Response: data1$y

Df Sum Sq Mean Sq F value Pr(>F)

data1$x 1479462873 479462873 171112 < 2.2e-16***

Residuals 997 2793641 2802

Signif. codes: 0‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

>abline(data1.reg, col = 2, lty = 2) #拟合直线