R与数据分析旧笔记（十）非线性模型

非线性模型

非线性模型

例子：销售额x与流通费率y

> x=c(1.5,2.8,4.5,7.5,10.5,13.5,15.1,16.5,19.5,22.5,24.5,26.5)
> y=c(7.0,5.5,4.6,3.6,2.9,2.7,2.5,2.4,2.2,2.1,1.9,1.8)
> plot(x,y)

1.直线回归

> lm.1=lm(y~x)
> summary(lm.1)

Call:
lm(formula = y ~ x)

Residuals:
    Min      1Q  Median      3Q     Max 
-0.9179 -0.5537 -0.1628  0.3953  1.6519 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  5.60316    0.43474  12.889 1.49e-07 ***
x           -0.17003    0.02719  -6.254 9.46e-05 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.7701 on 10 degrees of freedom
Multiple R-squared:  0.7964,    Adjusted R-squared:  0.776 
F-statistic: 39.11 on 1 and 10 DF,  p-value: 9.456e-05

，不理想

2.对数法，

> lm.log=lm(y~log(x))
> summary(lm.log)

Call:
lm(formula = y ~ log(x))

Residuals:
     Min       1Q   Median       3Q      Max 
-0.33291 -0.10133 -0.04693  0.16512  0.34844 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   7.3639     0.1688   43.64 9.60e-13 ***
log(x)       -1.7568     0.0677  -25.95 1.66e-10 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.2064 on 10 degrees of freedom
Multiple R-squared:  0.9854,    Adjusted R-squared:  0.9839 
F-statistic: 673.5 on 1 and 10 DF,  p-value: 1.66e-10
> plot(x,y);lines(x,fitted(lm.log))

比直线回归拟合效果要好，但还是有不足

3.指数法，

> lm.exp=lm(log(y)~x)
> summary(lm.exp)

Call:
lm(formula = log(y) ~ x)

Residuals:
     Min       1Q   Median       3Q      Max 
-0.18246 -0.10664 -0.01670  0.08079  0.25946 

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept)  1.759664   0.075101   23.43 4.54e-10 ***
x           -0.048809   0.004697  -10.39 1.12e-06 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.133 on 10 degrees of freedom
Multiple R-squared:  0.9153,    Adjusted R-squared:  0.9068 
F-statistic:   108 on 1 and 10 DF,  p-value: 1.116e-06

> plot(x,y);lines(x,exp(fitted(lm.exp)))

效果并无太大变化，反而更糟

4.幂函数法，

> lm.pow=lm(log(y)~log(x))
> summary(lm.pow)

Call:
lm(formula = log(y) ~ log(x))

Residuals:
      Min        1Q    Median        3Q       Max 
-0.054727 -0.020805  0.004548  0.024617  0.045896 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  2.19073    0.02951   74.23 4.81e-15 ***
log(x)      -0.47243    0.01184  -39.90 2.34e-12 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.0361 on 10 degrees of freedom
Multiple R-squared:  0.9938,    Adjusted R-squared:  0.9931 
F-statistic:  1592 on 1 and 10 DF,  p-value: 2.337e-12

> plot(x,y);lines(x,exp(fitted(lm.pow)))

检验结果来看上升了，也通过了假设检验，从图上来看拟合效果非常好