Question

我使用R中的betamix包来拟合beta回归的混合。如果我使用包中给出的示例代码：

data("ReadingSkills", package = "betareg")
set.seed(4040)
rs_mix <- betamix(accuracy ~ iq, data = ReadingSkills, k = 3,
              nstart = 10, extra_components = extraComponent(type = "uniform",
                                                             coef = 0.99, delta = 0.01))
summary(rs_mix)

运行拟合betamix对象的摘要会得到结果：

> summary(rs_mix)
$Comp.1
$Comp.1$mean
            Estimate Std. Error z value  Pr(>|z|)    
(Intercept)  1.40342    0.26332  5.3296  9.84e-08 ***
iq           0.82502    0.21630  3.8142 0.0001366 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

$Comp.1$precision
            Estimate Std. Error z value  Pr(>|z|)    
(Intercept)  2.68509    0.45435  5.9097 3.427e-09 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1


$Comp.2
$Comp.2$mean
             Estimate Std. Error z value  Pr(>|z|)    
(Intercept)  0.502523   0.082476  6.0930 1.108e-09 ***
iq          -0.048415   0.112923 -0.4287    0.6681    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

$Comp.2$precision
            Estimate Std. Error z value  Pr(>|z|)    
(Intercept)  4.25160    0.74737  5.6888 1.279e-08 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Component 2结果显示变量iq不重要。有没有办法从摘要结果中删除此变量？我尝试过使用summary(rs_mix)$Comp.1，它给了我错误：

Error in summary(rs_mix)$Comp.1 : 
  $ operator not defined for this S4 class

Answer 1

您可以使用@符号访问S4对象。对于S3对象，它的工作方式类似于$。

smr = summary(rs_mix)
comps = smr@components[[1]]
significance_level = 0.05

lapply(comps, function(x) {
  bool = x[['mean']][,4] < significance_level
  x[['mean']][bool, ]
})

$Comp.1
             Estimate Std. Error  z value     Pr(>|z|)
(Intercept) 1.4034184  0.2633230 5.329647 9.840410e-08
iq          0.8250193  0.2163036 3.814172 1.366404e-04

$Comp.2
    Estimate   Std. Error      z value     Pr(>|z|) 
5.025226e-01 8.247566e-02 6.092981e+00 1.108272e-09

从R中的摘要对象中删除无关紧要的变量

1 个答案: