2

有没有办法在 broom 包tidy函数的结果中添加一列，该函数可以将术语列与参数中使用的原始名称formula及其在参数中的列相关联data。

例如，如果我运行以下命令，我会得到：

library(ggplot2)
library(dplyr)

mod <- glm(mpg ~ wt + qsec + as.factor(carb), data = mtcars)

tidy(mod)

#               term     estimate std.error   statistic      p.value
# 1      (Intercept) 21.132995090 7.5756463  2.78959633 1.017187e-02
# 2               wt -4.916303175 0.6747590 -7.28601380 1.584408e-07
# 3             qsec  0.843355538 0.3930252  2.14580532 4.221188e-02
# 4 as.factor(carb)2  0.004133826 1.5321134  0.00269812 9.978695e-01
# 5 as.factor(carb)3 -0.755346006 2.3451222 -0.32209239 7.501715e-01
# 6 as.factor(carb)4 -0.489721798 2.0628564 -0.23739985 8.143615e-01
# 7 as.factor(carb)6 -0.886846134 3.4443957 -0.25747510 7.990068e-01
# 8 as.factor(carb)8 -0.894783610 3.7496630 -0.23863041 8.134180e-01

我正在寻找的是这样的：

#               term     estimate std.error   statistic      p.value   term_base
# 1      (Intercept) 21.132995090 7.5756463  2.78959633 1.017187e-02 
# 2               wt -4.916303175 0.6747590 -7.28601380 1.584408e-07          wt
# 3             qsec  0.843355538 0.3930252  2.14580532 4.221188e-02        qsec
# 4 as.factor(carb)2  0.004133826 1.5321134  0.00269812 9.978695e-01        carb
# 5 as.factor(carb)3 -0.755346006 2.3451222 -0.32209239 7.501715e-01        carb
# 6 as.factor(carb)4 -0.489721798 2.0628564 -0.23739985 8.143615e-01        carb
# 7 as.factor(carb)6 -0.886846134 3.4443957 -0.25747510 7.990068e-01        carb
# 8 as.factor(carb)8 -0.894783610 3.7496630 -0.23863041 8.134180e-01        carb

如果这个新列的第一行是空的，Intercept或者1. 只需要可以将术语列与传递给公式的原始变量名称相匹配的东西吗？

编辑

如果它不依赖于as.factor在公式中使用会很好，例如：

mod <- glm(mpg ~ wt + qsec + carb, data = mtcars %>% mutate(carb = factor(carb)))

tidy(mod)

#          term     estimate std.error   statistic      p.value
# 1 (Intercept) 21.132995090 7.5756463  2.78959633 1.017187e-02
# 2          wt -4.916303175 0.6747590 -7.28601380 1.584408e-07
# 3        qsec  0.843355538 0.3930252  2.14580532 4.221188e-02
# 4       carb2  0.004133826 1.5321134  0.00269812 9.978695e-01
# 5       carb3 -0.755346006 2.3451222 -0.32209239 7.501715e-01
# 6       carb4 -0.489721798 2.0628564 -0.23739985 8.143615e-01
# 7       carb6 -0.886846134 3.4443957 -0.25747510 7.990068e-01
# 8       carb8 -0.894783610 3.7496630 -0.23863041 8.134180e-01
4

1 回答 1

2

我们可以使用正则表达式来创建“term_base”列

tidy(mod) %>%
        mutate(term_base = sub("Intercept", "", gsub(".*\\(|\\).*", "", term)))
#              term     estimate std.error   statistic      p.value term_base
#1      (Intercept) 21.132995090 7.5756463  2.78959633 1.017187e-02          
#2               wt -4.916303175 0.6747590 -7.28601380 1.584408e-07        wt
#3             qsec  0.843355538 0.3930252  2.14580532 4.221188e-02      qsec
#4 as.factor(carb)2  0.004133826 1.5321134  0.00269812 9.978695e-01      carb
#5 as.factor(carb)3 -0.755346006 2.3451222 -0.32209239 7.501715e-01      carb
#6 as.factor(carb)4 -0.489721798 2.0628564 -0.23739985 8.143615e-01      carb
#7 as.factor(carb)6 -0.886846134 3.4443957 -0.25747510 7.990068e-01      carb
#8 as.factor(carb)8 -0.894783610 3.7496630 -0.23863041 8.134180e-01      carb

如果我们在步骤之前将“碳水化合物”添加到“术语”中，as.factor也可以将其从“术语”中删除mutatefactorglm

mtcars %>%
     mutate(carb = factor(carb)) %>% 
     glm(formula = mpg ~wt + qsec + carb, data = .) %>% 
     tidy(.) %>%
     mutate(term_base = sub("\\(.*\\)|\\d+", "", term))
#     term     estimate std.error   statistic      p.value term_base
#1 (Intercept) 21.132995090 7.5756463  2.78959633 1.017187e-02          
#2          wt -4.916303175 0.6747590 -7.28601380 1.584408e-07        wt
#3        qsec  0.843355538 0.3930252  2.14580532 4.221188e-02      qsec
#4       carb2  0.004133826 1.5321134  0.00269812 9.978695e-01      carb
#5       carb3 -0.755346006 2.3451222 -0.32209239 7.501715e-01      carb
#6       carb4 -0.489721798 2.0628564 -0.23739985 8.143615e-01      carb
#7       carb6 -0.886846134 3.4443957 -0.25747510 7.990068e-01      carb
#8       carb8 -0.894783610 3.7496630 -0.23863041 8.134180e-01      carb
于 2016-08-29T02:22:08.947 回答