R嵌套模型：创建模型公式列

Question

如何从模型的嵌套数据框创建一列公式（例如 y ~ x 或 y ~ log(x) 或...）？

在下面的尝试中，模型列包含具有最大 R 平方值的模型。创建一列模型公式的目的是确定每一行中使用了哪个模型。

library(tidyverse)
library(broom)

df <- gapminder::gapminder %>% 
  select(country, x = year, y = lifeExp) %>%
  group_by(country) %>%
  nest()

rsq_f <- function(model){summary(model)$r.squared}

best_model <- function(df){
  models <- list(
    lm(formula = y ~ x, data = df),
    lm(formula = y ~ log(x), data = df),
    lm(formula = log(y) ~ x, data = df),
    lm(formula = log(y) ~ log(x), data = df)
  )

  R_squared <- map_dbl(models, rsq_f)
  best_model_num <- which.max(R_squared)

  models[best_model_num][[1]]    
}

models <- df %>%
  mutate(
    model = map(data, best_model),
    rsq = map(model, broom::glance) %>% map_dbl("r.squared"),
    fun_call = map(model, formula)
  )

输出是

> models
# A tibble: 142 x 5
   country     data              model      rsq fun_call     
   <fct>       <list>            <list>   <dbl> <list>       
 1 Afghanistan <tibble [12 x 2]> <S3: lm> 0.949 <S3: formula>
 2 Albania     <tibble [12 x 2]> <S3: lm> 0.912 <S3: formula>
 3 Algeria     <tibble [12 x 2]> <S3: lm> 0.986 <S3: formula>
 4 Angola      <tibble [12 x 2]> <S3: lm> 0.890 <S3: formula>
 5 Argentina   <tibble [12 x 2]> <S3: lm> 0.996 <S3: formula>
 6 Australia   <tibble [12 x 2]> <S3: lm> 0.983 <S3: formula>
 7 Austria     <tibble [12 x 2]> <S3: lm> 0.994 <S3: formula>
 8 Bahrain     <tibble [12 x 2]> <S3: lm> 0.968 <S3: formula>
 9 Bangladesh  <tibble [12 x 2]> <S3: lm> 0.997 <S3: formula>
10 Belgium     <tibble [12 x 2]> <S3: lm> 0.995 <S3: formula>
# ... with 132 more rows

而不是 <S3: formula> 我想实际查看模型使用的公式。

Answer 1

为了让自己更清楚，我将 post 作为示例的答案，如果我理解正确的话，您会寻求包含公式的列，例如字符串 "y ~ x".

假设我们有一个简单的 lm:

x <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
y <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
my_lm <- lm(y~ x)

通过查看术语，您得到了公式，只是排列不正确：

as.character(my_lm[["terms"]])
# [1] "~" "y" "x"

您只需重新排列前两项：

paste(as.character(my_lm$terms)[2],as.character(my_lm$terms)[1], as.character(my_lm$terms)[-c(1:2)])
# [1] "y ~ x"

这可以用 mutate 分配给列。

Answer 2

根据 RLave 的评论，答案只是添加 as.character():

models <- df %>%
  mutate(
    model = map(data, best_model),
    rsq = map(model, broom::glance) %>% map_dbl("r.squared"),
    fun_call = map(model, formula) %>% as.character()
  )

给出：

# A tibble: 142 x 5
   country     data              model      rsq fun_call  
   <fct>       <list>            <list>   <dbl> <chr>     
 1 Afghanistan <tibble [12 x 2]> <S3: lm> 0.949 y ~ log(x)
 2 Albania     <tibble [12 x 2]> <S3: lm> 0.912 y ~ log(x)
 3 Algeria     <tibble [12 x 2]> <S3: lm> 0.986 y ~ log(x)
 4 Angola      <tibble [12 x 2]> <S3: lm> 0.890 y ~ log(x)
 5 Argentina   <tibble [12 x 2]> <S3: lm> 0.996 y ~ x     
 6 Australia   <tibble [12 x 2]> <S3: lm> 0.983 log(y) ~ x
 7 Austria     <tibble [12 x 2]> <S3: lm> 0.994 log(y) ~ x
 8 Bahrain     <tibble [12 x 2]> <S3: lm> 0.968 y ~ log(x)
 9 Bangladesh  <tibble [12 x 2]> <S3: lm> 0.997 log(y) ~ x
10 Belgium     <tibble [12 x 2]> <S3: lm> 0.995 log(y) ~ x
# ... with 132 more rows

R嵌套模型：创建模型公式列

R nested models: create column of model formulas

r

lm

purrr

broom