r - 为列表列数据框的每一行拟合不同的模型

Question

使用 tidyverse 中的列表列数据结构拟合因数据框的行而异的不同模型公式的最佳方法是什么？

在 R for Data Science 中，Hadley 提供了一个极好的示例，说明如何使用列表列数据结构并轻松拟合许多模型 ( http://r4ds.had.co.nz/many-models.html#gapminder )。我试图找到一种方法来拟合许多公式略有不同的模型。在下面改编自他的原始示例的示例中，为每个大陆拟合不同模型的最佳方法是什么？

library(gapminder)
library(dplyr)
library(tidyr)
library(purrr)
library(broom)

by_continent <- gapminder %>% 
  group_by(continent) %>% 
  nest()

by_continent <- by_continent %>% 
  mutate(model = map(data, ~lm(lifeExp ~ year, data = .)))

by_continent %>% 
  mutate(glance=map(model, glance)) %>% 
  unnest(glance, .drop=T)

## A tibble: 5 × 12
#  continent r.squared adj.r.squared     sigma statistic      p.value    df
#     <fctr>     <dbl>         <dbl>     <dbl>     <dbl>        <dbl> <int>
#1      Asia 0.4356350     0.4342026 8.9244419  304.1298 6.922751e-51     2
#2    Europe 0.4984659     0.4970649 3.8530964  355.8099 1.344184e-55     2
#3    Africa 0.2987543     0.2976269 7.6685811  264.9929 6.780085e-50     2
#4  Americas 0.4626467     0.4608435 6.8618439  256.5699 4.354220e-42     2
#5   Oceania 0.9540678     0.9519800 0.8317499  456.9671 3.299327e-16     2
## ... with 5 more variables: logLik <dbl>, AIC <dbl>, BIC <dbl>,
##   deviance <dbl>, df.residual <int>

我知道我可以通过迭代 by_continent 来做到这一点(效率不高，因为它估计每个大陆的每个模型:

formulae <- list(
  Asia=~lm(lifeExp ~ year, data = .),
  Europe=~lm(lifeExp ~ year + pop, data = .),
  Africa=~lm(lifeExp ~ year + gdpPercap, data = .),
  Americas=~lm(lifeExp ~ year - 1, data = .),
  Oceania=~lm(lifeExp ~ year + pop + gdpPercap, data = .)
)

for (i in 1:nrow(by_continent)) {
  by_continent$model[[i]] <- map(by_continent$data, formulae[[i]])[[i]]
}

by_continent %>% 
  mutate(glance=map(model, glance)) %>% 
  unnest(glance, .drop=T)

## A tibble: 5 × 12
#  continent r.squared adj.r.squared     sigma  statistic       p.value    df
#     <fctr>     <dbl>         <dbl>     <dbl>      <dbl>         <dbl> <int>
#1      Asia 0.4356350     0.4342026 8.9244419   304.1298  6.922751e-51     2
#2    Europe 0.4984677     0.4956580 3.8584819   177.4093  3.186760e-54     3
#3    Africa 0.4160797     0.4141991 7.0033542   221.2506  2.836552e-73     3
#4  Americas 0.9812082     0.9811453 8.9703814 15612.1901 4.227928e-260     1
#5   Oceania 0.9733268     0.9693258 0.6647653   243.2719  6.662577e-16     4
## ... with 5 more variables: logLik <dbl>, AIC <dbl>, BIC <dbl>,
##   deviance <dbl>, df.residual <int>

但是是否有可能在不返回基础 R 中循环的情况下执行此操作(并避免拟合我不需要的模型)？

我试过的是这样的:

by_continent <- by_continent %>% 
left_join(tibble::enframe(formulae, name="continent", value="formula"))

by_continent %>% 
   mutate(model=map2(data, formula, est_model))

但我似乎无法想出一个有效的 est_model 函数。我尝试了这个不起作用的函数 (h/t: https://gist.github.com/multidis/8138757 ):

  est_model <- function(data, formula, ...) {
  mc <- match.call()
  m <- match(c("formula","data"), names(mc), 0L)
  mf <- mc[c(1L, m)]
  mf[[1L]] <- as.name("model.frame")
  mf <- eval(mf, parent.frame())
  data.st <- data.frame(mf)

  return(data.st)
}

(诚​​然，这是一个人为的例子。我的实际情况是我的数据中有大量的观察缺少关键的自变量，所以我想用完整的观察中的所有变量拟合一个模型，而另一个只拟合一个变量子集的模型休息观察。)

更新

我想出了一个有效的 est_model 函数(虽然可能效率不高):

est_model <- function(data, formula, ...) {
  map(list(data), formula, ...)[[1]]
}

by_continent <- by_continent %>% 
   mutate(model=map2(data, formula, est_model))

by_continent %>% 
  mutate(glance=map(model, glance)) %>% 
  unnest(glance, .drop=T)

## A tibble: 5 × 12
#  continent r.squared adj.r.squared     sigma  statistic       p.value    df
#      <chr>     <dbl>         <dbl>     <dbl>      <dbl>         <dbl> <int>
#1      Asia 0.4356350     0.4342026 8.9244419   304.1298  6.922751e-51     2
#2    Europe 0.4984677     0.4956580 3.8584819   177.4093  3.186760e-54     3
#3    Africa 0.4160797     0.4141991 7.0033542   221.2506  2.836552e-73     3
#4  Americas 0.9812082     0.9811453 8.9703814 15612.1901 4.227928e-260     1
#5   Oceania 0.9733268     0.9693258 0.6647653   243.2719  6.662577e-16     4
## ... with 5 more variables: logLik <dbl>, AIC <dbl>, BIC <dbl>, deviance <dbl>,
##   df.residual <int>

Answer 1

我发现制作模型公式列表更容易。每个模型只适合一次对应的 continent .我添加了一个新列 formula到嵌套数据以确保 formula和 continent如果它们不是，它们的顺序相同。

formulae <- c(
    Asia= lifeExp ~ year,
    Europe= lifeExp ~ year + pop,
    Africa= lifeExp ~ year + gdpPercap,
    Americas= lifeExp ~ year - 1,
    Oceania= lifeExp ~ year + pop + gdpPercap
)

df <- gapminder %>%
    group_by(continent) %>%
    nest() %>%
    mutate(formula = formulae[as.character(continent)]) %>%
    mutate(model = map2(formula, data, ~ lm(.x, .y))) %>%
    mutate(glance=map(model, glance)) %>%
    unnest(glance, .drop=T)

# # A tibble: 5 × 12
#   continent r.squared adj.r.squared     sigma  statistic       p.value    df      logLik        AIC        BIC
#      <fctr>     <dbl>         <dbl>     <dbl>      <dbl>         <dbl> <int>       <dbl>      <dbl>      <dbl>
# 1      Asia 0.4356350     0.4342026 8.9244419   304.1298  6.922751e-51     2 -1427.65947 2861.31893 2873.26317
# 2    Europe 0.4984677     0.4956580 3.8584819   177.4093  3.186760e-54     3  -995.41016 1998.82033 2014.36475
# 3    Africa 0.4160797     0.4141991 7.0033542   221.2506  2.836552e-73     3 -2098.46089 4204.92179 4222.66639
# 4  Americas 0.9812082     0.9811453 8.9703814 15612.1901 4.227928e-260     1 -1083.35918 2170.71836 2178.12593
# 5   Oceania 0.9733268     0.9693258 0.6647653   243.2719  6.662577e-16     4   -22.06696   54.13392   60.02419
# # ... with 2 more variables: deviance <dbl>, df.residual <int>