Question

在tidyverse中使用list-columns数据结构来拟合不同数据框行的不同模型公式的最佳方法是什么？

在R for Data Science中，Hadley提供了一个很好的例子，说明如何使用列表 - 列数据结构并轻松地适应许多模型（http://r4ds.had.co.nz/many-models.html#gapminder）。我试图找到一种方法来适应许多模型与略有不同的公式。在下面的示例中，根据他的原始示例改编，适合每个大陆的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>

我知道我可以通过遍历by_continent来做到这一点（因为它估算每个大陆的每个模型都没有效率：

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>

但是有可能在不遵循基础R的循环（并避免我不需要的拟合模型）的情况下这样做吗？

我尝试的是这样的：

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)
}

（不可否认，这是一个人为的例子。我的实际情况是我的实质性观察中缺少关键的独立变量，因此我希望将一个模型与所有变量拟合在完整的观察上，另一个模型只有一个子集的关于其余观察的变量。）

更新

我提出了一个有效的est_model函数（尽管效率可能不高）：

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。每个模型仅适用于相应的continent。我在嵌套数据中添加了一个新列formula，以确保formula和continent处于相同的顺序，以防它们不相同。

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>

Answer 2

我发现purrr::at_depth()在原始问题中执行了我想要对est_model()执行的操作。这是我现在满意的解决方案：

library(gapminder)
library(tidyverse)
library(purrr)
library(broom)

fmlas <- tibble::tribble(
  ~continent, ~formula,
  "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 = .)
)

by_continent <- gapminder %>% 
  nest(-continent) %>%
  left_join(fmlas) %>%
  mutate(model=map2(data, formula, ~at_depth(.x, 0, .y)))

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

为列表 - 列数据框的每一行安装不同的模型

2 个答案: