Question

重新排列列表：我正在进行类似于以下代码的交叉验证，其中cv_chunk的数量是任意的：

library(purrr)
# randomly assign to a cross validation chunk
set.seed(11)
mtcars$cv_chunk <- sample(rep(1:3), nrow(mtcars), 1)

model_confint <- mtcars %>% 
  split(.$cv_chunk) %>% 
  map(~lm(mpg ~ cyl*qsec + gear - cv_chunk, data = .)) %>% 
  map(confint, levels = 0.95) %>%
  map(t)

  names(model_confint) <- paste0("CV_", names(model_confint))

# first element of the list
$CV_1
       (Intercept)        cyl      qsec      gear   cyl:qsec
2.5 %     -54.8983 -25.691233 -8.958490 -5.175215 -0.7161008
97.5 %    215.2629   9.901694  4.322784  5.185608  1.2804372

为了进一步处理每个这样的模型的置信水平，我需要重新排列model_confint，以便获得模型中每个系数的data.frame / list。例如。对于(Intercept)（依据cyl，qsec，...）：

$`(Intercept)`
       2.5 %    97.5 %
CV_1  -54.8983 215.26290
CV_2 -193.2070  84.48072
CV_3 -361.1489 545.04010

我确信使用apply函数或purrr包有一个很好的方法。但我被卡住了。

感谢您的帮助。

Answer 1

谢谢你的好榜样！我有两个直觉可以使这个问题更容易：您可以使用purrr::set_names在循环中更早地命名列表，并且可以通过先前映射到数据帧来避免转置。

library(purrr)

# randomly assign to a cross validation chunk
set.seed(11)
mtcars$cv_chunk <- sample(rep(1:3), nrow(mtcars), 1)

mtcars %>% 
  split(.$cv_chunk) %>% 
# Name the list here to use it as an argument later
  set_names(paste0("CV_", names(.))) %>% 
# Map the actual list and the names of the list to create one dataframe
  map(~lm(mpg ~ cyl*qsec + gear - cv_chunk, data = .)) %>% 
  map2_dfr(.x = ., .y = names(.), function(x, y) {
    df <- as.data.frame(confint(x, levels = .95))
    df$coeffs <- rownames(df)
    df$cv_chunk <- y
    df
    }
  ) %>%
 # Split the dataframe on the "coeffs" column
  split(.$coeffs) %>%
 # Remove the "coeffs" column from each dataframe
  map(function(x) x[colnames(x) != 'coeffs'])
#> $`(Intercept)`
#>        2.5 %    97.5 % cv_chunk
#> 1   -54.8983 215.26286     CV_1
#> 6  -193.2069  84.48072     CV_2
#> 11 -361.1489 545.04010     CV_3
#> 
#> $cyl
#>        2.5 %    97.5 % cv_chunk
#> 2  -25.69123  9.901694     CV_1
#> 7  -11.44012 24.073391     CV_2
#> 12 -26.36288 16.589356     CV_3
#> 
#> $`cyl:qsec`
#>         2.5 %    97.5 % cv_chunk
#> 5  -0.7161008 1.2804372     CV_1
#> 10 -1.4870954 0.5295074     CV_2
#> 15         NA        NA     CV_3
#> 
#> $gear
#>         2.5 %    97.5 % cv_chunk
#> 4   -5.175215  5.185608     CV_1
#> 9   -3.975621  8.092960     CV_2
#> 14 -30.785104 33.533967     CV_3
#> 
#> $qsec
#>         2.5 %    97.5 % cv_chunk
#> 3   -8.958490  4.322784     CV_1
#> 8   -2.257827 11.261528     CV_2
#> 13 -17.741244 12.929339     CV_3

Answer 2

可以使用broom：

来完成

library(purrr)
library(dplyr)
library(tidyr)
# randomly assign to a cross validation chunk
set.seed(11)
mtcars$cv_chunk <- sample(seq(3), nrow(mtcars), replace = TRUE)

mtcars %>% 
  split(.$cv_chunk) %>% 
  map(~lm(mpg ~ cyl*qsec + gear - cv_chunk, data = .)) %>% 
  # the following will work uder different seed. I will report a bug to `broom``
  #map_dfr(~broom::tidy(.x, conf.int=TRUE), .id="cv_chunk")
  map_dfr(~bind_cols(broom::tidy(.x), drop_na(broom::confint_tidy(.x))), .id="cv_chunk") %>% 
  select(cv_chunk, term, conf.low, conf.high) %>% 
  split(.$term)

正确的tidyverse做事方式将会使用 group_by(term) %>% nest() %>% pull(data)，但base函数split正在提供您想要的内容

$`(Intercept)`
   cv_chunk        term  conf.low conf.high
1         1 (Intercept)  -54.8983 215.26286
6         2 (Intercept) -193.2069  84.48072
11        3 (Intercept) -361.1489 545.04010

$cyl
   cv_chunk term  conf.low conf.high
2         1  cyl -25.69123  9.901694
7         2  cyl -11.44012 24.073391
12        3  cyl -26.36288 16.589356

$`cyl:qsec`
   cv_chunk     term   conf.low conf.high
5         1 cyl:qsec -0.7161008 1.2804372
10        2 cyl:qsec -1.4870954 0.5295074

$gear
   cv_chunk term   conf.low conf.high
4         1 gear  -5.175215  5.185608
9         2 gear  -3.975621  8.092960
14        3 gear -30.785104 33.533967

$qsec
   cv_chunk term   conf.low conf.high
3         1 qsec  -8.958490  4.322784
8         2 qsec  -2.257827 11.261528
13        3 qsec -17.741244 12.929339

R：重新排列清单，purrr

2 个答案: