将参数向量传递给 map 函数

Question

我正在尝试创建一个函数来映射嵌套的 tibble。此函数需要采用每行不同的参数向量。

当我对嵌套数据调用 purrr:map2() 时，purrr 尝试遍历参数向量的所有值和数据集中的所有行。我该怎么做才能将整个向量作为单个参数传递？

library(tidyverse)

myf <- function(x, params) {
  print(params)
  x %>%
    mutate(new_mpg = mpg + rnorm(n(), params[1], params[2])) %>%
    summarise(old = mean(mpg), new = mean(new_mpg)) %>%
    as.list()
}

# Calling function with params defined is great!
myf(mtcars, params = c(5, 10))
#> [1]  5 10
#> $old
#> [1] 20.09062
#> 
#> $new
#> [1] 25.62049

# Cannot work in purr as vector, tries to loop over param
mtcars %>%
  group_by(cyl) %>% # from base R
  nest()   %>%
  mutate(
    newold = map2(data, c(5, 10), myf),
  )
#> [1] 5
#> Warning in rnorm(n(), params[1], params[2]): NAs produced
#> [1] 10
#> Warning in rnorm(n(), params[1], params[2]): NAs produced
#> Error: Problem with `mutate()` column `newold`.
#> ℹ `newold = map2(data, c(5, 10), myf)`.
#> ℹ `newold` must be size 1, not 2.
#> ℹ The error occurred in group 1: cyl = 4.

# New function wrapper with hard-coded params
myf2 <- function(x){
  myf(x, c(5, 10))
}

# works great! but not what I need
mtcars %>%
  group_by(cyl) %>% # from base R
  nest()   %>%
  mutate(
    mean = 5, 
    sd = 10,
    newold = map(data, myf2),
  )
#> [1]  5 10
#> [1]  5 10
#> [1]  5 10
#> # A tibble: 3 × 5
#> # Groups:   cyl [3]
#>     cyl data                mean    sd newold          
#>   <dbl> <list>             <dbl> <dbl> <list>          
#> 1     6 <tibble [7 × 10]>      5    10 <named list [2]>
#> 2     4 <tibble [11 × 10]>     5    10 <named list [2]>
#> 3     8 <tibble [14 × 10]>     5    10 <named list [2]>

^{由 reprex package (v2.0.0)}

于 2021-11-29 创建

Answer 1

跳过 group_by() 步骤，只需使用 nest() - 否则您的数据在嵌套后将保持分组状态，需要取消分组。要使您的函数正常工作，只需将参数作为列表传递即可。

library(tidyverse)

mtcars %>%
  nest(data = -cyl) %>%
  mutate(
    newold = map2_df(data, list(c(5, 10)), myf)
  ) %>%
  unpack(newold)

# A tibble: 3 x 4
    cyl data                 old   new
  <dbl> <list>             <dbl> <dbl>
1     6 <tibble [7 x 10]>   19.7  30.7
2     4 <tibble [11 x 10]>  26.7  31.1
3     8 <tibble [14 x 10]>  15.1  17.0

Answer 2

你不需要map2。我想你需要的是 map.

mtcars %>%
  group_by(cyl) %>% # from base R
  nest()   %>%
  mutate(
    newold = map(data, myf, params = c(5, 10)),
  )
# [1]  5 10
# [1]  5 10
# [1]  5 10
# # A tibble: 3 x 3
# # Groups:   cyl [3]
# cyl data               newold          
# <dbl> <list>             <list>          
#   1     6 <tibble [7 x 10]>  <named list [2]>
#   2     4 <tibble [11 x 10]> <named list [2]>
#   3     8 <tibble [14 x 10]> <named list [2]>

如果你有多套params。您可以取消分组数据框，使用 params 添加列表列，然后使用 map2.

mtcars %>%
  group_by(cyl) %>%
  nest()   %>%
  ungroup() %>%
  # Add different sets of params
  mutate(Params = list(a = c(5, 10), b = c(6, 11), c = c(7, 12))) %>%
  mutate(
    newold = map2(data, Params, myf)
  )
# [1]  5 10
# [1]  6 11
# [1]  7 12
# # A tibble: 3 x 4
# cyl data               Params       newold          
# <dbl> <list>             <named list> <list>          
#   1     6 <tibble [7 x 10]>  <dbl [2]>    <named list [2]>
#   2     4 <tibble [11 x 10]> <dbl [2]>    <named list [2]>
#   3     8 <tibble [14 x 10]> <dbl [2]>    <named list [2]>

将参数向量传递给 map 函数

Pass a vector of arguments to map function

r

purrr