使用谓词函数整齐地替换缺少的值

推荐的整洁的更换方法是什么 NA 与使用 谓词函数?

我希望利用 tidyr::replace_na() (或类似的预定义缺少值处理程序)但我似乎无法使它与 purrr 或 dplyr 使用谓词函数的方法。

library(magrittr)

# Example data:
df <- tibble::tibble(
  id = c(rep("A", 3), rep("B", 3)),
  x = c(1, 2, NA, 10, NA, 30),
  y = c("a", NA, "c", NA, NA, "f")
)

# Works, but needs manual spec of columns that should be handled:
df %>% 
  tidyr::replace_na(list(x = 0))  
#> # A tibble: 6 x 3
#>   id        x y    
#>   <chr> <dbl> <chr>
#> 1 A         1 a    
#> 2 A         2 <NA> 
#> 3 A         0 c    
#> 4 B        10 <NA> 
#> 5 B         0 <NA> 
#> 6 B        30 f

# Doesn't work (at least not in the intended way):
df %>% 
  dplyr::mutate_if(
    function(.x) inherits(.x, c("integer", "numeric")),
    ~tidyr::replace_na(0)  
  )
#> # A tibble: 6 x 3
#>   id        x y    
#>   <chr> <dbl> <chr>
#> 1 A         0 a    
#> 2 A         0 <NA> 
#> 3 A         0 c    
#> 4 B         0 <NA> 
#> 5 B         0 <NA> 
#> 6 B         0 f

# Works, but uses an inline def of the replacement function:
df %>% 
  dplyr::mutate_if(
    function(.x) inherits(.x, c("integer", "numeric")),
    function(.x) dplyr::if_else(is.na(.x), 0, .x)
  )
#> # A tibble: 6 x 3
#>   id        x y    
#>   <chr> <dbl> <chr>
#> 1 A         1 a    
#> 2 A         2 <NA> 
#> 3 A         0 c    
#> 4 B        10 <NA> 
#> 5 B         0 <NA> 
#> 6 B        30 f

# Works, but uses an inline def of the replacement function:
df %>% 
  purrr::modify_if(
    function(.x) inherits(.x, c("integer", "numeric")),
    function(.x) dplyr::if_else(is.na(.x), 0, .x)
  )
#> # A tibble: 6 x 3
#>   id        x y    
#>   <chr> <dbl> <chr>
#> 1 A         1 a    
#> 2 A         2 <NA> 
#> 3 A         0 c    
#> 4 B        10 <NA> 
#> 5 B         0 <NA> 
#> 6 B        30 f

由Reprex软件包(v0.2.1)于2019-01-21创建