5

我正在尝试编写一个更复杂的自定义函数，因此为了简单起见，我创建了玩具示例。

假设我想写一个函数——

  1. 自动决定要运行的适当函数：例如，t 检验或方差分析。
  2. 接受"quoted"和unquoted参数

所以我编写了一个函数来运行 t 检验（按预期工作）：

set.seed(123)
library(rlang)
library(tidyverse)

# t-test function
fun_t <- function(data, x, y) {
  # make sure both quoted and unquoted arguments work
  x <- rlang::ensym(x)
  y <- rlang::ensym(y)

  # t-test
  broom::tidy(stats::t.test(
    formula = rlang::new_formula({{ y }}, {{ x }}),
    data = data
  ))
}

# works fine
fun_t(mtcars, am, wt)
#> # A tibble: 1 x 10
#>   estimate estimate1 estimate2 statistic p.value parameter conf.low
#>      <dbl>     <dbl>     <dbl>     <dbl>   <dbl>     <dbl>    <dbl>
#> 1     1.36      3.77      2.41      5.49 6.27e-6      29.2    0.853
#> # ... with 3 more variables: conf.high <dbl>, method <chr>,
#> #   alternative <chr>

fun_t(mtcars, "am", "wt")
#> # A tibble: 1 x 10
#>   estimate estimate1 estimate2 statistic p.value parameter conf.low
#>      <dbl>     <dbl>     <dbl>     <dbl>   <dbl>     <dbl>    <dbl>
#> 1     1.36      3.77      2.41      5.49 6.27e-6      29.2    0.853
#> # ... with 3 more variables: conf.high <dbl>, method <chr>,
#> #   alternative <chr>

然后我编写了一个函数来运行方差分析（按预期工作）：

# anova function
fun_anova <- function(data, x, y) {
  # make sure both quoted and unquoted arguments work
  x <- rlang::ensym(x)
  y <- rlang::ensym(y)

  # t-test
  broom::tidy(stats::aov(
    formula = rlang::new_formula({{ y }}, {{ x }}),
    data = data
  ))
}

# works fine
fun_anova(mtcars, cyl, wt)
#> # A tibble: 2 x 6
#>   term         df sumsq meansq statistic      p.value
#>   <chr>     <dbl> <dbl>  <dbl>     <dbl>        <dbl>
#> 1 cyl           1  18.2 18.2        47.4  0.000000122
#> 2 Residuals    30  11.5  0.384      NA   NA

fun_anova(mtcars, "cyl", "wt")
#> # A tibble: 2 x 6
#>   term         df sumsq meansq statistic      p.value
#>   <chr>     <dbl> <dbl>  <dbl>     <dbl>        <dbl>
#> 1 cyl           1  18.2 18.2        47.4  0.000000122
#> 2 Residuals    30  11.5  0.384      NA   NA

然后我写了一个元函数来从上面选择合适的函数——

fun_meta <- function(data, x, y) {
  # make sure both quoted and unquoted arguments work
  x <- rlang::ensym(x)
  y <- rlang::ensym(y)

  # which test to run?
  if (nlevels(data %>% dplyr::pull({{ x }})) == 2L) {
    .f <- fun_t
  } else {
    .f <- fun_anova
  }

  # executing the appropriate function
  rlang::exec(
    .fn = .f,
    data = data,
    x = x,
    y = y
  )
}

# using the meta-function
fun_meta(mtcars, am, wt)
#> Only strings can be converted to symbols

fun_meta(mtcars, "am", "wt")
#> Only strings can be converted to symbols

但这似乎不起作用。关于我在这里做错了什么以及如何让它发挥作用的任何想法？

4

1 回答 1

4

问题似乎源于通过在元函数中传递的内容，例如，传递x = rlang::ensym(am)给您的个人函数。rlang::exec()

该ensym()函数只接受字符串或符号，因此这样做会导致错误消息。鉴于此，将您的x和y参数转换为字符串应该会有所帮助。

所以元函数可以是：

fun_meta <- function(data, x, y) {
     # make sure both quoted and unquoted arguments work
     x <- rlang::ensym(x)
     y <- rlang::ensym(y)

     # which test to run?
     if (dplyr::n_distinct(data %>% dplyr::pull({{ x }})) == 2L) {
          .f <- fun_t
     } else {
          .f <- fun_anova
     }

     # executing the appropriate function
     rlang::exec(
          .fn = .f,
          data = data,
          x = rlang::as_string(x),
          y = rlang::as_string(y)
     )
}

（我切换到n_distinct()fromnlevels因为am并且cyl不是因素，所以我没有得到正确的结果来与你的原始结果进行比较。）

现在使用裸符号和字符串工作：

fun_meta(mtcars, am, wt)
    # A tibble: 1 x 10
  estimate estimate1 estimate2 statistic p.value parameter conf.low conf.high
     <dbl>     <dbl>     <dbl>     <dbl>   <dbl>     <dbl>    <dbl>     <dbl>
1     1.36      3.77      2.41      5.49 6.27e-6      29.2    0.853      1.86
# ... with 2 more variables: method <chr>, alternative <chr>
> fun_meta(mtcars, "am", "wt")

fun_meta(mtcars, "am", "wt")
# A tibble: 1 x 10
  estimate estimate1 estimate2 statistic p.value parameter conf.low conf.high
     <dbl>     <dbl>     <dbl>     <dbl>   <dbl>     <dbl>    <dbl>     <dbl>
1     1.36      3.77      2.41      5.49 6.27e-6      29.2    0.853      1.86
# ... with 2 more variables: method <chr>, alternative <chr>
于 2019-08-02T15:20:48.247 回答