为了简单起见,我正在尝试编写一个更复杂的自定义函数,因此我创建了一些玩具示例。
假设我想写一个函数-
"quoted"和unquoted参数因此,我编写了一个函数来运行t-test (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>然后编写一个函数来运行anova (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然后我编写了一个元函数来从上面选择适当的函数-
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但这似乎行不通。对于我在这里做错了什么,以及如何让它发挥作用,有什么想法吗?
发布于 2019-08-02 15:20:48
问题似乎源于将相当于什么的内容(例如,通过元函数中的x = rlang::ensym(am)通过rlang::exec()传递给您的单个函数)。
ensym()函数只接受字符串或符号,这样做会导致错误消息。有鉴于此,将x和y参数转换为字符串应该会有所帮助。
因此,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)
)
}(我从nlevels转到了nlevels,因为am和cyl不是因素,所以我没有得到与原始结果相比较的正确结果。)
现在,使用裸符号和字符串可以工作:
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>https://stackoverflow.com/questions/57328124
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