r语言 - 使用"rlang"评估用户参数时出错

Question

我正在尝试编写一个显示 Bartlett 测试结果的函数，但在管道操作链和公式中使用rlang正确评估用户参数时遇到麻烦。我做错了什么，我怎样才能让它工作？

library(tidyverse)
# function body
tryfn <- function(data, x, y) {
# creating a dataframe (works)
data <-
dplyr::select(
.data = data,
!!rlang::enquo(x),
!!rlang::enquo(y)
)
print(head(data))
# convert the grouping variable to factor (doesn't work)
data %<>%
stats::na.omit(.) %>%
dplyr::mutate_at(
.tbl = .,
.vars = !!rlang::enquo(x),
.funs = ~base::droplevels(x = base::as.factor(x = .))
)
print(head(data))
# running the test (doesn't work)
bartlett <- stats::bartlett.test(
formula = !!rlang::enquo(y) ~ !!rlang::enquo(x),
data = data,
na.action = na.omit
)
print(summary(bartlett))
}
# using the function
tryfn(
data = mtcars,
x = am,
y = wt
)
#>                   am    wt
#> Mazda RX4          1 2.620
#> Mazda RX4 Wag      1 2.875
#> Datsun 710         1 2.320
#> Hornet 4 Drive     0 3.215
#> Hornet Sportabout  0 3.440
#> Valiant            0 3.460
#> Error in !rlang::enquo(x): invalid argument type

^{创建于 2018-10-07 由 reprex 软件包 (v0.2.1(}

Answer 1

我们可以将字符串对象传递到.var中。 因此，将商数转换为quo_name并使用它

tryfn <- function(data, x, y) {
# creating a dataframe (works)
x <- enquo(x)
y <- enquo(y)
x1 <- quo_name(x)
y1 <- quo_name(y)
data <-
dplyr::select(
.data = data,
!!x,
!!y
)
print(head(data))
fml <- formula(paste0(y1, " ~ ", x1))

# convert the grouping variable to factor (doesn't work)
data <- data %>%
stats::na.omit(.) %>%
dplyr::mutate_at(
.var = x1,
.funs = ~base::droplevels(x = base::as.factor(x = .x))
)
bartlett <- stats::bartlett.test(
formula = fml,
data = data,
na.action = na.omit
)
bartlett    
}

-测试

# using the function
out <- tryfn(
data = mtcars,
x = am,
y = wt
)
out
#   Bartlett test of homogeneity of variances
#data:  wt by am
#Bartlett's K-squared = 0.71483, df = 1, p-value = 0.3978