借助 Cet - 您可能正在寻找这样的东西:
df <- data.frame(x = c(1:5), y = c(6:10), z = LETTERS[1:5])
my_fxn <- function (aaa, bbb, ccc, data) {
if (!missing(data)) {
aaa = as.numeric(data[[aaa]])
bbb = as.numeric(data[[bbb]])
ccc = as.character(data[[ccc]])
}
print(aaa[1])
}
my_fxn("x", "y", "z", df)
#> [1] 1
随着使用enquo()
from library(dplyr)
,我们不再需要输入字符作为函数变量:
library(dplyr)
my_fxn <- function (aaa, bbb, ccc, data) {
aaa <- enquo(aaa)
bbb <- enquo(bbb)
ccc <- enquo(ccc)
if (!missing(data)) {
aaa = as.numeric(pull(data, !!aaa))
bbb = as.numeric(pull(data, !!bbb))
ccc = as.character(pull(data, !!ccc))
}
print(aaa[1])
}
my_fxn(x, y, z, df)
#> [1] 1
有关功能构建的更多信息enquo()
and !!
可以在这里找到:https://dplyr.tidyverse.org/articles/programming.html#programming-recipes
最后,使用基本 R 解决方案deparse()
and substitute()
:
my_fxn <- function (aaa, bbb, ccc, data) {
aaa <- deparse(substitute(aaa))
bbb <- deparse(substitute(bbb))
ccc <- deparse(substitute(ccc))
if (!missing(data)) {
aaa = as.numeric(data[[aaa]])
bbb = as.numeric(data[[bbb]])
ccc = as.character(data[[ccc]])
}
print(aaa[1])
}
my_fxn(x, y, z, df)
#> [1] 1