我想使用...
表示我想从data.table
对象的自定义函数返回的变量。这是一个最小的可复制示例:
library(data.table)
d = data.table(mtcars)
getvar = function(...){
return(d[,.(xyz = mean(hp), ...), cyl])
}
getvar(mpg, cyl, disp)
[.data.table
(d,,。(N = .N,...),cyl)中的错误:找不到对象'cyl'
我希望得到的是:
d[,.(xyz = mean(hp), mpg, cyl, disp), cyl]
# cyl xyz mpg cyl disp
# 1: 6 122.28571 21.0 6 160.0
# 2: 6 122.28571 21.0 6 160.0
# 3: 6 122.28571 21.4 6 258.0
# 4: 6 122.28571 18.1 6 225.0
# 5: 6 122.28571 19.2 6 167.6
任何人都可以分享他们的解决方案吗?
[一种可能的解决方案是在函数中使用mget
,但返回一个列表,然后将xyz
与带有c
的列表合并。要进行此工作,需要将要添加的列指定为字符向量:
getvar = function(...){
return(d[, c(xyz = mean(hp), mget(...)), cyl])
}
getvar(c("mpg", "cyl", "disp"))
给出:
> getvar(c("mpg", "cyl", "disp")) cyl xyz mpg cyl disp 1: 6 122.28571 21.0 6 160.0 2: 6 122.28571 21.0 6 160.0 3: 6 122.28571 21.4 6 258.0 4: 6 122.28571 18.1 6 225.0 5: 6 122.28571 19.2 6 167.6 6: 6 122.28571 17.8 6 167.6 7: 6 122.28571 19.7 6 145.0 8: 4 82.63636 22.8 4 108.0 9: 4 82.63636 24.4 4 146.7 10: 4 82.63636 22.8 4 140.8 ....
要使它在不引用列名的情况下起作用,您必须使用一些非标准的评估策略:
getvar = function(...){
vars <- substitute(list(xyz = mean(hp), ...))
return(d[, eval(vars), cyl])
}
getvar(mpg, cyl, disp)
cyl xyz mpg cyl disp
1: 6 122.28571 21.0 6 160.0
2: 6 122.28571 21.0 6 160.0
3: 6 122.28571 21.4 6 258.0
4: 6 122.28571 18.1 6 225.0
5: 6 122.28571 19.2 6 167.6
...etc...
基于@Konrad Rudolph here的答案,我们可以编写函数
getvar = function(...){
dots = match.call(expand.dots = FALSE)$...
cols = sapply(dots, deparse)
d[, c(xyz = mean(hp), mget(cols)), cyl]
#thanks to @Jaap for simplified version
}
getvar(mpg, cyl, disp)
# cyl xyz mpg cyl disp
# 1: 6 122.29 21.0 6 160.0
# 2: 6 122.29 21.0 6 160.0
# 3: 6 122.29 21.4 6 258.0
# 4: 6 122.29 18.1 6 225.0
# 5: 6 122.29 19.2 6 167.6
# 6: 6 122.29 17.8 6 167.6
#....