我有问题以保留非标准评估的方式重构dplyr.让我们说我想创建一个总是选择和重命名的函数.
library(lazyeval)
library(dplyr)
df <- data.frame(a = c(1,2,3), f = c(4,5,6), lm = c(7, 8 , 9))
select_happy<- function(df, col){
col <- lazy(col)
fo <- interp(~x, x=col)
select_(df, happy=fo)
}
f <- function(){
print('foo')
}
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select_happy()当库函数使用非标准求值时,根据此帖子Refactor R代码的答案编写.select_happy()适用于未定义或在全局环境中定义的列名.但是,当列名也是另一个名称空间中的函数名称时,它会遇到问题.
select_happy(df, a)
# happy
# 1 1
# 2 2
# 3 3
select_happy(df, f)
# happy
# 1 4
# 2 5
# 3 6
select_happy(df, lm)
# Error in eval(expr, envir, enclos) (from #4) : …Run Code Online (Sandbox Code Playgroud)