当引用变量名时,NSE lazyeval :: lazy与substitute

2023年6月22日 172次阅读

我仍然试图围绕非标准评估以及如何在dplyr中使用它.当函数参数是变量名时,我无法理解为什么惰性求值很重要,因此原始上下文的环境看起来并不重要.

在下面的代码中,函数select3()使用惰性求值,但失败(我相信),因为它试图遵循变量名称顺序一直到base :: order.

可以在我的select4()中使用替换,或者是否有其他方法我应该实现此功能？什么时候保存原始环境实际上很重要,当我真的希望这些参数引用变量时？

谢谢！

library(dplyr)
library(lazyeval)
# Same as dplyr::select
select2 <- function(.data, ...) {
  select_(.data, .dots = lazy_dots(...))
}

# I want to have two capture groups of variables, so I need named arguments.
select3 <- function(.data, group1, group2) {
  out1 <- select_(.data, .dots = lazy(group1))
  out2 <- select_(.data, .dots = lazy(group2))

  list(out1, out2)
}


df <- data.frame(x = 1:2, y = 3:4, order = 5:6)

# select3 seems okay at first...
df %>% select2(x, y)
df %>% select3(x, y)


# But fails when the variable is a function defined in the namespace
df %>% select2(x, order)
df %>% select3(x, order)
# Error in eval(expr, envir, enclos) : object 'datafile' not found


# Using substitute instead of lazy works. But I'm not sure I understand the 
# implications of doing this.
select4 <- function(.data, group1, group2) {
  out1 <- select_(.data, .dots = substitute(group1))
  out2 <- select_(.data, .dots = substitute(group2))

  list(out1, out2)
}

df %>% select4(x, order)

PS相关说明,这是一个错误或预期的行为？

select(df, z)
# Error in eval(expr, envir, enclos) : object 'z' not found

# But if I define z as a numeric variable it works.
z <- 1
select(df, z)

更新

A. Webb在下面指出环境对于选择很重要,因为像one_of这样的特殊函数可以使用它的对象.

更新2

我曾经有一个丑陋的黑客作为修复,但这是一个更好的方式;我应该知道,甚至懒惰都有一个标准的评估对应部分lazy_

select6 <- function(.data, group1, group2) {
  g1 <- lazy_(substitute(group1), env = parent.frame())
  g2 <- lazy_(substitute(group2), env = parent.frame())

  out1 <- select_(.data, .dots = g1)
  out2 <- select_(.data, .dots = g2)

  list(out1, out2)
}

# Or even more like the original...

lazy_parent <- function(expr) {
  # Need to go up twice, because lazy_parent creates an environment for itself
  e1 <- substitute(expr)
  e2 <- do.call("substitute", list(e1), envir = parent.frame(1))

  lazy_(e2, parent.frame(2))
}

select7 <- function(.data, group1, group2) {
  out1 <- select_(.data, .dots = lazy_parent(group1))
  out2 <- select_(.data, .dots = lazy_parent(group2))

  list(out1, out2)
}

最佳答案 这里的问题是默认情况下延迟遵循promises,而order是由于延迟加载包而产生的承诺.

library(pryr)
is_promise(order)
#> TRUE

select中使用的lazy_dots的默认值是相反的.

但是这里还有其他的东西,其中特殊的……的性质用于提取未评估的表达.虽然您对替代品的使用在许多情况下都有效,但尝试通过select重命名将失败.

select4(df,foo=x,bar=order)
#> Error in select4(df, foo = x, bar = order) : 
#>   unused arguments (foo = x, bar = order)

但是,这有效

select5 <- function(.data, ...) {
  dots<-lazy_dots(...)
  out1 <- select_(.data, .dots=dots[1])
  out2 <- select_(.data, .dots=dots[2])
  list(out1, out2)
}


select5(df,foo=x,bar=order)
#> [[1]]  
#>   foo
#> 1   1
#> 2   2
#> 
#> [[2]]
#>   bar
#> 1   5
#> 2   6

另一个例子是,由于缺乏携带环境,替代品更直接失败,请考虑

vars<-c("x","y") 

select4(df,one_of(vars),order)
#>Error in one_of(vars, ...) : object 'vars' not found

select5(df,one_of(vars),order)
#> [[1]]
#>   x y
#> 1 1 3
#> 2 2 4
#> 
#> [[2]]
#>   order
#> 1     5
#> 2     6

select4版本失败,因为它找不到变量,其中select5由于lazy_dots携带环境而成功.注意select4(df,one_of(c(“x”,“y”)),order)是可以的,因为它使用文字.

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