R - 评估环境中的嵌套函数 [英] R - Evaluate a nested function in an environment

问题描述

我试图以沙盒编辑的方式运行大量R代码，方法是将所有必需的依赖项（函数和数据）加载到新环境中并评估该环境中的表达式。但是，我在调用环境中其他函数的函数时遇到了麻烦。这里有一个简单的例子：

I am trying to run a chunk of R code in a sandbox-ed fashion, by loading all the necessary dependencies (functions and data) into a new environment and evaluating an expression within that environment. However, I'm running into trouble with functions calling other functions in the environment. Here's a simple example:

jobenv <- new.env(parent=globalenv())
assign("f1", function(x) x*2, envir=jobenv)
assign("f2", function(y) f1(y) + 1, envir=jobenv)
expr <- quote(f2(3))

使用 eval on expr 因为 f2 找不到 f1

Using eval on expr fails since f2 can't find f1

> eval(expr, envir=jobenv)
Error in f2(3) : could not find function "f1"

显式附加环境工作

> attach(jobenv)
> eval(expr)
[1] 7

我可能错过了一些明显的东西，但我无法找到可用的 eval 调用的任何排列组合。有没有一种方法可以在不附加环境的情况下获得相同的效果？

I'm probably missing something obvious, but I couldn't find any permutation of the eval call that works. Is there a way to get the same effect without attaching the environment?

推荐答案

有很多方法可以做到这一点，但我有点像这样：

There are a number of ways of doing this, but I kind of like this one:

jobenv <- new.env(parent=globalenv())

local({
    f1 <- function(x) x*2
    f2 <- function(y) f1(y) + 1
}, envir=jobenv)

## Check that it works
ls(jobenv)
# [1] "f1" "f2"
local(f2(3), envir=jobenv)
# [1] 7
eval(quote(f2(3)), envir=jobenv)
# [1] 7

这篇关于R - 评估环境中的嵌套函数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！