在非本地列表中替换值的赋值

[[<- behaves differently for lists and environments when used on non-local objects:

[<-在非本地对象上使用时，列表和环境的行为不同:

lst = list()
env = new.env()

(function () lst[['x']] = 1)()
(function () env[['x']] = 1)()

lst
# list()

as.list(env)
# $x
# [1] 1

In other words, if the target of [[<- is an environment, it modifies the (nonlocal) environment, but if it’s a vector/list, it creates a new, local object.

换句话说，如果[<-的目标是一个环境，它将修改(非本地)环境，但是如果它是一个向量/列表，它将创建一个新的本地对象。

I would like to know two things:

我想知道两件事:

Why this difference in behaviour?
为什么会有这种行为上的差异?
Is there a way of achieving the same result for lists as for environments, without using <<-?
是否有一种方法可以在不使用<<-的情况下，为列表和环境实现相同的结果?

Regarding (1), I’m aware of multiple differences between lists and environments (in particular, I know that environments don’t get copied) but the documentation does not mention why the semantics of [[<- differ between the two — in particular, why the operator would operate on different scope. Is this a bug? It’s counter-intuitive at least, and requires some non-trivial implementation shenanigans.¹

关于(1)，我知道列表和环境之间有很多不同之处(特别是，我知道环境不会被复制)，但是文档中没有提到为什么[[<-]的语义在这两者之间存在差异——特别是为什么操作符会在不同的范围内操作。这是一个错误吗?至少这是违反直觉的，并且需要一些非平凡的实现诡计

Regarding (2), the obvious solution is of course to use <<-:

对于(2)，显然的解决方案当然是使用<<-:

(function () lst[['x']] <<- 1)()

However, I prefer understanding the difference rigorously rather than just working around them. Furthermore, I’ve so far used assign instead of <<- and I prefer this, as it allows me greater control over the scope of the assignment (in particular since I can specify inherits = FALSE. <<- is too much voodoo for my taste.

然而，我更喜欢严格地理解差异，而不是仅仅围绕它们工作。此外，到目前为止，我使用了assign而不是<-我更喜欢这种方法，因为它允许我对赋值范围有更大的控制(特别是因为我可以指定inherits = FALSE)。<-对我的口味来说是太多的巫术了。

However, the above cannot be solved (as far as I know) using assign because assign only works on environments, not lists. In particular, while assign('x', 1, env) works (and does the same as above), assign('x', 1, lst) doesn’t work.

但是，使用assign不能解决上述问题(据我所知)，因为assign只能在环境中工作，而不是在列表中。特别是，尽管assign('x'， 1, env)可以工作(并与上面一样)，assign('x'， 1, lst)不能工作。

¹ To elaborate, it’s of course expected that R does different thing for different object types using dynamic dispatch (e.g. via S3). However, this is not the case here (at least not directly): the distinction in scope resolution happens before the object type of the assignment target is known — otherwise the above would operate on the global lst, rather than creating a new local object. So internally [[<- has to do the equivalent of:

要详细说明，当然希望R对不同对象类型使用动态调度(例如通过S3)做不同的事情。但是，这里不是这种情况(至少不是直接):范围解析的区别发生在已知分配目标的对象类型之前——否则，上面的操作将在全局lst上进行，而不是创建新的本地对象。所以在内部[<-必须做等价的:

`[[<-` = function (x, i, value) {
    if (exists(x, mode = 'environment', inherits = TRUE))
        assign(i, value, pos = x, inherits = FALSE)
    else if (exists(x, inherits = FALSE)
        internal_assign(x, i, value)
    else
        assign(x, list(i = value), pos = parent.frame(), inherits = FALSE)
}

1 个解决方案

#1

The R-language definition (section 2.1.10) says:

R-language定义(2.1.10节)说:

Unlike most other R objects, environments are not copied when passed to functions or used in assignments.

与大多数其他R对象不同，环境在传递给函数或在赋值中使用时不会被复制。

Section "6.3 More on evaluation" also gives a slightly relevant hint:

“6.3更多的评估”也给出了一个稍微相关的提示:

Notice that evaluation in a given environment may actually change that environment, most obviously in cases involving the assignment operator, such as

请注意，给定环境中的评估实际上可能会改变该环境，最明显的是涉及赋值操作符的情况，例如
eval(quote(total <- 0), environment(robert$balance)) # rob Rob
This is also true when evaluating in lists, but the original list does not change because one is really working on a copy.

当在列表中求值时，这也是正确的，但是原始列表不会改变，因为一个人实际上正在处理一个副本。

So, the answer to your first question is that lists need to be copied to assign into them, but environments can be modified in place (which has huge performance implications).

因此，第一个问题的答案是列表需要被复制来分配给它们，但是环境可以被适当地修改(这具有巨大的性能影响)。

Regarding your second question:

关于你提到的第二个问题:

If you are working with a list, the only option seems to be to

如果你正在处理一个列表，唯一的选择似乎是

copy the list into the local scope (using get),
将列表复制到本地范围(使用get)，
assign into the list,
分配到列表中,
use assign to copy the modified list back into the original environment.
使用assign将修改后的列表复制回原始环境。

#1