在for循环中包含变量 [英] Enclosing variables within for loop
问题描述
#cartoon示例
a < - c（3,7,11）
f < - list（）
#manual initialization
f [[1]] <-function（x） （x）a [2] + x
f [[3]]< - 函数（x）a [3] + x
#desired result for the other examples
f [[1]]（1）
＃[1] 4
f [[3]]（1）
＃[1] 12
#attempted automation
for（i in 1：3）{
f [[i]] < - function（x）a （1）
$ b $ [$ 1]
$ [1] b＃[1] 12
请注意，在我们尝试自动化之后，当然，问题在于 i 没有包含在函数的私有环境中。所有的函数在全局环境（只能有一个值）中引用相同的 i ，因为for循环似乎并没有为每次迭代创建不同的环境。 b
$ b $ pre $ $ $ $ $ $
$<环境：R_GlobalEnv>
＃[[2]]
＃< environment：R_GlobalEnv>
＃[[3]]
＃< environment：R_GlobalEnv>
所以我尽管可以使用 local（） 和 force（）来捕捉 i 值
for（i in 1：3）{
f [[i]]< - local（{force（i）; function（x）a （1）
$
$ bf [[1]]（1）
＃[1] 12
f [[3]]（1）
＃[1] 12
但是这仍然不起作用。我可以看到它们都有不同的环境（通过 sapply（f，environment）），但它们看起来是空的（ ls.str（envir = environment （F [[1]]）））。 （1：3）{
f [[i]]< - local（$） {ai <-i; function（x）a [ai] + x}）
}
f [[1]]（1）
＃[1] 4
f [[3]]（1）
＃[1] 12
ls.str（envir = environment（f [[1]]））
＃ai： int 1
ls.str（envir = environment（f [[3]]））
＃ai：int 3
显然， force（）不能像我所期望的那样工作。我假设它会捕获 i 当前值到当前环境中。它在
#bad
f < - lapply（1：3，函数（i）函数（x）a [i] + x）
＃good
f < - lapply（1：3，function（i）{force（i）; function（x）a [i] + x} ）
其中 i 作为参数传递/ promise，但是这肯定不是在for循环中发生的事情。
所以我的问题是：可以创建这个函数列表，而不需要 local（）和变量重命名？有没有比 force（）更适合的函数，它将从父框架的变量捕获到本地/当前环境中？
这不是一个完整的答案，部分是因为我不确定问题到底是什么（即使我发现它很有趣！
相反，我只提供了两个替代的 - 循环。他们已经帮助澄清了我脑海中的问题（特别是帮助我首次明白为什么 force（）在调用 lapply（））。我希望他们也会帮助你。
首先，这是一个更接近您的功能 lapply（） call，它的工作原理和它一样：
a < - c（3， 7,11）
f < - list（）
##`for`loop等效于：
## f < - lapply（1：3，function（i ）{{（i）; function（x）a [i] + x}）
for（i in 1：3）{
f [[i]] < - {X < - function （i）{force（i）;函数（x）a [i] + x}; X（i）}
}
f [[1]]（1）
＃[1] 4
其次，这是一个使用 local（）但不会（严格地或字面地说）重命名我。但是，它通过将它的一个副本添加到当地环境中来重新它。从某种意义上来说，它与你的对于 -loop只有平凡的不同，但是通过把注意力集中在 i 的作用域，而不是它的名字，我认为这有助于揭露你的问题背后的真实问题。
a < - c 3,7,11）
f < - list（）
（i in 1：3）{
f [[i]]< - local（{i< -i;函数（x）a [i] + x}）
}
f [[1]]（1）
＃[1] 4
So consider the following chunk of code which does not work as most people might expect it to
#cartoon example
a <- c(3,7,11)
f <- list()
#manual initialization
f[[1]]<-function(x) a[1]+x
f[[2]]<-function(x) a[2]+x
f[[3]]<-function(x) a[3]+x
#desired result for the rest of the examples
f[[1]](1)
# [1] 4
f[[3]](1)
# [1] 12
#attempted automation
for(i in 1:3) {
f[[i]] <- function(x) a[i]+x
}
f[[1]](1)
# [1] 12
f[[3]](1)
# [1] 12
Note that we get 12 both times after we attempt to "automate". The problem is, of course, that i isn't being enclosed in the function's private environment. All the functions refer to the same i in the global environment (which can only have one value) since a for loop does not seem to create different environment for each iteration.
sapply(f, environment)
# [[1]]
# <environment: R_GlobalEnv>
# [[2]]
# <environment: R_GlobalEnv>
# [[3]]
# <environment: R_GlobalEnv>
So I though I could get around with with the use of local() and force() to capture the i value
for(i in 1:3) {
f[[i]] <- local({force(i); function(x) a[i]+x})
}
f[[1]](1)
# [1] 12
f[[3]](1)
# [1] 12
but this still doesn't work. I can see they all have different environments (via sapply(f, environment)) however they appear to be empty (ls.str(envir=environment(f[[1]]))). Compare this to
for(i in 1:3) {
f[[i]] <- local({ai<-i; function(x) a[ai]+x})
}
f[[1]](1)
# [1] 4
f[[3]](1)
# [1] 12
ls.str(envir=environment(f[[1]]))
# ai : int 1
ls.str(envir=environment(f[[3]]))
# ai : int 3
So clearly the force() isn't working like I was expecting. I was assuming it would capture the current value of i into the current environment. It is useful in cases like
#bad
f <- lapply(1:3, function(i) function(x) a[i]+x)
#good
f <- lapply(1:3, function(i) {force(i); function(x) a[i]+x})
where i is passed as a parameter/promise, but this must not be what's happening in the for-loop.
So my question is: is possible to create this list of functions without local() and variable renaming? Is there a more appropriate function than force() that will capture the value of a variable from a parent frame into the local/current environment?
This isn't a complete answer, partly because I'm not sure exactly what the question is (even though I found it quite interesting!).
Instead, I'll just present two alternative for-loops that do work. They've helped clarify the issues in my mind (in particular by helping me to understand for the first time why force() does anything at all in a call to lapply()). I hope they'll help you as well.
First, here is one that's a much closer equivalent of your properly function lapply() call, and which works for the same reason that it does:
a <- c(3,7,11)
f <- list()
## `for` loop equivalent of:
## f <- lapply(1:3, function(i) {force(i); function(x) a[i]+x})
for(i in 1:3) {
f[[i]] <- {X <- function(i) {force(i); function(x) a[i]+x}; X(i)}
}
f[[1]](1)
# [1] 4
Second, here is one that does use local() but doesn't (strictly- or literally-speaking) rename i. It does, though, "rescope" it, by adding a copy of it to the local environment. In one sense, it's only trivially different from your functioning for-loop, but by focusing attention on i's scope, rather than its name, I think it helps shed light on the real issues underlying your question.
a <- c(3,7,11)
f <- list()
for(i in 1:3) {
f[[i]] <- local({i<-i; function(x) a[i]+x})
}
f[[1]](1)
# [1] 4
这篇关于在for循环中包含变量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！
