递归地将列表变量发送到全局环境

时间:2021-03-04 19:26:08

Consider the following nested list, nest.

考虑以下嵌套列表,nest。

nest <- list(
    A = list(w = 1:5, x = letters[1:5]),
    B = list(y = 6:10, z = LETTERS[1:5])
)

I'd like to send all the individual variables in nest to the global environment. That is, the lists A and B, and the vectors w, x, y, and z should all go to the global environment. Here are a few of my tries along with their results. Notice that all of these only send some of the variables to the global environment.

我想将nest中的所有单个变量发送到全局环境。也就是说,列表A和B以及向量w,x,y和z都应该进入全局环境。以下是我的一些尝试以及他们的结果。请注意,所有这些只会将一些变量发送到全局环境。

list2env(nest, .GlobalEnv)
ls()
# [1] "A"    "B"    "nest"
list2env(unlist(nest, recursive = FALSE), .GlobalEnv)
ls()
# [1] "A.w"  "A.x"  "B.y"  "B.z"  "nest"
lapply(nest, list2env, envir = .GlobalEnv)
ls()
# [1] "nest" "w"    "x"    "y"    "z"   
with(nest, list2env(mget(ls()), .GlobalEnv))
ls()
# [1] "A"    "B"    "nest"

I also tried other recursive possibilities and got errors because when list2env hits the bottom of the list, it finds that x is not a list.

我还尝试了其他递归的可能性并得到了错误,因为当list2env命中列表的底部时,它发现x不是列表。

rapply(nest, list2env, envir = .GlobalEnv)
# Error in (function (x, envir = NULL, parent = parent.frame(), 
#   hash = (length(x) >  : first argument must be a named list
with(nest, {
    obj <- mget(ls())
    do.call(list2env, c(obj, envir = .GlobalEnv))
})
# Error in (function (x, envir = NULL, parent = parent.frame(), 
#     hash = (length(x) >  : unused arguments (A = list(w = 1:5, 
#     x = c("a", "b", "c", "d", "e")), B = list(y = 6:10, 
#     z = c("A", "B", "C", "D", "E")))

How can I recursively call list2env so that all variables go to the global environment? From a fresh R session, ls() would result in

如何递归调用list2env以便所有变量都转到全局环境?从一个新的R会话,ls()将导致

# [1] "A" "B" "nest" "w" "x" "y" "z"

I've also experimented with local and am having the same problem.

我也尝试了本地,我遇到了同样的问题。

1 个解决方案

#1


7  

Use a recursive function. Not elegant but it appears to work:

使用递归函数。不优雅,但似乎工作:

nest <- list(A = list(w = 1:5, x = letters[1:5]),
             B = list(y = 6:10, z = LETTERS[1:5]))

test <- function(x) {
    if(is.list(x)) { 
        list2env(x, envir = .GlobalEnv)
        lapply(x, test)
    }
}

test(nest)
ls()
# [1] "A"    "B"    "nest" "test" "w"    "x"    "y"    "z"   

#1


7  

Use a recursive function. Not elegant but it appears to work:

使用递归函数。不优雅,但似乎工作:

nest <- list(A = list(w = 1:5, x = letters[1:5]),
             B = list(y = 6:10, z = LETTERS[1:5]))

test <- function(x) {
    if(is.list(x)) { 
        list2env(x, envir = .GlobalEnv)
        lapply(x, test)
    }
}

test(nest)
ls()
# [1] "A"    "B"    "nest" "test" "w"    "x"    "y"    "z"