将（几乎）工作区中的所有对象存储在列表中

Let's say that I have many objects in my workspace (global environment) and I want to store most of those in a list. Here's a simplified example:

假设我的工作空间(全局环境)中有很多对象,我希望将大多数对象存储在列表中。这是一个简化的例子:

# Put some objects in the workspace
A <- 1
B <- 2
C <- 3

I would like to store objects A and C in a list. Of course, I can do that explicitly:

我想将对象A和C存储在列表中。当然,我可以明确地做到这一点:

mylist <- list(A,C)

However, when the number of objects in the workspace is very large, this would become rather cumbersome. Hence, I would like to do this differently and attempted the following:

但是,当工作空间中的对象数量非常大时,这将变得相当麻烦。因此,我想以不同的方式做到这一点并尝试以下方法:

mylist <- list(setdiff(ls(),B))

But this obviously is not what I want, as it only stores the names of the objects in the workspace.

但这显然不是我想要的,因为它只存储工作区中对象的名称。

Any suggestions on how I can do this?

有关如何做到这一点的任何建议?

Many thanks!

3 个解决方案

#1

Another option is to use mget:

另一种选择是使用mget:

mget(setdiff(ls(),"B"))

#2

EDIT : I think using lapply / sapply here raises too many problems. You should definitely use the mget answer.

编辑:我认为在这里使用lapply / sapply会引发太多问题。你绝对应该使用mget答案。

You can try :

你可以试试 :

mylist <- sapply(setdiff(ls(),"B"), get)

In certain cases, ie if all the objects in your workspace are of the same type, sapply will return a vector. For example :

在某些情况下,即如果工作空间中的所有对象属于同一类型,则sapply将返回向量。例如 :

sapply(setdiff(ls(),"B"), get)
# A C 
# 1 3

Otherwise, it will return a list :

否则,它将返回一个列表:

v <- list(1:2)
sapply(setdiff(ls(),"B"), get)
# $A
# [1] 1
# 
# $C
# [1] 3
# 
# $v
# [1] 1 2

So using lapply instead of sapply here could be safer, as Josh O'Brien pointed out.

因此,正如Josh O'Brien指出的那样,在这里使用lapply而不是sapply可能更安全。

#3

mget is definitely the easiest to use in this situation. However, you can achieve the same with as.list.environment and eapply:

在这种情况下,mget绝对是最容易使用的。但是,您可以使用as.list.environment和eapply实现相同的功能:

e2l <- as.list(.GlobalEnv)
# or: e2l <- as.list(environment()) 
# using environment() within a function returns the function's env rather than .GlobalEnv
e2l[! names(e2l) %in "B"]

# the following one sounds particularly manly with `force`
e2l <- eapply(environment(), force)
e2l[! names(e2l) %in "B"]

And one-liners:

 (function(x) x[!names(x)%in%"B"])(eapply(environment(), force))
 (function(x) x[!names(x)%in%"B"])(as.list(environment()))

#1