我有一个如下所示的函数,它生成具有相似名称的各种变量。在这个函数内部,我想将所有变量放入一个列表中。我知道这个例子看起来不太有效,因为我可以从一开始就使用列表,但在我的函数中,我有必要这样做。
test_function <- function(x) {
myenv <- new.env()
myenv$hello1 = "hello1string"
myenv$hello2 = "hello2string"
myenv$cello2 = "hello2string"
mylist <- lapply(ls(name = myenv, pattern = "hello"), get)
print(class(mylist))
}
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如何将所有以“hello”开头的变量放入列表中而不给出错误:Error in FUN(X[[i]], ...) : object 'hello1' not found when running test_function()。当将变量放入共享环境时甚至会发生这种情况。
我希望决赛mylist是班级名单,而不是角色。
谢谢!
您可以创建一个环境,然后在其中创建变量。然后使用ls()具有环境名称和正确模式的函数,您可以看到环境中与给定模式匹配的变量列表。
test_function <- function(x) {
myenv <- new.env()
myenv$hello1 = "hello1"
myenv$hello2 = "hello2"
myenv$cello2 = "hello2"
mylist <- ls(name = myenv, pattern = "hello")
print(mylist)
}
test_function(1)
# [1] "hello1" "hello2"
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您可以使用它mget来提取环境内变量列表的值。
test_function <- function(x, y, z, pattern) {
myenv <- new.env()
ls_vars <- list( hello1 = x,
hello2 = y,
cello2 = z)
list2env( ls_vars, myenv ) # add list of variables to myenv environment
newvar <- "hello3"
assign(newvar, value = "dfsfsf", envir = myenv) # assign new variable
mylist <- ls(name = myenv, pattern = pattern)
return(mget(mylist, envir = myenv))
}
test_function(x = "hello1", y = "hello2", z = "sdfsd", pattern = "hello")
# $hello1
# [1] "hello1"
#
# $hello2
# [1] "hello2"
#
# $hello3
# [1] "dfsfsf"
test_function(x = "hello1", y = "hello2", z = "sdfsd", pattern = "cello")
# $cello2
# [1] "sdfsd"
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