就像标题说的，我想更新一个函数列表，而不需要创建另一个对象。我有一个这样的代码：

add.filter <- function(filters.list, ..., after) {
  filters.list <- append(filters.list, ..., after={{after}})
  filters.list <<- filters.list
}

filters.list <- list(a = 1, b = 2, c = 3)
add.filter(filters.list, "d", after = 2)

filters.list
$a
[1] 1

$b
[1] 2

[[3]]
[1] "d"

$c
[1] 3

这工作正常，但我希望函数add.filter能够与函数列表一起工作。
例如：

filters.list <- list(function(x) replace(x, x == 3 , 300),
                     function(x) replace(x, x == "l" , 400)
                  )

如果参数是：我如何让函数add.filters（）工作？

add.filter(filters.list, function(x) replace(x, x == 300 , 5000), after = 1)

我希望add.filters（）的输出如下所示：

filters.list
[[1]]
function(x) replace(x, x == 3 , 300)

[[2]]
function(x) replace(x, x == 300 , 5000)

[[3]]
function(x) replace(x, x == "l" , 400)

总而言之，我想创建一个add.filters函数，它可以向现有的函数列表添加函数，而不必创建新的列表。我想用添加的新函数（filter）更新原始列表。我想让这样的东西工作：

add.filter <- function(filters.list, ..., after) {
              filters.list <- append(filters.list, ..., after={{after}})
              filters.list <<- filters.list
 }

g<-list(function(x) replace(x, x == 3 , 300),
        function(x) replace(x, x == "l" , 400)
)

add.filter(g, function(x) replace(x, x == 300 , 5000), after = 1)

g
[[1]]
function(x) replace(x, x == 3 , 300)

[[2]]
function(x) replace(x, x == "l" , 400)` 

this don't work well, so the question is how do I make the function add.filter to go for this type of list.

Thanks a lot