如何使用循环访问列表中的所有列表

Question

这是来自 one 的后续问题，我想在列表中创建一个列表，现在我希望能够访问嵌套列表并在其中执行函数。

我有一个 3 个列表，其中包含两个列表：

    my_list

    structure(list(a = structure(list(a = structure(list(value = c(1L, 
4L), class = structure(c(1L, 1L), .Label = c("a", "b", "c"), class = "factor")), .Names = c("value", 
"class"), row.names = c(1L, 4L), class = "data.frame"), b = structure(list(
    value = c(2, 1), class = structure(c(2L, 2L), .Label = c("a", 
    "b", "c"), class = "factor")), .Names = c("value", "class"
), row.names = c(2L, 5L), class = "data.frame")), .Names = c("a", 
"b")), b = structure(list(a = structure(list(value = c(6, 1), 
    class = structure(c(1L, 1L), .Label = c("a", "b", "c", "d"
    ), class = "factor")), .Names = c("value", "class"), row.names = c(1L, 
5L), class = "data.frame"), b = structure(list(value = c(5, 1
), class = structure(c(2L, 2L), .Label = c("a", "b", "c", "d"
), class = "factor")), .Names = c("value", "class"), row.names = c(2L, 
6L), class = "data.frame")), .Names = c("a", "b")), c = structure(list(
    a = structure(list(value = 5:6, class = structure(c(1L, 1L
    ), .Label = c("a", "b", "c", "d", "e"), class = "factor")), .Names = c("value", 
    "class"), row.names = 5:6, class = "data.frame"), b = structure(list(
        value = c(4, 1), class = structure(c(2L, 2L), .Label = c("a", 
        "b", "c", "d", "e"), class = "factor")), .Names = c("value", 
    "class"), row.names = c(4L, 7L), class = "data.frame")), .Names = c("a", 
"b"))), .Names = c("a", "b", "c"))

    $a
    $a$a
      value class
        1     a
        4     a

    $a$b
      value class
       2     b
       1     b


    $b
    $b$a
      value class
        6     a
        1     a

    $b$b
      value class
        5     b
        1     b


    $c
    $c$a
      value class
        5     a
        6     a

    $c$b
      value class
        4     b
        1     b

我想通过value != 1 subset 每个嵌套列表使用循环，以便生成的列表如下所示：

my_list
$a
        $a$a
          value class
            4     a

        $a$b
          value class
           2     b



        $b
        $b$a
          value class
            6     a


        $b$b
          value class
            5     b



        $c
        $c$a
          value class
            5     a
            6     a

        $c$b
          value class
            4     b

使用下面的代码，我只能访问list [[1]][[1]]、list[[2]][[2]] 和list[[3]][[3]]。如何通过循环访问其他人并应用subset 函数？

for (i in 1:length(my_list)) {
  my_list[[i]][[i]] <- subset(my_list[[i]][[i]],my_list[[i]][[i]]$value=1) 
 }

Answer 1

对于那些想知道如何为这个问题做一个嵌套循环的人——我得到了帮助来解决这个问题：

for (i in 1:length(my_list)) {
  for (j in 1:2) {
  my_list[[i]][[j]] <- subset(my_list[[i]][[j]],my_list[[i]][[j]]$value!=1) 
  }
}

Answer 2

我们可以使用嵌套的lapply 或rapply

lapply(my_list, function(x) lapply(x, function(y) subset(y, value!=1)))
#$a
#$a$a
#  value class
#4     4     a

#$a$b
#  value class
#2     2     b


#$b
#$b$a
#  value class
#1     6     a

#$b$b
#  value class
#2     5     b


#$c
#$c$a
#  value class
#5     5     a
#6     6     a

#$c$b
#  value class
#4     4     b