R 如何在嵌套列表的叶上轻松迭代

假设我有一个清单：

a <- list(a = 1, b = list(a = list(b = 1, c = 2), c = 1), c = 1)

a这个怎么样

a <- list(a = 1, b = list(a = list(b = 1, c = 2), c = 1), c = 1)

aunls <- unlist(a)
aunls[!grepl('c', names(aunls))]

a这个怎么样

a <- list(a = 1, b = list(a = list(b = 1, c = 2), c = 1), c = 1)

aunls <- unlist(a)
aunls[!grepl('c', names(aunls))]

a您可以使用递归lappy函数来执行此操作，例如：

foo <- function(x, name) {
  if(any(idx <- names(x) == name)) x <- x[!idx]
  if(is.list(x)) lapply(x, foo, name) else x
}

foo(a, "c")
# $a
# [1] 1
# 
# $b
# $b$a
# $b$a$b
# [1] 1

foo您可以使用递归lappy函数来实现这一点，例如：

foo <- function(x, name) {
  if(any(idx <- names(x) == name)) x <- x[!idx]
  if(is.list(x)) lapply(x, foo, name) else x
}

foo(a, "c")
# $a
# [1] 1
# 
# $b
# $b$a
# $b$a$b
# [1] 1

foo