重构任意嵌套列表

Question

我有一个结构如下的列表：

  l <- list(condition = "AND", 
            rules = list(list(id = "sex", 
                              field = "sex"),
                         list(condition = "AND", 
                              rules = list(
                                list(id = "species",
                                     field = "species"), 
                                list(condition = "AND",
                                     rules = list(
                                       list(id = "bill_length_mm", 
                                            field = "bill_length_mm")))))), 
            valid = TRUE)

我想转换成这个结构

  goal <- list(
    list(id = "sex",
         field = "sex"),
    list(id = "species",
         field = "species"),
    list(id = "bill_length_mm", 
         field = "bill_length_mm")
    )

解决方案需要足够灵活以处理 more/less rules 元素。非常感谢任何关于我如何做到这一点的建议！

Answer 1

使用递归函数 rrapply

library(rrapply)
out <- rrapply(l, condition = function(x, .xname)
         .xname %in% c('id', 'field'), how = "flatten" )

out2 <- unname(split(as.list(out), cumsum(names(out) == "id")) )

-检查预期输出

> identical(goal, out2)
[1] TRUE

或者如评论中提到的@Joris C.

out <- rrapply(l, classes = "list", condition = 
    function(x) "id" %in% names(x), how = "flatten")
names(out) <- NULL

Answer 2

看来你可以在这里使用递归策略。这将查找具有 id 元素的列表，并将 return 它们作为列表。

get_fields <- function (x) {
  if (is.list(x)) {
    if (!is.null(x$id)) {
      return(list(x))
    } else {
      unname(do.call("c", lapply(x, get_fields)))
    }
  } else {
    NULL
  }
}
dput(result<-get_fields(l))
# list(list(id = "sex", field = "sex"), list(id = "species", field = "species"), 
#    list(id = "bill_length_mm", field = "bill_length_mm"))

重构任意嵌套列表

Restructuring an arbitrarily nested list

r

lapply

purrr