如何在使用R保留结构的同时从嵌套列表创建所有组合?

本文介绍了如何在使用R保留结构的同时从嵌套列表创建所有组合?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

给出一个嵌套列表，如何在保留嵌套列表结构的同时从其元素创建所有可能的列表?

Given a nested list, how to create all possible lists from its elements, while preserving the structure of the nested list?

嵌套列表:

l = list( a = list( b = 1:2 ), c = list( d = list( e = 3:4, f = 5:6 ) ), g = 7 )

所需的输出:l的所有可能组合，同时保留结构，例如:

Desired output: all possible combinations of the elements of l, while preserving the structure, e.g.:

# One possible output: list( a = list( b = 1 ), c = list( d = list( e = 3, f = 5 ) ), g = 7 ) # Another possible output: list( a = list( b = 1 ), c = list( d = list( e = 4, f = 5 ) ), g = 7 )

到目前为止，我的方法是:

My approach so far is to:

拉平列表(例如，如此答案中所述)

expand.grid() 并得到一个矩阵，其中每一行代表一个唯一的组合

解析所得矩阵的每一行，并从 names() 使用正则表达式

flatten the list (e.g., as discussed in this answer)

expand.grid() and get a matrix where each row represents a unique combination

parse every row of the resulting matrix and reconstruct the structure from the names() using regular expressions

我正在寻找一种不太麻烦的方法，因为我不能保证列表元素的名称不会更改.

I am looking for a less cumbersome approach because I have no guarantee that the names of the list elements will not change.

推荐答案

utils中的relist函数似乎是为该任务而设计的:

The relist function from utils seems to be designed for this task:

rl <- as.relistable(l) r <- expand.grid(data.frame(rl), KEEP.OUT.ATTRS = F) > head(r, 5) b c.d.e c.d.f g 1 1 3 5 7 2 2 3 5 7 3 1 4 5 7 4 2 4 5 7 5 1 3 6 7

它保存列表的结构(skeleton).这意味着现在可以操纵嵌套列表中的数据，然后将其重新分配到结构中(flesh).这里是扩展矩阵的第一行.

It saves the structure of the list (skeleton). This means one can now manipulate the data within the nested list and re-assign it into the structure (flesh). Here with the first row of the expanded matrix.

r <- rep(unname(unlist(r[1,])),each = 2) l2 <- relist(r, skeleton = rl) > l2 $a $a$b [1] 1 1 $c $c$d $c$d$e [1] 3 3 $c$d$f [1] 5 5 $g [1] 7 attr(,"class") [1] "relistable" "list"

请注意，由于结构保持不变，因此我需要提供与原始列表中相同数量的元素.这就是为什么使用rep将元素重复两次的原因.我猜也可以用NA填充它.

Note that since the structure stays the same, I need to supply the same amount of elements as in the original list. This is why used rep to repeat the element twice. One could also fill it with NA, I guess.

对于每种可能的组合，请通过r进行迭代(扩展):

For every possible combination iterate through r (expanded):

lapply(1:nrow(r), function(x) relist(rep(unname(unlist(r[x,])),each = 2), skeleton = rl))