转置嵌套列表

本文介绍了转置嵌套列表的处理方法，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

我有一个列表结构，它表示像这样递给我的表

I have a list structure which represents a table being handed to me like this

> l = list(list(1, 4), list(2, 5), list(3, 6))
> str(l)
List of 3
 $ :List of 2
  ..$ : num 1
  ..$ : num 4
 $ :List of 2
  ..$ : num 2
  ..$ : num 5
 $ :List of 2
  ..$ : num 3
  ..$ : num 6

我想将其转换为此

> lt = list(x = c(1, 2, 3), y = c(4, 5, 6))
> str(lt)
List of 2
 $ x: num [1:3] 1 2 3
 $ y: num [1:3] 4 5 6

我编写了一个使用Reduce的非常简单的方法来执行此操作，但是我觉得必须有一种更聪明的方法来实现它.

I've written a function that does it in a really simple manner which uses Reduce, but I feel like there must be a smarter way to do it.

任何帮助表示赞赏，谢谢

Any help appreciated,Thanks

谢谢！非常感激.对答案进行基准测试，并在较大的测试案例中选择最快的答案:

Thanks all! Much appreciated. Benchmarked the answers and picked the fastest for a larger test case:

f1 = function(l) {
  k <- length(unlist(l)) / length(l)
  lapply(seq_len(k), function(i) sapply(l, "[[", i))
}

f2 = function(l) {
  n <- length(l[[1]])
  split(unlist(l, use.names = FALSE), paste0("x", seq_len(n)))
}

f3 = function(l) {
  split(do.call(cbind, lapply(l, unlist)), seq(unique(lengths(l))))
}

f4 = function(l) {
  l %>%
    purrr::transpose() %>%
    map(unlist)
}

f5 = function(l) {
  # bind lists together into a matrix (of lists)
  temp <- Reduce(rbind, l)
  # split unlisted values using indices of columns
  split(unlist(temp), col(temp))
}

f6 = function(l) {
  data.table::transpose(lapply(l, unlist))
}

microbenchmark::microbenchmark(
  lapply     = f1(l),
  split_seq  = f2(l),
  unique     = f3(l),
  tidy       = f4(l),
  Reduce     = f5(l),
  dt         = f6(l),
  times      = 10000
)

Unit: microseconds
      expr     min       lq     mean   median       uq      max neval
    lapply 165.057 179.6160 199.9383 186.2460 195.0005 4983.883 10000
 split_seq  85.655  94.6820 107.5544  98.5725 104.1175 4609.378 10000
    unique 144.908 159.6365 182.2863 165.9625 174.7485 3905.093 10000
      tidy  99.547 122.8340 141.9482 129.3565 138.3005 8545.215 10000
    Reduce 172.039 190.2235 216.3554 196.8965 206.8545 3652.939 10000
        dt  98.072 106.6200 120.0749 110.0985 116.0950 3353.926 10000

推荐答案

对于特定示例，您可以使用以下非常简单的方法:

For the specific example, you can use this pretty simple approach:

split(unlist(l), c("x", "y"))
#$x
#[1] 1 2 3
#
#$y
#[1] 4 5 6

它回收x-y向量并对其进行分割.

It recycles the x-y vector and splits on that.

要将其概括为每个列表中的"n"个元素，可以使用:

To generalize this to "n" elements in each list, you can use:

l = list(list(1, 4, 5), list(2, 5, 5), list(3, 6, 5)) # larger test case

split(unlist(l, use.names = FALSE), paste0("x", seq_len(length(l[[1L]]))))
# $x1
# [1] 1 2 3
#
# $x2
# [1] 4 5 6
#
# $x3
# [1] 5 5 5

这假定l顶层的所有列表元素的长度都与您的示例相同.

This assumes, that all the list elements on the top-level of l have the same length, as in your example.

这篇关于转置嵌套列表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！

thanks