lapply 创建的重塑列表

Question

我有带有纬度和经度的向量：

longDim
[1] -79.65770 -79.21761 -78.77750

latiDim
[1] -39.70588 -39.26471 -38.82353

我想并行遍历它们的组合。为此，我首先使用 expand.grid 创建一个包含所有可能组合的数据框：

my.grid <- expand.grid(longDim, latiDim)

然后我在结果数据框的行上使用了mclapply()：

mclapply(1:nrow(my.grid), function(x){some_function})

其中 some_function returns 包含两个对象的列表，每个对象的长度为 139。

因此，我得到了一个尺寸为 9x2 的嵌套列表，如下所示：

str(l1)
List of 9
 $ :List of 2
  ..$ su.25: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  ..$ su.30: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
 $ :List of 2
  ..$ su.25: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  ..$ su.30: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
 $ :List of 2
  ..$ su.25: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  ..$ su.30: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
 $ :List of 2
  ..$ su.25: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  ..$ su.30: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
 $ :List of 2
  ..$ su.25: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  ..$ su.30: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
 $ :List of 2
  ..$ su.25: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  ..$ su.30: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
 $ :List of 2
  ..$ su.25: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  ..$ su.30: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
 $ :List of 2
  ..$ su.25: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  ..$ su.30: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
 $ :List of 2
  ..$ su.25: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  ..$ su.30: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...

现在，我需要再次将此列表从 9x2 维度重新调整为 3x3x2 维度。这是我正在寻找的格式：

str(l2)
List of 3
 $ :List of 3
  ..$ :List of 2
  .. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  .. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  ..$ :List of 2
  .. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  .. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  ..$ :List of 2
  .. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  .. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
 $ :List of 3
  ..$ :List of 2
  .. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  .. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  ..$ :List of 2
  .. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  .. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  ..$ :List of 2
  .. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  .. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
 $ :List of 3
  ..$ :List of 2
  .. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  .. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  ..$ :List of 2
  .. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  .. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  ..$ :List of 2
  .. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
  .. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...

我怎样才能做到这一点？

可在以下位置找到重现 l1 和 l2 的代码：https://pastebin.com/raw/LTyZi0mp（这里 post 太长了）

Answer 1

您可以使用 split().

split(lst, cut(1:length(lst), 3, labels = FALSE))

测试

lst <- replicate(9, list(list(x = 1:5, y = 1:5)))
result <- split(lst, cut(seq_along(lst), 3, labels = FALSE))
str(result)

# List of 3
#  $ 1:List of 3
#   ..$ :List of 2
#   .. ..$ x: int [1:5] 1 2 3 4 5
#   .. ..$ y: int [1:5] 1 2 3 4 5
#   ..$ :List of 2
#   .. ..$ x: int [1:5] 1 2 3 4 5
#   .. ..$ y: int [1:5] 1 2 3 4 5
#   ..$ :List of 2
#   .. ..$ x: int [1:5] 1 2 3 4 5
#   .. ..$ y: int [1:5] 1 2 3 4 5
# etc.

Answer 2

我们也可以使用gl为spliting

创建分组索引

split(lst, as.integer(gl(length(lst1), 3, length(lst1))))

数据

lst1 <- replicate(9, list(list(x = 1:5, y = 1:5)))

lapply 创建的重塑列表

Reshape list created by lapply

r

list

reshape

lapply

mclapply

数据