将数组列表重塑为 R 中的数据帧

Reshape list of array to dataframe in R

我有数组列表,需要将其重塑为数据框。

示例输入:

> box
[[1]]
, , 1
          [,1]      [,2]      [,3]      [,4]
[1,] -88.44636 -84.29698 -84.29698 -88.44636

, , 2
         [,1]     [,2]    [,3]    [,4]
[1,] 32.28459 32.28459 41.7449 41.7449

[[2]]
NULL

[[3]]
, , 1
         [,1]     [,2]     [,3]     [,4]
[1,] 108.3619 108.4818 108.4818 108.3619

, , 2
          [,1]      [,2]      [,3]      [,4]
[1,] -6.537015 -6.537015 -6.439103 -6.439103

[[4]]
, , 1

         [,1]     [,2]     [,3]     [,4]
[1,] 108.5949 114.2009 114.2009 108.5949

, , 2

         [,1]     [,2]    [,3]    [,4]
[1,] -3.03971 -3.03971 2.08105 2.08105

输入结构:

> str(box)
List of 4
 $ : num [1, 1:4, 1:2] -88.4 -84.3 -84.3 -88.4 32.3 ...
 $ : NULL
 $ : num [1, 1:4, 1:2] 108.36 108.48 108.48 108.36 -6.54 ...
 $ : num [1, 1:4, 1:2] 108.59 114.2 114.2 108.59 -3.04 ...

期望的输出:

> bound
         X1        X2        X3        X4        X5        X6        X7        X8
1 -88.44636 -84.29698 -84.29698 -88.44636 32.284593 32.284593 41.744901 41.744901
2        NA        NA        NA        NA        NA        NA        NA        NA
3 108.36186 108.48179 108.48179 108.36186 -6.537015 -6.537015 -6.439103 -6.439103
4 108.59490 114.20087 114.20087 108.59490 -3.039710 -3.039710  2.081050  2.081050

我写代码。它工作并给了我想要的输出,但执行速度很慢:(

bound = NULL
for ( i in 1:length(box) ) {
  if ( !is.null(dim(box[i][[1]])) ) {
    bound = rbind(bound, data.frame(matrix(as.vector(box[i][[1]][1,,]),nrow=1))) 
  } else
    bound = rbind(bound,rep(NA,8))

  if ( i == 1) {
    colnames(bound) = paste("X",1:8,sep="")
  }
}

有人能给我更好的解决方案吗?既简单又快速,可以解决这个问题?从this discussion,apply一家人看合适。但我不知道该怎么做。

通过 list 循环 (lapply),我们用 NA 替换 NULL 元素,连接 (c) 非 NULL 元素,并且 rbind 以获得 matrix 输出。

m1 <- do.call(rbind,lapply(box, function(x) if(is.null(x)) NA else c(x)))

我们可以将其转换为 data.frame

as.data.frame(m1)

数据

box <- list(structure(c(-88.44636, -84.29698, -84.29698, -88.44636, 
32.28459, 32.28459, 41.7449, 41.7449), .Dim = c(1L, 4L, 2L)), 
NULL, structure(c(108.3619, 108.4818, 108.4818, 108.3619, 
-6.537015, -6.537015, -6.439103, -6.439103), .Dim = c(1L, 
4L, 2L)), structure(c(108.5949, 114.2009, 114.2009, 108.5949, 
-3.03971, -3.03971, 2.08105, 2.08105), .Dim = c(1L, 4L, 2L)))