将数组列表重新整形为R中的数据帧

时间:2015-09-16 08:49:47

标签: r apply reshape

我有数组列表,需要将其重新整理为数据帧。

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outpur,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="")
  }
}

有人可以给我更简单快速的解决方案来处理问题吗?来自this discussionapply家庭看起来合适。但我不知道该怎么做。

1 个答案:

答案 0 :(得分:2)

通过lapply循环(list),我们用NA替换NULL元素,连接(c)非NULL元素,{{1}获得rbind输出。

matrix

我们可以将其转换为m1 <- do.call(rbind,lapply(box, function(x) if(is.null(x)) NA else c(x)))

data.frame

数据

as.data.frame(m1)
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