Question

在上一篇关于coercion of variables into their appropriate format的帖子之后，我意识到问题是由于unlist()：ing，这似乎会扼杀变量的对象类。

考虑以下结构的嵌套列表（myList）

> str(myList)
List of 2
 $ lst1:List of 3
  ..$ var1: chr [1:4] "A" "B" "C" "D"
  ..$ var2: num [1:4] 1 2 3 4
  ..$ var3: Date[1:4], format: "1999-01-01" "2000-01-01" "2001-01-01" "2002-01-01"
 $ lst2:List of 3
  ..$ var1: chr [1:4] "Q" "W" "E" "R"
  ..$ var2: num [1:4] 11 22 33 44
  ..$ var3: Date[1:4], format: "1999-01-02" "2000-01-03" "2001-01-04" "2002-01-05"

包含最低级别的不同对象类型（character，numeric和Date）。我一直在使用

myNewLst <- lapply(myList, function(x) unlist(x,recursive=FALSE))
result <- do.call("rbind", myNewLst)

获得我所得矩阵的所需结构。但是，这会对所有变量产生强制转换为character，如下所示：

> str(result)
 chr [1:2, 1:12] "A" "Q" "B" "W" "C" "E" "D" "R" "1" "11" "2" "22" "3" "33" "4" "44" "10592" "10593" "10957" "10959" "11323" "11326" ...
 - attr(*, "dimnames")=List of 2
  ..$ : chr [1:2] "lst1" "lst2"
  ..$ : chr [1:12] "var11" "var12" "var13" "var14" ...

在阅读post on a similar issue后，我尝试使用do.call("c", x)

myNewLst <- lapply(myList, function(x) do.call("c", x))
result <- do.call("rbind", myNewLst)

不幸的是，这也导致所有变量都是character s，这是我的第一次尝试。所以我的问题是：如何在不丢失低级变量的对象类的情况下取消嵌套列表？是否有可以达到预期效果的替代方案？

myList的可重现代码：

myList <- list(
  "lst1" = list(
    "var1" = c("A","B","C","D"),
    "var2" = c(1,2,3,4),
    "var3" = c(as.Date('1999/01/01'),as.Date('2000/01/01'),as.Date('2001/01/01'),as.Date('2002/01/01'))
  ),
  "lst2" = list(
    "var1" = c("Q","W","E","R"),
    "var2" = c(11,22,33,44),
    "var3" = c(as.Date('1999/01/02'),as.Date('2000/01/03'),as.Date('2001/01/4'),as.Date('2002/01/05'))
  )
)

Answer 1

如果您的目标是将此列表列表转换为单个数据框，则以下代码应该有效：

result <- data.frame(var1 = unlist(lapply(myList, function(e) e[1]), use.names = FALSE), 
                     var2 = unlist(lapply(myList, function(e) e[2]), use.names = FALSE),
                     var3 = as.Date(unlist(lapply(myList, function(e) e[3]), use.names = FALSE), origin = "1970-01-01"))

这给出了：

> result
  var1 var2       var3
1    A    1 1999-01-01
2    B    2 2000-01-01
3    C    3 2001-01-01
4    D    4 2002-01-01
5    Q   11 1999-01-02
6    W   22 2000-01-03
7    E   33 2001-01-04
8    R   44 2002-01-05

当然，如果每个列表中有多个变量，您可以使用for循环使代码更简洁。

Answer 2

您可以使用Reduce()或do.call()将所有数据帧合并到一起。下面的代码应该可以使用

      Reduce(rbind,lapply(myList,data.frame,stringsAsFactors=F))

        var1 var2       var3
      1    A    1 1999-01-01
      2    B    2 2000-01-01
      3    C    3 2001-01-01
      4    D    4 2002-01-01
      5    Q   11 1999-01-02
      6    W   22 2000-01-03
      7    E   33 2001-01-04
      8    R   44 2002-01-05

此类课程也得到了维护：

  mapply(class,Reduce(rbind,lapply(myList,data.frame,stringsAsFactors=F)))
        var1        var2        var3 
 "character"   "numeric"      "Date"

取消列出嵌套列表并且不会丢失对象类

2 个答案: