展平列表和将列表键推到第二层的向量 [英] Flatten list and push list key to vector on second level

问题描述

我想这很简单，但是我似乎无法弄清楚. 我需要展平第二层结构，并将列表名称/键推到与其他向量相同级别的向量. myList的当前结构是

I suppose this is simple, but I just can't seem to figure it out. I need to flatten the second level structure and push the list name/key to a vector on the same level as the other vectors. The current structure of myList is

$ 13454:List of 30
  ..$ subjectId        : num 187
  ..$ procedureId      : num 3
  ..$ procedureSampleId: num 3
  ..$ timestamp        : chr "2017-04-21T17:15:10.911Z"
  ..$ n001             : num -999
  ..$ n002             : num -999
  ..$ gender           : num 1
  ..$ age              : num 18

 $ 13455:List of 30
  ..$ subjectId        : num 188
  ..$ procedureId      : num 3
  ..$ procedureSampleId: num 3
  ..$ timestamp        : chr "2017-04-21T17:15:10.913Z"
  ..$ n001             : num -999
  ..$ n002             : num -999      
  ..$ gender           : num -999
  ..$ age              : num 28

这是我正在寻找的结构

 $ ID               : chr  '13455' '13455'
 $ subjectId        : num 187 188
 $ procedureId:     : num  3 3

以此类推

我试图通过以下方式实现这一目标:

I've tried to achieve this by:

  myList2 <- sapply(names(myList), function(y){
    y <- unlist(c('ID' = y, myList[[y]]), use.names = TRUE)
  })

但是我最终得到了我所需要的全部转置结果.我可以去t(myList2)，但是我想了解如何正确执行此操作.谢谢！

But I end up with the full transposed result of what I need. I could go t(myList2) but I want to understand how to do this correctly. Thank you!

可复制的数据:

myList <- list('13454' = list('subjectId' = 187, 'procedureId' = 3, 'procedureSampleId' = 3, 'timestamp' = "2017-04-21T17:15:10.911Z", 'n001' = -999, 'n002' = -999, 'gender' = 1, 'age' = 18), '13455' = list('subjectId' = 188, 'procedureId' = 3, 'procedureSampleId' = 3, 'timestamp' = "2017-04-21T17:15:10.913Z", 'n001' = -999, 'n002' = -999, 'gender' = -999, 'age' = 28))

可以使用data.table包中的lapply()和rbindlist()将

推荐答案

myList转换为data.frame:

myList can be turned into a data.frame using lapply() and rbindlist() from the data.table package:

result <- data.table::rbindlist(lapply(myList, as.data.frame), idcol = "ID")
result[["ID"]] <- names(myList)
result
#      ID subjectId procedureId procedureSampleId                timestamp n001 n002 gender age
#1: 13454       187           3                 3 2017-04-21T17:15:10.911Z -999 -999      1  18
#2: 13455       188           3                 3 2017-04-21T17:15:10.913Z -999 -999   -999  28

编辑:可以更加简化:

library(data.table)
rbindlist(myList, idcol = "ID")[, ID := names(myList)][]

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