Adding dimensions to a list by ID column

Question

As a follow-up question to Flatten list and push list key to vector on second level , I'm now looking for an efficient way to reverse the procedure earlier described. In summary, I need to restructure a flat list from this:

> str(myList)
List of 9
 $ ID               : num [1:2] 13454 13455
 $ subjectId        : num [1:2] 187 188
 $ procedureId      : num [1:2] 3 3
 $ procedureSampleId: num [1:2] 3 3
 $ timestamp        : chr [1:2] "2017-04-21T17:15:10.911Z" "2017-04-21T17:15:10.913Z"
 $ n001             : num [1:2] -999 -999
 $ n002             : num [1:2] -999 -999
 $ gender           : num [1:2] 1 -999
 $ age              : num [1:2] 18 28

to this:

$ 13454:List of 8
  ..$ 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 8
  ..$ 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

Whereas each ID column should be a list of its own. So far I've been using for-loops to accomplish this, but it turned out to be highly inefficient.

Reproducible code:

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

In your expected output, the list of 30 seems odd when there are only 8 elements — akrun, May 07 '17 at 13:09

score 2 · Answer 1 · answered May 07 '17 at 13:03

2

One option would be transpose from purrr

library(tidyverse)
res <- myList %>%
             transpose %>%
             map(~.[-1])
names(res) <- myList[[1]]

answered May 07 '17 at 13:03

akrun

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1

This works and appears to be pretty efficient! Thank you:) – Comfort Eagle May 07 '17 at 13:21

score 1 · Accepted Answer · answered May 07 '17 at 15:17

A base R alternative would be to use a nested lapply and setNames to add the ID as outer level list names.

myNestList <- setNames(lapply(1:2,
                       function(i) lapply(myList[-1], "[", i)), myList[[1]])

which results in

str(myList)
List of 2
 $ :List of 8
  ..$ 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
 $ :List of 8
  ..$ 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

Works fine, thank you! I marked this answer as solution since it's not dependant on packages — Comfort Eagle, May 07 '17 at 17:44

Adding dimensions to a list by ID column

2 Answers2