Question

我有data.table，其中一些观测值包含data.frame。例如：

data.table(colA=   c('A1','A2','A3'), 
                  colB=list(data.frame(), 
                            data.frame(colsubB1=c('B2a','B2b'),colsubB2=c('B2c', 'B2d')), 
                            data.frame(colsubB1=c('A3a','A3b'),colsubB2=c('A3c', 'A3d'))),
                  colC=   c('C1','C2','C3'), 
                  colD=   c('D1','D2','D3')
                  )

返回

   colA         colB colC colD
1:   A1 <data.frame>   C1   D1
2:   A2 <data.frame>   C2   D2
3:   A3 <data.frame>   C3   D3

我希望得到以下结果：

    colA colsubB1 colsubB2 colC colD
1:   A1     <NA>     <NA>   C1   D1
2:   A2      B2a      B2c   C2   D2
3:   A2      B2b      B2d   C2   D2
4:   A3      A3a      A3c   C3   D3
5:   A3      A3b      A3d   C3   D3

请给我一些建议吗？

Answer 1

使用by：

DT[, if(nrow(colB[[1L]]) > 0) 
        colB[[1L]] 
    else 
        data.frame(colsubB1=NA_character_, colsubB2=NA_character_), 
    by=setdiff(names(DT), "colB")]

输出：

   colA colC colD colsubB1 colsubB2
1:   A1   C1   D1     <NA>     <NA>
2:   A2   C2   D2      B2a      B2c
3:   A2   C2   D2      B2b      B2d
4:   A3   C3   D3      A3a      A3c
5:   A3   C3   D3      A3b      A3d

数据：

library(data.table)
DT <- data.table(colA=   c('A1','A2','A3'), 
    colB=list(data.frame(), 
        data.frame(colsubB1=c('B2a','B2b'),colsubB2=c('B2c', 'B2d')), 
        data.frame(colsubB1=c('A3a','A3b'),colsubB2=c('A3c', 'A3d'))),
    colC=   c('C1','C2','C3'), 
    colD=   c('D1','D2','D3')

Answer 2

一种方法：

DT[, {
  lens  = sapply(colB, nrow)
  empty = data.table(colB[[which.max(lens > 0)]])[NA_integer_]
  unnested = rbindlist(replace(colB, lens == 0, list(empty)))
  repped   = .SD[rep(.I, pmax(lens, 1L))]

  # figure out column positions
  s = order(c(
    match(names(repped), names(DT)), 
    rep(match("colB", names(DT)), ncol(unnested))
  ))

  setcolorder(cbind(repped, unnested), s)
}, .SDcols=!"colB"]

   colA colsubB1 colsubB2 colC colD
1:   A1     <NA>     <NA>   C1   D1
2:   A2      B2a      B2c   C2   D2
3:   A2      B2b      B2d   C2   D2
4:   A3      A3a      A3c   C3   D3
5:   A3      A3b      A3d   C3   D3

如果colB的任何元素都不包含任何内容，则所需的输出不清楚。我猜您应该手动写出empty的值，就像在这种情况下@chinsoon的答案一样。

Answer 3

您实际上可以使用tidyr::unnest()：

library(data.table)
library(tidyr)
# data edited to avoid warnings
DT <- data.table(colA=   c('A1','A2','A3'), 
           colB=list(data.frame(row.names = 1), 
                     data.frame(colsubB1=c('B2a','B2b'),colsubB2=c('B2c', 'B2d'),
                                stringsAsFactors = FALSE), 
                     data.frame(colsubB1=c('A3a','A3b'),colsubB2=c('A3c', 'A3d'),
                                stringsAsFactors = FALSE)),
           colC=   c('C1','C2','C3'), 
           colD=   c('D1','D2','D3')
)
# with current version of tidyr, unnest can remove rows, should be fixed in next
# version, but for now we need to add a row to our empty data.frames
# no idea why the which was needed
DT[which(!lengths(colB)), colB := list(list(data.frame(row.names = 1)))]
# then just unnest
DT[, unnest(.SD,colB)]
#>    colA colC colD colsubB1 colsubB2
#> 1:   A1   C1   D1     <NA>     <NA>
#> 2:   A2   C2   D2      B2a      B2c
#> 3:   A2   C2   D2      B2b      B2d
#> 4:   A3   C3   D3      A3a      A3c
#> 5:   A3   C3   D3      A3b      A3d
# or 
unnest(DT, colB)
#>    colA colC colD colsubB1 colsubB2
#> 1:   A1   C1   D1     <NA>     <NA>
#> 2:   A2   C2   D2      B2a      B2c
#> 3:   A2   C2   D2      B2b      B2d
#> 4:   A3   C3   D3      A3a      A3c
#> 5:   A3   C3   D3      A3b      A3d

^{由reprex package（v0.3.0）于2019-07-11创建}

如何操作data.table中的data.frame

3 个答案: