Question

[[29]]
                type              lineNum                  cat                  sel 
                 "I"                 "25"                "DOR"               "MN-A" 
                 act                 desc          descChanged                 calc 
                 "+" "Door labor minimum"                  "1"                  "1" 
                 qty                 unit         coverageName              replace 
                 "1"                 "EA"           "Dwelling"              "58.58" 
               total                  acv          recoverable             acvTotal 
             "58.58"              "58.58"                  "1"              "58.58" 
            rcvTotal   isPartOfInitSettle           laborTotal            laborBase 
             "58.58"                  "0"              "58.58"              "58.58" 

[[30]]
                  type                lineNum                    cat                    sel 
                   "I"                   "26"                  "WDW"                 "MN-A" 
                   act                   desc            descChanged                   calc 
                   "+" "Window labor minimum"                    "1"                    "1" 
                   qty                   unit           coverageName                replace 
                   "1"                   "EA"             "Dwelling"                 "69.5" 
                 total                    acv            recoverable               acvTotal 
                "69.5"                 "69.5"                    "1"                 "69.5" 
              rcvTotal     isPartOfInitSettle                          laborBase 
                "69.5"                    "0"                                 "69.5"

这是我的数据样本。列数不相同。我正在尝试从中创建一个数据框架。我本质上想要两行，并按列名进行匹配。

我尝试使用plyr，但由于列数不相同，因此无法使用。

有什么建议吗？

Answer 1

我们可以使用map中的purrr

library(dplyr)
library(purrr)
map_df(lst, as.list)
# A tibble: 3 x 5
#      A     B     C     D     E
#  <int> <int> <int> <int> <int>
#1     1     2     3    NA    NA
#2     1     2    NA    NA    NA
#3     1     2     3     4     5

数据

lst <- list(setNames(1:3, LETTERS[1:3]), setNames(1:2, LETTERS[1:2]), 
            setNames(1:5, LETTERS[1:5]))

Answer 2

我刚发现

ID

完成工作

从列表到保留列名的数据框

2 个答案:

数据