r - 从列表中获取长格式数据框

Question

我有一个包含字符串的列表列表。每个子列表的第一个字符串描述了以下字符串所属的类别。我想获得一个(长格式)数据框，其中一列用于类别，一列用于内容。
如何从此列表中获取长格式的数据框:

mylist <- list(
  c("A","lorem","ipsum"),
  c("B","sed", "eiusmod", "tempor" ,"inci"),
  c("C","aliq", "ex", "ea"))

> mylist
[[1]]
[1] "A"     "lorem" "ipsum"

[[2]]
[1] "B"        "sed"      "eiusmod"  "tempor"   "incidunt"

[[3]]
[1] "C"       "aliquid" "ex"      "ea" 

它应该看起来像这个数据框

mydf <- data.frame(cate= c("A","A","B","B","B","B","C","C","C"),
               cont= c("lorem","ipsum","sed", "eiusmod", "tempor","inci","aliq", "ex", "ea"))

> mydf
  cate    cont
1   A    lorem
2   A    ipsum
3   B      sed
4   B  eiusmod
5   B   tempor
6   B incidunt
7   C  aliquid
8   C       ex
9   C       ea

我已经把类别和内容分开了。

cate <- sapply(mylist, "[[",1)
cont <- sapply(mylist, "[", -(1))

如何继续获得mydf？

Answer 1

使用您的原始列表而不是您创建的拆分对象，您可以尝试以下操作:

library(data.table)
setorder(melt(as.data.table(transpose(mylist)), 
              id.vars = "V1", na.rm = TRUE), V1, variable)[]
#    V1 variable   value
# 1:  A       V2   lorem
# 2:  A       V3   ipsum
# 3:  B       V2     sed
# 4:  B       V3 eiusmod
# 5:  B       V4  tempor
# 6:  B       V5    inci
# 7:  C       V2    aliq
# 8:  C       V3      ex
# 9:  C       V4      ea

为了好玩，您还可以尝试以下方法之一:

library(dplyr)
library(tidyr)

data_frame(id = seq_along(mylist), mylist) %>%
  unnest %>%
  group_by(id) %>%
  mutate(ind = mylist[1]) %>%
  slice(2:n())

library(purrr)
data_frame(
  value = mylist %>% map(~ .x[-1]) %>% unlist,
  ind = mylist %>% map(~ rep(.x[1], length(.x)-1)) %>% unlist
)

请注意，“purrr”也有 transpose 的事实会让您感到恼火。函数，这意味着如果你也加载了“data.table”，你将不得不养成使用 data.table::transpose 之类的习惯。或 purrr::transpose如果您正在使用这些功能(就像我在原始答案中所做的那样)。我还没有测试过，但我的猜测是“data.table”仍然是原始列表中最快的。