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.我想得到一个(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架

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

我已经分开了类别和内容.

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

如何进行获取mydf？

Answer 1

使用原始列表而不是您创建的拆分对象,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,你会对"purrr"也有一个transpose功能这一事实感到恼火,这意味着如果你也加载了"data.table",你将不得不习惯使用类似的东西,data.table::transpose或者purrr::transpose如果你正在使用那些功能(就像我在原始答案中所做的那样).我没有测试过,但我的猜测是"data.table"仍然是原始列表中最快的.

Answer 2

我们还可以rep结合OP的帖子中已经创建的变量.

dat <- data.frame(cat=rep(cate, lengths(cont)),
                  cont=unlist(cont))

因为有一些关于什么是"最佳"答案的讨论(如果有一个,我怀疑),这里有一些基准(如果性能很重要),基于要处理的100000个向量的列表:

Unit: milliseconds
   expr       min        lq     mean    median       uq      max neval cld
 heroka  56.24516  67.98583 122.1209  82.35606 117.6017 391.8297    50  a 
  akrun 258.86939 283.10408 363.5425 331.50263 448.9134 578.1818    50   b
 ananda  47.72320  61.05269 132.2678  76.22913 218.8286 385.5709    50  a 

基准测试代码假设变量cate和cont已创建的,因为这两个解决方案中使用它们.

heroka <- function(){
 data.frame(cat=rep(cate, lengths(cont)), cont=unlist(cont))
}

akrun <- function(){
  setNames(stack(setNames(cont, cate))[2:1], c('cate', 'cont'))
}

ananda <- function(){
  setorder(melt(as.data.table(transpose(mylist)), 
                id.vars = "V1", na.rm = TRUE), V1, variable)[]
}


mylist <- replicate(100000,c(sample(LETTERS[1:10],1),sample(LETTERS[1:10],sample(5))))
cate <- sapply(mylist, "[[",1)
cont <- sapply(mylist, "[", -(1))

tests <- microbenchmark(
  heroka = heroka(),
  akrun=akrun(),ananda=ananda(),
  times=50
)

Answer 3

我们可以stack在list用'cape' 命名'cont' 的元素之后使用.

setNames(stack(setNames(cont, cate))[2:1], c('cate', 'cont'))
#  cate    cont
#1    A   lorem
#2    A   ipsum
#3    B     sed
#4    B eiusmod
#5    B  tempor
#6    B    inci
#7    C    aliq
#8    C      ex
#9    C      ea