基于列表结构模式创建新列表

Question

我有一些数据如下所示：

   dat <- c("Sales","Jim","Halpert","","",
            "Reception","Pam","Beasley","","",
            "Not.Manager","Dwight","Schrute","Bears","Beets","BattlestarGalactica","","",
            "Manager","Michael","Scott","","")

每个数据“块”都是连续的，中间有一些空白。我想将数据转换成如下所示的列表列表：

iwant <- c(
           c("Sales","Jim","Halpert"),
           c("Reception","Pam","Beasley"),
           c("Not.Manager","Dwight","Schrute","Bears","Beets","BattlestarGalactica"),
           c("Manager","Michael","Scott")
           )

建议？我正在使用 rvest 和 stringi。我不想添加更多包。

Answer 1

我建议采用下一种方法。您最终会得到一个数据框，其中的变量格式与您想要的格式类似：

#Split chains
L1 <- strsplit(paste0(dat,collapse = " "),split = "  ")
#Split vectors from each chain
L2 <- lapply(L1[[1]],function(x) strsplit(trimws(x),split = " "))
#Format
L2 <- lapply(L2,as.data.frame)
#Remove zero dim data
L2[which(lapply(L2,nrow)==0)]<-NULL
#Format names
L2 <- lapply(L2,function(x) {names(x)<-'v';return(x)})
#Transform to dataframe
D1 <- as.data.frame(do.call(cbind,L2))
#Rename
names(D1) <- paste0('V',1:dim(D1)[2])
#Remove recycled values
D1 <- as.data.frame(apply(D1,2,function(x) {x[duplicated(x)]<-NA;return(x)}))

输出：

       V1        V2                  V3      V4
1   Sales Reception         Not.Manager Manager
2     Jim       Pam              Dwight Michael
3 Halpert   Beasley             Schrute   Scott
4    <NA>      <NA>               Bears    <NA>
5    <NA>      <NA>               Beets    <NA>
6    <NA>      <NA> BattlestarGalactica    <NA>

Answer 2

您可以将 rle、split 与 lapply 一起使用：

lapply(split(dat, with(rle(dat != ''), 
             rep(cumsum(values), lengths))), function(x) x[x!= ''])

#$`1`
#[1] "Sales"   "Jim"     "Halpert"

#$`2`
#[1] "Reception" "Pam"       "Beasley"  

#$`3`
#[1] "Not.Manager"         "Dwight"    "Schrute"     "Bears"   "Beets"            
#[6] "BattlestarGalactica"

#$`4`
#[1] "Manager" "Michael" "Scott"

rle 部分在 :

上创建组到 split

with(rle(dat != ''), rep(cumsum(values), lengths))
#[1] 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 3 3 3 4 4 4 4 4

在 split 之后，我们使用 lapply 从每个列表中删除所有空元素。

基于列表结构模式创建新列表

Create New Lists Based on List Structure Pattern

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stringi