[英]Create New Lists Based on List Structure Pattern
我有一些看起來像這樣的數據:
dat <- c("Sales","Jim","Halpert","","",
"Reception","Pam","Beasley","","",
"Not.Manager","Dwight","Schrute","Bears","Beets","BattlestarGalactica","","",
"Manager","Michael","Scott","","")
每個“塊”數據都是連續的,中間有一些空白。 我想將數據轉換為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")
)
建議? 我正在使用 rvest 和 stringi。 我不想添加更多包。
我會建議下一種方法。 您最終將得到一個 dataframe ,其變量格式與您想要的類似:
#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)}))
Output:
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>
您可以使用lapply
rle
split
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
部分創建要split
的組:
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
split
后,我們使用lapply
從每個列表中刪除任何空元素。
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