分隔数据框中的箭头分隔值，以使用R？分隔不相等的列？

我有一个具有以下示例值的数据帧

[1] "entry.cei"                                                                               
[2] "entry.lifecycle->hist.open.personal demand chequing account->exit.lifecycle->entry.cei"  
[3] "entry.lifecycle->hist.open.personal demand savings account->exit.lifecycle->entry.cei"   
[4] "entry.transaction->txn.no source available->exit.transaction->entry.cei"                 
[5] "entry.branch->exit.branch->entry.transaction->txn.in-branch->exit.transaction->entry.cei"

我需要用“->”将它们拆分成不同的列，比如V1、V2等。 例如：

           V1                             V2               V3             V4           V5     V6    V7
1   entry.cei   
2   entry.lifecycle hist.open.personal demand chequing account  exit.lifecycle  entry.cei   
3   entry.lifecycle hist.open.personal demand savings account   exit.lifecycle  entry.cei   

如何在R中实现这一点？

---

我尝试将rbind与strsplit（）一起使用，但我认为它需要相同数量的列。

最简单的方法是使用

gsub

用逗号替换

->

，然后使用

read.csv

。如果数据中有逗号，则只需使用

而不是逗号即可

read.csv(text = gsub("->", ",", x, fixed = TRUE), header = FALSE)
#                  V1                                         V2                V3            V4               V5        V6
# 1         entry.cei                                                                                                      
# 2   entry.lifecycle hist.open.personal demand chequing account    exit.lifecycle     entry.cei                           
# 3   entry.lifecycle  hist.open.personal demand savings account    exit.lifecycle     entry.cei                           
# 4 entry.transaction                    txn.no source available  exit.transaction     entry.cei                           
# 5      entry.branch                                exit.branch entry.transaction txn.in-branch exit.transaction entry.cei

或者

read.table(text = gsub("->", ",", x, fixed = TRUE), sep = ",", fill = TRUE)

---

您还可以使用

rbind

和

strsplit

，只要您首先使所有列表元素的长度相同。

长度看看我的最新答案<代码>读取.csv

更容易

s <- strsplit(x, "->", fixed = TRUE)
data.frame(do.call(rbind, lapply(s, `length<-`, max(sapply(s, length)))))
#                  X1                                         X2                X3            X4               X5        X6
# 1         entry.cei                                       <NA>              <NA>          <NA>             <NA>      <NA>
# 2   entry.lifecycle hist.open.personal demand chequing account    exit.lifecycle     entry.cei             <NA>      <NA>
# 3   entry.lifecycle  hist.open.personal demand savings account    exit.lifecycle     entry.cei             <NA>      <NA>
# 4 entry.transaction                    txn.no source available  exit.transaction     entry.cei             <NA>      <NA>
# 5      entry.branch                                exit.branch entry.transaction txn.in-branch exit.transaction entry.cei

x <- c("entry.cei", 
 "entry.lifecycle->hist.open.personal demand chequing account->exit.lifecycle->entry.cei", 
 "entry.lifecycle->hist.open.personal demand savings account->exit.lifecycle->entry.cei", 
 "entry.transaction->txn.no source available->exit.transaction->entry.cei", 
 "entry.branch->exit.branch->entry.transaction->txn.in-branch->exit.transaction->entry.cei")