使用R分隔数据框中的箭头分隔值以分隔不相等的列？答案

【问题标题】：Separating arrow separated values in data frame to separate unequal columns using R?使用R分隔数据框中的箭头分隔值以分隔不相等的列？
【发布时间】：2014-10-15 06:20:56
【问题描述】：

我有一个包含以下示例值的数据框。

[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“->”将它们分开，将它们放在不同的列中，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。

【问题讨论】：

看看我更新的答案。 read.csv 更容易

标签： r csv strsplit

【解决方案1】：

最简单的方法是使用gsub 将-> 替换为逗号，然后使用read.csv。如果数据中有逗号，那么只需使用> 而不是逗号就可以了。

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)

您仍然可以使用rbind 和strsplit，只要首先使所有列表元素的长度相同。 length<- 替换功能可以帮助解决这个问题。

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

原始x 向量在哪里

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")

【讨论】：