Question

我在R中有一个列如下：

Path Column
ag.1.4->ao.5.5->iv.9.12->ag.4.35
ao.11.234->iv.345.455.1.2->ag.9.531

我想将其转换为：

Path Column
ag->ao->iv->ag
ao->iv->ag

我该怎么做？

谢谢

以下是我的数据的全部输入：

structure(list(Rank = c(10394749L, 36749879L), Count = c(1L, 
1L), Percent = c(0.001011122, 0.001011122), Path = c("ao.legacy payment.not_completed->ao.legacy payment.not_completed->ao.legacy payment.completed", 
"ao.legacy payment.not_completed->agent.payment.completed")), .Names = c("Rank", 
"Count", "Percent", "Path"), class = "data.frame", row.names = c(NA, 
-2L))

Answer 1

您可以使用gsub来匹配.和.（\\.[0-9]+）后的数字，并将其替换为''。

 df1$Path.Column <- gsub('\\.[0-9]+', '', df1$Path.Column)
 df1
 #           Path.Column
 #1 ag -> ao -> iv -> ag
 #2       ao -> iv -> ag

更新

对于新数据集df2

gsub('\\.[^->]+(?=(->|\\b))', '', df2$Path, perl=TRUE)
#[1] "ao->ao->ao" "ao->agent"

和OP的帖子中显示的字符串

str2 <- c('ag.1.4->ao.5.5->iv.9.12->ag.4.35',
    'ao.11.234->iv.345.455.1.2->ag.9.531')

gsub('\\.[^->]+(?=(->|\\b))', '', str2, perl=TRUE)
 #[1] "ag->ao->iv->ag" "ao->iv->ag"

数据

df1 <- structure(list(Path.Column = c("ag.1 -> ao.5 -> iv.9 -> ag.4", 
"ao.11 -> iv.345 -> ag.9")), .Names = "Path.Column", 
class = "data.frame", row.names = c(NA, -2L))

df2  <- structure(list(Rank = c(10394749L, 36749879L), Count = c(1L, 
1L), Percent = c(0.001011122, 0.001011122), 
Path = c("ao.legacy payment.not_completed->ao.legacy payment.not_completed->ao.legacy payment.completed", 
"ao.legacy payment.not_completed->agent.payment.completed")), 
.Names = c("Rank", "Count", "Percent", "Path"), class = "data.frame", 
row.names = c(NA, -2L))

Answer 2

在'->'上拆分字符串并单独处理子字符串可能更容易

 # split the stirngs into parts
 subStrings <- strsplit(df$Path,'->')
 # remove eveything after **first** the dot
 subStrings<- lapply(subStrings,
                     function(x)gsub('\\..*','',x))
 # paste them back together.
 sapply(subStrings,paste0,collapse="->")
 #> "ao->ao->ao" "ao->agent"

或

 # split the stirngs into parts
 subStrings <- strsplit(df$Path,'->')
 # remove the parts of the identifiers after the dot
 subStrings<- lapply(subStrings,
                     function(x)gsub('\\.[^ \t]*','',x))
 # paste them back together.
 sapply(subStrings,paste0,collapse="->")
 #> "ao payment->ao payment->ao payment" "ao payment->agent"

R中的文字清理

2 个答案:

更新

数据