R中的文本清理

Question

我在 R 中有一列如下所示：

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

这是我的全部数据输出：

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 . 之后的数字 (\.[0-9]+) 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"

以及 OP 中显示的字符串 post

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

拆分 '->' 上的字符串并分别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中的文本清理

text cleaning in R

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r

data-cleaning

更新

数据