CODEWITHSUNDEEP
rtextdata-cleaning
blast00
blast00

Reputation: 559

text cleaning in R

I have a single column in R that looks like this:

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

I want to transform this into:

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

How can I do this?

Thank you

Here is my full dput from my data:

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))

Upvotes: 0

Views: 394

Answers (2)

akrun
akrun

Reputation: 887128

You could use gsub to match the . and numbers following the . (\\.[0-9]+) and replace it with ''.

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

Update

For the new dataset df2

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

and for the string showed in the OP's 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"

data

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))

Upvotes: 2

Jthorpe
Jthorpe

Reputation: 10167

It may be easeir to split the strings on '->' and process the substrings separately

 # 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"

or

 # 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"

Upvotes: 1

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