How to remove up to a certain punctuation when there are same punctuations

Asked Mar 20 '19 at 10:15

Active Mar 20 '19 at 10:15

Viewed 31 times

I have a dataframe with 1 column called filepath which contains file path:

filepath
C:\Windows\App\app1.exe
C:\Windows\App\folder1\app1.exe

I need to split up to the last backslash, and the exe file will be in another column. How should I split the directory path?

Below is the snapshot of the result I need to obtain.

I have tried regex expressions to obtain filepath1, eg:

df$filepath1 <- gsub("\\\\.*","",df$filepath)

but it doesnt work, because it will remove everything after the first match, and returning C: .

asked Mar 20 '19 at 10:15

R_abcdefg

1

You need `gsub("[^\\\\]+$", "", x)` – Wiktor Stribiżew Mar 20 '19 at 10:19
I guess you need the regex `(.*\\)[^\\]*$` and have to use the first matching group! :) – csabinho Mar 20 '19 at 10:21
@WiktorStribiżew Thanks! it works. :) – R_abcdefg Mar 20 '19 at 10:23
@csabinho, yours didnt work tho. But still thanks! – R_abcdefg Mar 20 '19 at 10:24
see ?`basename` and `dirname` – Richard Telford Mar 20 '19 at 10:25

0 Answers0