I have a data wherein one of the variable has a non-uniform pattern/format and I need to write a code in R which can remove that part of the string in the variable which follows a specific pattern.

There are links on replacement of patterns such as Extract a string between patterns/delimiters in R, Replace patterns separated by delimiter in R, and Remove part of a string but they haven't discussed the issue related to my data.

This is how the variable (c) looks like and below are the options I tried along with their results.

c <-  c("1998/123; 2001","181;2002/12","212")
c1 <- gsub("[0-9]/[0-9]", "", c) # returns 19923;2001, 181;2002, 212
c2 <- gsub("[0-9]/*", "", c) # returns ";",  " ", ";", ""
c3 <- gsub("[0-9][0-9][0-9][0-9]/", "", c) # returns 123;2001, 181;12, 212 
c4 <- gsub("*[0-9]/[0-9]*", "", c) # returns 199, 200, 212 
c5 <- gsub(" */* ", "", c) # no change
c6 <- str_replace_all(c,"/","") # returns 1998123, 200212, 212
c7 <- grep(fixed("/"), c, invert=TRUE, value = TRUE) # returns 212

a) There can be 3-8 digits after the forward slash. But there can only be 4 digits before the forward slash.

b) Each sub-string is separated by a semicolon delimiter.

c) I want to replace those substrings that contain the forward slash with blank. So, my result should be c(";2001", "181;" ,"212").

Kindly let me know where am I making the mistake. Any suggestions are very much welcome. Thanks.

CodePudding user response：

As the numbers before and after the forward slash have multiple digits you could use (1 or more) or * (0 and more) in your first approach to remove all of them:

c <-  c("1998/123; 2001","181;2002/12","212")

gsub("\\d \\/\\d ", "", c)
#> [1] "; 2001" "181;"   "212"