I am attempting to clean some data in R Studio.
Here's an example of my data.
LSOA name:
York 009A
Wychavon 014A
Bath and North East Somerset 001A
Aylesbury Vale 008C
Central Bedfordshire 030C
I want to be able to remove the code from the end of each. So that the resulting data looks like this:
LSOA name:
York
Wychavon
Bath and North East Somerset
Aylesbury Vale
Central Bedfordshire
I am quite new to regex so finding this quite difficult. From what I can tell, as there is a variable number of words before the code, a simple remove characters after a whitespace is not possible.
Any help would be hugely appreciated!
We can use sub
to match one or more spaces followed by one or more digits ( \\d+
) and an upper case letter ( [AZ]
) at the end ( $
) of the string and replace it with blank ( ""
)
df1$name <- sub("\\s+\\d+[A-Z]$", "", df1$name)
-output
df1
# name
#1 York
#2 Wychavon
#3 Bath and North East Somerset
#4 Aylesbury Vale
#5 Central Bedfordshire
df1 <- structure(list(name = c("York 009A", "Wychavon 014A",
"Bath and North East Somerset 001A",
"Aylesbury Vale 008C", "Central Bedfordshire 030C")), class = "data.frame",
row.names = c(NA,
-5L))
You can also use lookahead (?=\\s\\d+)
and backreference \\1
:
sub("(.*)(?=\\s\\d+).*", "\\1", df1$name, perl = T)
[1] "York" "Wychavon" "Bath and North East Somerset" "Aylesbury Vale"
[5] "Central Bedfordshire"
Another option is str_extract
and the nagative character class \\D
, which matches any char that is not a digit ( trimws
removes the whitespace).
library(stringr)
trimws(str_extract(df1$name, "\\D+"))
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.