I want to write a regex in R to remove all words of a string containing numbers

We could use gsub to match letters, digits, from a word boundary (\\b) to the next, and replace with blank ("")

gsub("\\s{2,}", " ", gsub("\\b[A-Z/0-9.-]+\\b", "", x))
#[1] "Saint Lucia"

Or using str_extract

library(stringr)
str_c(str_extract_all(x, "(?<= |^)[[:alpha:]]+(?= |$)")[[1]], collapse = " ")
#[1] "Saint Lucia"

You can use a replacing approach:

x<-"Saint  A/74/PV.46 12/12/2019 4/66 19-40538 Lucia"
gsub("\\s*(?<!\\S)(?!\\p{L}+(?!\\S))\\S+", "", x, perl=TRUE)
## => [1] "Saint Lucia"
library(stringr)
str_replace_all(x, "\\s*(?<!\\S)(?!\\p{L}+(?!\\S))\\S+", "")
## => [1] "Saint Lucia"

See the R demo. See the regex demo. Details:

• \s* - zero or more whitespaces
• (?<!\S) - start of string or a position immediately preceded with a whitespace
• (?!\p{L}+(?!\S)) - the next non-whitespace chunk cannot be a letter only word
• \S+ - one or more non-whitespace chars.

Or, you may match all letter only words in between whitespace boundaries and join the matches with a space:

paste(unlist(regmatches(x, gregexpr("(?<!\\S)\\p{L}+(?!\\S)", x, perl=TRUE))), collapse=" ")

See the R demo online. Also, see the regex demo, it matches

• (?<!\S) - a position at the start of string or right after a whitespace
• \p{L}+ - one or more Unicode letters
• (?!\S) - immediately on the right, there must be a whitespace or end of string.

You could use gsub to replace the characters starting from the first space(" ") to the last space with a single space.

x <- "Saint  A/74/PV.46 12/12/2019 4/66 19-40538 Lucia"
gsub(" .+ ", " ", x)
[1] "Saint Lucia"