Removing dates and all junks from texts using R - r

Question

I am cleaning a huge dataset made up of tens of thousands of texts using R. I know regular expression will do the job conveniently but I am poor in using it. I have combed stackoverflow but could not find solution. This is my dummy data:
foo_data <- c("03 / 05 / 2016 Education is good: WO0001982",
"04/02/2016 Health is a priority: WAI000553",
"09/ 08/2016 Economy is bad: 2031CE8D",
": : 21 / 05 / 13: Vehicle license is needed: DPH2790 ")
I want to remove all the dates, punctuations and IDs and want my result to be this:
[1] "Education is good"
[2] "Health is a priority"
[3] "Economy is bad"
[4] "Vehicle license is needed"
Any help in R will be appreciated.

Answer 1

I think specificity is in order here:
First, let's remove the date-like strings. I'll assume either mm/dd/yyyy or dd/mm/yyyy, where the first two can be 1-2 digits, and the third is always 4 digits. If this is variable, the regex can be changed to be a little more permissive:
foo_data2 <- gsub("\\d{1,2}\\s*/\\s*\\d{1,2}\\s*/\\s*\\d{4}", "", foo_data)
foo_data2
# [1] " Education is good: WO0001982" " Health is a priority: WO0002021" " Economy is bad: WO001999" " Vehicle license is needed: WO001050"
From here, the abbreviations seem rather easy to remove, as the other answers have demonstrated. You have not specified if the abbreviation is hard-coded to be anything after a colon, numbers prepended with "WO", or just some one-word combination of letters and numbers. Those could be:
gsub(":.*", "", foo_data2)
# [1] " Education is good" " Health is a priority" " Economy is bad" " Vehicle license is needed"
gsub("\\bWO\\S*", "", foo_data2)
# [1] " Education is good: " " Health is a priority: " " Economy is bad: " " Vehicle license is needed: "
gsub("\\b[A-Za-z]+\\d+\\b", "", foo_data2)
# [1] " Education is good: " " Health is a priority: " " Economy is bad: " " Vehicle license is needed: "
The : removal should be straight forward, and using trimws(.) will remove the leading/trailing spaces.
This can obviously be combined into a single regex (using the logical | with pattern grouping) or a single R call (nested gsub) without complication, I kept them broken apart for discussion.
I think https://stackoverflow.com/a/22944075/3358272 is a good reference for regex in general, note that while that page shows many regex things with single-backslashes, R requires all of those use double-backslashes (e.g., \d in regex needs to be \\d in R). The exception to this is if you use R-4's new raw-strings, where these two are identical:
"\\b[A-Za-z]+\\d+\\b"
r"(\b[A-Za-z]+\d+\b)"

Answer 2

Using stringr try this:
library(stringr)
library(magrittr)
str_remove_all(foo_data, "\\/|\\d+|\\: WO") %>%
str_squish()
#> [1] "Education is good" "Health is a priority"
#> [3] "Economy is bad" "Vehicle license is needed"
Created on 2021-04-22 by the reprex package (v2.0.0)
data
foo_data <- c("03 / 05 / 2016 Education is good: WO0001982", "04/02/2016 Health is a priority: WO0002021",
"09/ 08/2016 Economy is bad: WO001999", "09/08/ 2016 Vehicle license is needed: WO001050")

Answer 3

foo_data <- c("03 / 05 / 2016 Education is good: WO0001982", "04/02/2016 Health is a priority: WO0002021",
"09/ 08/2016 Economy is bad: WO001999", "09/08/ 2016 Vehicle license is needed: WO001050")
gsub(".*\\d{4}[[:space:]]+(.*):.*", "\\1", foo_data)
#> [1] "Education is good" "Health is a priority"
#> [3] "Economy is bad" "Vehicle license is needed"
Created on 2021-04-22 by the reprex package (v2.0.0)