Use grep to delete any string containing year less than 2014 - r

Question

Edited to add more context and data 5/12/2017
Using R version 3 on Windows
I have a data frame data2:
'data.frame': 1504 obs. of 14 variables:
$ Member.Name : chr "A" "B" "C"...
$ MSTATUS : Factor w/ 14 levels "","ACTIVE","ACTIVE;CHANGEDROLES;NONQUALIF",..: 13 2 2 2 2 4 13 13 2 13 ...
$ MCAT : Factor w/ 9 levels "","EDNEWCLASS",..: 5 4 9 6 6 6 9 9 4 4 ...
$ SALUTATION : Factor w/ 822 levels "","Aaron","Abigail",..: 285 2 2 2 4 4 4 4 5 5 ...
$ MEM_SUBCATEGORY : Factor w/ 22 levels "","AGENCYCEO",..: 22 6 8 15 8 6 8 1 6 6 ...
$ MEM_SUBTYPE : Factor w/ 25 levels "","AGENCY","AGENCYCEO",..: 24 6 6 20 6 6 6 6 6 6 ...
$ COUNTRY : Factor w/ 33 levels "","AE","AT","AU",..: 33 33 33 33 7 33 33 33 33 33 ...
$ F500 : Factor w/ 243 levels "","#1406 on Forbes Global 2000 ($11B)",..: 1 1 96 1 242 1 147 1 1 76 ...
$ OPT_LINE : Factor w/ 1467 levels "","(Formerly) Condé Nast",..: 1 1170 609 1333 251 1427 444 258 814 1207 ...
$ FLAGS : chr "2014PAGEJAMPARTICIPANT, \nPHOTO" "" "PUFOUNDINGMEMBER" "2014FLESPEAKER" ...
$ FLAGS_DESCR : chr "2014 Page Jam Participant, \nPhoto on File" "" "Page Up Founding Member" "2014 Future Leaders Experience Speaker" ...
$ Enroll.Date : Date, format: "2012-12-04" "2010-08-24" "2013-09-20" "2013-05-06" ...
$ Expiration.Date : Date, format: "2014-12-31" "2017-12-31" "2017-12-31" "2017-12-31" ...
$ Sponsorship.Amount: num 0 0 0 0 0 0 0 0 0 0 ...
For the FLAGS variable, I'd like to remove all row elements that contain a year less than 2014.
head(data2$FLAGS, n=3)
[1] "2011PRESIDENTS, \n2012CHAIRMANSCOUNCIL, \n2016CHAIRCOUNCIL" ""
[3] "2012COI"
So that FLAGS will look like:
head(data2$FLAGS, n=3)
[1] "\n2016CHAIRCOUNCIL" ""
[3] ""
The rows with no values can either be blank or NA, BUT if a row does contain an event with a year >=2014 and an event with a year <2014 than just delete the event less than 2014 and keep the other events in the row.

Answer 1

This regex works for your example. The idea is to match the first 3 characters of year for those elements that fail and drop them.
FLAGS[-grep("20(0|1[0123])", FLAGS)]
[1] "2014PAGEJAMPARTICIPANT, \nPHOTO" "\n2014PAGEJAMPARTICIPANT" "\n2014PUSPONSOR, \nPHOTO"
or, using invert, you'd have
FLAGS[grep("20(0|1[0123])", FLAGS, invert=TRUE)]
Note that it won't catch pre-2000s and you should be cautious if there are other "numeric" values in the vector.
To return a vector of the same length, with NAs replacing the earlier years, you could use is.na<- and grepl like this
is.na(FLAGS) <- grepl("20(0|1[0123])", FLAGS)
original data
FLAGS<-c("2014PAGEJAMPARTICIPANT, \nPHOTO", "2001ANNUALCONFERENCECOMM",
"\n2011GOVERNANCE", "\n2014PAGEJAMPARTICIPANT", "2013NEWMEMBERNOMINATOR",
"\n2014PUSPONSOR, \nPHOTO")
given OP's second question. The following more or less works:
sapply(strsplit(FLAGS, ","),
function(x) paste(gsub("(\\n)?20(0|1[0123]).*?(, |$)", "", trimws(x)), collapse=" "))
[1] " 2016CHAIRCOUNCIL" "" ""
Note that a "\n" is missing at the beginning and there is an additional (set of) space(s) at the beginning of the first element. The "\n" is removed be trimws. This makes the string a bit easier to work with. The additional spaces can be removed by wrapping the above expression in trimws, for example, trimws(sapply(strsplit(...))).
additional data
FLAGS <- c("2011PRESIDENTS, \n2012CHAIRMANSCOUNCIL, \n2016CHAIRCOUNCIL", "", "2012COI")

Answer 2

Here is one solution using stringr package:
library(stringr)
FLAGS[sapply(str_extract_all(FLAGS, '[0-9]{4}'),
function(x) !any(as.integer(x) < 2014))]
This solution assumes you may have more than one year in each value. If that is not the case, you can do something more simple like:
FLAGS[as.integer(str_extract(FLAGS, '[0-9]{4}')) >= 2014]
Assuming FLAGS is as follows:
FLAGS
[1] "2014PAGEJAMPARTICIPANT, \nPHOTO" "2001ANNUALCONFERENCECOMM"
[3] "\n2011GOVERNANCE" "\n2014PAGEJAMPARTICIPANT"
[5] "2013NEWMEMBERNOMINATOR" "\n2014PUSPONSOR, \nPHOTO"
You get result as:
[1] "2014PAGEJAMPARTICIPANT, \nPHOTO" "\n2014PAGEJAMPARTICIPANT"
[3] "\n2014PUSPONSOR, \nPHOTO"
EDITING ANSWER BASED ON QUESTION EDIT ABOVE
You can keep only values with 2014 or above and fill with NAs otherwise as follows:
data2$FLAGS <- ifelse(as.integer(str_extract(data2$FLAGS, '\\d+')) >= 2014,
data2$FLAGS, NA)
Result is as follows:
[1] "2014PAGEJAMPARTICIPANT, \nPHOTO" NA
[3] NA "\n2014PAGEJAMPARTICIPANT"
[5] NA "\n2014PUSPONSOR, \nPHOTO"