substring characters from a column in a data.table in R - regex

Question

Is there a more "r" way to substring two meaningful characters out of a longer string from a column in a data.table?
I have a data.table that has a column with "degree strings"... shorthand code for the degree someone got and the year they graduated.
> srcDT<- data.table(
alum=c("Paul Lennon","Stevadora Nicks","Fred Murcury"),
degree=c("W72","WG95","W88")
)
> srcDT
alum degree
1: Paul Lennon W72
2: Stevadora Nicks WG95
3: Fred Murcury W88
I need to extract the digits of the year from the degree, and put it in a new column called "degree_year"
No problem:
> srcDT[,degree_year:=substr(degree,nchar(degree)-1,nchar(degree))]
> srcDT
alum degree degree_year
1: Paul Lennon W72 72
2: Stevadora Nicks WG95 95
3: Fred Murcury W88 88
If only it were always that simple.
The problem is, the degree strings only sometimes look like the above. More often, they look like this:
srcDT<- data.table(
alum=c("Ringo Harrison","Brian Wilson","Mike Jackson"),
degree=c("W72 C73","WG95 L95","W88 WG90")
)
I am only interested in the 2 numbers next to the characters I care about: W & WG (and if both W and WG are there, I only care about WG)
Here's how I solved it:
x <-srcDT$degree ##grab just the degree column
z <-character() ## create an empty character vector
degree.grep.pattern <-c("WG[0-9][0-9]","W[0-9][0-9]")
## define a vector of regex's, in the order
## I want them
for(i in 1:length(x)){ ## loop thru all elements in degree column
matched=F ## at the start of the loop, reset flag to F
for(j in 1:length(degree.grep.pattern)){
## loop thru all elements of the pattern vector
if(length(grep(degree.grep.pattern[j],x[i]))>0){
## see if you get a match
m <- regexpr(degree.grep.pattern[j],x[i])
## if you do, great! grab the index of the match
y<-regmatches(x[i],m) ## then subset down. y will equal "WG95"
matched=T ## set the flag to T
break ## stop looping
}
## if no match, go on to next element in pattern vector
}
if(matched){ ## after finishing the loop, check if you got a match
yr <- substr(y,nchar(y)-1,nchar(y))
## if yes, then grab the last 2 characters of it
}else{
#if you run thru the whole list and don't match any pattern at all, just
# take the last two characters from the affilitation
yr <- substr(x[i],nchar(as.character(x[i]))-1,nchar(as.character(x[i])))
}
z<-c(z,yr) ## add this result (95) to the character vector
}
srcDT$degree_year<-z ## set the column to the results.
> srcDT
alum degree degree_year
1: Ringo Harrison W72 C73 72
2: Brian Wilson WG95 L95 95
3: Mike Jackson W88 WG90 90
This works. 100% of the time. No errors, no mis-matches.
The problem is: it doesn't scale. Given a data table with 10k rows, or 100k rows, it really slows down.
Is there a smarter, better way to do this? This solution is very "C" to me. Not very "R."
Thoughts on improvement?
Note: I gave a simplified example. In the actual data, there are about 30 different possible combinations of degrees, and combined with different years, there are something like 540 unique combinations of degree strings.
Also, I gave the degree.grep.pattern with only 2 patterns to match. In the actual work I'm doing, there are 7 or 8 patterns to match.

Answer 1

As it seem (per OPs) comments, there is no situation of "WG W", then a simple regex solution should do the job
srcDT[ , degree_year := gsub(".*WG?(\\d+).*", "\\1", degree)]
srcDT
# alum degree degree_year
# 1: Ringo Harrison W72 C73 72
# 2: Brian Wilson WG95 L95 95
# 3: Mike Jackson W88 WG90 90

Answer 2

Here's a solution based on the assumption that want the most recent degree with W in it:
regex <- "(?<=W|(?<=W)G)[0-9]{2}"
srcDT[ , degree_year :=
sapply(regmatches(degree,
gregexpr(regex, degree, perl = TRUE)),
function(x) max(as.integer(x)))]
> srcDT
alum degree degree_year
1: Ringo Harrison W72 C73 72
2: Brian Wilson WG95 L95 95
3: Mike Jackson W88 WG90 90
You said:
I gave the degree.grep.pattern with only 2 patterns to match. In the actual work I'm doing, there are 7 or 8 patterns to match.
But I'm not sure what this means. There are more options besides W and WG?

Answer 3

Here is one quick hack:
# split all words from degree and order so that WG is before W
words <- lapply(strsplit(srcDT$degree, " "), sort, decreasing=TRUE)
# obtain tags for each row (getting only first. But works since ordered)
tags <- mapply(Find, list(function(x) grepl("^WG|^W", x)), words)
# simple gsub to remove WG and W
(result <- gsub("^WG|^W", "", tags))
[1] "72" "95" "90"
Fast with 100k rows.

Answer 4

A solution without regular expressions, it's quite slow as it creates a sparse table... but it's clean and flexible so i leave it here.
First I split the degreeyears by space, then browse through them and build a clean structured table with one column per degree, that i fill it with years.
degreeyear_split <- sapply(srcDT$degree,strsplit," ")
for(i in 1:nrow(srcDT)){
for (degree_year in degreeyear_split[[i]]){
n <- nchar(degree_year)
degree <- substr(degree_year,1,n-2)
year <- substr(degree_year,n-1,n)
srcDT[i,degree] <- year
}}
Here I have my structure table, I paste W on the year i'm interested in, then paste WG on top of it.
srcDT$year <- srcDT$W
srcDT$year[srcDT$WG!=""]<-srcDT$WG[srcDT$WG!=""]
Then here's you result:
srcDT
alum degree W C WG L year
1: Ringo Harrison W72 C73 72 73 72
2: Brian Wilson WG95 L95 95 95 95
3: Mike Jackson W88 WG90 88 90 90