Looping grepl() through data.table (R)

Question

I have a dataset stored as a data.table DT that looks like this:

print(DT)
   category            industry
1: administration      admin
2: nurse         


        

Answer 1

As long as the match is always based on the start of the category string, then this works just fine:

dt[substring(category, 1, nchar(industry)) == industry]
#              category industry
# 1:     administration    admin
# 2:           trucking    truck
# 3:     administration    admin
# 4:           trucking    truck
# 5: nurse practitioner    nurse

Answer 2

Data.table is good at grouped operations; I think that's how it can help, assuming you have many rows with the same industry:

DT[ DT[, .I[grep(industry, category)], by = industry]$V1 ]

This uses the current idiom for subsetting by group, thanks to @eddi .

---

Comments. These might help further:

• If you have many rows with the same industry-category combo, try by=.(industry,category).

• Try something else in the place of grep (like the options in Ken and Richard's answers).

Answer 3

You could use stringi::stri_detect_fixed(). It is vectorized over both str and pattern.

DT[stringi::stri_detect_fixed(category, industry)]
#              category industry
# 1:     administration    admin
# 2:           trucking    truck
# 3:     administration    admin
# 4:           trucking    truck
# 5: nurse practitioner    nurse 

Alternatively, stringr::str_detect() can be used. It is also vectorized over both its arguments.

library(stringr)
DT[str_detect(category, fixed(industry))]

Or a base R option is to run grepl() through mapply()

DT[mapply(grepl, industry, category, fixed = TRUE)]

Or you could get the same result with Vectorize(grepl).

DT[Vectorize(grepl)(industry, category, fixed = TRUE)]

All of these produce the same result.

Data:

DT <- structure(list(category = c("administration", "nurse practitioner", 
"trucking", "administration", "warehousing", "warehousing", "trucking", 
"nurse practitioner", "nurse practitioner"), industry = c("admin", 
"truck", "truck", "admin", "nurse", "admin", "truck", "nurse", 
"truck")), .Names = c("category", "industry"), class = "data.frame", row.names = c(NA, 
-9L))
setDT(DT)