对照另一列检查一个数据框列中的值或模式以查看是否出现匹配项

Question

I have a list of brand name medications that I need to check whether or not they exist in a patients med list. The patients med list is primarily listed as generic form but I need to check if brand was entered and change it to generic. The patient's med list contains the drug plus directions in the column. My goal is to create a column that flags if brand shows up either with "yes" "no" or TRUE FALSE. My dataset contains about 5000 brand name entries and patient list contains about 60000 entries. I am not sure where to begin because of the difference in patterns from brand list to patient list. Capitalization is also inconsistent in the patients med list as well. Any help is appreciated.

Example dataset: MRN is patient ID

Brand <- c("Evista", "Rozerem", "Altace")

MRN <- c("121212", "121212", "231212", "432123", "432123", "542345", 
"323412", "242341", "412111", "642321")

MedList <- c("raloxifene 60mg daily", "Rozerem 8mg daily", "evista 60mg 
daily", "metoprolol tartate 25mg twice daily", "ramelteon 8mg daily", 
"ramipril 5mg daily", "omeprazole 20mg daily", "ALTACE 5mg nightly", 
"ramelteon 8mg daily", "imatinib 400mg daily")

Patients <- data.frame(MRN,MedList)

My goal is to end up with something like this

inlist <- c(FALSE, TRUE, TRUE, FALSE, FALSE, FALSE, FALSE, TRUE, FALSE, 
FALSE)

Patients <- cbind(Patients, inlist)

Thank you.

Answer 1

Try this:

grepl(paste(toupper(Brand), collapse = '|'), toupper(MedList))
 [1] FALSE  TRUE  TRUE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE

Answer 2

If only capitalization but not spelling is an issue, grepl should do what you want:

grepl(paste0(Brand, collapse = "|"), MedList, ignore.case = TRUE)

In case your pattern (ie, "Brand") is a really long vector, you may use str_detect() from stringr . It is much faster and supports longer patterns (but it has no ignore.case argument).

stringr::str_detect(tolower(MedList), paste0(tolower(Brand), collapse = "|"))