Question

I have the following R dataframe mydata

                       session      first     last            city
1  9cf571c8faa67cad2aa9ff41f3a26e38    cat+rat   biddix          fresno
2  e30f853d4e54604fd62858badb68113a      caleb     amos                
3  2ad41134cc285bcc06892fd68a471cd7 daniel+joe  folkers                
4  2ad41134cc285bcc06892fd68a471cd7 daniel+joe  folkers                
5  63a5e839510a647c1ff3b8aed684c2a5    charles   pierce           flint
6  691df47f2df12f14f000f9a17d1cc40e          j    franz prescott+valley
7  691df47f2df12f14f000f9a17d1cc40e          j    franz prescott+valley
8  b3a1476aa37ae4b799495256324a8d3d   carr%ie% mascorro            brea
9  bd9f1404b313415e7e7b8769376d2705       fred  morales       las+vegas
10 b50a610292803dc302f24ae507ea853a     aurora      lee                
11 fb74940e6feb0dc61a1b4d09fcbbcb37     andrew    price       yorkville

What is the proper line of code to remove all those rows where city is blank?

Answer 1

Maybe

subset(mydata,city!="")

? This assumes that the city column is stored in such a way that the blanks are zero-length strings. If they might be whitespace, then something like

grep("^[[:space:]]*$",mydata$city,invert=TRUE)

would find the elements you wanted. Since grepl doesn’t have an invert argument you could use (edit: thanks to @JoshO’Brien)

subset(mydata,!grepl("^[[:space:]]*$",city))

(No reproducible example given, so neither of these is tested.)