Delete last two characters in string if they match criteria - r

Question

I have 2 million names in a database. For example:
df <- data.frame(names=c("A ADAM", "S BEAN", "A APPLE A", "A SCHWARZENEGGER"))
> df
names
1 A ADAM
2 S BEAN
3 A APPLE A
4 A SCHWARZENEGGER
I want to delete ' A' (white space A) if these are the last two characters of the string.
I know that regex is our friend here. How do I efficiently apply a regex function to the last two characters of the string?
Desired output:
> output
names
1 A ADAM
2 S BEAN
3 A APPLE
4 A SCHWARZENEGGER

Answer 1

If you want good performance for millions of records, the stringi package is what you need. It even outperforms the base R functions:
require(stringi)
n <- 10000
x <- stri_rand_strings(n, 1:100)
ind <- sample(n, n/100)
x[ind] <- stri_paste(x[ind]," A")
baseR <- function(x){
sub("\\sA$", "", x)
}
stri1 <- function(x){
stri_replace_last_regex(x, "\\sA$","")
}
stri2 <- function(x){
ind <- stri_detect_regex(x, "\\sA$")
x[ind] <- stri_sub(x[ind],1, -3)
x
}
#if we assume that there can only be space, not any white character
#this is even faster (ca 200x)
stri3 <- function(x){
ind <- stri_endswith_fixed(x, " A")
x[ind] <- stri_sub(x[ind],1, -3)
x
}
head(stri2(x),44)
require(microbenchmark)
microbenchmark(baseR(x), stri1(x),stri2(x),stri3(x))
Unit: microseconds
expr min lq mean median uq max neval
baseR(x) 166044.032 172054.30 183919.6684 183112.1765 194586.231 219207.905 100
stri1(x) 36704.180 39015.59 41836.8612 40164.9365 43773.034 60373.866 100
stri2(x) 17736.535 18884.56 20575.3306 19818.2895 21759.489 31846.582 100
stri3(x) 491.963 802.27 918.1626 868.9935 1008.776 2489.923 100

Answer 2

We can use sub to match a space \\s followed by 'A' at the end ($) of the string and replace it with blank ("")
df$names <- sub("\\sA$", "", df$names)
df$names
#[1] "A ADAM" "S BEAN" "A APPLE" "A SCHWARZENEGGER"

Answer 3

The answer from #akrun is, of course, correct, but based on the comments I will just add one more thing when the column is factor.
Using the example of #vincentmajor in the comments:
df <- df2 <- data.frame(names = rep(c("A ADAM", "S BEAN", "A APPLE A", "A SCHWARZENEGGER"), length.out = 2000000))
# Probably we want the column to remain factor after substitution
system.time(
df$names <- factor(sub("\\sA$", "", df$names))
)
# user system elapsed
# 0.892 0.000 0.893
# Also if there are a lot of duplicates, like in this example,
# substituting the levels is way quicker
system.time(
levels(df2$names) <- sub("\\sA$", "", levels(df2$names))
)
# user system elapsed
# 0.052 0.000 0.053

Answer 4

Maybe not the fastest solution, but this will work too:
require(stringi)
x <- stri_rand_strings(10, 1:10)
ind <- sample(10, 5)
x[ind] <- stri_paste(x[ind]," A")
x
# [1] "z A" "hX" "uv0 A" "HQtD A" "kTNZh" "4SIVBh" "v28UrqS A" "uskxxNkl A"
# [9] "dKxloBsA6" "sRkCQp7sn4"
y <- stri_sub(x, -2,-1) == " A"
x[y] <- stri_sub(x[y], 1, -3)
x
# [1] "z" "hX" "uv0" "HQtD" "kTNZh" "4SIVBh" "v28UrqS" "uskxxNkl"
# [9] "dKxloBsA6" "sRkCQp7sn4"