CODEWITHSUNDEEP
rregexstringcharactergsub
wake_wake
wake_wake

Reputation: 1204

Delete last two characters in string if they match criteria

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

Upvotes: 4

Views: 1874

Answers (4)

rozowawoda
rozowawoda

Reputation: 27

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"

Upvotes: 2

bartektartanus
bartektartanus

Reputation: 16080

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

Upvotes: 6

alko989
alko989

Reputation: 7908

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

Upvotes: 3

akrun
akrun

Reputation: 887118

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"

Upvotes: 5

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