如果匹配条件，则删除字符串中的最后两个字符

Question

我的数据库中有 200 万个名字。例如：

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

如果这些是字符串的最后两个字符，我想删除 ' A'（白色 space A）。

我知道正则表达式是我们的朋友。如何有效地将正则表达式函数应用于字符串的最后两个字符？

期望的输出：

> output
             names
1           A ADAM
2           S BEAN
3          A APPLE
4 A SCHWARZENEGGER

Answer 1

我们可以用sub匹配一个space\s后跟[=17=]在字符串的末尾（$）并替换为空白 ("")

df$names <- sub("\sA$", "", df$names)
df$names
#[1] "A ADAM"           "S BEAN"           "A APPLE"          "A SCHWARZENEGGER"

Answer 2

@akrun 的回答当然是正确的，但根据评论，当列为 factor.

时，我将再添加一件事

在评论中使用@vincentmajor 的例子：

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 3

如果您希望对数百万条记录有良好的性能，stringi 包就是您所需要的。它甚至优于基本的 R 函数：

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 4

也许不是最快的解决方案，但这也行得通：

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"

如果匹配条件，则删除字符串中的最后两个字符

Delete last two characters in string if they match criteria

regex

string

r

character

gsub