Question

我在数据库中有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'（空格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

如果您想要数百万条记录的良好表现，stringi包就是您所需要的。它甚至优于基本R功能：

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

我们可以使用sub匹配字符串末尾（\\s）后跟空格$，后跟'A'，并将其替换为空白（""）

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

Answer 3

@akrun的答案当然是正确的，但根据评论，我会在列factor时再添加一个。

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 4

也许不是最快的解决方案，但这也会起作用：

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"

如果符合条件，则删除字符串中的最后两个字符

4 个答案: