如何在 R 中使用正则表达式替换字符串中的字符

Question

我试图替换字符串中的某个字符，但由于该模式是重复出现的，我想使用正则表达式进行迭代替换，如果可能的话。 例如，考虑一个链接http://www.bbc.co.uk/food/ingredients/by/letter/a 我想迭代地将最后一个'a'替换为'b'，然后是'c'、'd'等，直到'z'。 与此相关的正则表达式非常简单： http:\/\/www\.bbc\.co\.uk\/food\/ingredients\/by\/letter\/[a-z] 这将验证字符串，但我想自动化链接生成过程。我对'R'相当陌生，我想知道是否有类似的方法。

- shanky_thebearer

Answer 1

您可以使用paste0创建如下的链接列表：

l <- paste0("http://www.bbc.co.uk/food/ingredients/by/letter/", letters)

这导致以下列表的生成：

> l
 [1] "http://www.bbc.co.uk/food/ingredients/by/letter/a"
 [2] "http://www.bbc.co.uk/food/ingredients/by/letter/b"
 [3] "http://www.bbc.co.uk/food/ingredients/by/letter/c"
 [4] "http://www.bbc.co.uk/food/ingredients/by/letter/d"
 [5] "http://www.bbc.co.uk/food/ingredients/by/letter/e"
 [6] "http://www.bbc.co.uk/food/ingredients/by/letter/f"
 [7] "http://www.bbc.co.uk/food/ingredients/by/letter/g"
 [8] "http://www.bbc.co.uk/food/ingredients/by/letter/h"
 [9] "http://www.bbc.co.uk/food/ingredients/by/letter/i"
[10] "http://www.bbc.co.uk/food/ingredients/by/letter/j"
[11] "http://www.bbc.co.uk/food/ingredients/by/letter/k"
[12] "http://www.bbc.co.uk/food/ingredients/by/letter/l"
[13] "http://www.bbc.co.uk/food/ingredients/by/letter/m"
[14] "http://www.bbc.co.uk/food/ingredients/by/letter/n"
[15] "http://www.bbc.co.uk/food/ingredients/by/letter/o"
[16] "http://www.bbc.co.uk/food/ingredients/by/letter/p"
[17] "http://www.bbc.co.uk/food/ingredients/by/letter/q"
[18] "http://www.bbc.co.uk/food/ingredients/by/letter/r"
[19] "http://www.bbc.co.uk/food/ingredients/by/letter/s"
[20] "http://www.bbc.co.uk/food/ingredients/by/letter/t"
[21] "http://www.bbc.co.uk/food/ingredients/by/letter/u"
[22] "http://www.bbc.co.uk/food/ingredients/by/letter/v"
[23] "http://www.bbc.co.uk/food/ingredients/by/letter/w"
[24] "http://www.bbc.co.uk/food/ingredients/by/letter/x"
[25] "http://www.bbc.co.uk/food/ingredients/by/letter/y"
[26] "http://www.bbc.co.uk/food/ingredients/by/letter/z"

- Jaap

Answer 2

提取目录名称（即 /a 之前的所有内容），然后使用内置变量 letters 追加每个字母。

u <- "http://www.bbc.co.uk/food/ingredients/by/letter/a" # input string
file.path(dirname(u), letters)

这会得到一个字符向量：

 [1] "http://www.bbc.co.uk/food/ingredients/by/letter/a"
 [2] "http://www.bbc.co.uk/food/ingredients/by/letter/b"
 [3] "http://www.bbc.co.uk/food/ingredients/by/letter/c"
 [4] "http://www.bbc.co.uk/food/ingredients/by/letter/d"
 [5] "http://www.bbc.co.uk/food/ingredients/by/letter/e"
 [6] "http://www.bbc.co.uk/food/ingredients/by/letter/f"
 [7] "http://www.bbc.co.uk/food/ingredients/by/letter/g"
 [8] "http://www.bbc.co.uk/food/ingredients/by/letter/h"
 [9] "http://www.bbc.co.uk/food/ingredients/by/letter/i"
[10] "http://www.bbc.co.uk/food/ingredients/by/letter/j"
[11] "http://www.bbc.co.uk/food/ingredients/by/letter/k"
[12] "http://www.bbc.co.uk/food/ingredients/by/letter/l"
[13] "http://www.bbc.co.uk/food/ingredients/by/letter/m"
[14] "http://www.bbc.co.uk/food/ingredients/by/letter/n"
[15] "http://www.bbc.co.uk/food/ingredients/by/letter/o"
[16] "http://www.bbc.co.uk/food/ingredients/by/letter/p"
[17] "http://www.bbc.co.uk/food/ingredients/by/letter/q"
[18] "http://www.bbc.co.uk/food/ingredients/by/letter/r"
[19] "http://www.bbc.co.uk/food/ingredients/by/letter/s"
[20] "http://www.bbc.co.uk/food/ingredients/by/letter/t"
[21] "http://www.bbc.co.uk/food/ingredients/by/letter/u"
[22] "http://www.bbc.co.uk/food/ingredients/by/letter/v"
[23] "http://www.bbc.co.uk/food/ingredients/by/letter/w"
[24] "http://www.bbc.co.uk/food/ingredients/by/letter/x"
[25] "http://www.bbc.co.uk/food/ingredients/by/letter/y"
[26] "http://www.bbc.co.uk/food/ingredients/by/letter/z"

- G. Grothendieck

---

感谢 @g-grothendieck 的帮助。 - shanky_thebearer

Answer 3

另一个选项：

library(stringi)
v <- "http://www.bbc.co.uk/food/ingredients/by/letter/a"
stri_replace_last(v, letters, regex = "[a-z]")

- Steven Beaupré