R：如何在没有正则表达式的情况下使用 str_replace_all()

编程问答 2022-05-28

问题描述

我有一些文本数据，其中包含“[姓氏]”、“[女性姓名]”和“[男性姓名]”。例如，

c("I am [female name]. I am ten years old","My father is [male name][surname]","I went to school today")

我希望删除它们进行分析并期望得到

"I am . I am ten years old","My father is ","I went to school today"

但是当我运行下面的代码时，它返回的内容被破坏了。我认为 str_replace_all 可能会将 [ ] 的模式识别为正则表达式，但我不完全确定原因。

> str_replace_all(c("I am [female name]. I am ten years old","I went to school today"),"[surname]",'')

[1] "I  [fl ]. I  t y old" "My fth i [l ][]"      "I wt to chool tody"

有人知道怎么解决吗？ 提前致谢

解决方法

使用stringi::str_replace_all：

library(stringi)
data <- c("I am [female name]. I am ten years old","My father is [male name][surname]","I went to school today") 
remove_us <- c("[female name]","[male name]","[surname]")
stri_replace_all_fixed(data,remove_us,"",vectorize_all=FALSE)

结果

[1] "I am . I am ten years old" "My father is  "            "I went to school today"

见R proof。

但是，gsub 更简单：

gsub('\\[[^][]*]','',data)

见another R proof。

--------------------------------------------------------------------------------
  \[                       '['
--------------------------------------------------------------------------------
  [^][]*                   any character except: ']','[' (0 or more
                           times (matching the most amount possible))
--------------------------------------------------------------------------------
  ]                        ']'
rrtext-processing