如何使用str_replace在不删除字母的情况下分隔字符串中的单词？

n9vozmp4 于 2023-06-03 发布在其他

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我想把字符串中的单词分开，并在它们之间加一个空格。我怎么才能做到这一点，而不删除一些字母使用str_replace？

s1 <- c("Employee_Name", "EmpID", "MarriedID", "MaritalStatusID", "GenderID", 
        "EmpStatusID") |> print()
#> [1] "Employee_Name"   "EmpID"           "MarriedID"       "MaritalStatusID"
#> [5] "GenderID"        "EmpStatusID"
s1|> 
 stringr::str_remove_all("_") |> 
 # I want to separate the  words in  the string and add a space in between
 stringr::str_replace_all("([a-z][A-Z])", " ")
#> [1] "Employe ame"   "Em D"          "Marrie D"      "Marita tatu D"
#> [5] "Gende D"       "Em tatu D"

创建于2023-05-29带有reprex v2.0.2
我尝试了stringr：：str_replace_all（“（[a-z][A-Z]）"，““），但这会删除与模式匹配的字母。

来源：https://stackoverflow.com/questions/76360566/how-can-i-separate-words-in-a-string-without-deleting-letters-using-str-replace

3条答案

按热度按时间

cwtwac6a1#

使用OR操作符|和gsub，这只是一行代码。

gsub('_|(?<=[a-z])(?=[A-Z])', ' ', s1, perl=TRUE)
# [1] "Employee Name"     "Emp ID"            "Married ID"        "Marital Status ID"
# [5] "Gender ID"         "Emp Status ID"

赞(0）回复(0）举报 2023-06-03

np8igboo2#

您希望使用lookbehind和lookahead：

library(stringr)

s1 |>
  str_remove_all("_") |>
  str_replace_all("(?<=[a-z])(?=[A-Z])", " ")
# [1] "Employee Name"     "Emp ID"            "Married ID"       
# [4] "Marital Status ID" "Gender ID"         "Emp Status ID"

或者，使用反向引用捕获组：

s1 |>
  str_remove_all("_") |>
  str_replace_all("([a-z])([A-Z])", "\\1 \\2")
# [1] "Employee Name"     "Emp ID"            "Married ID"       
# [4] "Marital Status ID" "Gender ID"         "Emp Status ID"

赞(0）回复(0）举报 2023-06-03

ztmd8pv53#

s1 <- c("Employee_Name", "EmpID", "MarriedID", "MaritalStatusID", "GenderID", 
        "EmpStatusID")
str_replace_all(s1, '([A-Z]+)', ' \\1') |>
  str_replace_all('_', ' ') |>
  str_squish() |>
  print()

[1] "Employee Name"     "Emp ID"            "Married ID"        "Marital Status ID" "Gender ID"         "Emp Status ID"

赞(0）回复(0）举报 2023-06-03

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如何使用str_replace在不删除字母的情况下分隔字符串中的单词？

3条答案

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