如何提取包含在'$'和'/'（特殊字符）之间的子字符串，然后替换该子字符串的值？

Question

我有主字符串作为str和substring，需要命名为su_str：

su_str =“$ Region_Name”，

str <- " https://lglbw.pqr.xyz.com:58443/APG/lookup/All/Report%20Library/Amazon%20S3/Inventory/Regions/$Region_Name/Billing/report.csv"

以及其他文件中$ Region_Name = ap-southeast-1的值。

我试过了 ：

r <- unlist(stri_extract_all(p,"$ /"))

它会给出一个错误，如：

stri_extract_all（p，“$ /”）出错：你必须指定regex，fixed，coll或charclass

c_prop将是： 核心价值 $ DirectorName：DF-1C $ DirectorPortName：Ports，DF-1C $ MaskingViewName：000197801199，IS_LGLW9062_VIEW $ MaskingInitiatorPortName：启动器端口，IS_LGLW9062_VIEW $ MaskingAssDeviceName：相关设备，IS_LGLW9062_VIEW $ PoolName：000197801199，SRP_1 $ PoolBoundDevice：绑定设备，SRP_1 $ PortName：DF-1C：12 $ Region_Name：ap-southeast-1

如何解决这个问题，提出一些想法？提前致谢！！！

Answer 1

这适用于您的示例，它是否解决了您的问题？使用正则表达式（或正则表达式）时，必须使用两个反斜杠\\来转义R中的特殊字符。它看起来像使用stringi替换必须有特殊字符转义，但我不经常使用stringi所以希望有人可以使用stringi更好的方式来实现它

> library(stringi)
> 
> str <- " https://lglbw.pqr.xyz.com:58443/APG/lookup/All/Report%20Library/Amazon%20S3/Inventory/Regions/$Region_Name/Billing/report.csv"
> 
> # If you just want to extract the sequence of letters and underscores's after "$" and before the "/"
> unlist(stri_extract_all(str, regex = "\\$[[:alpha:]_]*\\b"))
[1] "$Region_Name"
> 
> # If you want to replace it with something else using base R
> 
> some_string <- "$Region_Name = ap-southeast-1"
> 
> gsub("\\$[[:alpha:]_]*\\b", some_string, str)
[1] " https://lglbw.pqr.xyz.com:58443/APG/lookup/All/Report%20Library/Amazon%20S3/Inventory/Regions/$Region_Name = ap-southeast-1/Billing/report.csv"
> 
> # Using stringi package
> 
> # Special characters have to be escaped
> some_string <- "\\$Region_Name \\= ap\\-southeast\\-1"
> 
> stri_replace_all(str, some_string, regex = "\\$[[:alpha:]_]*\\b")
[1] " https://lglbw.pqr.xyz.com:58443/APG/lookup/All/Report%20Library/Amazon%20S3/Inventory/Regions/$Region_Name = ap-southeast-1/"

编辑：如果你想要同一个子串的多个替换：

# If the substring will always be "$Region_Name"

su_str <- "$Region_Name"
replacements <- c("$Region_Name = ap-southeast-1/", "$Region_Name = ap-southeast-2/")

stri_replace_all(str, replacements, fixed = su_str)
[1] " https://lglbw.pqr.xyz.com:58443/APG/lookup/All/Report%20Library/Amazon%20S3/Inventory/Regions/$Region_Name = ap-southeast-1//Billing/report.csv"
[2] " https://lglbw.pqr.xyz.com:58443/APG/lookup/All/Report%20Library/Amazon%20S3/Inventory/Regions/$Region_Name = ap-southeast-2//Billing/report.csv"

Answer 2

你问题的标题和你要问的是两个不同的问题，但我会尝试解决它们。

关于使用stri_extract_all（）获得的错误，您需要指定要匹配的模式类型，我相信您正在尝试匹配固定模式，在这种情况下您可以使用

stri_extract_all_fixed() 

功能而不是。

但是，我不使用stri_extract_all（）来删除和替换您的子字符串。这是我的解决方案。

str <- " https://lglbw.pqr.xyz.com:58443/APG/lookup/All/Report%20Library/Amazon%20S3/Inventory/Regions/$Region_Name/Billing/report.csv"
reg<-"$Region_Name"
replce<-"ap-southeast-1"

# Custom function to return position of a sub string 
strpos_fixed<-function(x,y){
  a<-regexpr(y, x,fixed=T)
  b<-a[1]
  return(b)
}

part1<-substr(str,1,(strpos_fixed(str,reg)-1))
part2<-substr(str,(strpos_fixed(str,reg)+nchar(reg)),nchar(str))

part1 # Everything before "$Region_Name"
part2 # Everything after  "$Region_Name"

new<-paste(part1,replce,part2, sep ="")
new