Regular expression for anonymous emails

Question

i use regular expression in R

regexp <- "(^|[^([:alnum:]|.|_)]) [e-mail] ($|[^[:alnum:]])" 

to find the email address [e-mail] in the special text and replace it with anonym-mail-adress .

 tmp <- c(" [e-mail] ", ## <- should not be matched "aaaa [e-mail] bbbb", ## <- should be matched " [e-mail] ", ## <- should be matched " [e-mail] ", ## <- should not be matched " [e-mail] ", ## <- should not be matched "( [e-mail] )", ## <- should be matched "aaaa ( [e-mail] ) bbbb") ## <- should be matched replacement <- paste("\\1", " [e-mail] ", "\\2", sep="") gsub(regexp, replacement, tmp, ignore.case=TRUE) 

as a result I get

 > gsub(regexp, replacement, tmp, ignore.case=TRUE) [1] " [e-mail] " "aaaa [e-mail] bbbb" [3] " [e-mail] " " [e-mail] " [5] " [e-mail] " "( [e-mail] )" [7] "aaaa (abc.abc.de) bbbb" 

I do not know why the last two elements of the array do not match?

Thanks and best regards.

Answer 1

How about this?

 gsub("^( abc@abc )|(?<=[ (])( abc@abc )", " anonym@anonym ", tmp, perl=T) 

Sample before | : ^( abc@abc ) , of course, checks the beginning with abc@abc .

In the pattern after | uses a positive lookbehind and searches for abc@abc , which is preceded by space or ( (left), and if found, is replaced by anonym@anonym .

This is what I get: (Note: I replaced abc.abc in the last line abc@abc )

 [1] " [e-mail] " "aaaa [e-mail] bbbb" [3] " [e-mail] " " [e-mail] " [5] " [e-mail] " "( [e-mail] )" [7] "aaaa ( [e-mail] ) bbbb" 

Edit: To explain the problem with your regex, this seems like a problem with the part:

 [^([:alnum:]|.|_)] 

I think negation should be present in every expression | . In addition, you should use [.] Instead . , since the latter implies any character. Alternatively, instead of using negation for each character that you test, we can condense this part by removing all unnecessary | as:

 [^.[:alpha:]_] # not a . or _ or any alphanumeric # using gsub on it: gsub("(^|[^.[:alpha:]_]) abc@abc ", " anonym@anonym ", tmp) # [1] " [e-mail] " "aaaa [e-mail] bbbb" # [3] " [e-mail] " " [e-mail] " # [5] " [e-mail] " " [e-mail] )" # [7] "aaaa [e-mail] ) bbbb" 

You replace each abc@abc . But you will lose the character before abc@abc every time, because you also check it in the template. So you have to use a capture group. That is, if you complete the regular expression with () , you can refer to this “capture” using special variables such as \\1, \\2 etc.. Here we wrote (^|[^.[:alpha:]_]) , i.e. Part to abc@abc . Since this is the first capture, we will refer to it as \\1 to use it to recover the missing character in the previous result:

 gsub("(^|[^.[:alpha:]_]) abc@abc ", "\\ 1anonym@anonym ", tmp) # [1] " [e-mail] " "aaaa [e-mail] bbbb" # [3] " [e-mail] " " [e-mail] " # [5] " [e-mail] " "( [e-mail] )" # [7] "aaaa ( [e-mail] ) bbbb" 

This is the result you need. And this is the same as my original answer using a positive look. In this case, since it just checks that it is preceded by something, you do not need to capture anything. Only the abc@abc been replaced. Hope this helps.