[R] Need some help with regular expression

[Bert Gunter]

If I understand you correctly, I think you are making it more complex
than necessary. Using your example (thanks!!), the following should
get you started:


> x<- c("Name.MEMBER_TYPE: NMA -> STU ; CATEGORY:  -> 1 ; CITY: MISSISSAUGA -> Mississauga ; ZIP: L5N1H9 -> L5N 1H9 ; COUNTRY: CAN ->  ; MEMBER_STATUS:  -> N", "Name.MEMBER_TYPE: STU -> REG ; CATEGORY: 1 ->","Name.MEMBER_TYPE: -> STU")
>
> x
[1] "Name.MEMBER_TYPE: NMA -> STU ; CATEGORY:  -> 1 ; CITY:
MISSISSAUGA -> Mississauga ; ZIP: L5N1H9 -> L5N 1H9 ; COUNTRY: CAN ->
; MEMBER_STATUS:  -> N"

[2] "Name.MEMBER_TYPE: STU -> REG ; CATEGORY: 1 ->"
[3] "Name.MEMBER_TYPE: -> STU"
>
> sub(".*: *([[:alnum:]]* *-> *STU|STU *-> *[[:alnum:]]*).*","\\1",x)
[1] "NMA -> STU" "STU -> REG" "-> STU"


I am sure that you can get things to the form you desire in one go
with some fiddling of the above, but it was easier for me to write the
regex to pick out the pieces you wanted and leave the rest to you.
Others may have slicker ways to do it, of course.

HTH

Cheers,
Bert


Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Sat, Nov 19, 2016 at 8:06 PM, Steven Nagy <nstefi at gmail.com> wrote:
> I tried out a regular expression on this website:
>
> http://regexr.com/3en1m
>
>
>
> So the input text is:
>
> "Name.MEMBER_TYPE:  -> STU"
>
>
>
> The regular expression is: ((?:\w+|\s) -> STU|STU -> (?:\w+|\s))
>
> And it returns:
>
> "  -> STU"
>
>
>
> but when I use in R, it doesn't return the same result:
>
> strapply(c, "((?:\\w+|\\s) -> STU|STU -> (?:\\w+|\\s))", c, backref = -1,
> perl = TRUE)
>
> returns:
> "Name.MEMBER_TYPE: -> STU"
>
>
>
>
>
> Here is what I was trying to do:
>
>
>
> I need to extract some values from a log table, and I created a regular
> expression that helps me with that.
>
> The log table has cells with values like:
>
> a = "Name.MEMBER_TYPE: NMA -> STU ; CATEGORY:  -> 1 ; CITY: MISSISSAUGA ->
> Mississauga ; ZIP: L5N1H9 -> L5N 1H9 ; COUNTRY: CAN ->  ; MEMBER_STATUS:  ->
> N"
>
> or
> b = "Name.MEMBER_TYPE: STU -> REG ; CATEGORY: 1 ->"
>
> so I needed to extract the values that a STU member type is changing from
> and to, so I needed NMA, STU in the 1st case or STU, REG in the 2nd case.
>
> I came up with this expression which worked in both cases:
>
> strapply(strapply(a, "(\\w+ -> STU|STU -> \\w+)", c, backref = -1, perl =
> TRUE), "(\\w+) -> (\\w+)", c, backref = -2, perl = TRUE)
>
>
>
> But I had a 3rd case when the source member type was blank:
>
> c = "Name.MEMBER_TYPE: -> STU"
>
> and in that case it returned an error:
>
> strapply(strapply(c, "(\\w+ -> STU|STU -> \\w+)", c, backref = -1, perl =
> TRUE), "(\\w+) -> (\\w+)", c, backref = -2, perl = TRUE)
>
> Error: is.character(x) is not TRUE
>
>
>
> I found that the error is because this returns NULL:
>
> strapply(c, "(\\w+ -> STU|STU -> \\w+)", c, backref = -1, perl = TRUE)
>
>
>
>
>
> So I tried to modify the regular expression to match any word or blank
> space:
>
> strapply(c, "((?:\\w+|\\s) -> STU|STU -> (?:\\w+|\\s))", c, backref = -1,
> perl = TRUE)
>
>
>
> but this returned me the whole value of "c":
>
> "Name.MEMBER_TYPE:  -> STU"
>
> and I only needed "  -> STU" as it shows on the website regxr.com
>
>
>
> Is the result wrong on the regxr.com website or strapply returns the wrong
> result?
>
>
>
> Thanks,
>
> Steven
>
>
>         [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.