Need some help with regular expression

On Dec 15, 2016, at 8:46 AM, Steven Nagy <nstefi at gmail.com> wrote:

I tried to send this email, but it didn't go through. I guess pictures are
not allowed to send through HTML formatted emails?
I'm re-sending it again without the picture, just comment there instead as
placeholder.

Thanks,
Steven


From: Steven Nagy [mailto:nstefi at gmail.com] 
Sent: Monday, December 12, 2016 10:50 PM
To: 'Bert Gunter' <bgunter.4567 at gmail.com>
Cc: 'R-help' <r-help at r-project.org>
Subject: RE: [R] Need some help with regular expression

Hi Bert and all,

Sorry I was too busy at work and didn't have much time to continue this
until now.
So I studied "?regexp" and I can understand your regular expression now:
sub(".*: *([[:alnum:]]* *-> *STU|STU *-> *[[:alnum:]]*).*","\\1",x)

But I also wanted to split up these results in 2 columns, so your previous
command would give me this result:
[1] "NMA -> STU" "STU -> REG" "-> STU"

and I wanted to further split them up to show this:
From	To
NMA	STU
STU	REG
	STU

strsplit( sub(".*: *([[:alnum:]]* *-> *STU|STU *-> *[[:alnum:]]*).*","\\1",x), split="-> ")

sapply(  strsplit( sub(".*: *([[:alnum:]]* *-> *STU|STU *-> *[[:alnum:]]*).*","\\1",x), split="-> "), "[",1 )

sapply(  strsplit( sub(".*: *([[:alnum:]]* *-> *STU|STU *-> *[[:alnum:]]*).*","\\1",x), split="-> "), "[",2 )

I still don?t quite understand the backreferences, and how could I have 2
backreferences, one for the left side of the ?->? sign and one for the right
side?

So it seems like I need to apply the ?sub? function twice, similar how I
used the ?strapply? function twice in my original post:
strapply(strapply(a, "(file://w+ -> STU|STU -> file://w+)", c, backref = -1,
perl = TRUE), "(file://w+) -> (file://w+)", c, backref = -2, perl = TRUE)

or maybe there would be a more simple way of using only 1 ?sub? function and
2 backreferences?

Also I?m not sure what do I do after I get the data? How could I represent
the member type changes graphically? We need to analyze the behavior of
switching from STU to another type or from another type to STU.
Google Analytics has a nice chart under Behavior Flow, or Users Flow, and it
looks like this:
<here was my picture from Google Analytics - it's from Behavior Flow or
Users Flow showing flows from one category to another one and further to
another one>



Is there any graphical representation in R that is similar to this?

Thanks a lot,
Steven

-----Original Message-----
From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Bert Gunter
Sent: Sunday, November 20, 2016 10:05 PM
To: Aliz Csonka <mailto:lyzae.ro at gmail.com>
Cc: R-help <mailto:r-help at r-project.org>
Subject: Re: [R] Need some help with regular expression

Although others may respond, I think you will do much better studying
?regexp, which will answer all your questions. I believe the effort you will
make figuring it out will pay dividends for your future R/regular expression
usage that you cannot gain from my direct explanation.

Good luck.

Best,
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 Sun, Nov 20, 2016 at 6:40 PM, Steven Nagy <mailto:nstefi at gmail.com>
wrote:

Thanks a lot Bert. That's amazing. I am very new to both R and regular 
expressions. I don't really understand the regular expression that you 
used below.
And looks like I don't even need any special library, like the 
"gsubfn" for the strapply function.
I was trying to use the regexr.com website to analyze your regular 
expression, but it doesn't seem to match any text there.
Can you explain me the regular expression that you used?
".*: *([[:alnum:]]* *-> *STU|STU *-> *[[:alnum:]]*).*"
So the dot in the front means any character and the star after that 
means that it can repeat 0 or more times, right?
Then followed by a colon character ":" and a space, and what is the 
next star after that? It means that the sequence before that again can 
repeat 0 or more times?
And what are the double square brackets?
Is ":alnum:" specific to R? I don't think "regexr.com" understands 
that. Or maybe that site is for regular expressions in Javascript, and 
the syntax is different in R?

Thank you,
Steven

-----Original Message-----
From: Bert Gunter [mailto:bgunter.4567 at gmail.com]
Sent: Sunday, November 20, 2016 2:15 PM
To: Steven Nagy <mailto:nstefi at gmail.com>
Cc: R-help <mailto:r-help at r-project.org>
Subject: Re: [R] Need some help with regular expression

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:]]*).*","file://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 <mailto: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, "(file://w+ -> STU|STU -> file://w+)", c, backref =

-1, 

perl = TRUE), "(file://w+) -> (file://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, "(file://w+ -> STU|STU -> file://w+)", c, backref =

-1, 

perl = TRUE), "(file://w+) -> (file://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, "(file://w+ -> STU|STU -> file://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


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