Hi Jeff,
I was reviewing my old lecture notes and see that the professor did use \1 so I
think he was talking about regex in a non-platform specific context. But
obviously \\1 is the way to do it in R.
The examples you gave me to study really helped.
I was also going to ask how to identify empty strings AND blank character
strings but you will be happy to know that I figured it out on my own:
grep("^ *$",x)
Thank you.
Thank you Sarah, Bert and David too
Dan
-----Original Message-----
From: Jeff Newmiller [mailto:jdnew...@dcn.davis.ca.us]
Sent: Tuesday, October 29, 2013 11:08 AM
To: Lopez, Dan; R help (r-help@r-project.org)
Subject: Re: [R] Regular Expression returning unexpected results
Please read and follow the Posting Guide, in particular re plain text email.
You need to keep in mind that the characters in literal strings in R source
have to make it into RAM before the regex code can parse it. Since regex needs
a single backslash to escape normal parsing and interpret 1 as a back
reference, but the R parser also recognizes and removes backslashes in string
literals as escape characters, you need to escape the backslash with a
backslash in your R string literal.
nchar tells you how many characters are in the string. print renders the string
as it would need to be entered as R source code. cat sends the string directly
to the output (console). Study the output of the following commands at the R
prompt.
?Quotes
nchar("^([a-z]+) +\1 +[a-z]+ [0-9]")
print("^([a-z]+) +\1 +[a-z]+ [0-9]")
cat("^([a-z]+) +\1 +[a-z]+ [0-9]")
On most systems, a raw character code 1 is also known as Control-A, but the
effect it has on the terminal used as the console may vary according to your
setup, and it's effect on my system is not clear to me.
nchar("^([a-z]+) +\\1 +[a-z]+ [0-9]")
print("^([a-z]+) +\\1 +[a-z]+ [0-9]")
cat("^([a-z]+) +\\1 +[a-z]+ [0-9]")
grep("^([a-z]+) +\\1 +[a-z]+ [0-9]",lines)
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Sent from my phone. Please excuse my brevity.
"Lopez, Dan" <lopez...@llnl.gov> wrote:
>Hi,
>
>So I just took an intro to R programming class and one of the lectures
>was on Regular Expressions. I've been playing around with various R
>functions that use Regular Expressions.
>But this has me stumped. This was part of a quiz and I got it right
>through understanding the syntax. But when I try to run the thing it
>returns 'integer(0)'. Can you please tell me what I am doing wrong?
>
>#I copied and pasted this:
>going up and up and up
>night night at 8
>bye bye from up high
>heading, heading by 9
>
>#THEN
>lines<-readLines("clipboard")
>#This is what it looks like in R
>lines
>[1] "going up and up and up"
>[2] "night night at 8"
>[3] "bye bye from up high"
>[4] "heading, heading by 9"
>
>#THIS IS WHAT IS NOT WORKING THE WAY I THOUGHT. I was expecting it to
>return 2.
># "night night at 8" follows the pattern: Begins with a word then has
>at least one space then the same word then has at least one space then
>a word then a space then a single digit number.
>grep("^([a-z]+) +\1 +[a-z]+ [0-9]",lines)
>integer(0)
>
>#But simple examples DO work
>grep("[Hh]",lines)
>[1] 2 3 4
>grep('[0-9]',lines)
>[1] 2 4
>
> [[alternative HTML version deleted]]
>
>______________________________________________
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>PLEASE do read the posting guide
>http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.
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