Re: [R] sub/grep question: extract year

Quoting Marc Girondot via R-help : Hi everybody, I have some questions about the way that sub is working. I hope that someone has the answer: 1/ Why the second example does not return an empty string ? There is no match. subtext <- "-1980-" sub(".*(1980).*", "\\1", subtext) # return

Re: [R] sub/grep question: extract year

I answer myself to the third point: This pattern is better : pattern.year <- ".*\\b(18|19|20)([0-9][0-9])\\b.*" subtext <- "bla 1880 bla" sub(pattern.year, "\\1\\2", subtext) # return 1880 subtext <- "bla 1980 bla" sub(pattern.year, "\\1\\2", subtext) # return 1980 subtext <- "bla 2010 bla"

Re: [R] sub/grep question: extract year

So there is probably a command that resets the capture variables as I call them. No doubt someone will write what it is. On 9 Aug 2018 10:36, "john matthew" wrote: > Hi Marc. > For question 1. > I know in Perl that regular expressions when captured can be saved if not > overwritten. \\1 is the

Re: [R] sub/grep question: extract year

Hi Marc. For question 1. I know in Perl that regular expressions when captured can be saved if not overwritten. \\1 is the capture variable in your R examples. So the 2nd regular expression does not match but \\1 still has 1980 captured from the previous expression, hence the result. Maybe if

Re: [R] sub/grep question: extract year

I answer myself to the third point: This pattern is better to get a year: pattern.year <- ".*\\b(18|19|20)([0-9][0-9])\\b.*" subtext <- "bla 1880 bla" sub(pattern.year, "\\1\\2", subtext) # return 1880 subtext <- "bla 1980 bla" sub(pattern.year, "\\1\\2", subtext) # return 1980 subtext <- "bla

[R] sub/grep question: extract year

Hi everybody, I have some questions about the way that sub is working. I hope that someone has the answer: 1/ Why the second example does not return an empty string ? There is no match. subtext <- "-1980-" sub(".*(1980).*", "\\1", subtext) # return 1980 sub(".*(1981).*", "\\1", subtext) #