Hello,
Ok, try this, then.
(I've renamed the function and included some numbers in the text strings.)
x <- c("text, text, 2001, text, 1234", "text, 2000, 1234, text", "1999,
text, 1234, text, text")
extract.first.year <- function(x, n = 4){
pattern <- paste("\\D*(\\d{", n, "}).*$", sep="")
as.integer(sub(pattern, "\\1", x))
}
extract.first.year(x)
Also, thanks to arun for having reminded me of \\d and its negation,
\\D. It makes the code easier to read than [[:digit:]], which I
systematically use, and its negation in two places:
pattern <- paste("[^[:digit:]]*([[:digit:]]{", n, "}).*$", sep="")
Rui Barradas
Em 31-07-2012 19:36, jimi adams escreveu:
Thanks. Apparently my question wasn't quite specific enough, and my experience
with regular expressions is extremely limited.
This is doing *exactly* as expected for what i described, but i left out a few details.
The most important is that often #'s also appear in some of what i labeled as
"text" below.
This appears to be returning the LAST 4-digit number it finds in each of the
items over which it runs. How can i make it the first instead? That should do
the trick to make this exactly what i actually need, not just what i asked
about.
again, thanks!
jimi
On 31Jul, 2012, at 12:14 , Rui Barradas wrote:
Hello,
Try the following.
x <- c("text, text, 2001, text", "text, 2000, text", "1999, text, text, text")
extract.year <- function(x, n = 4){
pattern <- paste(".*([[:digit:]]{", n, "}).*", sep="")
as.integer(sub(pattern, "\\1", x))
}
extract.year(x)
The argument 'n' is the number of digits of year. Then use the function as you
want, within lapply, for instance, or directly as in
extract.year(foo$a)
Hope this helps,
Rui Barradas
Em 31-07-2012 16:33, jimi adams escreveu:
Hello,
I have a data frame, one element in that data frame is a LIST, with each
element being a character string. I am trying to extract the first year listed
in each of those character strings. The character elements are typically csv,
but the position of the year can vary (think citations with varying citation
standards). I.e.,
foo$a
[[1]]
[1] text, text, 2001, text
[2] text, 2000, text
[3] 1999, text, text, text, …
I'm trying to figure out how to create a new list such that each element is
that year, i.e., the result on the above would be:
foo$year
[[1]]
[1] 2001
[2] 2000
[3] 1999
…
For some reason i'm not figuring out how to properly get lapply and strsplit
(or other alternatives) to play nicely together. Any help greatly appreciated.
thanks,
jimi
jimi adams
Assistant Professor
Department of Sociology
American University
e: jad...@american.edu
w: jimiadams.com
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