On 04/07/2013 08:04, Dante.py wrote:
I have a method which is not so smart --use grep to match the pattern.
for example:
dates <- c("02/27/92", "02/27/92", "01/14/92", "02/28/92", "02/01/91")
day <- as.Date(dates, "%m/%d/%y")
day:
[1] "1992-02-27" "1992-02-27" "1992-01-14" "1992-02-28" "1991-02-01"
If I want to search for 1991, I can use:
grep("1991-*", day)
Hope for a better solution.
Here are 2. The first is quite obvious, the second is faster but needs
more knowledge.
x <- format(day, "%Y")
day[x >= "2007" & x <= "2009"]
x <- as.POSIXlt(day)$year + 1900
day[x >= 2007 & x <= 2009]
2013/7/4 Gallon Li <gallon...@gmail.com>
i have converted my data into date format like below:
day=as.Date(originaldate,"%m/%d/%Y")
day[1:5]
[1] "2008-04-12" "2011-07-02" "2011-09-02" "2008-04-12" "2008-04-12"
I wish to select only those observations from 2007 to 2009, how can I
select from this list?
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--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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