Ah! Thanks. ifelse() appears to do the job. note: subtracting 12 from chron rather than noon doesn't seem to work. I think (but am not totally sure) chron interprets 12 as fraction of a day - i.e. 12/1 days...
On Wed, Oct 15, 2008 at 1:33 PM, Greg Snow <[EMAIL PROTECTED]> wrote: > The if statement is for program flow control (do this bunch of code if this > condition is true, else do this), the vectorized version is the ifelse > function, that is the one that you want to use. > > Also note (this is for times in general, not sure about chron specifically) > subtracting noon from the times gives you time differences, subtracting a > time difference from a time will give you another time, so it may be simpler > to subtract just 12, rather than noon from your times. > > -- > Gregory (Greg) L. Snow Ph.D. > Statistical Data Center > Intermountain Healthcare > [EMAIL PROTECTED] > 801.408.8111 > > > > -----Original Message----- > > From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] > > project.org] On Behalf Of Joe Kaser > > Sent: Wednesday, October 15, 2008 2:19 PM > > To: r-help@r-project.org > > Subject: [R] Condition warning: has length > 1 and only the first > > element will be used > > > > Hello, > > > > I've been learning R functions recently and I've come across a problem > > that > > perhaps someone could help me with. > > > > # I have a a chron() object of times > > > > > hours=chron(time=c("01:00:00","18:00:00","13:00:00","10:00:00")) > > > > # I would like to subtract 12 hours from each time element, so I > > created a > > vector of 12 hours (actually, more like a vector of noons) > > > > > less.hours=chron(time=rep("12:00:00",4)) > > > > # If I just subtract the two vectors, I get some negative values, as > > fractions of a day > > > > > hours-less.hours > > [1] -0.45833333 0.25000000 0.04166667 -0.08333333 > > > > # I would like those negative values to cycle around and subtract the > > amount > > < 0 from midnight > > # That is to say, 01:00:00 - 12:00:00 should be 13:00:00 > > # because 01:00:00-12:00:00 = -11:00:00, and 24:00:00-11:00:00 = > > 13:00:00 > > # It's sort of like going back to the previous day, but without > > actually > > including information about which day it is > > > > # This is what I tried > > > > > test.function=function(x,y) { > > + sub = x-y > > + if(sub<0) x+y > > + } > > > > > test.function(hours,less.hours) > > Time in days: > > [1] 0.5416667 1.2500000 1.0416667 0.9166667 > > Warning message: > > In if (sub < 0) x + y : > > the condition has length > 1 and only the first element will be used > > > > > > # My questions are, why does it only use the first element?? Why does > > it not > > apply to all? Also, what happened to the elements where sub>= 0, it > > looks > > like they followed the rules > > # of if(sub<0). I feel like I must not be understanding something > > rather > > basic about how logical expressions operate within R > > # Help would be appreciated... > > > > [[alternative HTML version deleted]] > > > > ______________________________________________ > > R-help@r-project.org mailing list > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide http://www.R-project.org/posting- > > guide.html > > and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
