On Jan 3, 2011, at 11:09 AM, Muhammad Rahiz wrote:
Josh's recommendation to use predict works. At the same time, I'll
work on your suggestions, David.
Thanks.
Muhammad Rahiz
Researcher & DPhil Candidate (Climate Systems & Policy)
School of Geography & the Environment
University of Oxford
On Mon, 3 Jan 2011, David Winsemius wrote:
On Jan 3, 2011, at 9:52 AM, Muhammad Rahiz wrote:
Hi all,
I'm trying to get the value of y when x=203 by using the intersect
of three curves. The horizontal curve does not meet with the other
two. How can I rectify the code below?
Thanks
Muhammad
ts <- 1:10
dd <- 10:1
ts <- seq(200,209,1)
dd <- c(NA,NA,NA,NA,1.87,1.83,1.86,NA,1.95,1.96)
plot(ts,dd,ylim=c(1.5,2))
abline(lm(dd~ts),col="blue",lty=2)
abline(v=203,col="blue",lty=2)
xy <- lm(dd~ts)
fc <- function(x) coef(xy)[1] + x*coef(xy)[2]
val <- optimize(f=function(x) abs(fc(x)-203), c(1.5,2))
^???^
And a further observation: You should pay close attention to the
limits of optimization. If your answers keep coming out at the
extremes of the ranges then perhaps you have confused the x and y
coordinates there, as well as in your objective function.
--
David.
Joshua Wiley's answer made me realize that your question was
different
than I thought. At this point maybe you should have made your
objective function match on the coordinate difference to be
minimized.
You are at this point taking the absolute difference between an x-
value, 203. and a y-value. fc(x). That makes little sense. You can
use
predict() as Joshua showed you or you can change your objective
functions so it is more meaningful. Furthermore, I have had better
success with squared differences that with abs(differences). I'm not
sure whether that is due to the availability of a more informative
derivative. (Again, looks like homework and I refrain from posting
completed solutions.)
abline(h=val,col="blue",lty=2)
--
David Winsemius, MD
West Hartford, CT
David Winsemius, MD
West Hartford, CT
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