Thanks again David,
I've made the changes. The optimize() function works alright but it does
not give me the intersection. I suspect it must have to do with the
user-defined function which I am totally clueless and am inexperienced
with. Mind showing the way.
Thanks again and thanks for your patience...
* s1 cm
1 0 0.57223196
2 10 0.33110049
3 20 0.11163181
4 30 0.10242237
5 40 0.09254315
6 50 0.02739370
7 60 0.02567137
8 70 0.02492397
9 80 0.03206637
10 90 0.02487381
11 100 0.01747584
12 110 0.15977533
13 120 0.13317708
x<- read.table("test.data.txt",header=TRUE)
ds <- x[,2] ; cr <- x[,3]
plot(ds,cr)
fc <- function(x,a,b){a*exp(-b*x)}
fm <- nls(cr ~ fc(ds,a,b),start=c(a=1,b=0))
co <- coef(fm)
curve(fc(x,a=co[1],b=co[2]),add=TRUE,col="red",lwd=1.5)
n <- 1/2.71
int <- function(x) {coef(fm)[1] + x*coef(fm)[2]}
in1 <- optimize(f=function(x) abs(int(x)-n),c(0,120))
abline(v=in1$minimum,col="red",lty=2)
abline(h=n,col="red",lty=2)
Muhammad
David Winsemius wrote:
On Apr 23, 2010, at 8:06 PM, Muhammad Rahiz wrote:
Thanks David & Peter,
The locator() works but not practical as I have to repeat the
process many times.
Does the code works on linear regression only?
Should work for any process that can produce a function.
When i tried to find the intersection at a non-linear curve, i get
the following error
Error in optimize(f = function(x) abs(xyf(ds) - n), c(0, 13)) :
invalid function value in 'optimize'
Why did you did not change the name of the function?
And the code below would not have produced that error, so why did you
post it?
I'd like to know what the error means and how I can correct it.
I have my sample data and code as follows;
* s1 cm
1 0 0.57223196
2 10 0.33110049
3 20 0.11163181
4 30 0.10242237
5 40 0.09254315
6 50 0.02739370
7 60 0.02567137
8 70 0.02492397
9 80 0.03206637
10 90 0.02487381
11 100 0.01747584
12 110 0.15977533
13 120 0.13317708
rm(list=ls())
PLEASE, DON'T DO THAT!!!!. Or rather you can do it in your workspace
but don't post it. It's not fair to a person who may not read your
code line by line before pasting it into their workspace and having it
wiped out. Do you expect us to completely clear out our workspaces
just so we can answer your questions? At the very least comment it out
so it doesn't blow away half a days work for someone who is trying to
be helpful.
x<- read.table("test.data.txt",header=TRUE)
ds <- x[,2] # distance
cr <- x[,3] # correlation values
plot(ds,cr)
n <- 1/2.71
abline(h=n)
fc <- function(x,a,b){a*exp(-b*x)} # where a & b are constants
fm <- nls(cr ~ fc(ds,a,b),start=c(a=1,b=0))
co <- coef(fm)
curve(fc(x,a=co[1],b=co[2]),add=TRUE,col="red",lwd=1.5)
int <- function(x) coef(fm)[1] + x*coef(fm)[2]
in1 <- optimize(f=function(x) c(0,120), abs(int(ds)-n))
Huh? I don't really see anything in there that would be a function. I
think you would need to specify "fc" and coefficient values for "a"
and "b".
abline(v=in1$minimize) # ????
I think you would use in1$minimum once you get this far.
thanks,
Muhammad
On 04/23/2010 07:32 PM, Peter Ehlers wrote:
On 2010-04-23 11:46, David Winsemius wrote:
On Apr 23, 2010, at 1:06 PM, Muhammad Rahiz wrote:
Does anyone know of a method that I can get the intersection
where the
red and blue curves meet i.e. the value on the x-axis?
x<- 1:10
y<- 10:1
plot(x,y)
abline(lm(y~x),col="blue")
abline(h=2.5,col="red")
Two ways :
> xy<- lm(y~x)
> xyf<- function(x) coef(xy)[1] +x*coef(xy)[2]
# absolute difference
> optimise(f=function(x) abs(xyf(x)-2.5), c(1,10) )
$minimum
[1] 8.49998
$objective
(Intercept)
1.932015e-05
#N minimize squared difference
> optimise(f=function(x) (xyf(x)-2.5)^2, c(1,10) )
$minimum
[1] 8.5
$objective
(Intercept)
3.155444e-30
Another (crude) way is to use locator(). I usually maximize
the plot window for this.
David Winsemius, MD
West Hartford, CT
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