dear members,
I have two predictors, x1 and x2, and one response
variable, y. Moreover, a step function models the relationship between y and x1:
fx <- (x1 <= -2)*(x1^2) + (x1 > -2 && x1 < 2)*(x1^3) + (x1
> = 2)*(x1^4)
Can I include fx in an nls call to create something like this:
NLS1 <- nls(y ~ a*(sin(x2) + fx), start = list(a = 2)) ?
Will the above work? Or should I do something like this:
if(x1 <= -2) {
NLS2 <- nls(y[x1 <= -2] ~ a*(sin(x2[x1 <= -2]) + x1^2),start = list(a =
3)) }
if(x1 > -2 && x1 < 2) {
NLS3 <- nls(y[x1 > -2 && x1 < 2] ~ a*(sin(x2[x1 > -2
&& x1 < 2]) + x1^3),start = list(a = 4)) }
if(x1 > = 2) {
NLS4 <- nls(y[x1 >= 2] ~ a*(sin(x2[x1 >= 2]) + x1^4),start = list(a =
5)) }
If the first case doesn't work, is there any other method to include the
step function in the nls call without resorting to the if statements?
Also, the coefficient, a , gotten from the above two cases are different. Will
this affect the precision of the nonlinear fir between
y ~ (f(x1),f(x2)) ?
very many thanks for your time and effort
yours sincerely,
AKSHAY M KULKARNI
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On Thu, 18 Apr 2019 10:36:10 +0000 akshay kulkarni <akshay_e4 at hotmail.com> wrote:> fx <- (x1 <= -2)*(x1^2) + (x1 > -2 && x1 < 2)*(x1^3) + (x1 > > 2)*(x1^4) > > Can I include fx in an nls call to create something like this: > > NLS1 <- nls(y ~ a*(sin(x2) + fx), start = list(a = 2)) ?For now, you can, since fx does not depend on any of the parameters you optimize in the nls() call. (Actually, the model as presented should be solveable by lm(y ~ I(sin(x2) + fx) + 0), since it is linear in its only parameter.) If you make fx a function and use ifelse() to provide different outcomes depending on a condition in a vectorized fashion, you would make it easier to add new parameters later, should the need arise: fx <- function(x1, x2) ifelse(x1 <= -2, EXPR_IF_TRUE..., EXPR_IF_FALSE...) NLS1 <- nls(y ~ a*(sin(x2) + fx(x1, x2)), ...) -- Best regards, Ivan
dear Ivan,
THanks for the reply......But what do you mean by "since
fx does not depend on any of the parameters you
optimize in the nls() call."? Can you give an example?
very many thanks for your time and effort.....
Yours sincerely,
AKSHAY M KULKARNI
________________________________
From: Ivan Krylov <krylov.r00t at gmail.com>
Sent: Thursday, April 18, 2019 4:50 PM
To: akshay kulkarni
Cc: R help Mailing list
Subject: Re: [R] picewise function in nls....
On Thu, 18 Apr 2019 10:36:10 +0000
akshay kulkarni <akshay_e4 at hotmail.com> wrote:
> fx <- (x1 <= -2)*(x1^2) + (x1 > -2 && x1 < 2)*(x1^3)
+ (x1 > > 2)*(x1^4)
>
> Can I include fx in an nls call to create something like this:
>
> NLS1 <- nls(y ~ a*(sin(x2) + fx), start = list(a = 2)) ?
For now, you can, since fx does not depend on any of the parameters you
optimize in the nls() call. (Actually, the model as presented should be
solveable by lm(y ~ I(sin(x2) + fx) + 0), since it is linear in its
only parameter.)
If you make fx a function and use ifelse() to provide different
outcomes depending on a condition in a vectorized fashion, you would
make it easier to add new parameters later, should the need arise:
fx <- function(x1, x2)
ifelse(x1 <= -2, EXPR_IF_TRUE..., EXPR_IF_FALSE...)
NLS1 <- nls(y ~ a*(sin(x2) + fx(x1, x2)), ...)
--
Best regards,
Ivan
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