[R] Non-linear regression/Quantile regression
Ravi Varadhan
RVaradhan at jhmi.edu
Tue Jun 9 18:26:16 CEST 2009
Try `poly(Time, 2, raw=TRUE)'
Here is an example:
Time <- runif(100)
demand <- 1 + 0.5 * Time - 1.2 * Time^2 + rt(100, df=4)
rq1 <- rq(demand ~ Time + I(Time^2), tau = 1:3/4)
rq2 <- rq(demand ~ poly(Time, 2, raw=TRUE), tau = 1:3/4)
all.equal(c(rq1$coef), c(rq2$coef))
Ravi.
----------------------------------------------------------------------------
-------
Ravi Varadhan, Ph.D.
Assistant Professor, The Center on Aging and Health
Division of Geriatric Medicine and Gerontology
Johns Hopkins University
Ph: (410) 502-2619
Fax: (410) 614-9625
Email: rvaradhan at jhmi.edu
Webpage:
http://www.jhsph.edu/agingandhealth/People/Faculty_personal_pages/Varadhan.h
tml
----------------------------------------------------------------------------
--------
-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On
Behalf Of despaired
Sent: Tuesday, June 09, 2009 11:59 AM
To: r-help at r-project.org
Subject: Re: [R] Non-linear regression/Quantile regression
Hi,
thanks, it works :-)
But where is the difference between demand ~ Time + I(Time^2) and demand ~
poly(Time, 2) ?
Or: How do I have to interpret the results? (I get different results for the
two methods)
Thank you again!
Gabor Grothendieck wrote:
>
> Those are linear in the coefficients so try these:
>
> library(quantreg)
>
> rq1 <- rq(demand ~ Time + I(Time^2), data = BOD, tau= 1:3/4); rq1
>
> # or
> rq2 <- rq(demand ~ poly(Time, 2), data = BOD, tau = 1:3/4); rq2
>
>
> On Tue, Jun 9, 2009 at 10:55 AM, despaired<meyfarth at uni-potsdam.de> wrote:
>>
>> Hi,
>>
>> I'm relatively new to R and need to do a quantile regression. Linear
>> quantile regression works, but for my data I need some quadratic
>> function.
>> So I guess, I have to use a nonlinear quantile regression. I tried
>> the example on the help page for nlrq with my data and it worked. But
>> the example there was with a SSlogis model. Trying to write
>>
>> dat.nlrq <- nlrq(BM ~ I(Regen100^2), data=dat, tau=0.25, trace=TRUE)
>>
>> or
>>
>> dat.nlrq <- nlrq(BM ~ poly(Regen100^2), data=dat, tau=0.25,
>> trace=TRUE)
>>
>> (I don't know the difference) both gave me the following error message:
>>
>> error in getInitial.default(func, data, mCall =
>> as.list(match.call(func,
>> :
>> no 'getInitial' method found for "function" objects
>>
>> Looking in getInitial, it must have to do something with the starting
>> parameters or selfStart model. But I have no idea, what this is and
>> how I handle this problem. Can anyone please help?
>>
>> Thanks a lot in advance!
>> --
>> View this message in context:
>> http://www.nabble.com/Non-linear-regression-Quantile-regression-tp239
>> 44530p23944530.html Sent from the R help mailing list archive at
>> Nabble.com.
>>
>> ______________________________________________
>> R-help at 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.
>>
>
> ______________________________________________
> R-help at 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.
>
>
--
View this message in context:
http://www.nabble.com/Non-linear-regression-Quantile-regression-tp23944530p2
3945900.html
Sent from the R help mailing list archive at Nabble.com.
______________________________________________
R-help at 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.
More information about the R-help
mailing list