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fitting non linear data

5 messages · gregory lefebvre, Ko-Kang Kevin Wang, Jonathan Baron +2 more

Original
gregory lefebvre
# 
Hye every one, 

My question will certainly seem stupid as I am quite a beginner in R.

I would like to trace a curve which fits these two vectors:
 x<-c( 2,3,4,5,6,7,8,10 )
 y<-c( 20, 12, 8, 6, 5, 4.5, 4, 3.8)

It seems to follow a non linear model.

Could anyone help me because I could'nt find the answer I am looking for
in the FAQs.

In advance thank you for your time.

G. Lefebvre
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Ko-Kang Kevin Wang
# 
Hi,

----- Original Message -----
From: "gregory lefebvre" <greggory.lefebvre at univ-rouen.fr>
To: <r-help at stat.math.ethz.ch>
Sent: Wednesday, May 22, 2002 10:21 PM
Subject: [R] fitting non linear data
There are no stupid questions...:-)
Take a look at:
  library(help=MASS)
  library(help=nls)

In particular, the nls (non-linear regression) package.

The MASS book (Modern Applied Statistics with Splus) by Ripley and Venables
is a very good book to start.  In Chapter 8 (of the 3rd Edition), there are
quite an extensive explanations of how you can do non-linear models in
R/Splus.

MASS and nls packages are installed with R, you can open
$R_HOME/library/MASS/scripts3/ch08.R to look at ALL the R codes for that
chapter on non-linear models.

Cheers,

Ko-Kang Wang
------------------------------------------------
Ko-Kang Kevin Wang
Post Graduate PGDipSci Student
Department of Statistics
University of Auckland
New Zealand
www.stat.auckand.ac.nz/~kwan022


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Jonathan Baron
#
On 05/22/02 12:21, gregory lefebvre wrote:
The library nls (not included in the base) is one way to fit
nonlinear models.  You need to have some idea of the kind of
function you want to fit.  Without any idea of what these data
are, I thought an exponential function might do the trick, and I
took wild (and very wrong, because I forgot the minus sign for B)
guesses at the starting parameters:
Nonlinear regression model
  model:  y ~ A * exp(B * x) 
   data:  parent.frame 
         A          B 
35.6929652 -0.3290034 
 residual sum-of-squares:  16.28236 

Warning: In my experience nls does not always converge on a
solution (although it did this time).  Sometimes you need to fool
around with the starting parameters and other parameters, and
sometimes even then it won't work.  An alternative is optim(),
which seems to converge on a solution all the time (in my
experience) but which requires more study before you can use it.
Jesus Frias
# 
Gregory,

    Please have a look at the library nls (it comes with the base
distribution of R).

    You'll find a good example in

http://cran.r-project.org/doc/manuals/R-intro.pdf (pp 62 is for nonlinear
least squares).

    You can use that document and the rest of the manuals at

http://cran.r-project.org/

 to guide yourself up through R.

best regards,

IOsu


----- Original Message -----
From: "gregory lefebvre" <greggory.lefebvre at univ-rouen.fr>
To: <r-help at stat.math.ethz.ch>
Sent: Wednesday, May 22, 2002 11:21 AM
Subject: [R] fitting non linear data
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Frank E Harrell Jr
# 
To factor in model uncertainty you might bootstrap a semiparametric regression.   The Hmisc library areg.boot function, using the default method of avas, develops a semiparametric additive model and uses the bootstrap for confidence bands for the transformations on predictor and response variables.  As an example I tried

f <- areg.boot(I(y) ~ x)   # don't transform y, optimally transform x
# The default y ~ x shows a lot of uncertainty in transformation of y for
# large y

f

avas Additive Regression Model

areg.boot(x = I(y) ~ x)

n= 8   p= 1 

Apparent R2 on transformed Y scale: 0.886
Bootstrap validated R2            : 0.683

Coefficients of standardized transformations:

    Intercept             x 
-1.503799e-16  1.087857e+00 

summary(f)

summary.areg.boot(obj = f)

Estimates based on 100 resamples

Values to which predictors are set when estimating
effects of other predictors:

I(y)    x 
 5.5  5.5 

Estimates of differences of effects on Median Y (from first X value),
and bootstrap standard errors of these differences.
Settings for X are shown as row headings.


Predictor: x 
      
x      Differences      S.E Lower 0.95 Upper 0.95         Z      Pr(|Z|)
  3.75    0.000000       NA         NA         NA        NA           NA
  5.50   -4.614057 1.268541  -7.100353  -2.127762 -3.637294 0.0002755177
  7.25   -7.144868 2.030195 -11.123978  -3.165758 -3.519300 0.0004326864

plot(f)  # see attached

I had to fix a small bug in the function to make this work.
In the plot, confidence bands are in black, individual bootstrap estimates are in red.

Frank Harrell


On Wed, 22 May 2002 12:21:45 +0200
gregory lefebvre <greggory.lefebvre at univ-rouen.fr> wrote: