R help - May 2008 - Regarding anova result

 
  
Hi,

I fitted tree growth data with Chapman-Richards growth function using nls. 

summary(fit.nls) 

Formula: 

Parameters: 

  Estimate Std. Error t value Pr

Signif. codes:  0 ''***'' 0.001 ''**'' 0.01
''*'' 0.05 ''.'' 0.1 '' '' 1

Residual standard error: 1.879 on 713 degrees of freedom

Algorithm "port", convergence message: relative convergence (4) 

When I try to run the anova() function I get this:

anova(fit.nls)
Error in anova.nls(fit.nls) : anova is only defined for sequences of
"nls" objects

Could you tell me what the problem is?

Thanks & Regards,
Guru S 
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>>> "Guru S" <guru.rcom at rediffmail.com> 09/05/2008
08:18 >>>
>I fitted tree growth data with Chapman-Richards growth function using
nls. 
>When I try to run the anova() function I get this:
>Error in anova.nls(fit.nls) : anova is only defined for sequences of
"nls" objects

There seem to be two problems.
The first is that your summary is empty. I don't know why that is;
you'd have to supply the nls() command and data.
The second is exactly what the error message says; there is no anova
method for a single nls object. 
anova in R works two ways; on a single fit, in which case individual
term sums of squares are calculated, or on a sequence of fits, in which
case anova compares the SS for the different fits.

Using the example from the nls help:

 x <- 1:10
 y <- 2*x + 3                            # perfect fit
yeps <- y + rnorm(length(y), sd = 0.01) # added noise
y.nls1<-     nls(yeps ~ a + b*x, start = list(a = 0.12345, b 0.54321),
         trace = TRUE)

y.nls2<-     nls(yeps ~ a + b*x+c*x^2, start = list(a = 0.12345, b 0.54321,
c=0.01),
         trace = TRUE)

anova(y.nls1) #Returns your error message

anova(y.nls1, y.nls2) #Returns a test of one fit against the other

Hope that helps

Steve E



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Hi S Ellison,

 Thanks for your response,
 you said anova in R works two ways; 
 one is single fit, another one is sequence of fits.
Single fit is used to fit single nls() model result.  If  I am wrong plz advice
me.
 I want to use single fit. How to do that? In your example, why the
anova(y.nls1) is generating error? How to get anova result for
"anova(y.nls1)"
Thanks in advance,
With Regards,
Guru S

On Fri, 09 May 2008 S Ellison wrote :
>
>
> >>> "Guru S" <guru.rcom@rediffmail.com> 09/05/2008
08:18 >>>
> >I fitted tree growth data with Chapman-Richards growth function using
>nls.
> >When I try to run the anova() function I get this:
> >Error in anova.nls(fit.nls) : anova is only defined for sequences of
>"nls" objects
>
>There seem to be two problems.
>The first is that your summary is empty. I don't know why that is;
>you'd have to supply the nls() command and data.
>The second is exactly what the error message says; there is no anova
>method for a single nls object.
>anova in R works two ways; on a single fit, in which case individual
>term sums of squares are calculated, or on a sequence of fits, in which
>case anova compares the SS for the different fits.
>
>Using the example from the nls help:
>
>  x <- 1:10
>  y <- 2*x + 3                            # perfect fit
>yeps <- y + rnorm(length(y), sd = 0.01) # added noise
>y.nls1<-     nls(yeps ~ a + b*x, start = list(a = 0.12345, b
>0.54321),
>          trace = TRUE)
>
>y.nls2<-     nls(yeps ~ a + b*x+c*x^2, start = list(a = 0.12345, b
>0.54321, c=0.01),
>          trace = TRUE)
>
>anova(y.nls1) #Returns your error message
>
>anova(y.nls1, y.nls2) #Returns a test of one fit against the other
>
>Hope that helps
>
>Steve E
>
>
>
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