[R] Re: predict nlme syntax

[Petr Pikal]

Thank you. I somehow missed your answer and find it only after I 
went through archives. I had to have partial blindfoldness when 
searching in your book and in documentation for the answer.

Best regards
Petr Pikal 

On 10 May 2005 at 13:24, r-help at stat.math.ethz.ch wrote:

> Dear all
> 
> Please help me with correct syntax of predict.nlme.
> I would like to predict from nlme object for new data.
> I used predict(fit.nlme6, data=newdata) but I have always got 
> fitted values, no matter how I changed newdata.
> 
> I have
> 
> > summary(fit.nlme6)
> Nonlinear mixed-effects model fit by maximum likelihood
>   Model: konverze ~ SSfpl(tepl, A, B, xmid, scal) 
>  Data: limity.gr 
>        AIC      BIC    logLik
>   882.4939 907.6738 -433.2469
> 
> Random effects:
>  Formula: list(xmid ~ 1, scal ~ 1)
>  Level: spol.f
>  Structure: General positive-definite, Log-Cholesky 
> parametrization
>          StdDev    Corr 
> xmid     29.680114 xmid 
> scal      6.481679 0.249
> Residual  2.168191      
> 
> Fixed effects: list(A ~ 1, B ~ 1, xmid ~ 1, scal ~ 1) 
>         Value Std.Error  DF   t-value p-value
> A     36.1450  0.837050 154  43.18133       0
> B    101.0272  0.432074 154 233.81898       0
> xmid 735.3860  8.150964 154  90.22074       0
> scal  15.4453  2.201864 154   7.01466       0
>  Correlation: 
>      A      B      xmid  
> B    -0.088              
> xmid  0.057 -0.088       
> scal -0.089  0.469  0.036
> 
> Standardized Within-Group Residuals:
>           Min            Q1           Med            Q3           Max
> -3.7707629568 -0.3291628536  0.0005885683  0.4020944158  3.7911729382 
> 
> Number of Observations: 172
> Number of Groups: 15 
> 
> where **tepl** is independent variable and **spol.f** is grouping
> factor.
> 
> The newly constructed data frame newdata has the same structure 
> of spol.f levels as has the limity.gr data frame used for fitting.
> 
> > levels(limity.gr$spol.f)
>  [1] "1.8/3"   "4/3"     "6.3/3"   "10.8/3"  "1.8/7"   "1.8/12" 
> "1.8/30"  "6.3/30"  "10.8/30" "1.8/60"  "4/60"    "6.3/60"  "1.8/110"
> [14] "1.8/200" "1.8/300"
> 
> > levels(newdata$spol.f)
>  [1] "1.8/3"   "4/3"     "6.3/3"   "10.8/3"  "1.8/7"   "1.8/12" 
> "1.8/30"  "6.3/30"  "10.8/30" "1.8/60"  "4/60"    "6.3/60"  "1.8/110"
> [14] "1.8/200" "1.8/300" >
> 
> The only difference is in temperature.
> 
> Please advice how shall I change newdata to be able to use it in
> predict function.

> The argument's name is newdata, not data.

> Thank you.
> 
> Best regards
> 
> 

Petr Pikal
petr.pikal at precheza.cz