[R] Re: predict nlme syntax
Petr Pikal
petr.pikal at precheza.cz
Fri May 20 09:10:53 CEST 2005
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
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