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. Thank you. Best regards Petr Pikal petr.pikal at precheza.cz
Petr Pikal 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 haveThe argument's name is newdata, not data.
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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