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 BIClogLik
882.4939 907.6738 -433.2469
Random effects:
Formula: list(xmid ~ 1, scal ~ 1)
Level: spol.f
Structure: General positive-definite, Log-Cholesky
parametrization
StdDevCorr
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
B101.0272 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:
MinQ1 MedQ3 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$spol.f)
[1] 1.8/3 4/3 6.3/3 10.8/3 1.8/7 1.8/12
1.8/30 6.3/30 10.8/30 1.8/60 4/606.3/60
1.8/110
[14] 1.8/200 1.8/300
levels(newdata$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 4/606.3/60
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
[EMAIL PROTECTED]
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