On 29/11/10 11:57:31, Jude Ryan wrote: > Hi Georg, > > > The documentation (?nnet) says that y should be a matrix or data frame, > but in your case it is a vector. This is most likely the problem, if > you do not have other data issues going on. Convert y to a matrix (or > data frame) using ‘as.matrix’ and see if this solves your problem. > Library ‘nnet’ can do both classification and regression. I was able to > replicate your problem, using an example from Modern Applied Statistics > with S, Venables and Ripley, pages 246 and 247), by turning y into a > vector and verifying that all the predicted values are the same when y > is a vector. This is not the case when y is part of a data frame. You > can see this by running the code below. I tried about 4 neural network > packages in the past, including AMORE, but found ‘nnet’ to be the best > for my needs.
Hi Jude, thanks for the hint. I lately experimented both with the nnet(x,y, ...) and the nnet(formula, dataframe ...) interfaces to nnet and both yielded the same results. So changing the format of y from a vector to a matrix or a data frame didn't change anything at all. However, what _did_ change the outcome is to introduce the "decay" parameter (which I didn't have at all before). By default it is set to 0 which doesn't seem appropriate in my case. Setting it to "decay=1e-3" magically turned my output into an acceptable regression response instead of spitting out fixed values. I really love the predict interface for regression in each of the models I'm using. Clear code :-) So, for the record, the call for nnet for the regression problem is as follows: net.fitted <- nnet(formula, data = sp...@data[-testset,], decay=1e-3, size = 20, linout = TRUE) (where sp...@data is the data part of a SpatialPointsDataFrame. And yes, in selecting the [-testset,] data points I'm taking into account the existing spatial autocorrelation.) > # Neural Network model in Modern Applied Statistics with S, Venables > and Ripley, pages 246 and 247 Thanks for your help and the reference, I'm likely to order the book now :-) Leaving out the "decay" parameter changes the fitted.values in the "rock" example you mentioned as well, although not that much. Convergence speed does change as expected, so the parameter is working. I guess my problem is solved now, the rest is due to the specialties with my data sets. Georg. -- Research Assistant Otto-von-Guericke-Universität Magdeburg resea...@georgruss.de http://research.georgruss.de ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.