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I want to know if it is possible for a library in R to evaluate the association of independent variables and create a formula? I am trying to come up with a model to predict power consumption of a machine, using some hardware counters and performance attributes. When I use linear regression, I have no problem since I could represent my formula like power~lm(a1+a2+a3+a4), but for the non-linear case, I am not sure what would be the formula or which model should I choose. I would want to have a way to do this:

power ~ <some-non-linear-reg-pkg>(a1+a2+a3+non-linear(a4))

I reviewed some packages for non-linear regression such as nls and gnm, and they expect a formula to be provided by the user. I am however able to identify which variables have linear associations and which are non-linear (by performing correlation tests), the problem is building a formula out of them.

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  • $\begingroup$ I have plotted the data, and I don't see a straight relation; can you please elaborate on - "It might give you some inspiration as to the kinds of formula you'd want for a4"? $\endgroup$ – Sayan Mar 29 '13 at 18:29
  • $\begingroup$ Sorry, I thought it might work, but I just tried and it's not even close. I've deleted my previous comment. $\endgroup$ – Wayne Mar 29 '13 at 18:39
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    $\begingroup$ Because you appear to be in an exploratory mode ("trying to come up with a model"), why not first use exploratory methods like loess smoothing? $\endgroup$ – whuber Mar 29 '13 at 18:41
  • $\begingroup$ nls does non linear least squares in R. It is used differently from lm because the models aren't linear; you can't leave the parameters implicit. They appear in the formula. See ?nls. There's also the possibility of using glms if your model has a linear predictor (i.e. a transformation of the mean of the response is linear in predictors). But I would agree with @whuber - if you don't have a specific functional form in mind, start with some exploratory tools. $\endgroup$ – Glen_b Mar 30 '13 at 1:05
  • $\begingroup$ I tried gam, and I am getting better prediction accuracy. This might be a dumb question - but I am getting this vibe that a linear model is generally sufficient for most cases, may be this is a topic for another question, but comments are welcome. $\endgroup$ – Sayan Mar 30 '13 at 7:50
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Look at using linear regression but with a polynomial (poly) function or spline function on the predictors that you think may have a non-linear relationship. Then plotting and further examination of those results may suggest the form of a non-linear function (or the linear model may be sufficient for your purposes).

example:

library(TeachingDemos)  # for Predict.Plot
library(splines)

     fit.lm1 <- lm( Sepal.Width ~ ns(Petal.Width,3)*ns(Petal.Length,3)+Species,
             data=iris)

     Predict.Plot(fit.lm1, pred.var = "Petal.Width", Petal.Width = 1.22,
           Petal.Length = 4.3, Species = "versicolor",
             plot.args = list(ylim=range(iris$Sepal.Width), col='blue'),
           type = "response")
     Predict.Plot(fit.lm1, pred.var = "Petal.Width", Petal.Width = 1.22,
           Petal.Length = 4.3, Species = "virginica",
             plot.args = list(col='red'),
           type = "response", add=TRUE)
     Predict.Plot(fit.lm1, pred.var = "Petal.Width", Petal.Width = 1.22,
           Petal.Length = 4.4, Species = "virginica",
             plot.args = list(col='purple'),
           type = "response", add=TRUE)

This is actually the 1st example from the help page for Predict.Plot.

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A nonlinear regression function is never going to decide the functional form for you.

You might want to stick with a linear model and use a transformation of the variable with the nonlinear relationship. For example, check the correlation between power $a4^2$, $ln(a4)$, or some other transformation and use which ever one has the higher correlation.

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  • $\begingroup$ Thank you, I noticed that using ln in the formula yielded better prediction accuracy. I guess I need to explore the effects of transformations on the independent variables. $\endgroup$ – Sayan Mar 30 '13 at 7:52

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