Linear models in R with different combinations of variables - r

Question

I am new to R and I am stuck with a problem. I am trying to read a set of data in a table and I want to perform linear modeling. Below is how I read my data and my variables names:
>data =read.table(datafilename,header=TRUE)
>names(data)
[1] "price" "model" "size" "year" "color"
What I want to do is create several linear models using different combinations of the variables (price being the target ), such as:
> attach(data)
> model1 = lm(price~model+size)
> model2 = lm(price~model+year)
> model3 = lm(price~model+color)
> model4 = lm(price~model+size)
> model4 = lm(price~size+year+color)
#... and so on for all different combination...
My main aim is to compare the different models. Is there a more clever way to generate these models instead of hard coding the variables, especially that the number of my variables in some cases will increase to 13 or so.

Answer 1

If your goal is model selection there are several tools available in R which attempt to automate this process. Read the documentation on dredge(...) in the MuMIn package.
# dredge: example of use
library(MuMIn)
df <- mtcars[,c("mpg","cyl","disp","hp","wt")] # subset of mtcars
full.model <- lm(mpg ~ cyl+disp+hp+wt,df) # model for predicting mpg
dredge(full.model)
# Global model call: lm(formula = mpg ~ cyl + disp + hp + wt, data = df)
# ---
# Model selection table
# (Intrc) cyl disp hp wt df logLik AICc delta weight
# 10 39.69 -1.5080 -3.191 4 -74.005 157.5 0.00 0.291
# 14 38.75 -0.9416 -0.01804 -3.167 5 -72.738 157.8 0.29 0.251
# 13 37.23 -0.03177 -3.878 4 -74.326 158.1 0.64 0.211
# 16 40.83 -1.2930 0.011600 -0.02054 -3.854 6 -72.169 159.7 2.21 0.096
# 12 41.11 -1.7850 0.007473 -3.636 5 -73.779 159.9 2.37 0.089
# 15 37.11 -0.000937 -0.03116 -3.801 5 -74.321 161.0 3.46 0.052
# 11 34.96 -0.017720 -3.351 4 -78.084 165.6 8.16 0.005
# 9 37.29 -5.344 3 -80.015 166.9 9.40 0.003
# 4 34.66 -1.5870 -0.020580 4 -79.573 168.6 11.14 0.001
# 7 30.74 -0.030350 -0.02484 4 -80.309 170.1 12.61 0.001
# 2 37.88 -2.8760 3 -81.653 170.2 12.67 0.001
# 8 34.18 -1.2270 -0.018840 -0.01468 5 -79.009 170.3 12.83 0.000
# 6 36.91 -2.2650 -0.01912 4 -80.781 171.0 13.55 0.000
# 3 29.60 -0.041220 3 -82.105 171.1 13.57 0.000
# 5 30.10 -0.06823 3 -87.619 182.1 24.60 0.000
# 1 20.09 2 -102.378 209.2 51.68 0.000
You should consider these tools to help you make intelligent decisions. Do not let the tool make the decision for you!!!
For example, in this case dredge(...) suggests that the "best" model for predicting mpg, based on the AICc criterion, includes cyl and wt. But note that AICc for this model is 157.7 whereas the second best model has an AICc of 157.8, so these are basically the same. In fact, the first 5 models in this list are not significantly different in their ability to predict mpg. It does, however, narrow things down a bit. Among these 5, I would want to look at distribution of residuals (should be normal), trends in residuals (there should be none), and leverage (do some points have undue influence), before picking a "best" model.

Answer 2

Here's one way to get all of the combinations of variables using the combn function. It's a bit messy, and uses a loop (perhaps someone can improve on this with mapply):
vars <- c("price","model","size","year","color")
N <- list(1,2,3,4)
COMB <- sapply(N, function(m) combn(x=vars[2:5], m))
COMB2 <- list()
k=0
for(i in seq(COMB)){
tmp <- COMB[[i]]
for(j in seq(ncol(tmp))){
k <- k + 1
COMB2[[k]] <- formula(paste("price", "~", paste(tmp[,j], collapse=" + ")))
}
}
Then, you can call these formulas and store the model objects using a list or possibly give unique names with the assign function:
res <- vector(mode="list", length(COMB2))
for(i in seq(COMB2)){
res[[i]] <- lm(COMB2[[i]], data=data)
}

Answer 3

You can use stepwise multiple regression to determine what variables make sense to include. To get this started you write one lm() statement with all variables, such as:
library(MASS)
fit <- lm(price ~ model + size + year + color)
Then you continue with:
step <- stepAIC(model, direction="both")
Finally, you can use to following to show the results:
step$anova
Hope this gives you some inspiration for advancing your script.