Command for finding the best linear model in R - r

Question

Is there a way to get R to run all possible models (with all combinations of variables in a dataset) to produce the best/most accurate linear model and then output that model?
I feel like there is a way to do this, but I am having a hard time finding the information.

Answer 1

There are numerous ways this could be achieved, but for a simple way of doing this I would suggest that you have a look at the glmulti package, which is described in detail in this paper:
glmulti: An R Package for Easy Automated Model Selection with (Generalized) Linear Models
Alternatively, very simple example of the model selection as available on the Quick-R website:
# Stepwise Regression
library(MASS)
fit <- lm(y~x1+x2+x3,data=mydata)
step <- stepAIC(fit, direction="both")
step$anova # display results
Or to simplify even more, you can do more manual model comparison:
fit1 <- lm(y ~ x1 + x2 + x3 + x4, data=mydata)
fit2 <- lm(y ~ x1 + x2, data=mydata)
anova(fit1, fit2)

Answer 2

This should get you started. Although you should read my comment from above. This should build you a model based on all the data in your dataset and then compare all of the models with AIC and BIC.
# create a NULL vector called model so we have something to add our layers to
model=NULL
# create a vector of the dataframe column names used to build the formula
vars = names(data)
# remove variable names you don’t want to use (at least
# the response variable (if its in the first column)
vars = vars[-1]
# the combn function will run every different combination of variables and then run the glm
for(i in 1:length(vars)){
xx = combn(vars,i)
if(is.null(dim(xx))){
fla = paste("y ~", paste(xx, collapse="+"))
model[[length(model)+1]]=glm(as.formula(fla),data=data)
} else {
for(j in 1:dim(xx)[2]){
fla = paste("y ~", paste(xx[1:dim(xx)[1],j], collapse="+"))
model[[length(model)+1]]=glm(as.formula(fla),data=data)
}
}
}
# see how many models were build using the loop above
length(model)
# create a vector to extract AIC and BIC values from the model variable
AICs = NULL
BICs = NULL
for(i in 1:length(model)){
AICs[i] = AIC(model[[i]])
BICs[i] = BIC(model[[i]])
}
#see which models were chosen as best by both methods
which(AICs==min(AICs))
which(BICs==min(BICs))

Answer 3

I ended up running forwards, backwards, and stepwise procedures on data to select models and then comparing them based on AIC, BIC, and adj. R-sq. This method seemed most efficient. However, when I received the actual data to be used (the program I was writing was for business purposes), I was told to only model each explanatory variable against the response, so I was able to just call lm(response ~ explanatory) for each variable in question, since the analysis we ended up using it for wasn't worried about how they interacted with each other.

Answer 4

This is a very old question, but for those who are still encountering this discussion - the package olsrr and specifically the function ols_step_all_possible exhaustively produces an ols model for all possible subsets of variables, based on an lm object (such that by feeding it with a full model you will get all possible combinations), and returns a dataframe with R squared, adjusted R squared, aic, bic, etc. for all the models. This is very helpful in finding the best predictors but it is also very much time consuming.
see https://olsrr.rsquaredacademy.com/reference/ols_step_all_possible.html
I do not recommend just "cherry picking" the best performing model, rather I would actually look at the output and choose carefully for the most reasonable outcome. In case you would want to immediately get the best performing model (by some criteria, say number of predictors and R2) you may write a function that saves the dataframe, arranges it by number of predictors and orders it by descending R2 and spits out the top result.

Answer 5

The dredge() function in R also accomplishes this.