Question

I run a non-parallel randomForest object with no problem like this:

> rf <- randomForest(t2[,-c(1,2,7,12)],t2[,2],
+                    ,sampsize=c(10000),do.trace=F,importance=TRUE,ntree=1,,forest=TRUE)
Warning message:
In randomForest.default(t2[, -c(1, 2, 7, 12)], t2[, 2], , sampsize = c(10000),  :
  The response has five or fewer unique values.  Are you sure you want to do regression?
> rf

Call:
 randomForest(x = t2[, -c(1, 2, 7, 12)], y = t2[, 2], ntree = 1,      sampsize = c(10000), importance = TRUE, do.trace = F, forest = TRUE) 
               Type of random forest: regression
                     Number of trees: 1
No. of variables tried at each split: 2

          Mean of squared residuals: 0.07444926
                    % Var explained: -19.36
> rf$rsq
[1] -0.1936248

Now I run the same code using parallel stuff and get no MSE or % Var Explained:

> library("foreach")
> library("doSNOW")
> registerDoSNOW(makeCluster(2, type="SOCK"))
> 
> 
> 
> rf <- foreach(ntree = rep(1, 2), .combine = combine, .packages = "randomForest") %dopar%
+ randomForest(t2[,-c(1,2,7,12)],t2[,2],
+                    ,sampsize=c(10000),do.trace=F,importance=TRUE,ntree=1,,forest=TRUE)
> rf

Call:
 randomForest(x = t2[, -c(1, 2, 7, 12)], y = t2[, 2], ntree = 1,      sampsize = c(10000), importance = TRUE, do.trace = F, forest = TRUE) 
               Type of random forest: regression
                     Number of trees: 2
No. of variables tried at each split: 2

> rf$rsq
NULL

Any idea what I’m doing wrong? Thanks.

Answer 1

Its not the parallel bits, its the use of combine from the randomForest package. Quoting ?combine:

The confusion, err.rate, mse and rsq components (as well as the
corresponding components in the test compnent, if exist) of the
combined object will be NULL.

So you’ll have to calculate that stuff yourself using the predicted values.