Take the 2-minute tour ×

I'm doing some work with the randomForest package and while it works well, it can be time-consuming. Any one have any suggestions for speeding things up? I'm using a Windows 7 box w/ a dual core AMD chip. I know about R not being multi- thread/processor, but was curious if any of the parallel packages (rmpi, snow, snowfall, etc.) worked for randomForest stuff. Thanks.


I'm using rF for some classification work (0's and 1's). The data has about 8-12 variable columns and the training set is a sample of 10k lines, so it's decent size but not crazy. I'm running 500 trees and an mtry of 2, 3, or 4.

EDIT 2: Here's some output:

> head(t22)
  Id Fail     CCUse Age S-TFail         DR MonInc #OpenLines L-TFail RE M-TFail Dep
1  1    1 0.7661266  45       2 0.80298213   9120         13       0  6       0   2
2  2    0 0.9571510  40       0 0.12187620   2600          4       0  0       0   1
3  3    0 0.6581801  38       1 0.08511338   3042          2       1  0       0   0
4  4    0 0.2338098  30       0 0.03604968   3300          5       0  0       0   0
5  5    0 0.9072394  49       1 0.02492570  63588          7       0  1       0   0
6  6    0 0.2131787  74       0 0.37560697   3500          3       0  1       0   1
> ptm <- proc.time()
> RF<- randomForest(t22[,-c(1,2,7,12)],t22$Fail
+                    ,sampsize=c(10000),do.trace=F,importance=TRUE,ntree=500,,forest=TRUE)
Warning message:
In randomForest.default(t22[, -c(1, 2, 7, 12)], t22$Fail, sampsize = c(10000),  :
  The response has five or fewer unique values.  Are you sure you want to do regression?
> proc.time() - ptm
   user  system elapsed 
 437.30    0.86  450.97 
share|improve this question
Can you describe your data and modeling in a bit more detail? That will help a bit in making suggestions on how to address improvements. –  Iterator Oct 20 '11 at 1:54
You can often get rf models of equivalent performance only drawing a subsample of your data for each tree (sampsize), rather than a resample of size 10k for each tree. –  joran Oct 20 '11 at 2:31
I can run a rf model on 10k observations and 500 trees with 8 variables without using the sampsize argument and using the formula interface in under 15 seconds on my year old macbook air. I suspect there's something else going on. –  joran Oct 20 '11 at 2:47
It would really help if you posted code to create sample data and the code you're actually running (i.e. a reproducible example). –  Joshua Ulrich Oct 20 '11 at 6:11
Are those 12 variables stored as factors in the input data? If they are, are there many levels? As @joran says, don't use the formula interface; that is for convenience not speed. Are you sure you are not maxing out the RAM on your machine during fitting (i.e. are your discs swapping memory into/out of RAM)? 10K rows doesn;t seem like very much to me. –  Gavin Simpson Oct 20 '11 at 9:07

4 Answers 4

up vote 21 down vote accepted

The manual of the foreach package has a section on Parallel Random Forests (Using The foreach Package, Section 5.1):

> library("foreach")
> library("doSNOW")
> registerDoSNOW(makeCluster(4, type="SOCK"))

> x <- matrix(runif(500), 100)
> y <- gl(2, 50)

> rf <- foreach(ntree = rep(250, 4), .combine = combine, .packages = "randomForest") %dopar%
+    randomForest(x, y, ntree = ntree)
> rf
randomForest(x = x, y = y, ntree = ntree)
Type of random forest: classification
Number of trees: 1000

If we want want to create a random forest model with a 1000 trees, and our computer has four cores, we can split up the problem into four pieces by executing the randomForest function four times, with the ntree argument set to 250. Of course, we have to combine the resulting randomForest objects, but the randomForest package comes with a function called combine.

share|improve this answer
Unfortunately we loose some outputs when using parallel computations, such as the R² –  Stéphane Laurent Nov 17 at 13:13

There are two 'out of the box' options that address this problem. First, the caret package contains a method 'parRF' that handles this elegantly. I commonly use this with 16 cores to great effect. The randomShrubbery package also takes advantages of multiple cores for RF on Revolution R.

share|improve this answer
Hi Brent, in Revolution R does randomShrubbery take .xdf as input dataset or does have to be a dataframe. I have a 10GB .xdf training file. –  user1509107 Sep 10 '12 at 22:10
It's part of the train function though, so you won't find it with ?parRF, instead do ?train and read up. This answer is the better way forward really, cheers Brent –  N. McA. Dec 26 '12 at 16:36

Why don't you use an already parallelized and optimized implementation of Random Forest? Have a look to SPRINT using MPI. http://www.r-sprint.org/

share|improve this answer

Is there any particular reason why you're not using Python (namely the scikit-learn and multiprocessing modules) to implement this? Using joblib, I've trained random forests on datasets of similar size in a fraction of the time it takes in R. Even without multiprocessing, random forests are significantly faster in Python. Here's a quick example of training a RF classifier and cross validating in Python. You can also easily extract feature importances and visualize the trees.

import numpy as np
from sklearn.metrics import *
from sklearn.cross_validation import StratifiedKFold
from sklearn.ensemble import RandomForestClassifier

#assuming that you have read in data with headers
#first column corresponds to response variable 
y = data[1:, 0].astype(np.float)
X = data[1:, 1:].astype(np.float)

precision = np.array([])
accuracy = np.array([])
sensitivity = np.array([])
f1 = np.array([])
matthews = np.array([])

rf = RandomForestClassifier(n_estimators=100, max_features = 5, n_jobs = 2)

#divide dataset into 5 "folds", where classes are equally balanced in each fold
cv = StratifiedKFold(y, n_folds = 5)
for i, (train, test) in enumerate(cv):
        classes = rf.fit(X[train], y[train]).predict(X[test])
        precision = np.append(precision, (precision_score(y[test], classes)))
        accuracy = np.append(accuracy, (accuracy_score(y[test], classes)))
        sensitivity = np.append(sensitivity, (recall_score(y[test], classes)))
        f1 = np.append(f1, (f1_score(y[test], classes)))
        matthews = np.append(matthews, (matthews_corrcoef(y[test], classes)))
        cm = np.add(cm, (confusion_matrix(y[test], classes)))

print("Accuracy: %0.2f (+/- %0.2f)" % (accuracy.mean(), accuracy.std() * 2))
print("Precision: %0.2f (+/- %0.2f)" % (precision.mean(), precision.std() * 2))
print("Sensitivity: %0.2f (+/- %0.2f)" % (sensitivity.mean(), sensitivity.std() * 2))
print("F1: %0.2f (+/- %0.2f)" % (f1.mean(), f1.std() * 2))
print("Matthews: %0.2f (+/- %0.2f)" % (matthews.mean(), matthews.std() * 2))
share|improve this answer
+1 Not every screwdriver fits the screw. –  Brandon Bertelsen Sep 24 '13 at 7:56

Your Answer


