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Jan 8, 2012 at 23:46 comment added redcalx @joran. Yah, by forcing predictions to be 0 or 1 you lose the ability to make subtle adjustments to predictions (between 0 and 1) that can lower error (e.g. mean squared prediction error). As such I suspect that approach to be inferior. I tried it and most attempts at building a decision tree fail to find even a single split that improves error.
Jan 3, 2012 at 5:44 answer added topepo timeline score: 11
Jan 3, 2012 at 5:11 comment added joran I'm tempted to say that your description of the continuous case is correct (i.e. the standard way of doing things), but then your description of the binary variable case does not match up at all with my understanding of how random forests (or decision trees) work, so I'm worried that one of us is confused.
Jan 2, 2012 at 10:43 history edited redcalx CC BY-SA 3.0
minor rewording.
Jan 1, 2012 at 5:21 comment added Frank Harrell Very likely so. But a random forest is a mixture of trees (it is not a decision tree), so it approximates continuous relationships by making multiple splits, and in effect, using shrinkage. So I don't think your original question applies, if I understand it.
Dec 31, 2011 at 17:25 history tweeted twitter.com/#!/StackStats/status/153164918500241408
Dec 31, 2011 at 16:25 comment added redcalx The problem I'm working on right now has many predictor variables (a mix of continuous and binary) and a single target variable. Hence I believe RF is a reasonable approach to take.
Dec 31, 2011 at 15:56 comment added Frank Harrell Splitting on a continuous variable will be sure to make the resulting "model" not fit the data properly. If you have one continuous X and one continuous Y consider using the loess nonparametric smoother.
Dec 31, 2011 at 12:49 history edited chl CC BY-SA 3.0
edited tags; edited title
Dec 31, 2011 at 11:27 history asked redcalx CC BY-SA 3.0