similar to: Decision Tree and Random Forrest

Tjats great that you are familiar and thanks for responding. Have you ever done what I am referring to? I have alteady spent time going through links and tutorials about decision trees and random forrests and have even used them both before. Mike On Apr 13, 2016 5:32 PM, "Sarah Goslee" <sarah.goslee at gmail.com> wrote: It sounds like you want classification or regression trees.

Since you only have 3 predictors, each categorical with a small number of categories, you can use expand.grid to make a data.frame containing all possible combinations and give that the predict method for your model to get all possible predictions. Something like the following untested code. newdata <- expand.grid( Humidity = levels(Humidity), #(High, Medium,Low)

I need the output to have groups and the probability any given record in that group then has of being in the response class. Just like my email in the beginning i need the output that looks like if A and if B and if C then %77 it will be D. The examples you provided are just simply not similar. They are different and would take interpretation to get what i need. On Apr 14, 2016 1:26 AM,

Ah yes I will have to use the predict function. But the predict function will not get me there really. If I can take the example that I have a model predicting whether or not I will play golf (this is the dependent value), and there are three independent variables Humidity(High, Medium, Low), Pending_Chores(Taxes, None, Laundry, Car Maintenance) and Wind (High, Low). I would like rules like

I think you are missing the point of random forests. But if you just want to predict using the forest, there is a predict() method that you can use. Other than that, I certainly don't understand what you mean. Maybe someone else might. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka

Ok is there a way to do it with decision tree? I just need to make the decision rules. Perhaps I can pick one of the trees used with Random Forrest. I am somewhat familiar already with Random Forrest with respective to bagging and feature sampling and getting the mode from the leaf nodes and it being an ensemble technique of many trees. I am just working from the perspective that I need

Nope. Random forests are not decision trees -- they are ensembles (forests) of trees. You need to go back and read up on them so you understand how they work. The Hastie/Tibshirani/Friedman "The Elements of Statistical Learning" has a nice explanation, but I'm sure there are lots of good web resources, too. Cheers, Bert Bert Gunter "The trouble with having an open mind is

Hi I'm trying to get the top decision rules from a decision tree. Eventually I will like to do this with R and Random Forrest. There has to be a way to output the decsion rules of each leaf node in an easily readable way. I am looking at the randomforrest and rpart packages and I dont see anything yet. Mike [[alternative HTML version deleted]]

lapply(colordata2[ -1 ], f ) When you put the parentheses on, you are calling the function yourself before lapply gets a chance. The error pops up because you are giving a vector of numbers (the answer f gave you) to the second argument of lapply instead of a function. -- Sent from my phone. Please excuse my brevity. On April 7, 2016 7:31:18 AM PDT, Michael Artz <michaeleartz at

HI that did not work for me either. The value I got returned from that function was "<rounded mean> - <rounded mean>" :(. thanks for the reply through On Tue, Apr 19, 2016 at 10:34 AM, William Dunlap <wdunlap at tibco.com> wrote: > > That didn't work Jim! > > It always helps to say how the suggestion did not work. Jim's > function had a typo

If you are not using an anonymous function and say you had written the function out The below gives me the error > 'f(colordata2$color1)' is not a function, character or symbol' But then how is the anonymous function working? f <- function(col){ ifelse(col == 'blue', 1, 0) } responses <- lapply(colordata2[ -1 ], f(colordata2$color1) )

??? IQR returns a single number. > IQR(rnorm(10)) [1] 1.090168 To your 2nd response: "I could have used average, min, max, they all would have returned the same thing., " I can only respond: huh?? Are all your values identical? You really need to provide a small reproducible example as requested by the posting guide -- I certainly don't get it, and I'm done guessing.

If you show us, not just tell us about, a self-contained example someone might show you a non-hacky way of getting the job done. (I don't see an argument to plyr::ddply called 'transform'.) Bill Dunlap TIBCO Software wdunlap tibco.com On Tue, Apr 19, 2016 at 12:18 PM, Michael Artz <michaeleartz at gmail.com> wrote: > Oh thanks for that clarification Bert! Hope you enjoyed

Well, instead of your functions try: Mode <- function(x) { tabx <- table(x) tabx[which.max(tabx)] } and use R's IQR function instead of yours. ... so I still don't get why you want to return a character string instead of a value for the IQR; and the mode of a sample defined as above is generally a bad estimator of the mode of the distribution. To say more than that would

> That didn't work Jim! It always helps to say how the suggestion did not work. Jim's function had a typo in it - was that the problem? Or did you not change the call to ddply to use that function. Here is something that might "work" for you: library(plyr) data <- data.frame(groupColumn=rep(1:5,1:5), col1=2^(0:14)) myIqr <- function(x) {

Hi, Here is what I am doing notGroupedAll <- ddply(data ,~groupColumn ,summarise ,col1_mean=mean(col1) ,col2_mode=Mode(col2) #Function I wrote for getting the mode shown below ,col3_Range=myIqr(col3) ) groupedAll <- ddply(data ,~groupColumn ,summarise

Are you aware that there *already is* a function that does this? ?IQR (also your "function" iqr" is just a character string and would have to be parsed and evaluated to become a function. But this is a TERRIBLE way to do things in R as it completely circumvents R's central functional programming paradigm). Cheers, Bert Bert Gunter "The trouble with having an open mind

Again, IQR returns two both a .25 and a .75 value and it failed, which is why I didn't use it before. Also, the first function just returns tha same value repeating. Since they are the same, before the second call, using the mode function is just a way to grab one value. I could have used average, min, max, they all would have returned the same thing. Mike On Tue, Apr 19, 2016 at 7:24 PM,

Hi, Jumping into this thread mainly on the point of the mode of the distribution, while also supporting Bert's comments below on theory. If the vector 'x' that is being passed to this function is an integer vector, then a tabulation of the integers can yield a 'mode', presuming of course that there is only one unique mode. You may have to decide how you want to handle a

Lapply is not a vectorized function. It is compact to read, but it would not be worth using for this calculation. However, if your data frame had multiple color columns in your data frame that you wanted to make responses for then you might want to use lapply as a more compact version of a for loop to repeat this operation. colordata2 <- data.frame(id = c(1,2,3,4,5), color1 =