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...iven that the second and third panels of the first figure in the > first link I gave show a decision tree with decision rules at each split > and the number of samples at each direction, what _exactly_ is your > problem? > > > > On Wednesday, April 13, 2016, Michael Eugene <fartzy at hotmail.com> wrote: > >> I still need the output to match my requiremnt in my original post. With >> decision rules "clusters" and probability attached to them. The examples >> are sort of similar. You just provided links to general info about trees. >&gt...

...Pending_Chores = levels(Pending_Chores), #(Taxes, None, Laundry, Car Maintenance) Wind = levels(Wind)) # (High,Low) newdata$ProbabilityOfPlayingGolf <- predict(fittedModel, newdata=newdata) Bill Dunlap TIBCO Software wdunlap tibco.com On Fri, Apr 15, 2016 at 3:09 PM, Michael Artz <michaeleartz at gmail.com> wrote: > 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...

...t; wrote: > Hi Michael, > At a guess, try this: > > iqr<-function(x) { > return(paste(round(quantile(x,0.25),0),round(quantile(x,0.75),0),sep="-") > } > > .col3_Range=iqr(datat$tenure) > > Jim > > > > On Tue, Apr 19, 2016 at 11:15 AM, Michael Artz <michaeleartz at gmail.com> > wrote: > > Hi, > > I am trying to show an interquartile range while grouping values using > > the function ddply(). So my function call now is like > > > > groupedAll <- ddply(data > > ,~groupColumn &g...

...rtphone<div> </div><div> </div><!-- originalMessage --><div>-------- Original message --------</div><div>From: Sarah Goslee <sarah.goslee at gmail.com> </div><div>Date: 4/13/16 8:04 PM (GMT-06:00) </div><div>To: Michael Artz <michaeleartz at gmail.com> </div><div>Cc: "r-help at r-project.org" <R-help at r-project.org> </div><div>Subject: Re: [R] Decision Tree and Random Forrest </div><div> </div> On Wednesday, April 13, 2016, Michael Artz <michaeleartz...

...gt; names(responses) <- paste0( "response", 1:2 ) > > but you have to be careful to fix that code whenever you change the > colordata2 data frame to have more columns. > -- > Sent from my phone. Please excuse my brevity. > > On April 7, 2016 4:57:04 AM PDT, Michael Artz <michaeleartz at gmail.com> > wrote: >> >> Thaks so much! And how would you incorporate lapply() here? >> >> On Thu, Apr 7, 2016 at 6:52 AM, David Barron <dnbarron at gmail.com> wrote: >> >> ifelse is vectorised, so just use that without the loop...

...ou 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 gmail.com> wrote: >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...

...gt; Bert > > > Bert Gunter > > "The trouble with having an open mind is that people keep coming along > and sticking things into it." > -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) > > > On Wed, Apr 13, 2016 at 1:40 PM, Michael Artz <michaeleartz at gmail.com> > wrote: > > 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 >...

...portant point is that > paste(round(quantile(x,0.25),0),round(quantile(x,0.75),0),sep="-") > is not a function, it is an expression. ddplyr wants functions. > > > Bill Dunlap > TIBCO Software > wdunlap tibco.com > > On Tue, Apr 19, 2016 at 7:56 AM, Michael Artz <michaeleartz at gmail.com> > wrote: > >> That didn't work Jim! >> >> Thanks anyway >> >> On Mon, Apr 18, 2016 at 9:02 PM, Jim Lemon <drjimlemon at gmail.com> wrote: >> >> > Hi Michael, >> > At a guess, try this: >> &...

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 your coffee! I > ended up just using the transform argument in the ddply function. It > worked and it repeated, then I called a mode function in another call to > ddply that summarised...

...so, 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, Marc Schwartz <marc_schwartz at me.com> wrote: > 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 > vecto...

...to know why (if it's even relevant). Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Tue, Apr 19, 2016 at 4:25 PM, Michael Artz <michaeleartz at gmail.com> wrote: > Hi, > Here is what I am doing > > notGroupedAll <- ddply(data > ,~groupColumn > ,summarise > ,col1_mean=mean(col1) > ,col2_mode=Mode(col2) #Function I wrote fo...

...gt; Bert > > > Bert Gunter > > "The trouble with having an open mind is that people keep coming along > and sticking things into it." > -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) > > > On Wed, Apr 13, 2016 at 2:11 PM, Michael Artz <michaeleartz at gmail.com> > wrote: > > 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...

...#5 5 2048-8192 6144 The important point is that paste(round(quantile(x,0.25),0),round(quantile(x,0.75),0),sep="-") is not a function, it is an expression. ddplyr wants functions. Bill Dunlap TIBCO Software wdunlap tibco.com On Tue, Apr 19, 2016 at 7:56 AM, Michael Artz <michaeleartz at gmail.com> wrote: > That didn't work Jim! > > Thanks anyway > > On Mon, Apr 18, 2016 at 9:02 PM, Jim Lemon <drjimlemon at gmail.com> wrote: > > > Hi Michael, > > At a guess, try this: > > > > iqr<-function(x) { > >...

...mething useful. I clearly can't. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Tue, Apr 19, 2016 at 5:29 PM, Michael Artz <michaeleartz at gmail.com> wrote: > 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 fun...

...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 your coffee! I >> ended up just using the transform argument in the ddply function. It >> worked and it repeated, then I called a mode function in another call to >&...

...tp://www.r-bloggers.com/a-brief-tour-of-the-trees-and-forests/ You can get the same kind of information from random forests, but it's less straightforward. If you want a clear set of rules as in your golf example, then you need rpart or similar. Sarah On Wed, Apr 13, 2016 at 6:02 PM, Michael Artz <michaeleartz at gmail.com> wrote: > 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...

...ntral functional programming paradigm). Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Tue, Apr 19, 2016 at 7:56 AM, Michael Artz <michaeleartz at gmail.com> wrote: > That didn't work Jim! > > Thanks anyway > > On Mon, Apr 18, 2016 at 9:02 PM, Jim Lemon <drjimlemon at gmail.com> wrote: > >> Hi Michael, >> At a guess, try this: >> >> iqr<-function(x) { >> retu...

...ly() here? On Thu, Apr 7, 2016 at 6:52 AM, David Barron <dnbarron at gmail.com> wrote: > ifelse is vectorised, so just use that without the loop. > > colordata$response <- ifelse(colordata$color == 'blue', 1, 0) > > David > > On 7 April 2016 at 12:41, Michael Artz <michaeleartz at gmail.com> wrote: > >> Hi I'm not sure how to ask this, but its a very easy question to answer >> for >> an R person. >> >> What is an easy way to check for a column value and then assigne a new >> column a value based on that old co...

...9; + 0 Hope this helps, Rui Barradas ? Citando David Barron <dnbarron at gmail.com>: > ifelse is vectorised, so just use that without the loop. > > colordata$response <- ifelse(colordata$color == 'blue', 1, 0) > > David > > On 7 April 2016 at 12:41, Michael Artz <michaeleartz at gmail.com> wrote: >> Hi I'm not sure how to ask this, but its a very easy question to answer for >> an R person. >> >> What is an easy way to check for a column value and then assigne a new >> column a value based on that old column value? &g...

...D$x[which.max(D$y)] } where the second line of the function gets you the value of 'x' at the maximum of the density estimate. Of course, there is still the possibility of a multi-modal distribution and the nuances of which kernel is used, etc., etc. Food for thought. Regards, Marc Schwartz > On Apr 19, 2016, at 7:07 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > > 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. &...