cart-analysis

R randomForest - how to predict with a “getTree” tree

不羁的心 提交于 2020-01-24 14:23:08
问题 Background: I can make a random Forest in R: set.seed(1) library(randomForest) data(iris) model.rf <- randomForest(Species ~ ., data=iris, importance=TRUE, ntree=20, mtry = 2) I can predict values using the randomForest object that I just made: my_pred <- predict(model.rf) plot(iris$Species,my_pred) I can then peel off some random tree from the forest: idx <- sample(x = 1:20,size = 1,replace = F) single_tree <- getTree(model.rf,k=1) Questions: How do I predict from a single tree pulled from

R randomForest - how to predict with a “getTree” tree

自作多情 提交于 2020-01-24 14:22:25
问题 Background: I can make a random Forest in R: set.seed(1) library(randomForest) data(iris) model.rf <- randomForest(Species ~ ., data=iris, importance=TRUE, ntree=20, mtry = 2) I can predict values using the randomForest object that I just made: my_pred <- predict(model.rf) plot(iris$Species,my_pred) I can then peel off some random tree from the forest: idx <- sample(x = 1:20,size = 1,replace = F) single_tree <- getTree(model.rf,k=1) Questions: How do I predict from a single tree pulled from

Formulate data for rpart

笑着哭i 提交于 2020-01-17 00:44:11
问题 Concatenate columns name of a list to prepare a formula for rpart ? Just wanted to concatenate the names(log_data), log_data is a list of 60 vectors distinct vectors, so I just want their column names in a format so that I can put them in a formula of rpart in r..... like rpart(A ~ B + C + D + E ,log_data) , so here I just want to extract formula="A~B+C+D+E" as a whole string where A,B,C,D,E are the columns name which we have to extract from the log_data, or is there any better way to get a

What does the rpart “Error in as.character(x) : cannot coerce type 'builtin' to vector of type 'character' ” message mean?

核能气质少年 提交于 2019-12-23 19:53:03
问题 I've been banging my head against rpart for a few days now (trying to make classification trees for this dataset that I have), and I think it's time to ask a lifeline at this point :-) I'm sure it's something silly that I'm not seeing, but here's what I've been doing: EuropeWater <- read.csv(file=paste("/Users/artessaniccola/Documents/", "Magic Briefcase/CityTypology/Europe_water.csv",sep="")) library(rpart) attach(EuropeWater) names(EuropeWater) [1] "City" "waterpercapita_m3" "water_class"

How do I interpret rpart splits on factor variables when building classification trees in R?

无人久伴 提交于 2019-12-21 16:19:35
问题 If the factor variable is Climate, with 4 possible values: Tropical, Arid, Temperate, Snow, and a node in my rpart tree is labeled as "Climate:ab", what is the split? 回答1: I assume you use standard way to plot tree which is plot(f) text(f) As you can read in help to text.rpart , argument pretty on default factor variables are presented as letters, so a means levels(Climate)[1] and it means that on left node are observation with Climate==levels(Climate)[1] and on right the others. You could

Search for corresponding node in a regression tree using rpart

谁都会走 提交于 2019-12-18 04:15:33
问题 I'm pretty new to R and I'm stuck with a pretty dumb problem. I'm calibrating a regression tree using the rpart package in order to do some classification and some forecasting. Thanks to R the calibration part is easy to do and easy to control. #the package rpart is needed library(rpart) # Loading of a big data file used for calibration my_data <- read.csv("my_file.csv", sep=",", header=TRUE) # Regression tree calibration tree <- rpart(Ratio ~ Attribute1 + Attribute2 + Attribute3 + Attribute4

Splits and Root node of binary decision tree(CART)

旧时模样 提交于 2019-12-01 20:27:45
问题 How to find a split and root node in a regression tree, I made a regression tree from multiple vectors now I have to extract root node of rpart of multiple vectors.file contains numeric value of multiple vectors A , B , C , D , E , F , G , H ex. A vector contains 4,3,6,7,2,4,5,...and so on similarly others B,C,D,E,F,G,H .so want to extract F (which is a root node in my case) as an output from this input an after creating a tree .thank you.sorry unable to put any image :( Here's what I've done

Splits and Root node of binary decision tree(CART)

杀马特。学长 韩版系。学妹 提交于 2019-12-01 20:10:10
How to find a split and root node in a regression tree, I made a regression tree from multiple vectors now I have to extract root node of rpart of multiple vectors.file contains numeric value of multiple vectors A , B , C , D , E , F , G , H ex. A vector contains 4,3,6,7,2,4,5,...and so on similarly others B,C,D,E,F,G,H .so want to extract F (which is a root node in my case) as an output from this input an after creating a tree .thank you.sorry unable to put any image :( Here's what I've done so far log_data <- read.csv(file="C:\\Users\\AASHU\\Desktop\\CART\\syn.csv", header=T, as.is=T)

Search for corresponding node in a regression tree using rpart

谁说我不能喝 提交于 2019-11-29 04:31:40
I'm pretty new to R and I'm stuck with a pretty dumb problem. I'm calibrating a regression tree using the rpart package in order to do some classification and some forecasting. Thanks to R the calibration part is easy to do and easy to control. #the package rpart is needed library(rpart) # Loading of a big data file used for calibration my_data <- read.csv("my_file.csv", sep=",", header=TRUE) # Regression tree calibration tree <- rpart(Ratio ~ Attribute1 + Attribute2 + Attribute3 + Attribute4 + Attribute5, method="anova", data=my_data, control=rpart.control(minsplit=100, cp=0.0001)) After