#Introduction In this case we have an a dataset with four continuous columns and one categorical column. In such a classification problem we seek to classify the flowers with their known attributes on sepal and petal that is the width and length.

#Objective To understand the distribution of the flower continuous attributes from the descriptive statistics. Whether the data is normally distributed or skewed by the difference between mean and median.

To classify the flowers in to the three species category using the four attributes in decision tree algorithms with a section of the dataset and testing the model performance in the classification task with another section of the dataset.

loading required packages for the task

require(rpart)
## Loading required package: rpart
require(rpart.plot)
## Loading required package: rpart.plot
## Warning: package 'rpart.plot' was built under R version 4.2.1
require(mlbench)
## Loading required package: mlbench
require(caret)
## Loading required package: caret
## Loading required package: ggplot2
## Warning: package 'ggplot2' was built under R version 4.2.1
## Loading required package: lattice
require(pROC)
## Loading required package: pROC
## Type 'citation("pROC")' for a citation.
## 
## Attaching package: 'pROC'
## The following objects are masked from 'package:stats':
## 
##     cov, smooth, var
require(dplyr)
## Loading required package: dplyr
## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
require(ggplot2)

loading data and transforming variable of interest in to factors Extracting descriptive statistics of the variables

data(iris)
mydata <- iris
head(mydata)
mydata$Species<- as.factor(mydata$Species)
summary(mydata)
##   Sepal.Length    Sepal.Width     Petal.Length    Petal.Width   
##  Min.   :4.300   Min.   :2.000   Min.   :1.000   Min.   :0.100  
##  1st Qu.:5.100   1st Qu.:2.800   1st Qu.:1.600   1st Qu.:0.300  
##  Median :5.800   Median :3.000   Median :4.350   Median :1.300  
##  Mean   :5.843   Mean   :3.057   Mean   :3.758   Mean   :1.199  
##  3rd Qu.:6.400   3rd Qu.:3.300   3rd Qu.:5.100   3rd Qu.:1.800  
##  Max.   :7.900   Max.   :4.400   Max.   :6.900   Max.   :2.500  
##        Species  
##  setosa    :50  
##  versicolor:50  
##  virginica :50  
##                 
##                 
##

scatter plot of Sepal Length and Sepal width

ggplot(mydata,aes(Sepal.Width,(Sepal.Length),color=Species))+
  geom_point(position="jitter")

linear relationship between Petal Length and Sepal Length

ggplot(mydata, aes(Sepal.Length,Petal.Length , color=Species))+
  geom_point(position = "jitter")+
  geom_smooth()
## `geom_smooth()` using method = 'loess' and formula 'y ~ x'

Correlation test between Sepal width and Petal width

with(subset(mydata,mydata$Species=="setosa"), cor.test(Sepal.Width,Petal.Width))
## 
##  Pearson's product-moment correlation
## 
## data:  Sepal.Width and Petal.Width
## t = 1.6581, df = 48, p-value = 0.1038
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  -0.0487543  0.4800023
## sample estimates:
##      cor 
## 0.232752
with(subset(mydata,mydata$Species=="versicolor"), cor.test(Sepal.Width,Petal.Width))
## 
##  Pearson's product-moment correlation
## 
## data:  Sepal.Width and Petal.Width
## t = 6.1523, df = 48, p-value = 1.467e-07
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  0.4730884 0.7953482
## sample estimates:
##       cor 
## 0.6639987
with(subset(mydata,mydata$Species=="virginica"), cor.test(Sepal.Width,Petal.Width))
## 
##  Pearson's product-moment correlation
## 
## data:  Sepal.Width and Petal.Width
## t = 4.4187, df = 48, p-value = 5.648e-05
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  0.3050368 0.7098315
## sample estimates:
##      cor 
## 0.537728

##Decision Tree model

splitting in to train and test set

set.seed(1234)
ind <- sample(2, nrow(mydata), replace = T, prob = c(0.75, 0.25))
train <- mydata[ind == 1,]
test <- mydata[ind == 2,]

Training the decision tree model

tree <- rpart(Species ~., data = train)
rpart.plot(tree)

printcp(tree)
## 
## Classification tree:
## rpart(formula = Species ~ ., data = train)
## 
## Variables actually used in tree construction:
## [1] Petal.Length Petal.Width 
## 
## Root node error: 77/118 = 0.65254
## 
## n= 118 
## 
##        CP nsplit rel error   xerror     xstd
## 1 0.51948      0  1.000000 1.155844 0.060738
## 2 0.42857      1  0.480519 0.545455 0.067546
## 3 0.01000      2  0.051948 0.090909 0.033326

prediction and assessment on the train set

p <- predict(tree, train, type = 'class')
confusionMatrix(p, train$Species)
## Confusion Matrix and Statistics
## 
##             Reference
## Prediction   setosa versicolor virginica
##   setosa         40          0         0
##   versicolor      0         40         3
##   virginica       0          1        34
## 
## Overall Statistics
##                                           
##                Accuracy : 0.9661          
##                  95% CI : (0.9155, 0.9907)
##     No Information Rate : 0.3475          
##     P-Value [Acc > NIR] : < 2.2e-16       
##                                           
##                   Kappa : 0.9491          
##                                           
##  Mcnemar's Test P-Value : NA              
## 
## Statistics by Class:
## 
##                      Class: setosa Class: versicolor Class: virginica
## Sensitivity                  1.000            0.9756           0.9189
## Specificity                  1.000            0.9610           0.9877
## Pos Pred Value               1.000            0.9302           0.9714
## Neg Pred Value               1.000            0.9867           0.9639
## Prevalence                   0.339            0.3475           0.3136
## Detection Rate               0.339            0.3390           0.2881
## Detection Prevalence         0.339            0.3644           0.2966
## Balanced Accuracy            1.000            0.9683           0.9533

prediction and assessment on the test set

p1 <- predict(tree, test, type = 'class')
confusionMatrix(p1, test$Species)
## Confusion Matrix and Statistics
## 
##             Reference
## Prediction   setosa versicolor virginica
##   setosa         10          0         0
##   versicolor      0          9         2
##   virginica       0          0        11
## 
## Overall Statistics
##                                           
##                Accuracy : 0.9375          
##                  95% CI : (0.7919, 0.9923)
##     No Information Rate : 0.4062          
##     P-Value [Acc > NIR] : 3.355e-10       
##                                           
##                   Kappa : 0.9062          
##                                           
##  Mcnemar's Test P-Value : NA              
## 
## Statistics by Class:
## 
##                      Class: setosa Class: versicolor Class: virginica
## Sensitivity                 1.0000            1.0000           0.8462
## Specificity                 1.0000            0.9130           1.0000
## Pos Pred Value              1.0000            0.8182           1.0000
## Neg Pred Value              1.0000            1.0000           0.9048
## Prevalence                  0.3125            0.2812           0.4062
## Detection Rate              0.3125            0.2812           0.3438
## Detection Prevalence        0.3125            0.3438           0.3438
## Balanced Accuracy           1.0000            0.9565           0.9231

#conclusion The attributes are continuous and normally distributed this is from the descriptive statistics of the dataset. The machine learning model Decision Tree classifier performs we both on the train set and test dataset after assessment.