Beginner ML Project
Demo
February 23, 2021
Flower Classification
Project steps: 1. Define problem. 2. Prepare data. 3. Evaluate algorithms. 4. Improve results. 5. Present results.
Load Packages
library(caret)## Loading required package: lattice## Loading required package: ggplot2Load Data
data(iris)
datatset <- irisCreate a Test Dataset
test_index <- createDataPartition(datatset$Species, p=0.80, list=FALSE)
test <- datatset[-test_index, ]
train <- datatset[test_index, ]Summarize Dataset
#dimensions of train & test dataset
dim(train)## [1] 120 5dim(test)## [1] 30 5#Types of attributes
sapply(train, class)## Sepal.Length Sepal.Width Petal.Length Petal.Width Species
## "numeric" "numeric" "numeric" "numeric" "factor"#Peek at the data first 6 rows of the data
head(train)## Sepal.Length Sepal.Width Petal.Length Petal.Width Species
## 1 5.1 3.5 1.4 0.2 setosa
## 2 4.9 3.0 1.4 0.2 setosa
## 3 4.7 3.2 1.3 0.2 setosa
## 4 4.6 3.1 1.5 0.2 setosa
## 5 5.0 3.6 1.4 0.2 setosa
## 6 5.4 3.9 1.7 0.4 setosa#levels for the class (target)
levels(train$Species)## [1] "setosa" "versicolor" "virginica"#summarize the class distribution
percentage <- prop.table(table(train$Species)) * 100
cbind(freq = table(train$Species),percentage=percentage)## freq percentage
## setosa 40 33.33333
## versicolor 40 33.33333
## virginica 40 33.33333#Statistical Summary
summary(train)## Sepal.Length Sepal.Width Petal.Length Petal.Width
## Min. :4.300 Min. :2.200 Min. :1.100 Min. :0.100
## 1st Qu.:5.100 1st Qu.:2.800 1st Qu.:1.575 1st Qu.:0.300
## Median :5.800 Median :3.000 Median :4.300 Median :1.300
## Mean :5.847 Mean :3.057 Mean :3.772 Mean :1.197
## 3rd Qu.:6.400 3rd Qu.:3.400 3rd Qu.:5.100 3rd Qu.:1.800
## Max. :7.900 Max. :4.200 Max. :6.900 Max. :2.500
## Species
## setosa :40
## versicolor:40
## virginica :40
##
##
## Visualize Dataset
#Split into features and target
X_train <- train[,1:4]
y_train <- train[,5]#Boxplot for each attribute on one image
par(mfrow=c(1,4))
for(i in 1:4){
boxplot(X_train[,i], main=names(train)[i])
}
#Barplot for class breakdown
plot(y_train)
#Scatterplot Matrix
featurePlot(X_train, y_train, plot="box")
#Scatterplot Matrix
featurePlot(X_train, y_train, plot="pairs", auto.key= list(columns = 3))
#Density Plots for each attribute by class value
scales <- list(x=list(relation="free"), y=list(relation="free"))
featurePlot(X_train, y_train, plot="density", scales=scales, auto.key= list(columns = 3))
Evaluate Some Algorithms
#Run algorithms using 10-fold cross validation
control <- trainControl(method = "cv", number = 10)
metric <- "Accuracy"#Build models
#(a) linear algorithms
set.seed(7)
fit.lda <- train(Species~., data=train, method="lda", metric=metric, trControl=control)
#(b) nonlinear algorithms
#CART
set.seed(7)
fit.cart <- train(Species~., data=train, method="rpart", metric=metric, trControl=control)
#kNN
set.seed(7)
fit.knn <- train(Species~., data=train, method="knn", metric=metric, trControl=control)
#(c) advanced algorithms
#SVM
set.seed(7)
fit.svm <- train(Species~., data=train, method="svmRadial", metric=metric, trControl=control)
#Random Forest
set.seed(7)
fit.rf <- train(Species~., data=train, method="rf", metric=metric, trControl=control)#Select best model
# summarize accuracy of models
results <- resamples(list(lda=fit.lda, cart=fit.cart, knn=fit.knn, svm=fit.svm, rf=fit.rf))
summary(results)##
## Call:
## summary.resamples(object = results)
##
## Models: lda, cart, knn, svm, rf
## Number of resamples: 10
##
## Accuracy
## Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
## lda 0.8333333 1.0000000 1.0000000 0.9750000 1.0000000 1 0
## cart 0.8333333 0.9166667 0.9166667 0.9166667 0.9166667 1 0
## knn 0.9166667 0.9166667 0.9583333 0.9583333 1.0000000 1 0
## svm 0.8333333 0.9166667 0.9166667 0.9333333 0.9791667 1 0
## rf 0.8333333 0.9166667 0.9583333 0.9500000 1.0000000 1 0
##
## Kappa
## Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
## lda 0.750 1.000 1.0000 0.9625 1.00000 1 0
## cart 0.750 0.875 0.8750 0.8750 0.87500 1 0
## knn 0.875 0.875 0.9375 0.9375 1.00000 1 0
## svm 0.750 0.875 0.8750 0.9000 0.96875 1 0
## rf 0.750 0.875 0.9375 0.9250 1.00000 1 0# compare accuracy of models
dotplot(results)
# summarize Best Model
print(fit.knn)## k-Nearest Neighbors
##
## 120 samples
## 4 predictor
## 3 classes: 'setosa', 'versicolor', 'virginica'
##
## No pre-processing
## Resampling: Cross-Validated (10 fold)
## Summary of sample sizes: 108, 108, 108, 108, 108, 108, ...
## Resampling results across tuning parameters:
##
## k Accuracy Kappa
## 5 0.9500000 0.9250
## 7 0.9583333 0.9375
## 9 0.9500000 0.9250
##
## Accuracy was used to select the optimal model using the largest value.
## The final value used for the model was k = 7.Make Predictions
# estimate skill of kNN on the validation dataset
predictions <- predict(fit.knn, test)
confusionMatrix(predictions, test$Species)## Confusion Matrix and Statistics
##
## Reference
## Prediction setosa versicolor virginica
## setosa 10 0 0
## versicolor 0 10 0
## virginica 0 0 10
##
## Overall Statistics
##
## Accuracy : 1
## 95% CI : (0.8843, 1)
## No Information Rate : 0.3333
## P-Value [Acc > NIR] : 4.857e-15
##
## Kappa : 1
##
## Mcnemar's Test P-Value : NA
##
## Statistics by Class:
##
## Class: setosa Class: versicolor Class: virginica
## Sensitivity 1.0000 1.0000 1.0000
## Specificity 1.0000 1.0000 1.0000
## Pos Pred Value 1.0000 1.0000 1.0000
## Neg Pred Value 1.0000 1.0000 1.0000
## Prevalence 0.3333 0.3333 0.3333
## Detection Rate 0.3333 0.3333 0.3333
## Detection Prevalence 0.3333 0.3333 0.3333
## Balanced Accuracy 1.0000 1.0000 1.0000Reference
Brownlee, J. (2019, October 07). Your first machine learning project in R Step-By-Step. Retrieved February 15, 2021, from https://machinelearningmastery.com/machine-learning-in-r-step-by-step/