Practical ML Project - Exercise Prediction
Avijit
03rd March 2021
Synopsis
Using devices such as Jawbone Up, Nike FuelBand, and Fitbit it is now possible to collect a large amount of data about personal activity relatively inexpensively. This project uses data from accelerometers on the belt, forearm, arm, and dumbell of 6 participants who were asked to perform barbell lifts correctly and incorrectly in 5 different ways. The goal of your project is to predict the manner in which they did the exercise using the collected data. This feature is reflected in the “classe” variable in the training set.
Importing libraries
setwd("D:/R working directory/Practical ML Project")
library(RCurl)
library(caret)
library(rpart)
library(rpart.plot)
library(randomForest)
library(corrplot)Dataset
Loading data
trainFile0 <- "./pml-training.csv"
testFile0 <- "./pml-testing.csv"
if (!file.exists(trainFile0)) {
download.file("https://d396qusza40orc.cloudfront.net/predmachlearn/pml-training.csv",
destfile=trainFile0, method="curl")}
pml.data <- read.csv(trainFile0)
if (!file.exists(testFile0)) {
download.file("https://d396qusza40orc.cloudfront.net/predmachlearn/pml-testing.csv",
destfile=testFile0, method="curl")}
pml.validation <- read.csv(testFile0)Viewing data
dim(pml.data)## [1] 19622 160dim(pml.validation)## [1] 20 160w <- which(names(pml.data)!=names(pml.validation));w #160## [1] 160names(pml.data)[w]## [1] "classe"names(pml.validation)[w]## [1] "problem_id"table(pml.data[,w])##
## A B C D E
## 5580 3797 3422 3216 3607pml.validation[,w]## [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20Cleaning data
Columns containing more than 90% missing values and those with user info and timestamp are removed.
w <- which(colSums(is.na(pml.data)|pml.data=="")>0.9*dim(pml.data)[1])
as.vector(w)## [1] 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
## [19] 30 31 32 33 34 35 36 50 51 52 53 54 55 56 57 58 59 69
## [37] 70 71 72 73 74 75 76 77 78 79 80 81 82 83 87 88 89 90
## [55] 91 92 93 94 95 96 97 98 99 100 101 103 104 105 106 107 108 109
## [73] 110 111 112 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139
## [91] 141 142 143 144 145 146 147 148 149 150pml.data.c <- pml.data[,-w]
pml.data.c <- pml.data.c[,-c(1:7)] #user info and timestamp
pml.data.c$classe <- as.factor(pml.data.c$classe)
dim(pml.data); dim(pml.data.c)## [1] 19622 160## [1] 19622 53Same process is performed on validation dataset.
w <- which(colSums(is.na(pml.validation)|pml.validation=="")>
0.9*dim(pml.validation)[1])
pml.validation.c <- pml.validation[,-w]
pml.validation.c <- pml.validation.c[,-c(1:7)] #user info and timestamp
dim(pml.validation); dim(pml.validation.c)## [1] 20 160## [1] 20 53After cleaning the data, number of columns are narrowed down from 160 to 53 relevant ones
Splitting data
Data is split into training and testing datasets with a ratio of 70:30
set.seed(666)
inTrain <- createDataPartition(y=pml.data.c$classe, p=0.7, list=F)
training <- pml.data.c[inTrain,]
testing <- pml.data.c[-inTrain,]
dim(training)## [1] 13737 53dim(testing)## [1] 5885 53Prediction Models
1. Random forest
controlRf <- trainControl(method="cv", 5)
mod.rf <- train(classe ~ ., data=training, method="rf",
trControl=controlRf, ntree=250)
mod.rf$finalModel##
## Call:
## randomForest(x = x, y = y, ntree = 250, mtry = param$mtry)
## Type of random forest: classification
## Number of trees: 250
## No. of variables tried at each split: 27
##
## OOB estimate of error rate: 0.71%
## Confusion matrix:
## A B C D E class.error
## A 3903 1 1 0 1 0.0007680492
## B 23 2628 7 0 0 0.0112866817
## C 0 14 2371 11 0 0.0104340568
## D 0 0 29 2220 3 0.0142095915
## E 0 1 4 3 2517 0.0031683168plot(mod.rf)
Random Forest Testing
pred.rf <- predict(mod.rf, testing)
cm.rf <- confusionMatrix(testing$classe, pred.rf)
cm.rf$table## Reference
## Prediction A B C D E
## A 1672 2 0 0 0
## B 9 1128 2 0 0
## C 0 5 1016 4 1
## D 0 0 11 952 1
## E 0 0 3 0 1079cm.rf$overall[1] # Accuracy of model = 99.35%## Accuracy
## 0.99354291 - as.numeric(cm.rf$overall[1]) # out-of-sample error is 0.65%## [1] 0.006457094Accuracy of Random Forest Model = 99.35%
2. Regression Trees
mod.rpart <- rpart(classe ~ ., data=training, method="class")
prp(mod.rpart)
Regression Trees Testing
pred.rpart <- predict(mod.rpart, testing, type="class")
cm.rpart <- confusionMatrix(testing$classe, pred.rpart)
cm.rpart$table## Reference
## Prediction A B C D E
## A 1513 56 47 25 33
## B 185 625 109 92 128
## C 21 50 825 63 67
## D 57 79 150 603 75
## E 17 63 121 64 817cm.rpart$overall[1] # Accuracy of model = 74.5%## Accuracy
## 0.74477491 - as.numeric(cm.rpart$overall[1]) # out-of-sample error is 25.5%## [1] 0.2552251Accuracy of Regression Trees Model = 74.5%
3. Gradient Boosting
mod.gbm <- train(classe ~ ., data=training, verbose=F, method="gbm")
mod.gbm$finalModel## A gradient boosted model with multinomial loss function.
## 150 iterations were performed.
## There were 52 predictors of which 52 had non-zero influence.Gradient Boosting Testing
pred.gbm <- predict(mod.gbm, testing)
cm.gbm <- confusionMatrix(testing$classe, pred.gbm)
cm.gbm$table## Reference
## Prediction A B C D E
## A 1651 12 4 6 1
## B 48 1058 30 3 0
## C 1 36 978 10 1
## D 0 3 37 919 5
## E 2 8 17 9 1046cm.gbm$overall[1] # Accuracy of model = 96%## Accuracy
## 0.96040781 - as.numeric(cm.gbm$overall[1]) # out-of-sample error is 4%## [1] 0.03959218Accuracy of Gradient Boosting Model = 96%
Results
Out of the tested three machine learning prediction models, the random forest model stands out with the highest accuracy of prediction of 99.35%. So, we deploy random forest model to the validation dataset and obtain the following results.
predict(mod.rf, pml.validation.c)## [1] B A B A A E D B A A B C B A E E A B B B
## Levels: A B C D EAfter submitting to prediction assignment part of the project, we find that all 20/20 predicted values were correct.