### Introduction

In a study, six participants were asked to perform dumbbell lifts correctly and incorrectly in five different ways. Data were gathered from accelerometers on the belt, forearm, arm, and dumbbell of each participant. The test subjects were all males aged 20-28 years and the dumbbells were light, 1.25kg.

This report describes a method for predicting the way in which the participants did the exercise (the classe variable in the data). Class A indicates the correct execution of the exercise while the other four values of this variable represent common mistakes.

To begin the analysis and prediction, relevant packages were installed. Then, the data were downloaded and an exploratory analysis was performed. Finally, the machine learning algorithm was created and testing on a training set of data. The prediction model was also used to predict 20 test cases.

The training and test sets of data were first downloaded (using this function) and assigned to dataframes (the training data set was assigned to train and the testing data set was assigned to test). Packages required for this analysis (caret, doParallel, and randomForest) were loaded into memory using this function.

## download the data if it is not currently in the working directory
## check if the expected directory exists in the working directory.
dir <- paste("./", dir, sep = "")
if (!file.exists(dir)) {
## if not, check whether the expected file exists in the working
## directory
if (!file.exists(file)) {
}
}
}

## download the training data and assign it to a dataframe called "train," converting
## blank observations to NA

## download the testing data and assign it to a dataframe called "test"

## load relevant packages into the R environment, installing them if necessary
pkgInst <- function(x) {
for (i in x) {
## "require" returns TRUE invisibly if it was able to load package
if (!require(i, character.only = TRUE)) {
## if package was not able to be loaded, install it
install.packages(i, dependencies = TRUE,
repos="http://cran.r-project.org/")
require (i, character.only = TRUE)
}
}
}

## assign names of required packages to "pkgs"
pkgs <- c("R2HTML", "caret", "doParallel", "randomForest")

pkgInst(pkgs)
## Loading required package: R2HTML
## randomForest 4.6-12
## Type rfNews() to see new features/changes/bug fixes.
##
## Attaching package: 'randomForest'
##
## The following object is masked from 'package:ggplot2':
##
##     margin

### Exploratory Data Analysis

First, a summary of the training data was displayed. For the sake of brevity, this is available here.

## assign the summary table of the training data set to the variable summ
summ <- summary(train)

## use the R2HTML package's "HTML" function to write the output to an html file
HTML(summ, file="summary.html")

For purposes of this exercise, the row number (variable X) is irrelevant and is thus dropped from the analysis. Also, several columns in the dataframe have 19,216 observations with value of NA (e.g., max_roll_belt). These are removed from consideration as having complete data sets is likely to increase prediction accuracy.

## remove the column with the row number
train$X <- NULL ## remove the columns with 19,216 NA observations train <- train[,colSums(is.na(train)) < 19216] The remaining fifty-eight (58) variables were used as predictors (there are (fifty-nine (59) remaining columns including classe). A histogram showing the frequency of the values of classe was created to see if there is a large skew in the data toward one of the possible ways of performing the exercise. The frequency histogram was created using the method described in this Stack Overflow answer ## create a variable to represent the x-axis as "classe" is a factor and ## histograms can only be drawn for numeric data classnum <- as.numeric(train$classe)

## create the histogram and change its density to percentages
h <- hist(classnum)
h$density <- h$counts / sum(h$counts) * 100 ## plot the histogram, leaving the x-axis blank plot(h, freq = F, main = "Histogram of Classe for the Training Set", ylab = "Percentage", xlab="Classe", xaxt="n", col="blue") ## add the axis axis(1,at=1:5,labels = c("A","B","C","D","E")) Based on the histogram, it does not seem that any of the values is much more likely than any of the others. ### Prediction Algorithm A random forest was chosen as the prediction algorithm, despite its relatively slow computational speed, due to the likelihood that accuracy would be increased. The caret package’s train function was used to fit the model, using cross validation with ten repeats (k = 10). Principal Components Analysis (PCA) was also used to preprocess the data in order to further reduce the number of predictors. Two cores (the total number of processors available on the machine fitting the model) were used to aid in decreasing computational time. The prediction algorithm’s code chunk was also cached to facilitate re-running the model. ## use two cores registerDoParallel(cores=2) ## set trainControl parameters ctrl <- trainControl(method = "cv", repeats = 10) ## fit the model modelFit <- train(classe ~ ., data = train, method = "rf", trControl = ctrl, preProcess = "pca") ### In-Sample Error: Accuracy with Training Set The prediction algorithm was then used to predict the training data set in order to determine its accuracy. ## first, display the model's output modelFit ## Random Forest ## ## 19622 samples ## 58 predictor ## 5 classes: 'A', 'B', 'C', 'D', 'E' ## ## Pre-processing: principal component signal extraction (80), centered ## (80), scaled (80) ## Resampling: Cross-Validated (10 fold) ## Summary of sample sizes: 17659, 17660, 17662, 17658, 17659, 17661, ... ## Resampling results across tuning parameters: ## ## mtry Accuracy Kappa Accuracy SD Kappa SD ## 2 0.9867494 0.9832400 0.001982079 0.002506222 ## 41 0.9815513 0.9766656 0.002694792 0.003407737 ## 80 0.9805827 0.9754408 0.002756402 0.003486487 ## ## Accuracy was used to select the optimal model using the largest value. ## The final value used for the model was mtry = 2. ## then predict the training set predTrain <- predict(modelFit, train) ## then display the accuracy of the model table(predTrain,train$classe)
##
## predTrain    A    B    C    D    E
##         A 5580    0    0    0    0
##         B    0 3797    0    0    0
##         C    0    0 3422    0    0
##         D    0    0    0 3216    0
##         E    0    0    0    0 3607

From the above table, the prediction algorithm’s in-sample error (the accuracy of its predictions for the training set) is 100%. This follows as the training set was used to create the prediction algorithm, though this accuracy is very high even for in-sample error. Of course, the actual predictive accuracy of the algorithm is the out of sample error, which was next calculated by predicting the values of the test set. This must be less than or equal to the in-sample error (regardless of the in-sample error’s actual value).

### Out of Sample Error: Accuracy with Testing Set

Per the assignment, the prediction algorithm created above is applied to the twenty (20) cases in the testing data set. For each test case, a text file is created with a single capital letter (A, B, C, D, or E) corresponding to the prediction for the corresponding problem in the test data set. These files were then manually submitted to Coursera for grading.

## predict the testing set and store the results in a character vector
predTest <- as.character(predict(modelFit, test))

## display the predictions
predTest
##  [1] "B" "A" "A" "A" "A" "E" "D" "B" "A" "A" "B" "C" "B" "A" "E" "E" "A"
## [18] "B" "B" "B"
## create a directory called "test_output," suppressing warnings if the
dir.create(file.path("test_output"), showWarnings = FALSE)

## create a function to write the files to be submitted
pml_write_files = function(x){
n = length(x)
for(i in 1:n){
filename = paste0("test_output/problem_id_", i, ".txt")
write.table(x[i], file = filename, quote = FALSE, row.names = FALSE,
col.names=FALSE)
}
}

## then create the files, one for each prediction
pml_write_files(predTest)

According to the “Prediction Assignment Submission” on Coursera, the algorithm described above accurately predicted nineteen (19) of the twenty (20) test cases, so its out of sample accuracy (accuracy with the test data) was 95% – only test problem 3 was incorrectly predicted. This level of accuracy makes sense due to out of sample error necessarily being less than or equal to in-sample error. Due to the small sample size of the test data set, it is possible that the algorithm would not perform as well for a larger data set.

### Citations

• Liaw, A. and M. Wiener (2002). Classification and Regression by randomForest. R News Vol. 2 N. 3, Vienna, Austria.
• Lecoutre, Eric (2003). The R2HTML Package. R News, Vol 3 N. 3, Vienna, Austria.
• Velloso, E., Bulling, A., Gellersen, H., Ugulino, W., Fuks, H. (2013). Qualitative Activity Recognition of Weight Lifting Exercises.
Proceedings of 4th International Conference in Cooperation with SIGCHI (Augmented Human ’13), ACM SIGCHI, Stuttgart, Germany.