Executive summary

One thing that people regularly do is quantify how much of a particular activity they do, but they rarely quantify how well they do it.
In this project, we have seen that using the data from accelerometers on the belt, forearm, arm, and dumbell we could predict with Random Forest method the manner in which 6 participants did an exercise with 99.3% accurancy.

Data processing

Load data

suppressWarnings(library(caret))
training <- read.csv("Data/pml-training.csv", header=T) #load training data
test <- read.csv("Data/pml-testing.csv", header=T) #load test data

Cleaning data

training <- subset(training, select=-c(1:6)) #remove row numbers, time stamp and user names
test <- subset(test, select=-c(1:6)) #remove row numbers, time stamp and user names
classe <- training$classe #save classe information
training <- training[ , colSums(is.na(training)) == 0] #Remove all columns containing at least one NA
test <- test[ , colSums(is.na(test)) == 0] #Remove all columns containing at least one NA
training <- training[,sapply(training,is.numeric)] # remove non-numeric variables
test <- test[,sapply(test,is.numeric)] # remove non-numeric variables
training$classe <- classe; rm(classe) #add classe information and delete variable

Prepare training, validate and test data

Training data set: - It includes the parameter to estimate: classe - We will split the training set on two subsets (50% observations each), train and validate
- Train to build the model and validate to validate it

Test data set: - Includes 20 observations, without the parameter classe but with problem_id

set.seed(33433)
inTraining <- createDataPartition(training$classe, p=0.5, list=FALSE)
train <- training[ inTraining,]
validate  <- training[-inTraining,]
rm(training, inTraining)

Fit model

We will follow the recommendation of the study and use a Random Forest method to model the train data.
Fore the model accurancy we will be using a 10 fold cross validation and each forest with 10 trees. If accuracy is low then we will try a second model.

set.seed(33433)
tr <- trainControl(method = "cv", number = 10) #setting parameters for the model
fit.rf <- train(classe~.,                      #fit model
              method = "rf", #random forest model
              data = train, 
              trControl = tr, #model parameters
              allowParallel = TRUE, #use parallel processing (if availible)
              ntree = 10, #number of trees
              verbose = FALSE)

print(fit.rf$finalModel)
## 
## Call:
##  randomForest(x = x, y = y, ntree = 10, mtry = param$mtry, allowParallel = TRUE,      verbose = FALSE) 
##                Type of random forest: classification
##                      Number of trees: 10
## No. of variables tried at each split: 27
## 
##         OOB estimate of  error rate: 2.47%
## Confusion matrix:
##      A    B    C    D    E class.error
## A 2726   21    5    7    4  0.01339124
## B   23 1814   24   15    4  0.03510638
## C    2   26 1651   17    0  0.02653302
## D    2   14   28 1533    8  0.03280757
## E    2   16    9   13 1749  0.02235886

Estimating model accuracy

We evaluate the model with the validation data

set.seed(33433)
pred.rf <- predict(fit.rf, newdata = validate) #Predict with validation data
confusionMatrix(pred.rf, validate$classe)$overall[1] #Measure accuracy with validation data
## Accuracy 
## 0.993578

We get 99.3% accuracy!!
There is no need to improve the model

Most relevant variables

Here we plot the variables that have a higher impact on the fitted model

VarImportance <- varImp(fit.rf)
plot(VarImportance, main = "Most relevant variables", top = 10)

Predict classe

Now we will predit classe for each problem_id of the test data set

pred.rf.test<-predict(fit.rf, newdata = test[,-which(names(test) %in% "problem_id")])
t(data.frame(problem_id = test$problem_id, prediction = pred.rf.test))
##            [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
## problem_id " 1" " 2" " 3" " 4" " 5" " 6" " 7" " 8" " 9" "10"  "11"  "12" 
## prediction "B"  "A"  "B"  "A"  "A"  "E"  "D"  "B"  "A"  "A"   "B"   "C"  
##            [,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20]
## problem_id "13"  "14"  "15"  "16"  "17"  "18"  "19"  "20" 
## prediction "B"   "A"   "E"   "E"   "A"   "B"   "B"   "B"

Data source