machinelearninglifting.Rmd

---
title: "Machine Learning for Weight Lifting Exercises"
author: "Jawad ALAOUI"
date: "23 October 2015"
output: html_document
---

First of all, we download an read the data:

```{r, echo=TRUE, eval=TRUE}
train <- read.csv("training.csv")
test <- read.csv("testing.csv")
```

#Model selection

Knowing that we are in front of a classification problem, we suggest to select 3 classification models that we will compare using Cross-validation. Here are suggested models:**Bagging**, **Random Forest** and **Boosting**.

#Features selection

Let's remove variables with negligible variance and that contain no added value for our classification model. We will also remove columns that contain information about the features extraction process.

```{r, echo=TRUE, eval=TRUE, warning=FALSE, results='hide'}
library(caret)
num_features <- dim(train)[2]
train_class <- train[,num_features]
trainfeatures <- train[, -c(1:7, num_features)]
trainfeatures <- trainfeatures[,-nearZeroVar(trainfeatures, saveMetrics = FALSE)]
num_features <- dim(trainfeatures)[2] 
```

Also, we'll remove all covariates were we have more the 50% N.A values.

```{r, echo=TRUE, eval=TRUE, warning=FALSE}
var_na_ratio <- apply(is.na(trainfeatures),2,mean)
trainfeatures <- trainfeatures[,var_na_ratio<0.5]
```

Now, let's remove covariates that introduce too much variance. To do so we need to calculate the features correlation and use *findCorrelation* function to spot the covariates to remove in order to reduce pair-wise correlations.

```{r, echo=TRUE, eval=TRUE, warning=FALSE}
trainfeatures <- trainfeatures[,-findCorrelation(cor(trainfeatures), cutoff = 0.8)]
num_features <- dim(trainfeatures)[2]
print(num_features)
```

$39$ features remain. We can start testing the models. To compare the models, we are going to use **train** function and cross-validation.

As the computation will be time consuming, we are going to use k-fold method with $k = 3$ instead of $10$ (default value). Cross-validation in train function will tune the parameters and select the ones that most perform for each model.

We will also split the training set to 2 different parts. One part will be used for building the model an the second part to make a final evaluation of the 3 models.

```{r, echo=TRUE, eval=TRUE, warning=FALSE}
set.seed(7575)

iTrain <- createDataPartition(y=train_class, p=0.5, list=FALSE)
training_Feats <- trainfeatures[iTrain,]
training_Class <- train_class[iTrain]
eval_Feats <- trainfeatures[-iTrain,]
eval_Class <- train_class[-iTrain]
```

##Bagging Model:

```{r, echo=TRUE, eval=TRUE, warning=FALSE}
control <- trainControl(method = "cv", number = 3)
set.seed(7676)
treebag <- train(training_Class ~ ., data=training_Feats, method="treebag", trControl=control)
print(treebag)
confusionMatrix(eval_Class, predict(treebag, eval_Feats))

```

##Random Forest Model:

```{r, echo=TRUE, eval=TRUE, warning=FALSE}
set.seed(7777)
rf <- train(training_Class~ ., data=training_Feats, method="rf", importance = T, trControl=control)
print(rf)
confusionMatrix(eval_Class, predict(rf, eval_Feats))
```

##Boosting Model:

```{r, echo=TRUE, eval=TRUE, warning=FALSE}
set.seed(7878)
gbm <- train(training_Class~ .,data=training_Feats,method="gbm",verbose = F, trControl=control)
print(gbm)
confusionMatrix(eval_Class, predict(gbm, eval_Feats))
```

#Conclusion

Random forest has the best Accuracy on the out of sample evaluation $98.93 \%$.
the calculated Accuracy with Cross-validation is $98.09 \%$. It's also interesting to look at the Out Of Bag estimate of error rate : $1.39 \%$.

```{r, echo=TRUE, eval=TRUE, warning=FALSE}
rf$finalModel$err.rate[rf$finalModel$ntree, "OOB"]*100

```

All these results show that we have a good performance for the Random forest Model. Therefore, this is the model that we'll keep to predict the outcomes for the test set.

```{r, echo=TRUE, eval=TRUE, warning=FALSE}
predict(rf,test)
```