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<h1>Machine Learning Project</h1>
<p>This is an R Markdown document. Or is it?</p>
<p>The purpose of this porject is to develop a model for predicting a type of activity through supervised learning based on aquired data.</p>
<p>The subject in question is of multiple test subjects performing exercises with gyroscopes and accelerametor sensors attached to them. There are 5 exercises labled A to E that describe the exercise they performed. Using this data it should be possible to predict the exercise type based soley on the data.</p>
<p>However, none of this matters! By using feature selection and random forests, we can build a model that predicts the type of activity without knowing much about the data at all. As long as we have an accuracy of 95% or more, we can be confident that the algorithm works.</p>
<p>First we load the data. We are given a testing set of 19622 rows and a testing set of 20 rows. We will eventually split the training set up into another training and testing set in order to validate our model before applying it.</p>
<pre><code>## Loading required package: caret
## Loading required package: lattice
## Loading required package: ggplot2
</code></pre>
<pre><code class="r">training <- read.csv('pml-training.csv',na.strings = c("NA",""))
testing <- read.csv('pml-testing.csv',na.strings = c("NA",""))
</code></pre>
<p>Next, we need to remove columns where we didn't receive any data. Remember that any transformation we make to the training set, we have to make to the testing set.</p>
<pre><code class="r">keep <- colSums(training !=0) != 0
keep <- keep[keep == TRUE]
training2<- training[,which(!is.na(names(keep)))]
testing <- testing[,which(!is.na(names(keep)))]
</code></pre>
<p>Then we remove columns with low variance. If a column has little variance, then it is a bad predictor since every activity type will have similar values.</p>
<pre><code class="r">nsv <- nearZeroVar(training2,saveMetrics=TRUE)
training2 <- training2[,which(nsv$freqRatio > 1.1)]
testing <- testing[,which(nsv$freqRatio > 1.1)]
</code></pre>
<p>There are also columns that relate to timestamps that don't make any sense to use.</p>
<pre><code class="r">testing <- testing[,-c(1,3,4,5,6)]
training2 <- training2[,-c(1,3,4,5,6)]
</code></pre>
<p>Since we want to know how well our algorithm is performing, we split the training data up again:</p>
<pre><code class="r">inTrain= createDataPartition(y=training2$classe,p=0.75,list=FALSE)
training3 <- training2[inTrain,]
subTrain <- training2[-inTrain,]
</code></pre>
<p>Finally, we run a random forest algorithm on the training3 dataset to build a model using random forests</p>
<pre><code class="r">modelFit <- train(classe ~ .,method="rf",data=training3)
</code></pre>
<pre><code>## Loading required package: randomForest
</code></pre>
<pre><code>## Warning: package 'randomForest' was built under R version 3.1.1
</code></pre>
<pre><code>## randomForest 4.6-10
## Type rfNews() to see new features/changes/bug fixes.
</code></pre>
<p>We can then run a prediction on the subTrain dataset using this model and compare the results</p>
<pre><code class="r">testPC <- predict(modelFit,subTrain)
dazed <- confusionMatrix(subTrain$classe,testPC)
testAccuracy = dazed$overall['Accuracy']
</code></pre>
<p>We find that we have an accuracy of 0.9802 ! That's better than our accepted accuracy of 95%.</p>
<p>All that's left to do is run the prediction on our test set:</p>
<pre><code class="r">testTC <- predict(modelFit,testing)
</code></pre>
<p>And the answers are:
B, A, B, A, A, E, D, B, A, A, B, C, B, A, E, E, A, B, B, B</p>
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