primaryobjects · May 5, 2016 16:25
diff --git a/boston.csv b/boston.csv
diff --git a/boston.R b/boston.R
 library(rpart)
 library(rpart.plot)
 library(caTools)
 library(caret)
 library(e1071)

 boston <- read.csv('boston.csv')

 plot(boston$LON, boston$LAT)

 # Plot points on map along river.
 points(boston$LON[boston$CHAS==1], boston$LAT[boston$CHAS==1], col='blue', pch=19)
 # Plot MIT on map.
 points(boston$LON[boston$TRACT==3531], boston$LAT[boston$TRACT==3531], col='red', pch=19)
 # Plot higher-priced real-estate.
 points(boston$LON[boston$MEDV>=21.2], boston$LAT[boston$MEDV>=21.2], col='red', pch=1)

 # Try building a linear model to predict house prices by lat and lon.
 latlonlm <- lm(MEDV ~ LAT + LON, data=boston)

 plot(boston$LON, boston$LAT)
 points(boston$LON[boston$MEDV>=21.2], boston$LAT[boston$MEDV>=21.2], col='red', pch=19)

 # Where does the linear model think high priced houses are?
 # The linear regression model has plotted blue '$' towards the left-side of the map only, missing all the higher priced houses on the right.
 # You can see a well defined line where the linear regression cuts off at. This is the problem that using a regression tree can solve.
 points(boston$LON[latlonlm$fitted.values>=21.2], boston$LAT[latlonlm$fitted.values>=21.2], col='blue', pch='$')

 # Build a regression tree model.
 latlontree <- rpart(MEDV ~ LAT + LON, data=boston)
 prp(latlontree)

 # Plot the original house layout again.
 plot(boston$LON, boston$LAT)
 points(boston$LON[boston$MEDV>=21.2], boston$LAT[boston$MEDV>=21.2], col='red', pch=19)

 # Get the predicted values from the tree.
 fittedValues <- predict(latlontree)

 # Where does the regression tree think high priced houses are?
 # This looks much better. Now the predictions are spread across the map.
 points(boston$LON[fittedValues>=21.2], boston$LAT[fittedValues>=21.2], col='blue', pch='$')

 # Try simplifying the tree.
 latlontree <- rpart(MEDV ~ LAT + LON, data=boston, minbucket=50)
 plot(latlontree)
 text(latlontree)

 # Draw the first few divisions from the tree on the map. Notice how the sections are grouped around thicker clusters.
 plot(boston$LON, boston$LAT)
 abline(v=-71.07)
 abline(h=42.21)
 abline(h=42.17)
 points(boston$LON[boston$MEDV>=21.2], boston$LAT[boston$MEDV>=21.2], col='red', pch=19)

 set.seed(123)

 # Build a training and test set.
 split <- sample.split(boston$MEDV, SplitRatio=0.7)
 train <- subset(boston, split==TRUE)
 test <- subset(boston, split==FALSE)

 # Build a linear regression model on more features.
 linreg <- lm(MEDV ~ LAT + LON + CRIM + ZN + INDUS + CHAS + NOX + RM + AGE + DIS + RAD + TAX + PTRATIO, data=train)

 # Calculate sum of squared errors: 3037.088
 linreg.pred <- predict(linreg, newdata=test)
 linreg.sse <- sum((linreg.pred - test$MEDV)^2)

 # Plot this linear regression predicted prices.
 plot(test$LON, test$LAT)
 points(test$LON[test$MEDV>=21.2], test$LAT[test$MEDV>=21.2], col='red', pch=19)
 points(test$LON[linreg.pred>=21.2], test$LAT[linreg.pred>=21.2], col='blue', pch='!')

 # Build a regression tree.
 tree <- rpart(MEDV ~ LAT + LON + CRIM + ZN + INDUS + CHAS + NOX + RM + AGE + DIS + RAD + TAX + PTRATIO, data=train)
 prp(tree)

 # Calculate sum of squared errors: 4328.988
 # Regression trees are not as good as linear regression for this problem.
 tree.pred <- predict(tree, newdata=test)
 tree.sse <- sum((tree.pred - test$MEDV)^2)

 # Use cross-validation to determine the optimal minimum number of branches for our tree model.
 tr.control <- trainControl(method='cv', number=10)
 cp.grid <- expand.grid(.cp=(0:10)*0.001)

 # Use caret to train the model with cross validation and rpart.
 # The result is cp = 0.001 for the best model.
 tr <- train(MEDV ~ LAT + LON + CRIM + ZN + INDUS + CHAS + NOX + RM + AGE + DIS + RAD + TAX + PTRATIO, data=train, method='rpart', trControl=tr.control, tuneGrid=cp.grid)

 # Notice the best tree is a pretty complex tree.
 best.tree <- tr$finalModel
 prp(best.tree)

 # Calculate sum of squared errors: 3675.766
 # This tree is better, but the linear regression model was better.
 best.tree.pred <- predict(best.tree, newdata=test)
 best.tree.sse <- sum((best.tree.pred - test$MEDV)^2)

 # Plot original map with highest house prices.
 plot(test$LON, test$LAT)
 points(test$LON[test$MEDV>=21.2], test$LAT[test$MEDV>=21.2], col='red', pch=19)
 # Plot best tree's predictions for highest house prices (even though not as good as linear regression model).
 points(test$LON[best.tree.pred >=21.2], test$LAT[best.tree.pred >=21.2], col='blue', pch='!')
diff --git a/plot-best-tree.png b/plot-best-tree.png
diff --git a/plot-linear-regression.png b/plot-linear-regression.png
	library(rpart)
	library(rpart.plot)
	library(caTools)
	library(caret)
	library(e1071)

	boston <- read.csv('boston.csv')

	plot(boston$LON, boston$LAT)

	# Plot points on map along river.
	points(boston$LON[boston$CHAS==1], boston$LAT[boston$CHAS==1], col='blue', pch=19)
	# Plot MIT on map.
	points(boston$LON[boston$TRACT==3531], boston$LAT[boston$TRACT==3531], col='red', pch=19)
	# Plot higher-priced real-estate.
	points(boston$LON[boston$MEDV>=21.2], boston$LAT[boston$MEDV>=21.2], col='red', pch=1)

	# Try building a linear model to predict house prices by lat and lon.
	latlonlm <- lm(MEDV ~ LAT + LON, data=boston)

	plot(boston$LON, boston$LAT)
	points(boston$LON[boston$MEDV>=21.2], boston$LAT[boston$MEDV>=21.2], col='red', pch=19)

	# Where does the linear model think high priced houses are?
	# The linear regression model has plotted blue '$' towards the left-side of the map only, missing all the higher priced houses on the right.
	# You can see a well defined line where the linear regression cuts off at. This is the problem that using a regression tree can solve.
	points(boston$LON[latlonlm$fitted.values>=21.2], boston$LAT[latlonlm$fitted.values>=21.2], col='blue', pch='$')

	# Build a regression tree model.
	latlontree <- rpart(MEDV ~ LAT + LON, data=boston)
	prp(latlontree)

	# Plot the original house layout again.
	plot(boston$LON, boston$LAT)
	points(boston$LON[boston$MEDV>=21.2], boston$LAT[boston$MEDV>=21.2], col='red', pch=19)

	# Get the predicted values from the tree.
	fittedValues <- predict(latlontree)

	# Where does the regression tree think high priced houses are?
	# This looks much better. Now the predictions are spread across the map.
	points(boston$LON[fittedValues>=21.2], boston$LAT[fittedValues>=21.2], col='blue', pch='$')

	# Try simplifying the tree.
	latlontree <- rpart(MEDV ~ LAT + LON, data=boston, minbucket=50)
	plot(latlontree)
	text(latlontree)

	# Draw the first few divisions from the tree on the map. Notice how the sections are grouped around thicker clusters.
	plot(boston$LON, boston$LAT)
	abline(v=-71.07)
	abline(h=42.21)
	abline(h=42.17)
	points(boston$LON[boston$MEDV>=21.2], boston$LAT[boston$MEDV>=21.2], col='red', pch=19)

	set.seed(123)

	# Build a training and test set.
	split <- sample.split(boston$MEDV, SplitRatio=0.7)
	train <- subset(boston, split==TRUE)
	test <- subset(boston, split==FALSE)

	# Build a linear regression model on more features.
	linreg <- lm(MEDV ~ LAT + LON + CRIM + ZN + INDUS + CHAS + NOX + RM + AGE + DIS + RAD + TAX + PTRATIO, data=train)

	# Calculate sum of squared errors: 3037.088
	linreg.pred <- predict(linreg, newdata=test)
	linreg.sse <- sum((linreg.pred - test$MEDV)^2)

	# Plot this linear regression predicted prices.
	plot(test$LON, test$LAT)
	points(test$LON[test$MEDV>=21.2], test$LAT[test$MEDV>=21.2], col='red', pch=19)
	points(test$LON[linreg.pred>=21.2], test$LAT[linreg.pred>=21.2], col='blue', pch='!')

	# Build a regression tree.
	tree <- rpart(MEDV ~ LAT + LON + CRIM + ZN + INDUS + CHAS + NOX + RM + AGE + DIS + RAD + TAX + PTRATIO, data=train)
	prp(tree)

	# Calculate sum of squared errors: 4328.988
	# Regression trees are not as good as linear regression for this problem.
	tree.pred <- predict(tree, newdata=test)
	tree.sse <- sum((tree.pred - test$MEDV)^2)

	# Use cross-validation to determine the optimal minimum number of branches for our tree model.
	tr.control <- trainControl(method='cv', number=10)
	cp.grid <- expand.grid(.cp=(0:10)*0.001)

	# Use caret to train the model with cross validation and rpart.
	# The result is cp = 0.001 for the best model.
	tr <- train(MEDV ~ LAT + LON + CRIM + ZN + INDUS + CHAS + NOX + RM + AGE + DIS + RAD + TAX + PTRATIO, data=train, method='rpart', trControl=tr.control, tuneGrid=cp.grid)

	# Notice the best tree is a pretty complex tree.
	best.tree <- tr$finalModel
	prp(best.tree)

	# Calculate sum of squared errors: 3675.766
	# This tree is better, but the linear regression model was better.
	best.tree.pred <- predict(best.tree, newdata=test)
	best.tree.sse <- sum((best.tree.pred - test$MEDV)^2)

	# Plot original map with highest house prices.
	plot(test$LON, test$LAT)
	points(test$LON[test$MEDV>=21.2], test$LAT[test$MEDV>=21.2], col='red', pch=19)
	# Plot best tree's predictions for highest house prices (even though not as good as linear regression model).
	points(test$LON[best.tree.pred >=21.2], test$LAT[best.tree.pred >=21.2], col='blue', pch='!')
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