Machine Learning Engineer Nanodegree¶

Model Evaluation & Validation¶

Project: Predicting Boston Housing Prices¶

By Michael Eryan¶

Welcome to the first project of the Machine Learning Engineer Nanodegree! In this notebook, some template code has already been provided for you, and you will need to implement additional functionality to successfully complete this project. You will not need to modify the included code beyond what is requested. Sections that begin with 'Implementation' in the header indicate that the following block of code will require additional functionality which you must provide. Instructions will be provided for each section and the specifics of the implementation are marked in the code block with a 'TODO' statement. Please be sure to read the instructions carefully!

In addition to implementing code, there will be questions that you must answer which relate to the project and your implementation. Each section where you will answer a question is preceded by a 'Question X' header. Carefully read each question and provide thorough answers in the following text boxes that begin with 'Answer:'. Your project submission will be evaluated based on your answers to each of the questions and the implementation you provide.

Note: Code and Markdown cells can be executed using the Shift + Enter keyboard shortcut. In addition, Markdown cells can be edited by typically double-clicking the cell to enter edit mode.

Getting Started¶

In this project, you will evaluate the performance and predictive power of a model that has been trained and tested on data collected from homes in suburbs of Boston, Massachusetts. A model trained on this data that is seen as a good fit could then be used to make certain predictions about a home — in particular, its monetary value. This model would prove to be invaluable for someone like a real estate agent who could make use of such information on a daily basis.

The dataset for this project originates from the UCI Machine Learning Repository. The Boston housing data was collected in 1978 and each of the 506 entries represent aggregated data about 14 features for homes from various suburbs in Boston, Massachusetts. For the purposes of this project, the following preprocessing steps have been made to the dataset:

16 data points have an 'MEDV' value of 50.0. These data points likely contain missing or censored values and have been removed.
1 data point has an 'RM' value of 8.78. This data point can be considered an outlier and has been removed.
The features 'RM', 'LSTAT', 'PTRATIO', and 'MEDV' are essential. The remaining non-relevant features have been excluded.
The feature 'MEDV' has been multiplicatively scaled to account for 35 years of market inflation.

Run the code cell below to load the Boston housing dataset, along with a few of the necessary Python libraries required for this project. You will know the dataset loaded successfully if the size of the dataset is reported.

# Import libraries necessary for this project
import warnings
warnings.simplefilter(action = "ignore", category = FutureWarning)
#added because was still giving me warnings

import numpy as np
import pandas as pd
from sklearn.cross_validation import ShuffleSplit

# Import supplementary visualizations code visuals.py
import visuals as vs

# Pretty display for notebooks
%matplotlib inline

# ME: some other libraries I need
import sklearn
print('The scikit-learn version is {}.'.format(sklearn.__version__))

from sklearn.metrics import r2_score
from sklearn.cross_validation import train_test_split
from sklearn.tree import DecisionTreeRegressor
from sklearn.metrics import make_scorer
from sklearn.grid_search import GridSearchCV

import matplotlib.pyplot as plt
import seaborn as sns

The scikit-learn version is 0.19.0.

# Load the Boston housing dataset
data = pd.read_csv('housing.csv')
prices = data['MEDV']
features = data.drop('MEDV', axis = 1)
    
print ("Boston housing dataset has {} data points with {} variables each.".format(*data.shape))
print ("The data has already been cleaned up, some observations were dropped and only 3 relevant features kept.") 
print ("\n")
print (data.head(n=1))

Boston housing dataset has 489 data points with 4 variables each.
The data has already been cleaned up, some observations were dropped and only 3 relevant features kept.


      RM  LSTAT  PTRATIO      MEDV
0  6.575   4.98     15.3  504000.0

Data Exploration¶

In this first section of this project, you will make a cursory investigation about the Boston housing data and provide your observations. Familiarizing yourself with the data through an explorative process is a fundamental practice to help you better understand and justify your results.

Since the main goal of this project is to construct a working model which has the capability of predicting the value of houses, we will need to separate the dataset into features and the target variable. The features, 'RM', 'LSTAT', and 'PTRATIO', give us quantitative information about each data point. The target variable, 'MEDV', will be the variable we seek to predict. These are stored in features and prices, respectively.

Implementation: Calculate Statistics¶

For your very first coding implementation, you will calculate descriptive statistics about the Boston housing prices. Since numpy has already been imported for you, use this library to perform the necessary calculations. These statistics will be extremely important later on to analyze various prediction results from the constructed model.

In the code cell below, you will need to implement the following:

Calculate the minimum, maximum, mean, median, and standard deviation of 'MEDV', which is stored in prices.
- Store each calculation in their respective variable.

#Summary statistics for the target variable - using Numpy

# TODO: Minimum price of the data
minimum_price = np.min(prices)

# TODO: Maximum price of the data
maximum_price = np.max(prices)

# TODO: Mean price of the data
mean_price = np.mean(prices)

# TODO: Median price of the data
median_price = np.median(prices)

# TODO: Standard deviation of prices of the data
std_price = np.std(prices)

# Show the calculated statistics
print ("Statistics for Boston housing dataset:\n")
print ("Minimum price: ${:,.2f}".format(minimum_price))
print ("Maximum price: ${:,.2f}".format(maximum_price))
print ("Mean price: ${:,.2f}".format(mean_price))
print ("Median price ${:,.2f}".format(median_price))
print ("Standard deviation of prices: ${:,.2f}".format(std_price))

Statistics for Boston housing dataset:

Minimum price: $105,000.00
Maximum price: $1,024,800.00
Mean price: $454,342.94
Median price $438,900.00
Standard deviation of prices: $165,171.13

#ME some additional stats 
print ("See all the statistics for the target variable together using Pandas. \n")
pd.options.display.float_format = '${:,.2f}'.format
print (prices.describe(percentiles=[0.01,.25, .5, .75, 0.99]))
pd.reset_option("display.float_format")

See all the statistics for the target variable together using Pandas. 

count         $489.00
mean      $454,342.94
std       $165,340.28
min       $105,000.00
1%        $147,000.00
25%       $350,700.00
50%       $438,900.00
75%       $518,700.00
99%       $954,912.00
max     $1,024,800.00
Name: MEDV, dtype: float64

#ME: 
plt.title("Boston Housing Prices Boxplot")
plt.ylabel("$")
plt.boxplot(prices)
plt.show()

#ME: Additional goodlooking EDA
#Scatterplot matrix using Seaborn
sns.pairplot(data, size=2.5)

plt.title("Scatterplot Matrix")
plt.tight_layout()
plt.show()
print ("\n Look especially at the target variable MEDV charts: RM (rooms) and LSTAT (status of population) have clear relationships with MEDV, while PTRATIO is noisier.")

#Pearson correlation heatmap 
corr = np.corrcoef(data.values.T)
heatm = sns.heatmap(corr,
                 cbar=True,
                 annot=True,
                 square=True,
                 fmt='.2f',
                 annot_kws={'size': 15},
                 yticklabels=list(data),
                 xticklabels=list(data))

plt.title("Correlation Heat Map")
plt.tight_layout()
plt.show()

print ("\n Correlation heat map nicely complements the scatterplot matrix. RM and LSTAT are strongly correlated with MEDV, while PTRATIO (pupil-teacher ratio) is weakly correlated with MEDV.")

 Look especially at the target variable MEDV charts: RM (rooms) and LSTAT (status of population) have clear relationships with MEDV, while PTRATIO is noisier.

 Correlation heat map nicely complements the scatterplot matrix. RM and LSTAT are strongly correlated with MEDV, while PTRATIO (pupil-teacher ratio) is weakly correlated with MEDV.

#Also, thanks to a recommendation, can visualize these relationships by plotting the best fit line
plt.figure(figsize=(20, 5))
for i, col in enumerate(features.columns):
    plt.subplot(1, 3, i+1)
    plt.plot(data[col], prices, 'o')
    fit = np.polyfit(data[col], prices, 1)
    plt.plot(data[col], data[col] * fit[0] + fit[1], lw=3)
    plt.title(col + ' Pearson correlation ' + str(np.round(np.corrcoef(data[col], prices)[1][0], 3)))
    plt.xlabel(col)
    plt.ylabel('prices')

Question 1 - Feature Observation¶

As a reminder, we are using three features from the Boston housing dataset: 'RM', 'LSTAT', and 'PTRATIO'. For each data point (neighborhood):

'RM' is the average number of rooms among homes in the neighborhood.
'LSTAT' is the percentage of homeowners in the neighborhood considered "lower class" (working poor).
'PTRATIO' is the ratio of students to teachers in primary and secondary schools in the neighborhood.

Using your intuition, for each of the three features above, do you think that an increase in the value of that feature would lead to an increase in the value of 'MEDV' or a decrease in the value of 'MEDV'? Justify your answer for each.

Hint: This problem can phrased using examples like below.

Would you expect a home that has an 'RM' value(number of rooms) of 6 be worth more or less than a home that has an 'RM' value of 7?
Would you expect a neighborhood that has an 'LSTAT' value(percent of lower class workers) of 15 have home prices be worth more or less than a neighborhood that has an 'LSTAT' value of 20?
Would you expect a neighborhood that has an 'PTRATIO' value(ratio of students to teachers) of 10 have home prices be worth more or less than a neighborhood that has an 'PTRATIO' value of 15?

Answer 1:

'RM'

Increase in RM leads to an increase in MEDV. RM is directly proportional (positively correlated) to MEDV - higher values of RM are associated with higher values of MEDV. Intuition: house price is directly related to the square footage, and more rooms typically means larger square footage. Therefore, more rooms is associated with higher housing prices.

'LSTAT'

Increase in LSTAT leads to a decrease in MEDV. LSAT is inversely proportional (negatively correlated) to MEDV - meaning higher values of LSTAT are associated with lower values of MEDV. Intuition: LSTAT is the percentage of the residents that are poor, and poor people live in lower priced housing. So, the more poor residents, the more cheaper houses in the area which drags the median house value lower.

'PTRATIO'

Increase in PTRATIO leads to a decrease in MEDV. PTRATIO is inversely proportional (negatively correlated) to MEDV - meaning higher values of PTRATIO is associated with lower values of MEDV. Intuition: pupil to teacher ratio is another proxy variable for population's affluence. Higher PTRATIO means the population is poorer, and poor residents live in cheaper housing which drags the median house value lower.

Developing a Model¶

In this second section of the project, you will develop the tools and techniques necessary for a model to make a prediction. Being able to make accurate evaluations of each model's performance through the use of these tools and techniques helps to greatly reinforce the confidence in your predictions.

Implementation: Define a Performance Metric¶

It is difficult to measure the quality of a given model without quantifying its performance over training and testing. This is typically done using some type of performance metric, whether it is through calculating some type of error, the goodness of fit, or some other useful measurement. For this project, you will be calculating the coefficient of determination, R², to quantify your model's performance. The coefficient of determination for a model is a useful statistic in regression analysis, as it often describes how "good" that model is at making predictions.

The values for R² range from 0 to 1, which captures the percentage of squared correlation between the predicted and actual values of the target variable. A model with an R² of 0 is no better than a model that always predicts the mean of the target variable, whereas a model with an R² of 1 perfectly predicts the target variable. Any value between 0 and 1 indicates what percentage of the target variable, using this model, can be explained by the features. A model can be given a negative R² as well, which indicates that the model is arbitrarily worse than one that always predicts the mean of the target variable.

For the performance_metric function in the code cell below, you will need to implement the following:

Use r2_score from sklearn.metrics to perform a performance calculation between y_true and y_predict.
Assign the performance score to the score variable.

# Developing a Model
# R^2 is a measure of goodness of fit
# One way to calculate R^2 is to square the Pearson correlation of the target variable and predicted values. 
# Create a simple wrapper for sklearn's procedure.

# TODO: Import 'R^2_score'
def performance_metric(y_true, y_predict):
    """ Calculates and returns the performance score between 
        true and predicted values based on the metric chosen. """
    
    # TODO: Calculate the performance score between 'y_true' and 'y_predict'
    score = r2_score(y_true, y_predict)
    # Return the score
    return score

#Can also calculate the R2 score manually:
#y_true_mean = np.mean(y_true)
#SSres = sum(np.square(np.subtract(y_true, y_predict)))
#SStot = sum(np.square(np.subtract(y_true, y_true_mean)))
#score = 1.0 - SSres/SStot

Question 2 - Goodness of Fit¶

Assume that a dataset contains five data points and a model made the following predictions for the target variable:

True Value	Prediction
3.0	2.5
-0.5	0.0
2.0	2.1
7.0	7.8
4.2	5.3

Run the code cell below to use the performance_metric function and calculate this model's coefficient of determination.

# Calculate the performance of this model
score = performance_metric([3, -0.5, 2, 7, 4.2], [2.5, 0.0, 2.1, 7.8, 5.3])
print("Model has a coefficient of determination, R^2, of {:.3f}.".format(score))

Model has a coefficient of determination, R^2, of 0.923.

Would you consider this model to have successfully captured the variation of the target variable?
Why or why not?

Hint: The R2 score is the proportion of the variance in the dependent variable that is predictable from the independent variable. In other words:

R2 score of 0 means that the dependent variable cannot be predicted from the independent variable.
R2 score of 1 means the dependent variable can be predicted from the independent variable.
R2 score between 0 and 1 indicates the extent to which the dependent variable is predictable. An
R2 score of 0.40 means that 40 percent of the variance in Y is predictable from X.

Answer 2:

Yes, R^2 = 0.923 is very close to 1!

R^2 of 0.923 is a high value meaning the target variable is strongly correlated with the predicted variable.

Another way to look at is that the model explains the target variable's variation very well - 92.3% of the variation in the true value has been explained. Whatever model produced these predicted values was a good model.

Implementation: Shuffle and Split Data¶

Your next implementation requires that you take the Boston housing dataset and split the data into training and testing subsets. Typically, the data is also shuffled into a random order when creating the training and testing subsets to remove any bias in the ordering of the dataset.

For the code cell below, you will need to implement the following:

Use train_test_split from sklearn.cross_validation to shuffle and split the features and prices data into training and testing sets.
- Split the data into 80% training and 20% testing.
- Set the random_state for train_test_split to a value of your choice. This ensures results are consistent.
Assign the train and testing splits to X_train, X_test, y_train, and y_test.

# TODO: Import 'train_test_split' - ME: I prefer to import everything on top
# TODO: Shuffle and split the data into training and testing subsets
X_train, X_test, y_train, y_test = train_test_split(features, prices, test_size=0.2, random_state=3)

# Success
print ("Training and testing split was successful.")
print ("Train shapes (X,y): ", X_train.shape, y_train.shape)
print ("Test shapes (X,y): ", X_test.shape, y_test.shape)

Training and testing split was successful.
Train shapes (X,y):  (391, 3) (391,)
Test shapes (X,y):  (98, 3) (98,)

Question 3 - Training and Testing¶

What is the benefit to splitting a dataset into some ratio of training and testing subsets for a learning algorithm?

Hint: Think about how overfitting or underfitting is contingent upon how splits on data is done.

Answer 3:

Splitting the data into training and test samples is vital to produce useful generalizable models.

To understand how good our model is, we must test its performance on the test sample that was not used to build the model. This test sample represents completely new data that was unknown to us when we built the model. So, testing the performance on this new data gives us more accurate goodness of fit metrics.

If we evaluate the model on the same data that was used for training, we may be mislead by overly optimistic performance. We might think the model will perform well on new data but actually it overfits - it performs well only on the training data set but will perform poorly on brand new data. If there is enough data, it might also be beneficial to have a separate validation sample as well - to fine tune the model before testing on the test sample. If there is not enough data, the same set may be used for validation and testing though.

Using more data for training might help to reduce the underfitting of the model, but this really depends on the usefulness of the explanatory variables. So, there is a trade-off - we must have a test sample to avoid overfitting but it cannot proportionally too large because we want more training data for the model.

Analyzing Model Performance¶

In this third section of the project, you'll take a look at several models' learning and testing performances on various subsets of training data. Additionally, you'll investigate one particular algorithm with an increasing 'max_depth' parameter on the full training set to observe how model complexity affects performance. Graphing your model's performance based on varying criteria can be beneficial in the analysis process, such as visualizing behavior that may not have been apparent from the results alone.

Learning Curves¶

The following code cell produces four graphs for a decision tree model with different maximum depths. Each graph visualizes the learning curves of the model for both training and testing as the size of the training set is increased. Note that the shaded region of a learning curve denotes the uncertainty of that curve (measured as the standard deviation). The model is scored on both the training and testing sets using R², the coefficient of determination.

Run the code cell below and use these graphs to answer the following question.

# Produce learning curves for varying training set sizes and maximum depths
vs.ModelLearning(features, prices)

Question 4 - Learning the Data¶

Choose one of the graphs above and state the maximum depth for the model.
What happens to the score of the training curve as more training points are added? What about the testing curve?
Would having more training points benefit the model?

Hint: Are the learning curves converging to particular scores? Generally speaking, the more data you have, the better. But if your training and testing curves are converging with a score above your benchmark threshold, would this be necessary? Think about the pros and cons of adding more training points based on if the training and testing curves are converging.

Answer 4:

Chosen graph: max_depth=3

I chose the graph produced by the model with max_depth=3 because the training and testing scores converge on a pretty high value of R^2=0.8.

What I see in the graph is that as more training points are added, the training score slowly declines and eventually levels off at about 0.8. This tells me that as more data is added, the model does a worse job in fitting the variety of data.

At the same time by looking at the test learning curve, I see that the model's score on the test (unseen) data first jumps and then gradually increases and levels off close to 0.8 as well. This is good news - with at least 300 observations, the models performs well at R^2=0.8 much overfitting (small gap between the two curves). I would pick this model as my winner.

Having more than 300 training observations for model with max_depth=3 would not hurt the model but I doubt it would improve the model very much giving the existing convergence. Plus, it might not always be feasible or cost effective to collect more data for marginal improvements in the model performance.

Complexity Curves¶

The following code cell produces a graph for a decision tree model that has been trained and validated on the training data using different maximum depths. The graph produces two complexity curves — one for training and one for validation. Similar to the learning curves, the shaded regions of both the complexity curves denote the uncertainty in those curves, and the model is scored on both the training and validation sets using the performance_metric function.

Run the code cell below and use this graph to answer the following two questions Q5 and Q6.

vs.ModelComplexity(X_train, y_train)

Question 5 - Bias-Variance Tradeoff¶

When the model is trained with a maximum depth of 1, does the model suffer from high bias or from high variance?
How about when the model is trained with a maximum depth of 10? What visual cues in the graph justify your conclusions?

Hint: High bias is a sign of underfitting(model is not complex enough to pick up the nuances in the data) and high variance is a sign of overfitting(model is by-hearting the data and cannot generalize well). Think about which model(depth 1 or 10) aligns with which part of the tradeoff.

Answer 5:

The model with maximum depth=1 suffers from high bias because R^2 is pretty low at about 0.4 on the graph which means that only 40% of the variation is explained. The model fits the data poorly (aka underfits) because it misses the underlying relationship between features and the target.

The model with maximum depth=10 suffers from high variance because training data R^2 is high at 0.99 while for validation data it is lower at 0.7. This is easily seen by the wide gap between the two curves which gets wider as we increase the depth.

As we increase the maximum depth hyper-parameter, the model overfits the training data and fits the validation data poorer meaning that the model memorizes the pecularities of the training data which hurts its performance on the validation set. An overfitted model does not generalize well because it fits new data poorly.

Question 6 - Best-Guess Optimal Model¶

Which maximum depth do you think results in a model that best generalizes to unseen data?
What intuition lead you to this answer?

Hint: Look at the graph above Question 5 and see where the validation scores lie for the various depths that have been assigned to the model. Does it get better with increased depth? At what point do we get our best validation score without overcomplicating our model? And remember, Occams Razor states "Among competing hypotheses, the one with the fewest assumptions should be selected."

Answer 6:

I would stick to the choice I made previously: at max_depth=3 the model generalizes better than at the other depths.

The model's R^2 is about 0.8 on both training and validation data which is a pretty high goodness of fit in itself (low bias) and there is not much overfit (low variance) as seen in the very small gap between the testing and validaton scores.

Plus with the max_depth=3 it is a simpler model than the runner-up max_depth=4, so the rule of parsimony (aka "simpler is better") is also its favor.

Evaluating Model Performance¶

In this final section of the project, you will construct a model and make a prediction on the client's feature set using an optimized model from fit_model.

Question 7 - Grid Search¶

What is the grid search technique?
How it can be applied to optimize a learning algorithm?

Hint: When explaining the Grid Search technique, be sure to touch upon why it is used, what the 'grid' entails and what the end goal of this method is. To solidify your answer, you can also give an example of a parameter in a model that can be optimized using this approach.

Answer 7:

Grid search is a brute force method to optimize the model's performance. The way I imagine the "grid" is a hyper-dimensional matrix with specified range of values for each margin.

Say, a model has three different hyper-parameters ("levers") A,B,C each with own range of discrete values - A(range), B(range), C(range). What I want is to find the optimal set of these parameters that makes the model produces the best results. I am looking for a single solution out of the A(range) B(range) C(range) grid of possible solutions. To accomplish this, I need to search through this whole grid and find the best solution.

To apply it to any algorithm we need to know which hyper-parameters we can tinker with to find the best the solution. We then write a program to try all the combinations of these hyper-parameters, calculate the performance metric for each and pick the best performing one.

Question 8 - Cross-Validation¶

What is the k-fold cross-validation training technique?
What benefit does this technique provide for grid search when optimizing a model?

Hint: When explaining the k-fold cross validation technique, be sure to touch upon what 'k' is, how the dataset is split into different parts for training and testing and the number of times it is run based on the 'k' value.

When thinking about how k-fold cross validation helps grid search, think about the main drawbacks of grid search which are hinged upon using a particular subset of data for training or testing and how k-fold cv could help alleviate that. You can refer to the docs for your answer.

Answer 8:

Much like with the holdout method in which we split the data into training, test and, optionally, validation sets, the k-fold cross-validation is another a clever way to avoid the problem of overfitting in order to produce a reliable and generalizable model.

To implement k-fold cross-validation we start as usual by splitting the original data into training and test samples. We leave this original test sample until the very end of our modeling process.

Then we need to randomly split the training data set into k number of folds (chunks) - say 10 such folds. Then we treat one chunk as the test set and the remaining k-1 chunks as the training set. We then train the model again but now using another chunk as the test set. This way every chunk and its underlying observations will be used to evaluate the performance only once and all of the training data will be used both for training the model and for measuring its performance.

We end up training the same model on different pairs of data sets k times. When all the k-folds have been used we calculate the average performance metrics across all the iterations. These average metrics are much more reliable than the metrics calculated on a single set of training/validation/test data which can suffer from pure dumb luck.

Needless to say, we must be careful to pick the k-number so that each chunk has enough observations in it to be useful as the test sample.

K-folding technique is beneficial to grid search because it gives us the average model performance for each set of hyper-parameters in the grid. This minimizes the risk of overfitting due to a particular "lucky" pair of training and test sets because we use all of the training data as both training and test data to get an all-around measure of model performance. K-folding allows us to avoid setting aside data for model validation (we still need to set aside test data as always) which gives us more data for training the model. At the end of training we still evaluate the final model's performance on the original test data which represents "never before seen" data for the model.

Implementation: Fitting a Model¶

Your final implementation requires that you bring everything together and train a model using the decision tree algorithm. To ensure that you are producing an optimized model, you will train the model using the grid search technique to optimize the 'max_depth' parameter for the decision tree. The 'max_depth' parameter can be thought of as how many questions the decision tree algorithm is allowed to ask about the data before making a prediction. Decision trees are part of a class of algorithms called supervised learning algorithms.

In addition, you will find your implementation is using ShuffleSplit() for an alternative form of cross-validation (see the 'cv_sets' variable). While it is not the K-Fold cross-validation technique you describe in Question 8, this type of cross-validation technique is just as useful!. The ShuffleSplit() implementation below will create 10 ('n_splits') shuffled sets, and for each shuffle, 20% ('test_size') of the data will be used as the validation set. While you're working on your implementation, think about the contrasts and similarities it has to the K-fold cross-validation technique.

Please note that ShuffleSplit has different parameters in scikit-learn versions 0.17 and 0.18. For the fit_model function in the code cell below, you will need to implement the following:

Use DecisionTreeRegressor from sklearn.tree to create a decision tree regressor object.
- Assign this object to the 'regressor' variable.
Create a dictionary for 'max_depth' with the values from 1 to 10, and assign this to the 'params' variable.
Use make_scorer from sklearn.metrics to create a scoring function object.
- Pass the performance_metric function as a parameter to the object.
- Assign this scoring function to the 'scoring_fnc' variable.
Use GridSearchCV from sklearn.grid_search to create a grid search object.
- Pass the variables 'regressor', 'params', 'scoring_fnc', and 'cv_sets' as parameters to the object.
- Assign the GridSearchCV object to the 'grid' variable.

# ME: Decision tree regressor
# First, let's do a simple decision tree to get an idea about it. 
# Decision trees do not necessarily require any transformation of the data and can be applied to both categorical and continuous 
# target variables. 
# The difference is that while for classification we can use Gini or entropy for regression 
#we need to MSE or R^2 as the impurity measure to split the nodes. 

dtree = DecisionTreeRegressor(max_depth=2, random_state=3) #best looking graph at 2, random state to make model reproducible
X1 = data[['RM']].values
Y1 = data[['MEDV']].values
dtree.fit(X1,Y1 )

sort_idx = X1.flatten().argsort()

plt.scatter(X1[sort_idx], Y1[sort_idx], c='steelblue', edgecolor='white', s=70)
plt.plot(X1[sort_idx], dtree.predict(X1[sort_idx]), color='black', lw=2)    
plt.xlabel('RM')
plt.ylabel('MEDV')
plt.show()

print ("Here is how the decision tree looks like when using a single feature - RM. Not very pretty but it captures the general relationship in a step function.")

Here is how the decision tree looks like when using a single feature - RM. Not very pretty but it captures the general relationship in a step function.

# TODO: Import 'make_scorer', 'DecisionTreeRegressor', and 'GridSearchCV'

# Fitting a Model using decision tree regressor and grid search
# ME: replaced hard-coded by parameters for iterations and test size for prototyping
def fit_model(X, y, v_iter=10, v_size=0.20, v_depth=11):
    """ Performs grid search over the 'max_depth' parameter for a 
        decision tree regressor trained on the input data [X, y]. """
    
    # Create cross-validation sets from the training data
    cv_sets = ShuffleSplit(X.shape[0], n_iter = v_iter, test_size = v_size, random_state = 0)

    # TODO: Create a decision tree regressor object
    regressor = DecisionTreeRegressor()

    # TODO: Create a dictionary for the parameter 'max_depth' with a range from 1 to 10
    params = {'max_depth':np.arange(1,v_depth)} # P3 needs this np fn instead of range

    # TODO: Transform 'performance_metric' into a scoring function using 'make_scorer' - which was defined above as R^2_score
    scoring_fnc = make_scorer(performance_metric)

    # TODO: Create the grid search object
    grid = GridSearchCV(regressor, param_grid=params, scoring=scoring_fnc, cv=cv_sets)

    # Fit the grid search object to the data to compute the optimal model
    grid = grid.fit(X, y)

    # Return the optimal model after fitting the data
    return grid.best_estimator_

Making Predictions¶

Once a model has been trained on a given set of data, it can now be used to make predictions on new sets of input data. In the case of a decision tree regressor, the model has learned what the best questions to ask about the input data are, and can respond with a prediction for the target variable. You can use these predictions to gain information about data where the value of the target variable is unknown — such as data the model was not trained on.

Question 9 - Optimal Model¶

What maximum depth does the optimal model have? How does this result compare to your guess in Question 6?

Run the code block below to fit the decision tree regressor to the training data and produce an optimal model.

# Fit the training data to the model using grid search
reg = fit_model(X_train, y_train, v_iter=10)

# Produce the value for 'max_depth'
print ("Parameter 'max_depth' is {} for the optimal model.".format(reg.get_params()['max_depth']))

Parameter 'max_depth' is 4 for the optimal model.

Hint: The answer comes from the output of the code snipped above.

Answer 9:

Winner: max_depth=4.

For a single re-shuffling & splitting iteration the max_depth of the winner is 3 but for n_iter=10 - max_depth=4.

Interesting, where does it change? At n_iter=4 the best max_depth becomes 4 and stays the same up to n_iter=10.

So, grid search identified max_depth=4 as the optimal model while my initial guess was 3. Must have been pretty close competition though as it was not easily visible to me.

Question 10 - Predicting Selling Prices¶

Imagine that you were a real estate agent in the Boston area looking to use this model to help price homes owned by your clients that they wish to sell. You have collected the following information from three of your clients:

Feature	Client 1	Client 2	Client 3
Total number of rooms in home	5 rooms	4 rooms	8 rooms
Neighborhood poverty level (as %)	17%	32%	3%
Student-teacher ratio of nearby schools	15-to-1	22-to-1	12-to-1

What price would you recommend each client sell his/her home at?
Do these prices seem reasonable given the values for the respective features?

Hint: Use the statistics you calculated in the Data Exploration section to help justify your response. Of the three clients, client 3 has has the biggest house, in the best public school neighborhood with the lowest poverty level; while client 2 has the smallest house, in a neighborhood with a relatively high poverty rate and not the best public schools.

Run the code block below to have your optimized model make predictions for each client's home.

# Produce a matrix for client data
# features: rooms, poverty, p-t ratio
client_data = [[5, 17, 15], # Client 1
               [4, 32, 22], # Client 2
               [8, 3, 12]]  # Client 3

# Show predictions
for i, price in enumerate(reg.predict(client_data)):
    print("Predicted selling price for Client {}'s home: ${:,.2f}".format(i+1, price))

Predicted selling price for Client 1's home: $420,622.22
Predicted selling price for Client 2's home: $235,122.22
Predicted selling price for Client 3's home: $896,280.00

#To understand if the model makes sense, we can quickly eyeball the descriptive stats for the features
features.describe()

Answer 10:

Model recommended prices are:

Client 1's home: \$420,622.22
Client 2's home: \$235,122.22
Client 3's home: \$896,280.00

The first two of predicted prices are below and the last one is way above the average MEDV of $454K that we found above during exploratory data analysis. This makes sense as the first two houses have a lower number of rooms (5 and 4 vs 8), higher level of poverty (17% and 32% vs 3%) and high student to teach ratio (15-to-1 and 22-to-1 vs 12-to-1).

Another way to look at it is to compare each house's features to the averages from the whole data set. The reasoning is similar though - first two houses have below average number of rooms (mean=6.24), above average level of poverty (mean=12.94%). Client 2 has above average level of student to teach ratio (mean=18.52), but Client 1 has slightly below average though Client 3 is even lower.

The factors that are worse than average for Client 1 and Client 2 drag the predicted prices down in comparison to Client 3's house which is the largest house, located in an area with very little poverty and schools which can afford more teachers per student.

For our purposes, using regression to predict house prices seems to make sense.

How reliable are these estimates is another question though.

#per recommendation, additional fancy plots to show where these houses fit in the data
#note how easy it is to see when house is above or below the averages

plt.figure(figsize=(20, 5))
y_ax = [[3,9],[0,40],[11,23]]
for i, col in enumerate(features.columns):
    plt.subplot(1, 3, i+1)
    plt.boxplot(data[col])
    plt.title(col)
    for j in range(3):
        plt.plot(1, client_data[j][i], marker="o")
        plt.annotate('Client '+str(j+1), xy=(1,client_data[j][i]))
        plt.ylim(y_ax[i])

Sensitivity¶

An optimal model is not necessarily a robust model. Sometimes, a model is either too complex or too simple to sufficiently generalize to new data. Sometimes, a model could use a learning algorithm that is not appropriate for the structure of the data given. Other times, the data itself could be too noisy or contain too few samples to allow a model to adequately capture the target variable — i.e., the model is underfitted.

Run the code cell below to run the fit_model function ten times with different training and testing sets to see how the prediction for a specific client changes with respect to the data it's trained on.

vs.PredictTrials(features, prices, fit_model, client_data)

Trial 1: $391,183.33
Trial 2: $419,700.00
Trial 3: $415,800.00
Trial 4: $420,622.22
Trial 5: $418,377.27
Trial 6: $411,931.58
Trial 7: $399,663.16
Trial 8: $407,232.00
Trial 9: $351,577.61
Trial 10: $413,700.00

Range in prices: $69,044.61

ME: Yes, indeed, just by chance the same model can produce very different predictions when using different training and test data sets. Perhaps, this is the reasoning why house appraisals are not done using regressions.

Question 11 - Applicability¶

In a few sentences, discuss whether the constructed model should or should not be used in a real-world setting.

Hint: Take a look at the range in prices as calculated in the code snippet above. Some questions to answering:

How relevant today is data that was collected from 1978? How important is inflation?
Are the features present in the data sufficient to describe a home? Do you think factors like quality of apppliances in the home, square feet of the plot area, presence of pool or not etc should factor in?
Is the model robust enough to make consistent predictions?
Would data collected in an urban city like Boston be applicable in a rural city?
Is it fair to judge the price of an individual home based on the characteristics of the entire neighborhood?

Answer 11:

No, this model should not be used in a real-world setting.

I would not use this data set for anything other than practice. The data set is old and the underlying relationships between features and the house price may no longer hold. The data set has just a few hundred useable observations, so I doubt it represents well such a large city as Boston. Three features that we used for this exercise are hardly enough to explain all the variation in the data. Though R^2 of 0.8 was actually decent for this small data set.

More worrisome is the sensitivity of the model to the chancey selection of data for training and testing. As we have seen above, the predictions range widely entirely due to chance. This is further evidence that to get accurate predictions we need both more observations and more features.

Any model built on Boston data would, most likely, not be applicable to another city. Unless we had data representative of other cities as well, we can use this model to predict prices of the given area and at the given time.

While it is very attractive to use regression models to predict house prices. Given good data, itt would, probably, work really well for pre-sale appraisals, however, it is not allowed by law. Instead, the appraisers are to use actual sale prices of similiar houses allowing for some minor adjustments.

Note: Once you have completed all of the code implementations and successfully answered each question above, you may finalize your work by exporting the iPython Notebook as an HTML document. You can do this by using the menu above and navigating to
File -> Download as -> HTML (.html). Include the finished document along with this notebook as your submission.

	RM	LSTAT	PTRATIO
count	489.000000	489.000000	489.000000
mean	6.240288	12.939632	18.516564
std	0.643650	7.081990	2.111268
min	3.561000	1.980000	12.600000
25%	5.880000	7.370000	17.400000
50%	6.185000	11.690000	19.100000
75%	6.575000	17.120000	20.200000
max	8.398000	37.970000	22.000000