Machine Learning Engineer Nanodegree

Model Evaluation & Validation

Project: Predicting Boston Housing Prices

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.

In [2]:
# Import libraries necessary for this project
import numpy as np
import pandas as pd
from sklearn.model_selection import ShuffleSplit

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

# Pretty display for notebooks
%matplotlib inline

# Load the Boston housing dataset
data = pd.read_csv('housing.csv')
prices = data['MEDV']
features = data.drop('MEDV', axis = 1)
    
# Success
print("Boston housing dataset has {} data points with {} variables each.".format(*data.shape))

Boston housing dataset has 489 data points with 4 variables each.

/home/christian/anaconda2/lib/python2.7/site-packages/sklearn/cross_validation.py:41: DeprecationWarning: This module was deprecated in version 0.18 in favor of the model_selection module into which all the refactored classes and functions are moved. Also note that the interface of the new CV iterators are different from that of this module. This module will be removed in 0.20.
  "This module will be removed in 0.20.", DeprecationWarning)
/home/christian/anaconda2/lib/python2.7/site-packages/sklearn/learning_curve.py:22: DeprecationWarning: This module was deprecated in version 0.18 in favor of the model_selection module into which all the functions are moved. This module will be removed in 0.20
  DeprecationWarning)

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.
In [3]:
# TODO: Minimum price of the data
minimum_price = prices.min()

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

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

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

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

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

Statistics for Boston housing dataset:

Minimum price: $105000.0
Maximum price: $1024800.0
Mean price: $454342.944785
Median price $438900.0
Standard deviation of prices: $165340.277653

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:

  • RM: moderate positive correlation to MEDV: Space costs money in the real estate markets. Hence, more rooms in houses should lead to an increase in the value of houses.
  • LTAT: strong negative correlation to MEDV: People want to build their houses in the best places. "Lower class" neighborhoods are not considered to be the best places usually because they come at the expense of good restaurants, cultural events, and low crime rates. Hence, an increase in the percentage of homeowners in the neighborhood should decrease the value of the houses.
  • PTRATIO: small negative correlation to MEDV: A premise of good education is that teachers should be able to teach few students. In fact, a central goal of education is the 2 Sigma Problem coined by the educational psychologist Benjamin Bloom. How can we achieve the same outcomes that 1 to 1 tutoring yields with bigger classes?Again, small classes are deemed to be positive. A high ratio of students to teachers can be considered a negative condition for learning because many students have to be taught by a single teacher. A low ratio of students to teachers in turn should be considered positive because teachers might better be able to educate their students. Hence a higher students to teachers ratio should yield to a decrease in the value of the houses.

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.
In [4]:
# TODO: Import 'r2_score'
from sklearn.metrics import r2_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

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.

In [5]:
# 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:

According to the R^2 value the model is a good fit to the data. The model can explain 92.3% of the variance in Y and hence is almost a perfect fit to the data. In other words, the model is much better the predict the actual value than assuming the mean value of y as a baseline. However, R^2 should be interpreted with causion since splitting the dataset into a testing, and training dataset or checking for the number of features to not inflate the R^2, are crucial for its interpretation. You can have a high R^2 that heavily overfits to a testing set or you can have a high R^2 that is inflated by using hundreds of features to build the 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.
In [6]:
# TODO: Import 'train_test_split'
from sklearn.model_selection import train_test_split

# 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=0.33, random_state=42)

# Success
print("Training and testing split was successful.")

Training and testing split was successful.

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:

Splitting a dataset into a testing and training subset is necessary to check for over- and underfitting. Overfitting occurs if a model is closely adapted to the training data and does not generalize to unkown data (or testing data). Underfitting occurs if a model is too simple and hence doesn't caputure the patterns in the data. By splitting a dataset into a testing and training subset a machine learning engineer can test for over- and underfitting. If a model shows good performance on the training data and bad performance on the test data the model overfits. If a model shows bad performance on the training data and bad performance on the test data the model underfits. The size of the testing dataset is usually smaller than the training set. It is also a golden rule never to use the testing set for training.

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.

In [7]:
# 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:

  • Choose one of the graphs: According to the four learning curves the decision tree with a maximum depth of three (max_depth = 3) is the best model.
  • Adding training points: As more training points are added the training and testing score (here R^2) converge to a value of about 0.8. At first, with few training points, the model heavily overfits, since the model fits well to the training data but does not generalize to the training set. With an increase in data points the score (R^2) of the training and testing set becomes more and more similar until their absolute difference in scores doesn't change anymore. In contrast to the example from Lesson 8 (Detecting Errors) -> 5. Learning Curves, the high score on the y-axis means that we have a good model because a high R^2 indicates a good fit to the data. The example from the lecture depicted the error on the y-axis and thus should be interpreted differently.
  • More training points: Adding more training points does not seem to affect the quality of the model. If there still would be a variance in the absolute deviation between the training and testing score for the last 100 data points, adding testing points might positively affect the model. However, there is no variance in the absolute deviation between the training and testing score between data points ranging from 250 to 350. Thus, I would conclude that adding testing points would not affect the quality of the model. According to the learning curve about 200 training points should be enough to generate a good decision tree model with a maximum depth of three.

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.

In [8]:
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:

  • Maximum depth of 1: The model with a maximum depth of 1 seems to suffer from high bias. High bias occurs if a model is too simple to capture the underlying pattern in the data. For example, Linear regression might be a model that leads to high bias if the underlying data is not linear. I assume that the decision tree with a maximum depth of 1 suffers from high bias because both the training and the testing scores yield a medium value of R^2. Thus the model generated from the training data does not fit the data very well since it is too simple.
  • Maximum depth of 10: The complexity curve for a maximum depth of 10 clearly shows evidence for high variance. The R^2 for the training set is almost perfect, indicating a model that fits too well to the training set. On the other hand, the R^2 for the testing set is about .3 below the R^2 for the testing set. Thus, the model does not generalizes well to new data. Compared to a maximum depth of 3 the model with a maximum depth of 10 does worse. As a rule of thumb, the stronger the deviation between training and testing score, with the training score being higher, the more the model suffers from high variance.

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:

I think the decision tree model with a maximum depth of 3 best generalizes to unseen data. I chose a maximum depth of 3 because this model provides the highest score on the validation set. The R^2 for this model is about .80, meaning that the model can explain almost 80% of the variance in the testing set. Another reason to pick the model with a maximum depth of 3 is that this model does not suffer from high variance compared to the models with a higher maximum depth. Both, the score for the training and the testing set are similar. If the model would suffer from high variance both scores would deviate. Also, the model doesn't seem to show evidence for high bias since we obtain a very good R^2.

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:

  • What is grid search?: Supervised machine learning algorithms very often have multiple parameters that can be adjusted by a machine learning engineer. For example, support vector machines have the parameters gamma and C which change the behavior of the underlying algorithm. A machine learning engineer might train multiple such models and change these parameters from model to model. However, this approach would resemble a trial-and-error approach. A better way to formalize this problem is to use the grid search technique. With grid search you define the parameters of an algorithm before hand and then run the algorithm multiple times with the defined parameters. For example, if you would train a support vector machine and want to tune the parameters gamma and C you could first define two values of gamma and two values of C (see here). What you would get is a grid of two rows and two columns. The rows might indicate the values of gamma and the columns might indicate the values of C. With a two and two grid the grid search technique would have to run the support vector machine four times.
  • Optimize a learning algorithm: From running an algorithm multiple times with the a priori defined parameter values you are able to select the algorithm which provides the best evaluation metrics. In our example we might find that the support vector machine with a gamma of 2 and a C of 100 yields the highest F1-score. Thus, we could be fairly confident that this algorithm performs best on new data.

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:

  • What is k-fold cross-validation?: When evaluating an algorithm with multiple paramaters on a test set, a machine learning engineer might be tempted to compare the evaluation metrics on the test set and then decide upon the best model. I could, for example, tweak a parameter on an algorithm 4 times and then calculate my evaluation metric for each of the models. By doing so, knowledge about the test set "leaks" into the model and its evaluation. Such a "leak" would break the golden rule of splitting a dataset in a test and training set: Never use the testing set for training. A solution to this problem might be the use of a validation set. Instead of two datasets you would have three: a training set, a testing set and a validation set. In my example, I would have to train my model, test it on the validation set and then report the evaluation metric of my model from the testing set. However, this procedure would result in a reduced training set since I would have to split the dataset one more time. A solution to this problem is the k-fold cross-validation technqiue. In k-fold cross-validation the training set is split into k-bins of equal size. For example, I could split a training set of 100 datapoints into five bins of 20 datapoints. Each bin is used as a testing set once. So, if I have a training set of five bins I would generate five testing sets. The remaining bins are used for training my model. Hence, you can run your algorithm k times and then average the resulting evaluation metric of each iteration. The benefit of using k-fold cross-validation is that you are able to validate your model on an evaluation metric without touching the testing set. However, k-fold cross-validation is computationally more expensive and takes more time.
  • Benefit for grid search: The idea behind the grid search technique is to find the model that provides the best evaluation metric on the testing set. Again, doing so would break the golden rule of splitting a dataset into a test and training set. Information about the testing set could leak into the evaluation of our model. K-fold cross-validation can solve this problem because we can calculate the evaluation metric of a model and compare them without looking at the testing set. By averaging the evaluation metric through k-fold cross-validation we can be fairly confident, which of the tuned models is the best and calculate it's predicitve power using our evaluation metrics on the testing set.

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.
In [14]:
# TODO: Import 'make_scorer', 'DecisionTreeRegressor', and 'GridSearchCV'
from sklearn.metrics import make_scorer
from sklearn.tree import DecisionTreeRegressor
from sklearn.model_selection import GridSearchCV

def fit_model(X, y):
    """ 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
    # sklearn version 0.18: ShuffleSplit(n_splits=10, test_size=0.1, train_size=None, random_state=None)
    # sklearn versiin 0.17: ShuffleSplit(n, n_iter=10, test_size=0.1, train_size=None, random_state=None)
    # cv_sets = ShuffleSplit(X.shape[0], n_iter = 10, test_size = 0.20, random_state = 0)
    cv_sets = ShuffleSplit(n_splits = 10, test_size = 0.20, 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': range(1, 10)}

    # TODO: Transform 'performance_metric' into a scoring function using 'make_scorer' 
    scoring_fnc = make_scorer(performance_metric)

    # TODO: Create the grid search cv object --> GridSearchCV()
    # Make sure to include the right parameters in the object:
    # (estimator, param_grid, scoring, cv) which have values 'regressor', 'params', 'scoring_fnc', and 'cv_sets' respectively.
    grid = GridSearchCV(regressor, params, 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.

In [17]:
# Fit the training data to the model using grid search
reg = fit_model(X_train, y_train)

# 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: The optimal model has a maximum depth of 4. In question 6 I guessed that the optimal model has a maximum depth of 3. Hence, I was pretty close to the optimal model. The model with a maximum depth of 6 already showed evidence for high variance. Since I was not provided with the learning curves for a maximum depth between 3 and 6 a maximum depth of 3 was the best pick among the four graphs. But apparently, a depth of 4 is even better.

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.

In [18]:
# Produce a matrix for client data
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: $404,460.00
Predicted selling price for Client 2's home: $241,981.25
Predicted selling price for Client 3's home: $946,800.00

Answer: According to my decision tree regressor with a maximum depth of 4, I would recommend to sell the houses for the price depicted in the cell above. Compared to my intuitions from the data exploration section these results seem reasonable. I predicted that an increase in the number of rooms should incrase the value of an estate. A high neighberhood poverty level in turn should decrease the value of an estate. The same goes for the student-teacher ratio. A high student teacher-ratio in the nearby schools should decrease the value of an estate. According to these metrics the three clients can be arranged to the following order: Client3 > Client 1 > Client 2. In other words:

Rooms Poverty Student-teacher ratio
8 > 5 > 4
3% < 17% < 32%
12/1 < 15/1 < 22/1
Client 3 > Client 1 > Client 2

The predicted house prices match with these intuitions. Apparently the neighborhood poverty level has a big impact on the housing prices, given the difference between the predicted house for client 3 and client 1 (\$946,800.00 to \$404,460.00).

Looking at the descriptive statistics from the data exploration section each of the houses seem to be within the common range for the boston housing dataset. The maximum price for a house in that area was \$1,024,800.00. The house of client 3 approximates that value. The mean price for a house was \$454,342.94 and thus in between the house prices for our fictitious three clients.

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.

In [19]:
vs.PredictTrials(features, prices, fit_model, client_data)

Trial 1: $391,183.33
Trial 2: $424,935.00
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: $73,357.39

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: I would not propose the model for a real-word setting. First, the model does not seem to be robust. A range in \$73,357 is non-trivial and might affect the chances for people whether to buy a new house or not. Second, the model is based on an outdated dataset. One can assume that there has been a significant inflation in the housing prices from 1978 to now. But most importantly, the features to train the model might have changed dramatically since 1978. A region with a high poverty rate then might be prosperous nowadays. The same goes for the student-teacher ratio. Third, people base their decision whether to buy a house not only on these three features. I guess the specific location of a house is a very important feature to buy a house. I would like to buy a house with a nice view and a close connection to a supermarket. I would not like to buy a house that is engulfed by skyscrapers. Another feature would be the cost of building the house itself. There is a huge difference in the overall price whether I buy a prefabricated house or house build and designed by an architect. Rather than using the proposed decision tree model I would first try to find a new updated dataset and think carefully about features that might also be important for people buying houses.

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.