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Given the following text description, write Python code to implement the functionality described below step by step
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Description:
 Logistic Regression with L2 regularization
The goal of this second notebook is to implement your own logistic regression classifier with L2 regularization. You will do the following
Step1:  Load and process review dataset
For this assignment, we will use the same subset of the Amazon product review dataset that we used in Module 3 assignment. The subset was chosen to contain similar numbers of positive and negative reviews, as the original dataset consisted of mostly positive reviews.
Step2:  Just like we did previously, we will work with a hand-curated list of important words extracted from the review data. We will also perform 2 simple data transformations
Step3:  Now, let us take a look at what the dataset looks like (Note
Step4:  Train-Validation split
We split the data into a train-validation split with 80% of the data in the training set and 20% of the data in the validation set. We use seed=2 so that everyone gets the same result.
Note
Step5:  Convert SFrame to NumPy array
Just like in the second assignment of the previous module, we provide you with a function that extracts columns from an SFrame and converts them into a NumPy array. Two arrays are returned
Step6:  We convert both the training and validation sets into NumPy arrays.
Warning
Step7:  Are you running this notebook on an Amazon EC2 t2.micro instance? (If you are using your own machine, please skip this section)
It has been reported that t2.micro instances do not provide sufficient power to complete the conversion in acceptable amount of time. For interest of time, please refrain from running get_numpy_data function. Instead, download the binary file containing the four NumPy arrays you'll need for the assignment. To load the arrays, run the following commands
Step8:  Adding  L2 penalty
Let us now work on extending logistic regression with L2 regularization. As discussed in the lectures, the L2 regularization is particularly useful in preventing overfitting. In this assignment, we will explore L2 regularization in detail.
Recall from lecture and the previous assignment that for logistic regression without an L2 penalty, the derivative of the log likelihood function is
Step9:  Quiz question
Step11:  Quiz question
Step12:  Explore effects of L2 regularization
Now that we have written up all the pieces needed for regularized logistic regression, let's explore the benefits of using L2 regularization in analyzing sentiment for product reviews. As iterations pass, the log likelihood should increase.
Below, we train models with increasing amounts of regularization, starting with no L2 penalty, which is equivalent to our previous logistic regression implementation.
Step13:  Compare coefficients
We now compare the coefficients for each of the models that were trained above. We will create a table of features and learned coefficients associated with each of the different L2 penalty values.
Below is a simple helper function that will help us create this table.
Step14:  Now, let's run the function add_coefficients_to_table for each of the L2 penalty strengths.
Step15:  Using the coefficients trained with L2 penalty 0, find the 5 most positive words (with largest positive coefficients). Save them to positive_words. Similarly, find the 5 most negative words (with largest negative coefficients) and save them to negative_words.
Quiz Question. Which of the following is not listed in either positive_words or negative_words?
Step16:  Let us observe the effect of increasing L2 penalty on the 10 words just selected. We provide you with a utility function to  plot the coefficient path.
Step17:  Run the following cell to generate the plot. Use the plot to answer the following quiz question.
Step18:  Quiz Question
Step19:  Below, we compare the accuracy on the training data and validation data for all the models that were trained in this assignment.  We first calculate the accuracy values and then build a simple report summarizing the performance for the various models.
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Python Code:
from __future__ import division
import graphlab
Explanation: Logistic Regression with L2 regularization
The goal of this second notebook is to implement your own logistic regression classifier with L2 regularization. You will do the following:
Extract features from Amazon product reviews.
Convert an SFrame into a NumPy array.
Write a function to compute the derivative of log likelihood function with an L2 penalty with respect to a single coefficient.
Implement gradient ascent with an L2 penalty.
Empirically explore how the L2 penalty can ameliorate overfitting.
Fire up GraphLab Create
Make sure you have the latest version of GraphLab Create. Upgrade by
pip install graphlab-create --upgrade
See this page for detailed instructions on upgrading.
End of explanation
products = graphlab.SFrame('amazon_baby_subset.gl/')
Explanation: Load and process review dataset
For this assignment, we will use the same subset of the Amazon product review dataset that we used in Module 3 assignment. The subset was chosen to contain similar numbers of positive and negative reviews, as the original dataset consisted of mostly positive reviews.
End of explanation
# The same feature processing (same as the previous assignments)
# ---------------------------------------------------------------
import json
with open('important_words.json', 'r') as f: # Reads the list of most frequent words
    important_words = json.load(f)
important_words = [str(s) for s in important_words]
def remove_punctuation(text):
    import string
    return text.translate(None, string.punctuation) 
# Remove punctuation.
products['review_clean'] = products['review'].apply(remove_punctuation)
# Split out the words into individual columns
for word in important_words:
    products[word] = products['review_clean'].apply(lambda s : s.split().count(word))
Explanation: Just like we did previously, we will work with a hand-curated list of important words extracted from the review data. We will also perform 2 simple data transformations:
Remove punctuation using Python's built-in string functionality.
Compute word counts (only for the important_words)
Refer to Module 3 assignment for more details.
End of explanation
products
Explanation: Now, let us take a look at what the dataset looks like (Note: This may take a few minutes).
End of explanation
train_data, validation_data = products.random_split(.8, seed=2)
print 'Training set   : %d data points' % len(train_data)
print 'Validation set : %d data points' % len(validation_data)
Explanation: Train-Validation split
We split the data into a train-validation split with 80% of the data in the training set and 20% of the data in the validation set. We use seed=2 so that everyone gets the same result.
Note: In previous assignments, we have called this a train-test split. However, the portion of data that we don't train on will be used to help select model parameters. Thus, this portion of data should be called a validation set. Recall that examining performance of various potential models (i.e. models with different parameters) should be on a validation set, while evaluation of selected model should always be on a test set.
End of explanation
import numpy as np
def get_numpy_data(data_sframe, features, label):
    data_sframe['intercept'] = 1
    features = ['intercept'] + features
    features_sframe = data_sframe[features]
    feature_matrix = features_sframe.to_numpy()
    label_sarray = data_sframe[label]
    label_array = label_sarray.to_numpy()
    return(feature_matrix, label_array)
Explanation: Convert SFrame to NumPy array
Just like in the second assignment of the previous module, we provide you with a function that extracts columns from an SFrame and converts them into a NumPy array. Two arrays are returned: one representing features and another representing class labels. 
Note: The feature matrix includes an additional column 'intercept' filled with 1's to take account of the intercept term.
End of explanation
feature_matrix_train, sentiment_train = get_numpy_data(train_data, important_words, 'sentiment')
feature_matrix_valid, sentiment_valid = get_numpy_data(validation_data, important_words, 'sentiment') 
Explanation: We convert both the training and validation sets into NumPy arrays.
Warning: This may take a few minutes.
End of explanation
'''
produces probablistic estimate for P(y_i = +1 | x_i, w).
estimate ranges between 0 and 1.
'''
def predict_probability(feature_matrix, coefficients):
    # Take dot product of feature_matrix and coefficients  
    ## YOUR CODE HERE
    scores = np.dot(feature_matrix, coefficients)
    
    # Compute P(y_i = +1 | x_i, w) using the link function
    ## YOUR CODE HERE
    predictions = 1 / (1 + np.exp(-scores))
    
    return predictions
Explanation: Are you running this notebook on an Amazon EC2 t2.micro instance? (If you are using your own machine, please skip this section)
It has been reported that t2.micro instances do not provide sufficient power to complete the conversion in acceptable amount of time. For interest of time, please refrain from running get_numpy_data function. Instead, download the binary file containing the four NumPy arrays you'll need for the assignment. To load the arrays, run the following commands:
arrays = np.load('module-4-assignment-numpy-arrays.npz')
feature_matrix_train, sentiment_train = arrays['feature_matrix_train'], arrays['sentiment_train']
feature_matrix_valid, sentiment_valid = arrays['feature_matrix_valid'], arrays['sentiment_valid']
Building on logistic regression with no L2 penalty assignment
Let us now build on Module 3 assignment. Recall from lecture that the link function for logistic regression can be defined as:
$$
P(y_i = +1 | \mathbf{x}_i,\mathbf{w}) = \frac{1}{1 + \exp(-\mathbf{w}^T h(\mathbf{x}_i))},
$$
where the feature vector $h(\mathbf{x}_i)$ is given by the word counts of important_words in the review $\mathbf{x}_i$. 
We will use the same code as in this past assignment to make probability predictions since this part is not affected by the L2 penalty.  (Only the way in which the coefficients are learned is affected by the addition of a regularization term.)
End of explanation
def feature_derivative_with_L2(errors, feature, coefficient, l2_penalty, feature_is_constant): 
    
    # Compute the dot product of errors and feature
    ## YOUR CODE HERE
    derivative = np.dot(errors, feature)
    # add L2 penalty term for any feature that isn't the intercept.
    if not feature_is_constant: 
        ## YOUR CODE HERE
        derivative = derivative - 2 * l2_penalty * coefficient
        
    return derivative
Explanation: Adding  L2 penalty
Let us now work on extending logistic regression with L2 regularization. As discussed in the lectures, the L2 regularization is particularly useful in preventing overfitting. In this assignment, we will explore L2 regularization in detail.
Recall from lecture and the previous assignment that for logistic regression without an L2 penalty, the derivative of the log likelihood function is:
$$
\frac{\partial\ell}{\partial w_j} = \sum_{i=1}^N h_j(\mathbf{x}_i)\left(\mathbf{1}[y_i = +1] - P(y_i = +1 | \mathbf{x}_i, \mathbf{w})\right)
$$
 Adding L2 penalty to the derivative 
It takes only a small modification to add a L2 penalty. All terms indicated in red refer to terms that were added due to an L2 penalty.
Recall from the lecture that the link function is still the sigmoid:
$$
P(y_i = +1 | \mathbf{x}_i,\mathbf{w}) = \frac{1}{1 + \exp(-\mathbf{w}^T h(\mathbf{x}_i))},
$$
We add the L2 penalty term to the per-coefficient derivative of log likelihood:
$$
\frac{\partial\ell}{\partial w_j} = \sum_{i=1}^N h_j(\mathbf{x}_i)\left(\mathbf{1}[y_i = +1] - P(y_i = +1 | \mathbf{x}_i, \mathbf{w})\right) \color{red}{-2\lambda w_j }
$$
The per-coefficient derivative for logistic regression with an L2 penalty is as follows:
$$
\frac{\partial\ell}{\partial w_j} = \sum_{i=1}^N h_j(\mathbf{x}i)\left(\mathbf{1}[y_i = +1] - P(y_i = +1 | \mathbf{x}_i, \mathbf{w})\right) \color{red}{-2\lambda w_j }
$$
and for the intercept term, we have
$$
\frac{\partial\ell}{\partial w_0} = \sum{i=1}^N h_0(\mathbf{x}_i)\left(\mathbf{1}[y_i = +1] - P(y_i = +1 | \mathbf{x}_i, \mathbf{w})\right)
$$
Note: As we did in the Regression course, we do not apply the L2 penalty on the intercept. A large intercept does not necessarily indicate overfitting because the intercept is not associated with any particular feature.
Write a function that computes the derivative of log likelihood with respect to a single coefficient $w_j$. Unlike its counterpart in the last assignment, the function accepts five arguments:
 * errors vector containing $(\mathbf{1}[y_i = +1] - P(y_i = +1 | \mathbf{x}_i, \mathbf{w}))$ for all $i$
 * feature vector containing $h_j(\mathbf{x}_i)$  for all $i$
 * coefficient containing the current value of coefficient $w_j$.
 * l2_penalty representing the L2 penalty constant $\lambda$
 * feature_is_constant telling whether the $j$-th feature is constant or not.
End of explanation
def compute_log_likelihood_with_L2(feature_matrix, sentiment, coefficients, l2_penalty):
    indicator = (sentiment==+1)
    scores = np.dot(feature_matrix, coefficients)
    
    lp = np.sum((indicator-1)*scores - np.log(1. + np.exp(-scores))) - l2_penalty*np.sum(coefficients[1:]**2)
    
    return lp
Explanation: Quiz question: In the code above, was the intercept term regularized?
To verify the correctness of the gradient ascent algorithm, we provide a function for computing log likelihood (which we recall from the last assignment was a topic detailed in an advanced optional video, and used here for its numerical stability).
$$\ell\ell(\mathbf{w}) = \sum_{i=1}^N \Big( (\mathbf{1}[y_i = +1] - 1)\mathbf{w}^T h(\mathbf{x}_i) - \ln\left(1 + \exp(-\mathbf{w}^T h(\mathbf{x}_i))\right) \Big) \color{red}{-\lambda\|\mathbf{w}\|_2^2} $$
End of explanation
def sigmoid(x):
    return 1 / (1 + np.exp(-x))
def compute_derivatives_with_L2(features, indicator, weights, L2):
    
    indicator can only consists of 1(positive) and 0 (negative)
    
    scores = np.dot(features, weights)
    predictions = sigmoid(scores)
    differences = indicator - predictions
    return np.dot(differences, features) - 2 * L2 * weights
def my_logistic_regression_with_L2(feature_matrix, sentiment, initial_coefficients, step_size, l2_penalty, max_iter):
    coefficients = np.array(initial_coefficients) # make sure it's a numpy array
    for itr in xrange(max_iter):
        # Predict P(y_i = +1|x_i,w) using your predict_probability() function
        ## YOUR CODE HERE
        predictions = predict_probability(feature_matrix, coefficients)
        
        # Compute indicator value for (y_i = +1)
        indicator = (sentiment==+1)
        
        # Compute the derivatives
        derivatives = compute_derivatives_with_L2(feature_matrix, indicator, coefficients, l2_penalty)
        
        # Update weights
        coefficients = coefficients + derivatives * step_size
        
        # Checking whether log likelihood is increasing
        if itr <= 15 or (itr <= 100 and itr % 10 == 0) or (itr <= 1000 and itr % 100 == 0) \
        or (itr <= 10000 and itr % 1000 == 0) or itr % 10000 == 0:
            lp = compute_log_likelihood_with_L2(feature_matrix, sentiment, coefficients, l2_penalty)
            print 'iteration %*d: log likelihood of observed labels = %.8f' % \
                (int(np.ceil(np.log10(max_iter))), itr, lp)
    return coefficients
def logistic_regression_with_L2(feature_matrix, sentiment, initial_coefficients, step_size, l2_penalty, max_iter):
    coefficients = np.array(initial_coefficients) # make sure it's a numpy array
    for itr in xrange(max_iter):
        # Predict P(y_i = +1|x_i,w) using your predict_probability() function
        ## YOUR CODE HERE
        predictions = predict_probability(feature_matrix, coefficients)
        
        # Compute indicator value for (y_i = +1)
        indicator = (sentiment==+1)
        
        # Compute the errors as indicator - predictions
        errors = indicator - predictions
        for j in xrange(len(coefficients)): # loop over each coefficient
            is_intercept = (j == 0)
            # Recall that feature_matrix[:,j] is the feature column associated with coefficients[j].
            # Compute the derivative for coefficients[j]. Save it in a variable called derivative
            ## YOUR CODE HERE
            derivative = feature_derivative_with_L2(errors, feature_matrix[:,j], 
                                                    coefficients[j], l2_penalty, is_intercept)
            
            # add the step size times the derivative to the current coefficient
            ## YOUR CODE HERE
            coefficients[j] = coefficients[j] + derivative * step_size
        
        # Checking whether log likelihood is increasing
        if itr <= 15 or (itr <= 100 and itr % 10 == 0) or (itr <= 1000 and itr % 100 == 0) \
        or (itr <= 10000 and itr % 1000 == 0) or itr % 10000 == 0:
            lp = compute_log_likelihood_with_L2(feature_matrix, sentiment, coefficients, l2_penalty)
            print 'iteration %*d: log likelihood of observed labels = %.8f' % \
                (int(np.ceil(np.log10(max_iter))), itr, lp)
    return coefficients
Explanation: Quiz question: Does the term with L2 regularization increase or decrease $\ell\ell(\mathbf{w})$?
The logistic regression function looks almost like the one in the last assignment, with a minor modification to account for the L2 penalty.  Fill in the code below to complete this modification.
End of explanation
my_logistic_regression_with_L2(feature_matrix_train, sentiment_train,
                                                     initial_coefficients=np.zeros(194),
                                                     step_size=5e-6, l2_penalty=0, max_iter=501)
# run with L2 = 0
coefficients_0_penalty = logistic_regression_with_L2(feature_matrix_train, sentiment_train,
                                                     initial_coefficients=np.zeros(194),
                                                     step_size=5e-6, l2_penalty=0, max_iter=501)
# run with L2 = 4
coefficients_4_penalty = logistic_regression_with_L2(feature_matrix_train, sentiment_train,
                                                      initial_coefficients=np.zeros(194),
                                                      step_size=5e-6, l2_penalty=4, max_iter=501)
# run with L2 = 10
coefficients_10_penalty = logistic_regression_with_L2(feature_matrix_train, sentiment_train,
                                                      initial_coefficients=np.zeros(194),
                                                      step_size=5e-6, l2_penalty=10, max_iter=501)
# run with L2 = 1e2
coefficients_1e2_penalty = logistic_regression_with_L2(feature_matrix_train, sentiment_train,
                                                       initial_coefficients=np.zeros(194),
                                                       step_size=5e-6, l2_penalty=1e2, max_iter=501)
# run with L2 = 1e3
coefficients_1e3_penalty = logistic_regression_with_L2(feature_matrix_train, sentiment_train,
                                                       initial_coefficients=np.zeros(194),
                                                       step_size=5e-6, l2_penalty=1e3, max_iter=501)
# run with L2 = 1e5
coefficients_1e5_penalty = logistic_regression_with_L2(feature_matrix_train, sentiment_train,
                                                       initial_coefficients=np.zeros(194),
                                                       step_size=5e-6, l2_penalty=1e5, max_iter=501)
Explanation: Explore effects of L2 regularization
Now that we have written up all the pieces needed for regularized logistic regression, let's explore the benefits of using L2 regularization in analyzing sentiment for product reviews. As iterations pass, the log likelihood should increase.
Below, we train models with increasing amounts of regularization, starting with no L2 penalty, which is equivalent to our previous logistic regression implementation.
End of explanation
table = graphlab.SFrame({'word': ['(intercept)'] + important_words})
def add_coefficients_to_table(coefficients, column_name):
    table[column_name] = coefficients
    return table
Explanation: Compare coefficients
We now compare the coefficients for each of the models that were trained above. We will create a table of features and learned coefficients associated with each of the different L2 penalty values.
Below is a simple helper function that will help us create this table.
End of explanation
add_coefficients_to_table(coefficients_0_penalty, 'coefficients [L2=0]')
add_coefficients_to_table(coefficients_4_penalty, 'coefficients [L2=4]')
add_coefficients_to_table(coefficients_10_penalty, 'coefficients [L2=10]')
add_coefficients_to_table(coefficients_1e2_penalty, 'coefficients [L2=1e2]')
add_coefficients_to_table(coefficients_1e3_penalty, 'coefficients [L2=1e3]')
add_coefficients_to_table(coefficients_1e5_penalty, 'coefficients [L2=1e5]')
Explanation: Now, let's run the function add_coefficients_to_table for each of the L2 penalty strengths.
End of explanation
word_weights = zip(coefficients_0_penalty[1:],important_words)
sorted_word_weights = sorted(word_weights, reverse=True)
positive_words = [word for (weight, word) in sorted_word_weights[:5]]
negative_words = [word for (weight, word) in sorted_word_weights[-5:]]
positive_words
negative_words
Explanation: Using the coefficients trained with L2 penalty 0, find the 5 most positive words (with largest positive coefficients). Save them to positive_words. Similarly, find the 5 most negative words (with largest negative coefficients) and save them to negative_words.
Quiz Question. Which of the following is not listed in either positive_words or negative_words?
End of explanation
import matplotlib.pyplot as plt
%matplotlib inline
plt.rcParams['figure.figsize'] = 10, 6
def make_coefficient_plot(table, positive_words, negative_words, l2_penalty_list):
    cmap_positive = plt.get_cmap('Reds')
    cmap_negative = plt.get_cmap('Blues')
    
    xx = l2_penalty_list
    plt.plot(xx, [0.]*len(xx), '--', lw=1, color='k')
    
    table_positive_words = table.filter_by(column_name='word', values=positive_words)
    table_negative_words = table.filter_by(column_name='word', values=negative_words)
    del table_positive_words['word']
    del table_negative_words['word']
    
    for i in xrange(len(positive_words)):
        color = cmap_positive(0.8*((i+1)/(len(positive_words)*1.2)+0.15))
        plt.plot(xx, table_positive_words[i:i+1].to_numpy().flatten(),
                 '-', label=positive_words[i], linewidth=4.0, color=color)
        
    for i in xrange(len(negative_words)):
        color = cmap_negative(0.8*((i+1)/(len(negative_words)*1.2)+0.15))
        plt.plot(xx, table_negative_words[i:i+1].to_numpy().flatten(),
                 '-', label=negative_words[i], linewidth=4.0, color=color)
        
    plt.legend(loc='best', ncol=3, prop={'size':16}, columnspacing=0.5)
    plt.axis([1, 1e5, -1, 2])
    plt.title('Coefficient path')
    plt.xlabel('L2 penalty ($\lambda$)')
    plt.ylabel('Coefficient value')
    plt.xscale('log')
    plt.rcParams.update({'font.size': 18})
    plt.tight_layout()
Explanation: Let us observe the effect of increasing L2 penalty on the 10 words just selected. We provide you with a utility function to  plot the coefficient path.
End of explanation
make_coefficient_plot(table, positive_words, negative_words, l2_penalty_list=[0, 4, 10, 1e2, 1e3, 1e5])
Explanation: Run the following cell to generate the plot. Use the plot to answer the following quiz question.
End of explanation
def get_classification_accuracy(feature_matrix, sentiment, coefficients):
    scores = np.dot(feature_matrix, coefficients)
    apply_threshold = np.vectorize(lambda x: 1. if x > 0  else -1.)
    predictions = apply_threshold(scores)
    
    num_correct = (predictions == sentiment).sum()
    accuracy = num_correct / len(feature_matrix)    
    return accuracy
Explanation: Quiz Question: (True/False) All coefficients consistently get smaller in size as the L2 penalty is increased.
Quiz Question: (True/False) The relative order of coefficients is preserved as the L2 penalty is increased. (For example, if the coefficient for 'cat' was more positive than that for 'dog', this remains true as the L2 penalty increases.)
Measuring accuracy
Now, let us compute the accuracy of the classifier model. Recall that the accuracy is given by
$$
\mbox{accuracy} = \frac{\mbox{# correctly classified data points}}{\mbox{# total data points}}
$$
Recall from lecture that that the class prediction is calculated using
$$
\hat{y}_i = 
\left{
\begin{array}{ll}
      +1 & h(\mathbf{x}_i)^T\mathbf{w} > 0 \
      -1 & h(\mathbf{x}_i)^T\mathbf{w} \leq 0 \
\end{array} 
\right.
$$
Note: It is important to know that the model prediction code doesn't change even with the addition of an L2 penalty. The only thing that changes is the estimated coefficients used in this prediction.
Based on the above, we will use the same code that was used in Module 3 assignment.
End of explanation
train_accuracy = {}
train_accuracy[0]   = get_classification_accuracy(feature_matrix_train, sentiment_train, coefficients_0_penalty)
train_accuracy[4]   = get_classification_accuracy(feature_matrix_train, sentiment_train, coefficients_4_penalty)
train_accuracy[10]  = get_classification_accuracy(feature_matrix_train, sentiment_train, coefficients_10_penalty)
train_accuracy[1e2] = get_classification_accuracy(feature_matrix_train, sentiment_train, coefficients_1e2_penalty)
train_accuracy[1e3] = get_classification_accuracy(feature_matrix_train, sentiment_train, coefficients_1e3_penalty)
train_accuracy[1e5] = get_classification_accuracy(feature_matrix_train, sentiment_train, coefficients_1e5_penalty)
validation_accuracy = {}
validation_accuracy[0]   = get_classification_accuracy(feature_matrix_valid, sentiment_valid, coefficients_0_penalty)
validation_accuracy[4]   = get_classification_accuracy(feature_matrix_valid, sentiment_valid, coefficients_4_penalty)
validation_accuracy[10]  = get_classification_accuracy(feature_matrix_valid, sentiment_valid, coefficients_10_penalty)
validation_accuracy[1e2] = get_classification_accuracy(feature_matrix_valid, sentiment_valid, coefficients_1e2_penalty)
validation_accuracy[1e3] = get_classification_accuracy(feature_matrix_valid, sentiment_valid, coefficients_1e3_penalty)
validation_accuracy[1e5] = get_classification_accuracy(feature_matrix_valid, sentiment_valid, coefficients_1e5_penalty)
# Build a simple report
for key in sorted(validation_accuracy.keys()):
    print "L2 penalty = %g" % key
    print "train accuracy = %s, validation_accuracy = %s" % (train_accuracy[key], validation_accuracy[key])
    print "--------------------------------------------------------------------------------"
Explanation: Below, we compare the accuracy on the training data and validation data for all the models that were trained in this assignment.  We first calculate the accuracy values and then build a simple report summarizing the performance for the various models.
End of explanation
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