Implementing logistic regression from scratch

The goal of this notebook is to implement your own logistic regression classifier. You will:

  • Extract features from Amazon product reviews.
  • Convert an SFrame into a NumPy array.
  • Implement the link function for logistic regression.
  • Write a function to compute the derivative of the log likelihood function with respect to a single coefficient.
  • Implement gradient ascent.
  • Given a set of coefficients, predict sentiments.
  • Compute classification accuracy for the logistic regression model.

Let's get started!

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.


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import graphlab

Load review dataset

For this assignment, we will use a subset of the Amazon product review dataset. The subset was chosen to contain similar numbers of positive and negative reviews, as the original dataset consisted primarily of positive reviews.


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products = graphlab.SFrame('amazon_baby_subset.gl/')

One column of this dataset is 'sentiment', corresponding to the class label with +1 indicating a review with positive sentiment and -1 indicating one with negative sentiment.


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products['sentiment']

Let us quickly explore more of this dataset. The 'name' column indicates the name of the product. Here we list the first 10 products in the dataset. We then count the number of positive and negative reviews.


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products.head(10)['name']

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print '# of positive reviews =', len(products[products['sentiment']==1])
print '# of negative reviews =', len(products[products['sentiment']==-1])

Note: For this assignment, we eliminated class imbalance by choosing a subset of the data with a similar number of positive and negative reviews.

Apply text cleaning on the review data

In this section, we will perform some simple feature cleaning using SFrames. The last assignment used all words in building bag-of-words features, but here we limit ourselves to 193 words (for simplicity). We compiled a list of 193 most frequent words into a JSON file.

Now, we will load these words from this JSON file:


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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]

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print important_words

Now, we will perform 2 simple data transformations:

  1. Remove punctuation using Python's built-in string functionality.
  2. Compute word counts (only for important_words)

We start with Step 1 which can be done as follows:


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def remove_punctuation(text):
    import string
    return text.translate(None, string.punctuation) 

products['review_clean'] = products['review'].apply(remove_punctuation)

Now we proceed with Step 2. For each word in important_words, we compute a count for the number of times the word occurs in the review. We will store this count in a separate column (one for each word). The result of this feature processing is a single column for each word in important_words which keeps a count of the number of times the respective word occurs in the review text.

Note: There are several ways of doing this. In this assignment, we use the built-in count function for Python lists. Each review string is first split into individual words and the number of occurances of a given word is counted.


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for word in important_words:
    products[word] = products['review_clean'].apply(lambda s : s.split().count(word))

The SFrame products now contains one column for each of the 193 important_words. As an example, the column perfect contains a count of the number of times the word perfect occurs in each of the reviews.


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products['perfect']

Now, write some code to compute the number of product reviews that contain the word perfect.

Hint:

  • First create a column called contains_perfect which is set to 1 if the count of the word perfect (stored in column perfect) is >= 1.
  • Sum the number of 1s in the column contains_perfect.

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Quiz Question. How many reviews contain the word perfect?

Convert SFrame to NumPy array

As you have seen previously, NumPy is a powerful library for doing matrix manipulation. Let us convert our data to matrices and then implement our algorithms with matrices.

First, make sure you can perform the following import.


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import numpy as np

We now 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 that the feature matrix includes an additional column 'intercept' to take account of the intercept term.


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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)

Let us convert the data into NumPy arrays.


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# Warning: This may take a few minutes...
feature_matrix, sentiment = get_numpy_data(products, important_words, 'sentiment')

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feature_matrix.shape

Quiz Question: How many features are there in the feature_matrix?

Quiz Question: Assuming that the intercept is present, how does the number of features in feature_matrix relate to the number of features in the logistic regression model?

Now, let us see what the sentiment column looks like:


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sentiment

Recall from lecture that the link function is given by: $$ P(y_i = +1 | \mathbf{x}_i,\mathbf{w}) = \frac{1}{1 + \exp(-\mathbf{w}^T h(\mathbf{x}_i))}, $$

where the feature vector $h(\mathbf{x}_i)$ represents the word counts of important_words in the review $\mathbf{x}_i$. Complete the following function that implements the link function:


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'''
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
    ...
    
    # Compute P(y_i = +1 | x_i, w) using the link function
    # YOUR CODE HERE
    predictions = ...
    
    # return predictions
    return predictions

Aside. How the link function works with matrix algebra

Since the word counts are stored as columns in feature_matrix, each $i$-th row of the matrix corresponds to the feature vector $h(\mathbf{x}_i)$: $$ [\text{feature_matrix}] = \left[ \begin{array}{c} h(\mathbf{x}_1)^T \\ h(\mathbf{x}_2)^T \\ \vdots \\ h(\mathbf{x}_N)^T \end{array} \right] = \left[ \begin{array}{cccc} h_0(\mathbf{x}_1) & h_1(\mathbf{x}_1) & \cdots & h_D(\mathbf{x}_1) \\ h_0(\mathbf{x}_2) & h_1(\mathbf{x}_2) & \cdots & h_D(\mathbf{x}_2) \\ \vdots & \vdots & \ddots & \vdots \\ h_0(\mathbf{x}_N) & h_1(\mathbf{x}_N) & \cdots & h_D(\mathbf{x}_N) \end{array} \right] $$

By the rules of matrix multiplication, the score vector containing elements $\mathbf{w}^T h(\mathbf{x}_i)$ is obtained by multiplying feature_matrix and the coefficient vector $\mathbf{w}$. $$ [\text{score}] = [\text{feature_matrix}]\mathbf{w} = \left[ \begin{array}{c} h(\mathbf{x}_1)^T \\ h(\mathbf{x}_2)^T \\ \vdots \\ h(\mathbf{x}_N)^T \end{array} \right] \mathbf{w} = \left[ \begin{array}{c} h(\mathbf{x}_1)^T\mathbf{w} \\ h(\mathbf{x}_2)^T\mathbf{w} \\ \vdots \\ h(\mathbf{x}_N)^T\mathbf{w} \end{array} \right] = \left[ \begin{array}{c} \mathbf{w}^T h(\mathbf{x}_1) \\ \mathbf{w}^T h(\mathbf{x}_2) \\ \vdots \\ \mathbf{w}^T h(\mathbf{x}_N) \end{array} \right] $$

Checkpoint

Just to make sure you are on the right track, we have provided a few examples. If your predict_probability function is implemented correctly, then the outputs will match:


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dummy_feature_matrix = np.array([[1.,2.,3.], [1.,-1.,-1]])
dummy_coefficients = np.array([1., 3., -1.])

correct_scores      = np.array( [ 1.*1. + 2.*3. + 3.*(-1.),          1.*1. + (-1.)*3. + (-1.)*(-1.) ] )
correct_predictions = np.array( [ 1./(1+np.exp(-correct_scores[0])), 1./(1+np.exp(-correct_scores[1])) ] )

print 'The following outputs must match '
print '------------------------------------------------'
print 'correct_predictions           =', correct_predictions
print 'output of predict_probability =', predict_probability(dummy_feature_matrix, dummy_coefficients)

Compute derivative of log likelihood with respect to a single coefficient

Recall from lecture: $$ \frac{\partial\ell}{\partial w_j} = \sum_{i=1}^N h_j(\mathbf{x}_i)\left(\mathbf{1}[y_i = +1] - P(y_i = +1 | \mathbf{x}_i, \mathbf{w})\right) $$

We will now write a function that computes the derivative of log likelihood with respect to a single coefficient $w_j$. The function accepts two 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 $i$.

Complete the following code block:


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def feature_derivative(errors, feature):     
    # Compute the dot product of errors and feature
    derivative = ...
    
    # Return the derivative
    return derivative

In the main lecture, our focus was on the likelihood. In the advanced optional video, however, we introduced a transformation of this likelihood---called the log likelihood---that simplifies the derivation of the gradient and is more numerically stable. Due to its numerical stability, we will use the log likelihood instead of the likelihood to assess the algorithm.

The log likelihood is computed using the following formula (see the advanced optional video if you are curious about the derivation of this equation):

$$\ell\ell(\mathbf{w}) = \sum_{i=1}^N \Big( (\mathbf{1}[y_i = +1] - 1)\mathbf{w}^T h(\mathbf{x}_i) - \ln\left(1 + \exp(-\mathbf{w}^T h(\mathbf{x}_i))\right) \Big) $$

We provide a function to compute the log likelihood for the entire dataset.


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def compute_log_likelihood(feature_matrix, sentiment, coefficients):
    indicator = (sentiment==+1)
    scores = np.dot(feature_matrix, coefficients)
    logexp = np.log(1. + np.exp(-scores))
    
    # Simple check to prevent overflow
    mask = np.isinf(logexp)
    logexp[mask] = -scores[mask]
    
    lp = np.sum((indicator-1)*scores - logexp)
    return lp

Checkpoint

Just to make sure we are on the same page, run the following code block and check that the outputs match.


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dummy_feature_matrix = np.array([[1.,2.,3.], [1.,-1.,-1]])
dummy_coefficients = np.array([1., 3., -1.])
dummy_sentiment = np.array([-1, 1])

correct_indicators  = np.array( [ -1==+1,                                       1==+1 ] )
correct_scores      = np.array( [ 1.*1. + 2.*3. + 3.*(-1.),                     1.*1. + (-1.)*3. + (-1.)*(-1.) ] )
correct_first_term  = np.array( [ (correct_indicators[0]-1)*correct_scores[0],  (correct_indicators[1]-1)*correct_scores[1] ] )
correct_second_term = np.array( [ np.log(1. + np.exp(-correct_scores[0])),      np.log(1. + np.exp(-correct_scores[1])) ] )

correct_ll          =      sum( [ correct_first_term[0]-correct_second_term[0], correct_first_term[1]-correct_second_term[1] ] ) 

print 'The following outputs must match '
print '------------------------------------------------'
print 'correct_log_likelihood           =', correct_ll
print 'output of compute_log_likelihood =', compute_log_likelihood(dummy_feature_matrix, dummy_sentiment, dummy_coefficients)

Taking gradient steps

Now we are ready to implement our own logistic regression. All we have to do is to write a gradient ascent function that takes gradient steps towards the optimum.

Complete the following function to solve the logistic regression model using gradient ascent:


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from math import sqrt

def logistic_regression(feature_matrix, sentiment, initial_coefficients, step_size, 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 = ...
        
        # 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
            
            # 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 = ...
            
            # add the step size times the derivative to the current coefficient
            ## YOUR CODE HERE
            ...
        
        # 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(feature_matrix, sentiment, coefficients)
            print 'iteration %*d: log likelihood of observed labels = %.8f' % \
                (int(np.ceil(np.log10(max_iter))), itr, lp)
    return coefficients

Now, let us run the logistic regression solver.


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coefficients = logistic_regression(feature_matrix, sentiment, initial_coefficients=np.zeros(194),
                                   step_size=1e-7, max_iter=301)

Quiz question: As each iteration of gradient ascent passes, does the log likelihood increase or decrease?

Predicting sentiments

Recall from lecture that class predictions for a data point $\mathbf{x}$ can be computed from the coefficients $\mathbf{w}$ using the following formula: $$ \hat{y}_i = \left\{ \begin{array}{ll} +1 & \mathbf{x}_i^T\mathbf{w} > 0 \\ -1 & \mathbf{x}_i^T\mathbf{w} \leq 0 \\ \end{array} \right. $$

Now, we will write some code to compute class predictions. We will do this in two steps:

  • Step 1: First compute the scores using feature_matrix and coefficients using a dot product.
  • Step 2: Using the formula above, compute the class predictions from the scores.

Step 1 can be implemented as follows:


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# Compute the scores as a dot product between feature_matrix and coefficients.
scores = np.dot(feature_matrix, coefficients)

Now, complete the following code block for Step 2 to compute the class predictions using the scores obtained above:


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Quiz question: How many reviews were predicted to have positive sentiment?


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Measuring accuracy

We will now measure the classification accuracy of the model. Recall from the lecture that the classification accuracy can be computed as follows:

$$ \mbox{accuracy} = \frac{\mbox{# correctly classified data points}}{\mbox{# total data points}} $$

Complete the following code block to compute the accuracy of the model.


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num_mistakes = ... # YOUR CODE HERE
accuracy = ... # YOUR CODE HERE
print "-----------------------------------------------------"
print '# Reviews   correctly classified =', len(products) - num_mistakes
print '# Reviews incorrectly classified =', num_mistakes
print '# Reviews total                  =', len(products)
print "-----------------------------------------------------"
print 'Accuracy = %.2f' % accuracy

Quiz question: What is the accuracy of the model on predictions made above? (round to 2 digits of accuracy)

Which words contribute most to positive & negative sentiments?

Recall that in Module 2 assignment, we were able to compute the "most positive words". These are words that correspond most strongly with positive reviews. In order to do this, we will first do the following:

  • Treat each coefficient as a tuple, i.e. (word, coefficient_value).
  • Sort all the (word, coefficient_value) tuples by coefficient_value in descending order.

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coefficients = list(coefficients[1:]) # exclude intercept
word_coefficient_tuples = [(word, coefficient) for word, coefficient in zip(important_words, coefficients)]
word_coefficient_tuples = sorted(word_coefficient_tuples, key=lambda x:x[1], reverse=True)

Now, word_coefficient_tuples contains a sorted list of (word, coefficient_value) tuples. The first 10 elements in this list correspond to the words that are most positive.

Ten "most positive" words

Now, we compute the 10 words that have the most positive coefficient values. These words are associated with positive sentiment.


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Quiz question: Which word is not present in the top 10 "most positive" words?

Ten "most negative" words

Next, we repeat this exercise on the 10 most negative words. That is, we compute the 10 words that have the most negative coefficient values. These words are associated with negative sentiment.


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Quiz question: Which word is not present in the top 10 "most negative" words?