The goal of this notebook is to implement your own logistic regression classifier. You will:
Let's get started!
Make sure you have the latest version of GraphLab Create. Upgrade by
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import graphlab
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']
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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.
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:
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']
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Now, write some code to compute the number of product reviews that contain the word perfect.
Hint:
contains_perfect which is set to 1 if the count of the word perfect (stored in column perfect) is >= 1.contains_perfect.
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products['contains_perfect'] = products['perfect'] >= 1
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products['contains_perfect'].sum()
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Quiz Question. How many reviews contain the word perfect?
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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
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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
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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
scores = np.dot(feature_matrix, coefficients)
# Compute P(y_i = +1 | x_i, w) using the link function
# YOUR CODE HERE
predictions = map(lambda x: 1 / (1 + np.exp(- x)), scores)
# 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)
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 = np.dot(errors, feature)
# 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)
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 = 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
# 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(errors, feature_matrix[:,j])
# add the step size times the derivative to the current coefficient
## YOUR CODE HERE
coefficients[j] += step_size * derivative
# 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?
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 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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y_hat = scores > 0
Quiz question: How many reviews were predicted to have positive sentiment?
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y_hat.sum()
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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 = (y_hat == sentiment).sum() # YOUR CODE HERE
accuracy = 1. - float(num_mistakes) / len(products) # 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 = %.3f' % accuracy
Quiz question: What is the accuracy of the model on predictions made above? (round to 2 digits of accuracy)
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:
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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.
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sorted(word_coefficient_tuples, key=lambda x: x[1], reverse=True)[:10]
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Quiz question: Which word is not present in the top 10 "most positive" words?
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sorted(word_coefficient_tuples, key=lambda x: x[1], reverse=False)[:10]
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Quiz question: Which word is not present in the top 10 "most negative" words?