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import pandas as pd
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products = pd.read_csv('../../data/amazon_baby_subset.csv')
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products['sentiment']
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products['sentiment'].size
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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]))
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# The same feature processing (same as the previous assignments)
# ---------------------------------------------------------------
import json
with open('../../data/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
translator = str.maketrans('', '', string.punctuation)
return str(text).translate(translator)
# 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))
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.
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with open('../../data/module-4-assignment-train-idx.json', 'r') as f:
train_idx = json.load(f)
train_data = products.ix[train_idx]
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with open ('../../data/module-4-assignment-validation-idx.json', 'r') as f:
v_idx = json.load(f)
validation_data = products.ix[v_idx]
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.
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import numpy as np
def get_numpy_data(data_frame, features, label):
data_frame['intercept'] = 1
features = ['intercept'] + features
features_frame = data_frame[features]
feature_matrix = features_frame.as_matrix()
label_array = data_frame[label]
return(feature_matrix, label_array)
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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')
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.)
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def prediction(score):
return (1 / (1 + np.exp(-score)))
'''
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
scores = np.dot(feature_matrix, coefficients)
# Compute P(y_i = +1 | x_i, w) using the link function
predictions = np.apply_along_axis(prediction, 0, scores)
# return predictions
return predictions
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.
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.
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def feature_derivative_with_L2(errors, feature, coefficient, l2_penalty, feature_is_constant):
# Compute the dot product of errors and feature
derivative = np.dot(feature, errors)
# add L2 penalty term for any feature that isn't the intercept.
if not feature_is_constant:
derivative = derivative - 2 * l2_penalty * coefficient
return derivative
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).
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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
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.
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from math import sqrt
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 range(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 range(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_with_L2(errors, feature_matrix[:, j], coefficients[j], l2_penalty, j == 0)
# add the step size times the derivative to the current coefficient
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_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
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.
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# 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)
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# 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)
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# 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)
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# 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)
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# 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)
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# 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)
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important_words.insert(0, 'intercept')
data = np.array(important_words)
table = pd.DataFrame(columns = ['words'], data = data)
def add_coefficients_to_table(coefficients, column_name):
table[column_name] = coefficients
return table
important_words.remove('intercept')
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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]')
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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?
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def make_tuple(column_name):
word_coefficient_tuples = [(word, coefficient) for word, coefficient in zip( table['words'], table[column_name])]
return word_coefficient_tuples
positive_words = list(map(lambda x: x[0], sorted(make_tuple('coefficients [L2=0]'), key=lambda x:x[1], reverse=True)[:5]))
negative_words = list(map(lambda x: x[0], sorted(make_tuple('coefficients [L2=0]'), key=lambda x:x[1], reverse=False)[:5]))
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positive_words
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negative_words
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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.
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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[table['words'].isin(positive_words)]
table_negative_words = table[table['words'].isin(negative_words)]
del table_positive_words['words']
del table_negative_words['words']
for i in range(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].as_matrix().flatten(),
'-', label=positive_words[i], linewidth=4.0, color=color)
for i in range(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].as_matrix().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()
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make_coefficient_plot(table, positive_words, negative_words, l2_penalty_list=[0, 4, 10, 1e2, 1e3, 1e5])
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.)
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.
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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
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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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)
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# 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("--------------------------------------------------------------------------------")
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# Optional. Plot accuracy on training and validation sets over choice of L2 penalty.
import matplotlib.pyplot as plt
%matplotlib inline
plt.rcParams['figure.figsize'] = 10, 6
sorted_list = sorted(train_accuracy.items(), key=lambda x:x[0])
plt.plot([p[0] for p in sorted_list], [p[1] for p in sorted_list], 'bo-', linewidth=4, label='Training accuracy')
sorted_list = sorted(validation_accuracy.items(), key=lambda x:x[0])
plt.plot([p[0] for p in sorted_list], [p[1] for p in sorted_list], 'ro-', linewidth=4, label='Validation accuracy')
plt.xscale('symlog')
plt.axis([0, 1e3, 0.78, 0.786])
plt.legend(loc='lower left')
plt.rcParams.update({'font.size': 18})
plt.tight_layout
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