In this assignment, you'll be analyzing movie reviews from Rotten Tomatoes. This assignment will cover:
Useful libraries for this assignment
In [30]:
%matplotlib inline
import json
import requests
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
pd.set_option('display.width', 500)
pd.set_option('display.max_columns', 30)
# set some nicer defaults for matplotlib
from matplotlib import rcParams
#these colors come from colorbrewer2.org. Each is an RGB triplet
dark2_colors = [(0.10588235294117647, 0.6196078431372549, 0.4666666666666667),
(0.8509803921568627, 0.37254901960784315, 0.00784313725490196),
(0.4588235294117647, 0.4392156862745098, 0.7019607843137254),
(0.9058823529411765, 0.1607843137254902, 0.5411764705882353),
(0.4, 0.6509803921568628, 0.11764705882352941),
(0.9019607843137255, 0.6705882352941176, 0.00784313725490196),
(0.6509803921568628, 0.4627450980392157, 0.11372549019607843),
(0.4, 0.4, 0.4)]
rcParams['figure.figsize'] = (10, 6)
rcParams['figure.dpi'] = 150
rcParams['axes.color_cycle'] = dark2_colors
rcParams['lines.linewidth'] = 2
rcParams['axes.grid'] = False
rcParams['axes.facecolor'] = 'white'
rcParams['font.size'] = 14
rcParams['patch.edgecolor'] = 'none'
def remove_border(axes=None, top=False, right=False, left=True, bottom=True):
"""
Minimize chartjunk by stripping out unnecesary plot borders and axis ticks
The top/right/left/bottom keywords toggle whether the corresponding plot border is drawn
"""
ax = axes or plt.gca()
ax.spines['top'].set_visible(top)
ax.spines['right'].set_visible(right)
ax.spines['left'].set_visible(left)
ax.spines['bottom'].set_visible(bottom)
#turn off all ticks
ax.yaxis.set_ticks_position('none')
ax.xaxis.set_ticks_position('none')
#now re-enable visibles
if top:
ax.xaxis.tick_top()
if bottom:
ax.xaxis.tick_bottom()
if left:
ax.yaxis.tick_left()
if right:
ax.yaxis.tick_right()
Rotten Tomatoes gathers movie reviews from critics. An entry on the website typically consists of a short quote, a link to the full review, and a Fresh/Rotten classification which summarizes whether the critic liked/disliked the movie.
When critics give quantitative ratings (say 3/4 stars, Thumbs up, etc.), determining the Fresh/Rotten classification is easy. However, publications like the New York Times don't assign numerical ratings to movies, and thus the Fresh/Rotten classification must be inferred from the text of the review itself.
This basic task of categorizing text has many applications. All of the following questions boil down to text classification:
Language is incredibly nuanced, and there is an entire field of computer science dedicated to the topic (Natural Language Processing). Nevertheless, we can construct basic language models using fairly straightforward techniques.
You will be starting with a database of Movies, derived from the MovieLens dataset. This dataset includes information for about 10,000 movies, including the IMDB id for each movie.
Your first task is to download Rotten Tomatoes reviews from 3000 of these movies, using the Rotten Tomatoes API (Application Programming Interface).
Web APIs are a more convenient way for programs to interact with websites. Rotten Tomatoes has a nice API that gives access to it's data in JSON format.
To use this, you will first need to register for an API key. For "application URL", you can use anything -- it doesn't matter.
After you have a key, the documentation page shows the various data you can fetch from Rotten Tomatoes -- each type of data lives at a different web address. The basic pattern for fetching this data with Python is as follows (compare this to the Movie Reviews tab on the documentation page):
In [31]:
api_key = 'YOUR KEY HERE'
movie_id = '770672122' # toy story 3
url = 'http://api.rottentomatoes.com/api/public/v1.0/movies/%s/reviews.json' % movie_id
#these are "get parameters"
options = {'review_type': 'top_critic', 'page_limit': 20, 'page': 1, 'apikey': api_key}
data = requests.get(url, params=options).text
data = json.loads(data) # load a json string into a collection of lists and dicts
print json.dumps(data['reviews'][0], indent=2) # dump an object into a json string
In [32]:
from io import StringIO
movie_txt = requests.get('https://raw.github.com/cs109/cs109_data/master/movies.dat').text
movie_file = StringIO(movie_txt) # treat a string like a file
movies = pd.read_csv(movie_file, delimiter='\t')
#print the first row
movies[['id', 'title', 'imdbID', 'year']].irow(0)
Out[32]:
We'd like you to write a function that looks up the first 20 Top Critic Rotten Tomatoes reviews for a movie in the movies dataframe. This involves two steps:
Movie Alias API to look up the Rotten Tomatoes movie id from the IMDB idMovie Reviews API to fetch the first 20 top-critic reviews for this movieNot all movies have Rotten Tomatoes IDs. In these cases, your function should return None. The detailed spec is below. We are giving you some freedom with how you implement this, but you'll probably want to break this task up into several small functions.
Hint In some situations, the leading 0s in front of IMDB ids are important. IMDB ids have 7 digits
In [33]:
"""
Function
--------
fetch_reviews(movies, row)
Use the Rotten Tomatoes web API to fetch reviews for a particular movie
Parameters
----------
movies : DataFrame
The movies data above
row : int
The row of the movies DataFrame to use
Returns
-------
If you can match the IMDB id to a Rotten Tomatoes ID:
A DataFrame, containing the first 20 Top Critic reviews
for the movie. If a movie has less than 20 total reviews, return them all.
This should have the following columns:
critic : Name of the critic
fresh : 'fresh' or 'rotten'
imdb : IMDB id for the movie
publication: Publication that the critic writes for
quote : string containing the movie review quote
review_data: Date of review
rtid : Rotten Tomatoes ID for the movie
title : Name of the movie
If you cannot match the IMDB id to a Rotten Tomatoes ID, return None
Examples
--------
>>> reviews = fetch_reviews(movies, 0)
>>> print len(reviews)
20
>>> print reviews.irow(1)
critic Derek Adams
fresh fresh
imdb 114709
publication Time Out
quote So ingenious in concept, design and execution ...
review_date 2009-10-04
rtid 9559
title Toy story
Name: 1, dtype: object
"""
#your code here
def base_url():
return 'http://api.rottentomatoes.com/api/public/v1.0/'
def rt_id_by_imdb(imdb):
"""
Queries the RT movie_alias API. Returns the RT id associated with an IMDB ID,
or raises a KeyError if no match was found
"""
url = base_url() + 'movie_alias.json'
imdb = "%7.7i" % imdb
params = dict(id=imdb, type='imdb', apikey=api_key)
r = requests.get(url, params=params).text
r = json.loads(r)
return r['id']
def _imdb_review(imdb):
"""
Query the RT reviews API, to return the first page of reviews
for a movie specified by its IMDB ID
Returns a list of dicts
"""
rtid = rt_id_by_imdb(imdb)
url = base_url() + 'movies/{0}/reviews.json'.format(rtid)
params = dict(review_type='top_critic',
page_limit=20,
page=1,
country='us',
apikey=api_key)
data = json.loads(requests.get(url, params=params).text)
data = data['reviews']
data = [dict(fresh=r['freshness'],
quote=r['quote'],
critic=r['critic'],
publication=r['publication'],
review_date=r['date'],
imdb=imdb, rtid=rtid
) for r in data]
return data
def fetch_reviews(movies, row):
m = movies.irow(row)
try:
result = pd.DataFrame(_imdb_review(m['imdbID']))
result['title'] = m['title']
except KeyError:
return None
return result
Use the function you wrote to retrieve reviews for the first 3,000 movies in the movies dataframe.
In [34]:
"""
Function
--------
build_table
Parameters
----------
movies : DataFrame
The movies data above
rows : int
The number of rows to extract reviews for
Returns
--------
A dataframe
The data obtained by repeatedly calling `fetch_reviews` on the first `rows`
of `movies`, discarding the `None`s,
and concatenating the results into a single DataFrame
"""
#your code here
def build_table(movies, rows):
dfs = [fetch_reviews(movies, r) for r in range(rows)]
dfs = [d for d in dfs if d is not None]
return pd.concat(dfs, ignore_index=True)
In [35]:
#you can toggle which lines are commented, if you
#want to re-load your results to avoid repeatedly calling this function
#critics = build_table(movies, 3000)
#critics.to_csv('critics.csv', index=False)
critics = pd.read_csv('critics.csv')
#for this assignment, let's drop rows with missing quotes
critics = critics[~critics.quote.isnull()]
2.1 How many reviews, critics, and movies are in this dataset?
In [4]:
#your code here
n_reviews = len(critics)
n_movies = critics.rtid.unique().size
n_critics = critics.critic.unique().size
print "Number of reviews: %i" % n_reviews
print "Number of critics: %i" % n_critics
print "Number of movies: %i" % n_movies
2.2 What does the distribution of number of reviews per reviewer look like? Make a histogram
In [5]:
#Your code here
def histogram_style():
remove_border(left=False)
plt.grid(False)
plt.grid(axis='y', color='w', linestyle='-', lw=1)
critics.groupby('critic').rtid.count().hist(log=True, bins=range(20), edgecolor='white')
plt.xlabel("Number of reviews per critic")
plt.ylabel("N")
histogram_style()
2.3 List the 5 critics with the most reviews, along with the publication they write for
In [6]:
#Your code here
#note: there are a few valid ways to deal with critics that write for several publications.
counts = critics.groupby(['critic', 'publication']).critic.count()
counts.sort()
counts[-1:-6:-1]
Out[6]:
2.4 Of the critics with > 100 reviews, plot the distribution of average "freshness" rating per critic
In [8]:
#Your code here
df = critics.copy()
df['fresh'] = df.fresh == 'fresh'
grp = df.groupby('critic')
counts = grp.critic.count() # number of reviews by each critic
means = grp.fresh.mean() # average freshness for each critic
means[counts > 100].hist(bins=10, edgecolor='w', lw=1)
plt.xlabel("Average rating per critic")
plt.ylabel("N")
plt.yticks([0, 2, 4, 6, 8, 10])
histogram_style()
2.5
Using the original movies dataframe, plot the rotten tomatoes Top Critics Rating as a function of year. Overplot the average for each year, ignoring the score=0 examples (some of these are missing data). Comment on the result -- is there a trend? What do you think it means?
In [64]:
#Your code here
data = movies[['year', 'rtTopCriticsRating']]
data = data.convert_objects(convert_numeric=True)
data = data[(data['rtTopCriticsRating'] > 0)]
means = data.groupby('year').mean().dropna()
plt.plot(data['year'], data['rtTopCriticsRating'], 'o', mec='none', alpha=.2, label='Data')
plt.plot(means.index, means['rtTopCriticsRating'], '-', label='Yearly Average')
plt.legend(loc='lower left', frameon=False)
plt.xlabel("Year")
plt.ylabel("Average Score")
remove_border()
Answer This graph shows a trend towards a lower average score, as well as a greater abundance of low scores, with time. This is probably at least partially a selection effect -- Rotten Tomatoes probably doesn't archive reviews for all movies, especially ones that came out before the website existed. Thus, reviews of old movies are more often "the classics". Mediocre old movies have been partially forgotten, and are underrepresented in the data.
You will now use a Naive Bayes classifier to build a prediction model for whether a review is fresh or rotten, depending on the text of the review. See Lecture 9 for a discussion of Naive Bayes.
Most models work with numerical data, so we need to convert the textual collection of reviews to something numerical. A common strategy for text classification is to represent each review as a "bag of words" vector -- a long vector of numbers encoding how many times a particular word appears in a blurb.
Scikit-learn has an object called a CountVectorizer that turns text into a bag of words. Here's a quick tutorial:
In [11]:
from sklearn.feature_extraction.text import CountVectorizer
text = ['Hop on pop', 'Hop off pop', 'Hop Hop hop']
print "Original text is\n", '\n'.join(text)
vectorizer = CountVectorizer(min_df=0)
# call `fit` to build the vocabulary
vectorizer.fit(text)
# call `transform` to convert text to a bag of words
x = vectorizer.transform(text)
# CountVectorizer uses a sparse array to save memory, but it's easier in this assignment to
# convert back to a "normal" numpy array
x = x.toarray()
print
print "Transformed text vector is \n", x
# `get_feature_names` tracks which word is associated with each column of the transformed x
print
print "Words for each feature:"
print vectorizer.get_feature_names()
# Notice that the bag of words treatment doesn't preserve information about the *order* of words,
# just their frequency
3.1
Using the critics dataframe, compute a pair of numerical X, Y arrays where:
(nreview, nwords) array. Each row corresponds to a bag-of-words representation for a single review. This will be the input to your model.nreview-element 1/0 array, encoding whether a review is Fresh (1) or Rotten (0). This is the desired output from your model.Make sure to remove items with no review text
In [12]:
#hint: Consult the scikit-learn documentation to
# learn about what these classes do do
from sklearn.cross_validation import train_test_split
from sklearn.naive_bayes import MultinomialNB
"""
Function
--------
make_xy
Build a bag-of-words training set for the review data
Parameters
-----------
critics : Pandas DataFrame
The review data from above
vectorizer : CountVectorizer object (optional)
A CountVectorizer object to use. If None,
then create and fit a new CountVectorizer.
Otherwise, re-fit the provided CountVectorizer
using the critics data
Returns
-------
X : numpy array (dims: nreview, nwords)
Bag-of-words representation for each review.
Y : numpy array (dims: nreview)
1/0 array. 1 = fresh review, 0 = rotten review
Examples
--------
X, Y = make_xy(critics)
"""
def make_xy(critics, vectorizer=None):
#Your code here
if vectorizer is None:
vectorizer = CountVectorizer()
X = vectorizer.fit_transform(critics.quote)
X = X.tocsc() # some versions of sklearn return COO format
Y = (critics.fresh == 'fresh').values.astype(np.int)
return X, Y
In [13]:
X, Y = make_xy(critics)
3.2 Next, randomly split the data into two groups: a training set and a validation set.
Use the training set to train a MultinomialNB classifier,
and print the accuracy of this model on the validation set
Hint
You can use train_test_split to split up the training data
In [14]:
#Your code here
xtrain, xtest, ytrain, ytest = train_test_split(X, Y)
clf = MultinomialNB().fit(xtrain, ytrain)
print "Accuracy: %0.2f%%" % (100 * clf.score(xtest, ytest))
3.3:
We say a model is overfit if it performs better on the training data than on the test data. Is this model overfit? If so, how much more accurate is the model on the training data compared to the test data?
In [16]:
# Your code here. Print the accuracy on the test and training dataset
training_accuracy = clf.score(xtrain, ytrain)
test_accuracy = clf.score(xtest, ytest)
print "Accuracy on training data: %0.2f" % (training_accuracy)
print "Accuracy on test data: %0.2f" % (test_accuracy)
Interpret these numbers in a few sentences here
Answer Some overfitting seems to be happening here, since the error rate on the test data (23%) is more than twice as large as the error rate on the training data (10%). It's possible (though unlikely) that the accuracy difference is a product of chance, and not a symptom of overfitting. This could be tested with cross-validation, by repeatedly fitting and scoring the classifier on different train/test splits. If the performance on the training data is consistently better than the test data, then overfitting has occured. This is the case here.
3.4: Model Calibration
Bayesian models like the Naive Bayes classifier have the nice property that they compute probabilities of a particular classification -- the predict_proba and predict_log_proba methods of MultinomialNB compute these probabilities.
Being the respectable Bayesian that you are, you should always assess whether these probabilities are calibrated -- that is, whether a prediction made with a confidence of x% is correct approximately x% of the time.
Let's make a plot to assess model calibration. Schematically, we want something like this:
In words, we want to:
clf.predict_probaHints
The output of clf.predict_proba(X) is a (N example, 2) array. The first column gives the probability $P(Y=0)$ or $P(Rotten)$, and the second gives $P(Y=1)$ or $P(Fresh)$.
The above image is just a guideline -- feel free to explore other options!
In [17]:
"""
Function
--------
calibration_plot
Builds a plot like the one above, from a classifier and review data
Inputs
-------
clf : Classifier object
A MultinomialNB classifier
X : (Nexample, Nfeature) array
The bag-of-words data
Y : (Nexample) integer array
1 if a review is Fresh
"""
#your code here
def calibration_plot(clf, xtest, ytest):
prob = clf.predict_proba(xtest)[:, 1]
outcome = ytest
data = pd.DataFrame(dict(prob=prob, outcome=outcome))
#group outcomes into bins of similar probability
bins = np.linspace(0, 1, 20)
cuts = pd.cut(prob, bins)
binwidth = bins[1] - bins[0]
#freshness ratio and number of examples in each bin
cal = data.groupby(cuts).outcome.agg(['mean', 'count'])
cal['pmid'] = (bins[:-1] + bins[1:]) / 2
cal['sig'] = np.sqrt(cal.pmid * (1 - cal.pmid) / cal['count'])
#the calibration plot
ax = plt.subplot2grid((3, 1), (0, 0), rowspan=2)
p = plt.errorbar(cal.pmid, cal['mean'], cal['sig'])
plt.plot(cal.pmid, cal.pmid, linestyle='--', lw=1, color='k')
plt.ylabel("Empirical P(Fresh)")
remove_border(ax)
#the distribution of P(fresh)
ax = plt.subplot2grid((3, 1), (2, 0), sharex=ax)
plt.bar(left=cal.pmid - binwidth / 2, height=cal['count'],
width=.95 * (bins[1] - bins[0]),
fc=p[0].get_color())
plt.xlabel("Predicted P(Fresh)")
remove_border()
plt.ylabel("Number")
In [18]:
calibration_plot(clf, xtest, ytest)
3.5 We might say a model is over-confident if the freshness fraction is usually closer to 0.5 than expected (that is, there is more uncertainty than the model predicted). Likewise, a model is under-confident if the probabilities are usually further away from 0.5. Is this model generally over- or under-confident?
Answer This model is over-confident. For a properly calibrated model, we would expect ~10% of the P(Fresh)~0.1 reviews to actually be fresh. However, the actual freshness rate is closer to 30%. Likewise, for reviews where P(Fresh) ~0.9, the actuall freshness fraction is closer to 0.7. In other words, there is more uncertainty in the outcome than implied by the model.
Our classifier has a few free parameters. The two most important are:
The min_df keyword in CountVectorizer, which will ignore words which appear in fewer than min_df fraction of reviews. Words that appear only once or twice can lead to overfitting, since words which occur only a few times might correlate very well with Fresh/Rotten reviews by chance in the training dataset.
The alpha keyword in the Bayesian classifier is a "smoothing parameter" -- increasing the value decreases the sensitivity to any single feature, and tends to pull prediction probabilities closer to 50%.
As discussed in lecture and HW2, a common technique for choosing appropriate values for these parameters is cross-validation. Let's choose good parameters by maximizing the cross-validated log-likelihood.
3.6 Using clf.predict_logproba, write a function that computes the log-likelihood of a dataset
In [19]:
"""
Function
--------
log_likelihood
Compute the log likelihood of a dataset according to
a bayesian classifier.
The Log Likelihood is defined by
L = Sum_fresh(logP(fresh)) + Sum_rotten(logP(rotten))
Where Sum_fresh indicates a sum over all fresh reviews,
and Sum_rotten indicates a sum over rotten reviews
Parameters
----------
clf : Bayesian classifier
x : (nexample, nfeature) array
The input data
y : (nexample) integer array
Whether each review is Fresh
"""
#your code here
def log_likelihood(clf, x, y):
prob = clf.predict_log_proba(x)
rotten = y == 0
fresh = ~rotten
return prob[rotten, 0].sum() + prob[fresh, 1].sum()
Here's a function to estimate the cross-validated value of a scoring function, given a classifier and data
In [20]:
from sklearn.cross_validation import KFold
def cv_score(clf, x, y, score_func):
"""
Uses 5-fold cross validation to estimate a score of a classifier
Inputs
------
clf : Classifier object
x : Input feature vector
y : Input class labels
score_func : Function like log_likelihood, that takes (clf, x, y) as input,
and returns a score
Returns
-------
The average score obtained by randomly splitting (x, y) into training and
test sets, fitting on the training set, and evaluating score_func on the test set
Examples
cv_score(clf, x, y, log_likelihood)
"""
result = 0
nfold = 5
for train, test in KFold(y.size, nfold): # split data into train/test groups, 5 times
clf.fit(x[train], y[train]) # fit
result += score_func(clf, x[test], y[test]) # evaluate score function on held-out data
return result / nfold # average
3.7
Fill in the remaining code in this block, to loop over many values of alpha and min_df to determine
which settings are "best" in the sense of maximizing the cross-validated log-likelihood
In [21]:
#the grid of parameters to search over
alphas = [0, .1, 1, 5, 10, 50]
min_dfs = [1e-5, 1e-4, 1e-3, 1e-2, 1e-1]
#Find the best value for alpha and min_df, and the best classifier
best_alpha = None
best_min_df = None
max_loglike = -np.inf
for alpha in alphas:
for min_df in min_dfs:
vectorizer = CountVectorizer(min_df = min_df)
X, Y = make_xy(critics, vectorizer)
#your code here
clf = MultinomialNB(alpha=alpha)
loglike = cv_score(clf, X, Y, log_likelihood)
if loglike > max_loglike:
max_loglike = loglike
best_alpha, best_min_df = alpha, min_df
In [22]:
print "alpha: %f" % best_alpha
print "min_df: %f" % best_min_df
3.8 Now that you've determined values for alpha and min_df that optimize the cross-validated log-likelihood, repeat the steps in 3.1, 3.2, and 3.4 to train a final classifier with these parameters, re-evaluate the accuracy, and draw a new calibration plot.
In [23]:
#Your code here
vectorizer = CountVectorizer(min_df=best_min_df)
X, Y = make_xy(critics, vectorizer)
xtrain, xtest, ytrain, ytest = train_test_split(X, Y)
clf = MultinomialNB(alpha=best_alpha).fit(xtrain, ytrain)
calibration_plot(clf, xtest, ytest)
# Your code here. Print the accuracy on the test and training dataset
training_accuracy = clf.score(xtrain, ytrain)
test_accuracy = clf.score(xtest, ytest)
print "Accuracy on training data: %0.2f" % (training_accuracy)
print "Accuracy on test data: %0.2f" % (test_accuracy)
3.9 Discuss the various ways in which Cross-Validation has affected the model. Is the new model more or less accurate? Is overfitting better or worse? Is the model more or less calibrated?
Answer The new model is slightly less accurate on the test data (74% vs 77% on the original model). However, it is both better calibrated and less over-fit than before. In other words, while the classification accuracy is slightly worse, the probabilities themselves are more accurate. The model is still slightly over-confident when making low P(Fresh) predictions. However, the calibration plot shows the model is usually within 1 error bar of the expected performance where P(Fresh) >= 0.2. Finally, the new model makes less-conclusive predictions on average -- the histogram in the calibration plot is more uniformly distributed, with fewer predictions clustered around P(Fresh) = 0 or 1.
To think about/play with, but not to hand in: What would happen if you tried this again using a function besides the log-likelihood -- for example, the classification accuracy?
4.1
Using your classifier and the vectorizer.get_feature_names method, determine which words best predict a positive or negative review. Print the 10 words
that best predict a "fresh" review, and the 10 words that best predict a "rotten" review. For each word, what is the model's probability of freshness if the word appears one time?
Try computing the classification probability for a feature vector which consists of all 0s, except for a single 1. What does this probability refer to?
np.eye generates a matrix where the ith row is all 0s, except for the ith column which is 1.
In [22]:
## Your code here
words = np.array(vectorizer.get_feature_names())
x = np.eye(xtest.shape[1])
probs = clf.predict_log_proba(x)[:, 0]
ind = np.argsort(probs)
good_words = words[ind[:10]]
bad_words = words[ind[-10:]]
good_prob = probs[ind[:10]]
bad_prob = probs[ind[-10:]]
print "Good words\t P(fresh | word)"
for w, p in zip(good_words, good_prob):
print "%20s" % w, "%0.2f" % (1 - np.exp(p))
print "Bad words\t P(fresh | word)"
for w, p in zip(bad_words, bad_prob):
print "%20s" % w, "%0.2f" % (1 - np.exp(p))
4.2
One of the best sources for inspiration when trying to improve a model is to look at examples where the model performs poorly.
Find 5 fresh and rotten reviews where your model performs particularly poorly. Print each review.
In [23]:
#Your code here
x, y = make_xy(critics, vectorizer)
prob = clf.predict_proba(x)[:, 0]
predict = clf.predict(x)
bad_rotten = np.argsort(prob[y == 0])[:5]
bad_fresh = np.argsort(prob[y == 1])[-5:]
print "Mis-predicted Rotten quotes"
print '---------------------------'
for row in bad_rotten:
print critics[y == 0].quote.irow(row)
print
print "Mis-predicted Fresh quotes"
print '--------------------------'
for row in bad_fresh:
print critics[y == 1].quote.irow(row)
print
4.3 What do you notice about these mis-predictions? Naive Bayes classifiers assume that every word affects the probability independently of other words. In what way is this a bad assumption? In your answer, report your classifier's Freshness probability for the review "This movie is not remarkable, touching, or superb in any way".
In [25]:
clf.predict_proba(vectorizer.transform(['This movie is not remarkable, touching, or superb in any way']))
Out[25]:
Answer Many mis-predictions seem due to the fact that the quotes use more ambivalent language -- quotes along the lines of "this should have been a good movie, but it wasn't". Words like "but", "not", etc. act to negate the sentiment of words. However, because Naive Bayes treats each word separately, it isn't able to capture these kind of word interactions. Because the quote "this movie is not remarkable, touching, or superb in any way" contains typically positive words like remarkabke/touching/superb, the classifier gives it P(Fresh)=0.98.
4.4 If this was your final project, what are 3 things you would try in order to build a more effective review classifier? What other exploratory or explanatory visualizations do you think might be helpful?
There are many things worth trying. Some examples:
To submit your homework, create a folder named lastname_firstinitial_hw3 and place this notebook file in the folder. Double check that this file is still called HW3.ipynb, and that it contains your solutions. Please do not include the critics.csv data file, if you created one. Compress the folder (please use .zip compression) and submit to the CS109 dropbox in the appropriate folder. If we cannot access your work because these directions are not followed correctly, we will not grade your work.
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