In the mini-project, you'll learn the basics of text analysis using a subset of movie reviews from the rotten tomatoes database. You'll also use a fundamental technique in Bayesian inference, called Naive Bayes. This mini-project is based on Lab 10 of Harvard's CS109 class. Please free to go to the original lab for additional exercises and solutions.
In [1]:
%matplotlib inline
import numpy as np
import scipy as sp
import matplotlib as mpl
import matplotlib.cm as cm
import matplotlib.pyplot as plt
import pandas as pd
import seaborn as sns
from six.moves import range
# Setup Pandas
pd.set_option('display.width', 500)
pd.set_option('display.max_columns', 100)
pd.set_option('display.notebook_repr_html', True)
# Setup Seaborn
sns.set_style("whitegrid")
sns.set_context("poster")
In [2]:
critics = pd.read_csv('./critics.csv')
#let's drop rows with missing quotes
critics = critics[~critics.quote.isnull()]
critics.head()
Out[2]:
In [3]:
n_reviews = len(critics)
n_movies = critics.rtid.unique().size
n_critics = critics.critic.unique().size
print("Number of reviews: {:d}".format(n_reviews))
print("Number of critics: {:d}".format(n_critics))
print("Number of movies: {:d}".format(n_movies))
In [4]:
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("Number of Critics")
plt.yticks([0, 2, 4, 6, 8, 10]);
The histogram above displays the average rating per critic for critics that have more than 100 reviews. The shape seems like a positively skewed distribution with most ratings near 0.6. The histogram tells us that most of these critics have an average rating of 0.6 or above. Since this only accounts for critics who have posted over 100 reviews, we can assume that these critics watch movies often due to their love and enjoyment of movies. With this assumption it makes sense why most of the average ratings are positive.
All the diagrams here are snipped from Introduction to Information Retrieval by Manning et. al. which is a great resource on text processing. For additional information on text mining and natural language processing, see Foundations of Statistical Natural Language Processing by Manning and Schutze.
Also check out Python packages nltk, spaCy, pattern, and their associated resources. Also see word2vec.
Let us define the vector derived from document $d$ by $\bar V(d)$. What does this mean? Each document is treated as a vector containing information about the words contained in it. Each vector has the same length and each entry "slot" in the vector contains some kind of data about the words that appear in the document such as presence/absence (1/0), count (an integer) or some other statistic. Each vector has the same length because each document shared the same vocabulary across the full collection of documents -- this collection is called a corpus.
To define the vocabulary, we take a union of all words we have seen in all documents. We then just associate an array index with them. So "hello" may be at index 5 and "world" at index 99.
Suppose we have the following corpus:
A Fox one day spied a beautiful bunch of ripe grapes hanging from a vine trained along the branches of a tree. The grapes seemed ready to burst with juice, and the Fox's mouth watered as he gazed longingly at them.
Suppose we treat each sentence as a document $d$. The vocabulary (often called the lexicon) is the following:
$V = \left\{\right.$ a, along, and, as, at, beautiful, branches, bunch, burst, day, fox, fox's, from, gazed, grapes, hanging, he, juice, longingly, mouth, of, one, ready, ripe, seemed, spied, the, them, to, trained, tree, vine, watered, with$\left.\right\}$
Then the document
A Fox one day spied a beautiful bunch of ripe grapes hanging from a vine trained along the branches of a tree
may be represented as the following sparse vector of word counts:
$$\bar V(d) = \left( 4,1,0,0,0,1,1,1,0,1,1,0,1,0,1,1,0,0,0,0,2,1,0,1,0,0,1,0,0,0,1,1,0,0 \right)$$or more succinctly as
[(0, 4), (1, 1), (5, 1), (6, 1), (7, 1), (9, 1), (10, 1), (12, 1), (14, 1), (15, 1), (20, 2), (21, 1), (23, 1),
(26, 1), (30, 1), (31, 1)]
along with a dictionary
{
0: a, 1: along, 5: beautiful, 6: branches, 7: bunch, 9: day, 10: fox, 12: from, 14: grapes,
15: hanging, 19: mouth, 20: of, 21: one, 23: ripe, 24: seemed, 25: spied, 26: the,
30: tree, 31: vine,
}
Then, a set of documents becomes, in the usual sklearn style, a sparse matrix with rows being sparse arrays representing documents and columns representing the features/words in the vocabulary.
Notice that this representation loses the relative ordering of the terms in the document. That is "cat ate rat" and "rat ate cat" are the same. Thus, this representation is also known as the Bag-Of-Words representation.
Here is another example, from the book quoted above, although the matrix is transposed here so that documents are columns:
Such a matrix is also catted a Term-Document Matrix. Here, the terms being indexed could be stemmed before indexing; for instance, jealous and jealousy after stemming are the same feature. One could also make use of other "Natural Language Processing" transformations in constructing the vocabulary. We could use Lemmatization, which reduces words to lemmas: work, working, worked would all reduce to work. We could remove "stopwords" from our vocabulary, such as common words like "the". We could look for particular parts of speech, such as adjectives. This is often done in Sentiment Analysis. And so on. It all depends on our application.
From the book:
$$S_{12} = \frac{\bar V(d_1) \cdot \bar V(d_2)}{|\bar V(d_1)| \times |\bar V(d_2)|}$$The standard way of quantifying the similarity between two documents $d_1$ and $d_2$ is to compute the cosine similarity of their vector representations $\bar V(d_1)$ and $\bar V(d_2)$:
There is a far more compelling reason to represent documents as vectors: we can also view a query as a vector. Consider the query q = jealous gossip. This query turns into the unit vector $\bar V(q)$ = (0, 0.707, 0.707) on the three coordinates below.
$$\bar V(q) \cdot \bar V(d)$$The key idea now: to assign to each document d a score equal to the dot product:
Then we can use this simple Vector Model as a Search engine.
In [5]:
from sklearn.feature_extraction.text import CountVectorizer
text = ['Hop on pop', 'Hop off pop', 'Hop Hop hop']
print("Original text is\n{}".format('\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{}".format(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
In [6]:
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
X, y = make_xy(critics)
From Bayes' Theorem, we have that
$$P(c \vert f) = \frac{P(c \cap f)}{P(f)}$$where $c$ represents a class or category, and $f$ represents a feature vector, such as $\bar V(d)$ as above. We are computing the probability that a document (or whatever we are classifying) belongs to category c given the features in the document. $P(f)$ is really just a normalization constant, so the literature usually writes Bayes' Theorem in context of Naive Bayes as
$$P(c \vert f) \propto P(f \vert c) P(c) $$$P(c)$ is called the prior and is simply the probability of seeing class $c$. But what is $P(f \vert c)$? This is the probability that we see feature set $f$ given that this document is actually in class $c$. This is called the likelihood and comes from the data. One of the major assumptions of the Naive Bayes model is that the features are conditionally independent given the class. While the presence of a particular discriminative word may uniquely identify the document as being part of class $c$ and thus violate general feature independence, conditional independence means that the presence of that term is independent of all the other words that appear within that class. This is a very important distinction. Recall that if two events are independent, then:
$$P(A \cap B) = P(A) \cdot P(B)$$Thus, conditional independence implies
$$P(f \vert c) = \prod_i P(f_i | c) $$where $f_i$ is an individual feature (a word in this example).
To make a classification, we then choose the class $c$ such that $P(c \vert f)$ is maximal.
There is a small caveat when computing these probabilities. For floating point underflow we change the product into a sum by going into log space. This is called the LogSumExp trick. So:
$$\log P(f \vert c) = \sum_i \log P(f_i \vert c) $$There is another caveat. What if we see a term that didn't exist in the training data? This means that $P(f_i \vert c) = 0$ for that term, and thus $P(f \vert c) = \prod_i P(f_i | c) = 0$, which doesn't help us at all. Instead of using zeros, we add a small negligible value called $\alpha$ to each count. This is called Laplace Smoothing.
$$P(f_i \vert c) = \frac{N_{ic}+\alpha}{N_c + \alpha N_i}$$where $N_{ic}$ is the number of times feature $i$ was seen in class $c$, $N_c$ is the number of times class $c$ was seen and $N_i$ is the number of times feature $i$ was seen globally. $\alpha$ is sometimes called a regularization parameter.
Since we are modeling word counts, we are using variation of Naive Bayes called Multinomial Naive Bayes. This is because the likelihood function actually takes the form of the multinomial distribution.
$$P(f \vert c) = \frac{\left( \sum_i f_i \right)!}{\prod_i f_i!} \prod_{f_i} P(f_i \vert c)^{f_i} \propto \prod_{i} P(f_i \vert c)$$where the nasty term out front is absorbed as a normalization constant such that probabilities sum to 1.
There are many other variations of Naive Bayes, all which depend on what type of value $f_i$ takes. If $f_i$ is continuous, we may be able to use Gaussian Naive Bayes. First compute the mean and variance for each class $c$. Then the likelihood, $P(f \vert c)$ is given as follows
$$P(f_i = v \vert c) = \frac{1}{\sqrt{2\pi \sigma^2_c}} e^{- \frac{\left( v - \mu_c \right)^2}{2 \sigma^2_c}}$$Exercise: Implement a simple Naive Bayes classifier:
In [7]:
#your turn
In [8]:
from sklearn.naive_bayes import MultinomialNB
from sklearn.model_selection import train_test_split
xtrain, xtest, ytrain, ytest = train_test_split(X, y)
clf = MultinomialNB()
clf.fit(xtrain, ytrain)
print("Accuracy score for training set:", clf.score(xtrain, ytrain))
print("Accuracy score for test set:", clf.score(xtest, ytest))
The score for the training set (0.92) is much higher than the score for the testing set (0.77). Although the classifier does well, this indicates that the classifier may be overfit.
We need to know what value to use for $\alpha$, and we also need to know which words to include in the vocabulary. As mentioned earlier, some words are obvious stopwords. Other words appear so infrequently that they serve as noise, and other words in addition to stopwords appear so frequently that they may also serve as noise.
First, let's find an appropriate value for min_df for the CountVectorizer. min_df can be either an integer or a float/decimal. If it is an integer, min_df represents the minimum number of documents a word must appear in for it to be included in the vocabulary. If it is a float, it represents the minimum percentage of documents a word must appear in to be included in the vocabulary. From the documentation:
min_df: When building the vocabulary ignore terms that have a document frequency strictly lower than the given threshold. This value is also called cut-off in the literature. If float, the parameter represents a proportion of documents, integer absolute counts. This parameter is ignored if vocabulary is not None.
Exercise: Construct the cumulative distribution of document frequencies (df). The $x$-axis is a document count $x_i$ and the $y$-axis is the percentage of words that appear less than $x_i$ times. For example, at $x=5$, plot a point representing the percentage or number of words that appear in 5 or fewer documents.
Exercise: Look for the point at which the curve begins climbing steeply. This may be a good value for `min_df`. If we were interested in also picking `max_df`, we would likely pick the value where the curve starts to plateau. What value did you choose?
In [9]:
# Your turn.
In [10]:
df = list(sorted((X > 0).sum(axis=0).reshape(-1).tolist()[0]))
rows, features = X.shape
height, axis = np.histogram(df, bins=len(np.unique(df)))
cumhist = np.cumsum(height * 1, axis=0) / features
axis = np.insert(axis, 0, 0)
cumhist = np.insert(cumhist, 0, 0)
plt.plot(axis[:-1], cumhist)
plt.xlim(-1, 25)
plt.xlabel("Document Count")
plt.ylabel("CDF")
Out[10]:
Observing the plot above, the curve begins to climb steeply at near 0. The curve starts to plateau shortly after at around 1 or 2 documents. Thus our min_df and max_df values can be close to 0 and 2, respectively. We can experiment with different values of min_df.
The parameter $\alpha$ is chosen to be a small value that simply avoids having zeros in the probability computations. This value can sometimes be chosen arbitrarily with domain expertise, but we will use K-fold cross validation. In K-fold cross-validation, we divide the data into $K$ non-overlapping parts. We train on $K-1$ of the folds and test on the remaining fold. We then iterate, so that each fold serves as the test fold exactly once. The function cv_score performs the K-fold cross-validation algorithm for us, but we need to pass a function that measures the performance of the algorithm on each fold.
In [11]:
from sklearn.model_selection import KFold
def cv_score(clf, X, y, scorefunc):
result = 0.
nfold = 5
for train, test in KFold(nfold).split(X): # split data into train/test groups, 5 times
clf.fit(X[train], y[train]) # fit the classifier, passed is as clf.
result += scorefunc(clf, X[test], y[test]) # evaluate score function on held-out data
return result / nfold # average
We use the log-likelihood as the score here in scorefunc. The higher the log-likelihood, the better. Indeed, what we do in cv_score above is to implement the cross-validation part of GridSearchCV.
The custom scoring function scorefunc allows us to use different metrics depending on the decision risk we care about (precision, accuracy, profit etc.) directly on the validation set. You will often find people using roc_auc, precision, recall, or F1-score as the scoring function.
In [12]:
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()
We'll cross-validate over the regularization parameter $\alpha$.
Let's set up the train and test masks first, and then we can run the cross-validation procedure.
In [13]:
from sklearn.model_selection import train_test_split
_, itest = train_test_split(range(critics.shape[0]), train_size=0.7)
mask = np.zeros(critics.shape[0], dtype=np.bool)
mask[itest] = True
Exercise: What does using the function `log_likelihood` as the score mean? What are we trying to optimize for?
Exercise: Without writing any code, what do you think would happen if you choose a value of $\alpha$ that is too high?
Exercise: Using the skeleton code below, find the best values of the parameter `alpha`, and use the value of `min_df` you chose in the previous exercise set. Use the `cv_score` function above with the `log_likelihood` function for scoring.
What does using the function log_likelihood as the score mean? What are we trying to optimize for? The function, log_likelihood, uses the predict_log_proba method which returns the log-probability of the sample for each class. In other words, it returns how likely a single sample is fresh and how likely that sample is rotten. The log_likelihood function then takes the sum of the likelihood probabilities based on their correct classification. We are trying to optimize for the highest log-probabilities of correctly classified samples.
Without writing any code, what do you think would happen if you choose a value of α that is too high? When alpha is too high, it simplifies the model which results in underfitting. An alpha value that is too low will overfit the data.
In [14]:
from sklearn.naive_bayes import MultinomialNB
#the grid of parameters to search over
alphas = [.1, 1, 5, 10, 50]
min_dfs =[0, 0.01, 0.015, 0.001, 0.0005, 0.00055, 0.0001, 0.00015]
best_min_df = None # YOUR TURN: put your value of min_df here.
#Find the best value for alpha and min_df, and the best classifier
best_alpha = None
maxscore=-np.inf
for alpha in alphas:
for min_df in min_dfs:
vectorizer = CountVectorizer(min_df=min_df)
Xthis, ythis = make_xy(critics, vectorizer)
Xtrainthis = Xthis[mask]
ytrainthis = ythis[mask]
# your turn
clf = MultinomialNB(alpha = alpha)
score = cv_score(clf, Xtrainthis, ytrainthis, log_likelihood)
print("Score:", score, "| min_df:", min_df, "| alpha:", alpha)
if score > maxscore:
maxscore = score
best_min_df = min_df
best_alpha = alpha
In [15]:
print("best alpha: {}".format(best_alpha))
In [16]:
print("best min_df: {}".format(best_min_df))
Exercise: Using the best value of `alpha` you just found, calculate the accuracy on the training and test sets. Is this classifier better? Why (not)?
In [17]:
vectorizer = CountVectorizer(min_df=best_min_df)
X, y = make_xy(critics, vectorizer)
xtrain=X[mask]
ytrain=y[mask]
xtest=X[~mask]
ytest=y[~mask]
clf = MultinomialNB(alpha=best_alpha).fit(xtrain, ytrain)
#your turn. 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: {:2f}".format(training_accuracy))
print("Accuracy on test data: {:2f}".format(test_accuracy))
The accuracy on both the training data and test data is lower. However, this classifier is not overfit like the previous classifier.
In [18]:
from sklearn.metrics import confusion_matrix
print(confusion_matrix(ytest, clf.predict(xtest)))
We use a neat trick to identify strongly predictive features (i.e. words).
In [19]:
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("{:>20}".format(w), "{:.2f}".format(1 - np.exp(p)))
print("Bad words\t P(fresh | word)")
for w, p in zip(bad_words, bad_prob):
print("{:>20}".format(w), "{:.2f}".format(1 - np.exp(p)))
Exercise: Why does this method work? What does the probability for each row in the identity matrix represent
This method creates a dataset where each row contains one word (identity matrix). The trained classifier is then used to make predictions on each word. The probability on each row in the identity matrix represents the probability that it is classified as fresh. You can figure out which words have the highest and lowest probabilities. This tells you which words are the strongly predictive. Higher probabilities indicate a strong predictive feature/word for a fresh review, while lower probabilities indicate a strong predictive feature/word for a rotten review.
In [20]:
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.iloc[row])
print("")
print("Mis-predicted Fresh quotes")
print('--------------------------')
for row in bad_fresh:
print(critics[y == 1].quote.iloc[row])
print("")
In [21]:
#your turn
In [22]:
sentence = ['This movie is not remarkable, touching, or superb in any way']
print('Review:',sentence[0])
sentence = vectorizer.transform(sentence)
print('\nRotten Probability:', clf.predict_proba(sentence)[0, 0])
print('Fresh Probability:', clf.predict_proba(sentence)[0, 1])
if clf.predict(sentence)[0] == 1:
print('\nThis review is Fresh.')
else:
print('\nThis review is Rotten')
I would expect this sentence to be correctly classified as rotten. Since this classifier is using the bag-of-words approach, it only looks at individual words. With this approach it is difficult to correctly predict sentences with words that reverse its meaning such as 'not'. The sentence had many words that would be classified as fresh which resulted in a high probability that this is a fresh review.
TF-IDF stands for
Term-Frequency X Inverse Document Frequency.
In the standard CountVectorizer model above, we used just the term frequency in a document of words in our vocabulary. In TF-IDF, we weight this term frequency by the inverse of its popularity in all documents. For example, if the word "movie" showed up in all the documents, it would not have much predictive value. It could actually be considered a stopword. By weighing its counts by 1 divided by its overall frequency, we downweight it. We can then use this TF-IDF weighted features as inputs to any classifier. TF-IDF is essentially a measure of term importance, and of how discriminative a word is in a corpus. There are a variety of nuances involved in computing TF-IDF, mainly involving where to add the smoothing term to avoid division by 0, or log of 0 errors. The formula for TF-IDF in scikit-learn differs from that of most textbooks:
where $n_{td}$ is the number of times term $t$ occurs in document $d$, $\vert D \vert$ is the number of documents, and $\vert d : t \in d \vert$ is the number of documents that contain $t$
In [23]:
# http://scikit-learn.org/dev/modules/feature_extraction.html#text-feature-extraction
# http://scikit-learn.org/dev/modules/classes.html#text-feature-extraction-ref
from sklearn.feature_extraction.text import TfidfVectorizer
tfidfvectorizer = TfidfVectorizer(min_df=1, stop_words='english')
Xtfidf=tfidfvectorizer.fit_transform(critics.quote)
There are several additional things we could try. Try some of these as exercises:
In [24]:
# Your turn
1. Build a Naive Bayes model where the features are n-grams instead of words. N-grams are phrases containing n words next to each other: a bigram contains 2 words, a trigram contains 3 words, and 6-gram contains 6 words. This is useful because "not good" and "so good" mean very different things. On the other hand, as n increases, the model does not scale well since the feature set becomes more sparse.
In [25]:
alphas = [.1, 1, 5, 10, 50]
min_dfs =[0, 0.01, 0.015, 0.001, 0.0005, 0.00055, 0.0001, 0.00015]
best_min_df = None
best_alpha = None
maxscore=-np.inf
for alpha in alphas:
for min_df in min_dfs:
vectorizer = CountVectorizer(min_df=min_df, ngram_range=(2, 2))
Xthis, ythis = make_xy(critics, vectorizer)
Xtrainthis = Xthis[mask]
ytrainthis = ythis[mask]
clf = MultinomialNB(alpha = alpha)
score = cv_score(clf, Xtrainthis, ytrainthis, log_likelihood)
print("Score:", score, "| min_df:", min_df, "| alpha:", alpha)
if score > maxscore:
maxscore = score
best_min_df = min_df
best_alpha = alpha
In [26]:
print("best alpha: {}".format(best_alpha))
print("best min_df: {}".format(best_min_df))
In [27]:
vectorizer = CountVectorizer(min_df=best_min_df, ngram_range=(2, 2))
X, y = make_xy(critics, vectorizer)
xtrain=X[mask]
ytrain=y[mask]
xtest=X[~mask]
ytest=y[~mask]
clf = MultinomialNB(alpha=best_alpha).fit(xtrain, ytrain)
training_accuracy = clf.score(xtrain, ytrain)
test_accuracy = clf.score(xtest, ytest)
print("Accuracy on training data: {:2f}".format(training_accuracy))
print("Accuracy on test data: {:2f}".format(test_accuracy))
In [28]:
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("{:>20}".format(w), "{:.2f}".format(1 - np.exp(p)))
print("Bad words\t P(fresh | word)")
for w, p in zip(bad_words, bad_prob):
print("{:>20}".format(w), "{:.2f}".format(1 - np.exp(p)))
2. Try a model besides Naive Bayes, one that would allow for interactions between words -- for example, a Random Forest classifier.
In [29]:
from sklearn.ensemble import RandomForestClassifier
from sklearn.model_selection import GridSearchCV
min_dfs =[0.001, 0.0005, 0.00055, 0.0001, 0.00015]
best_gridsearch = None
best_min_df = None
maxscore=-np.inf
for min_df in min_dfs:
vectorizer = CountVectorizer(min_df=min_df)
Xthis, ythis = make_xy(critics, vectorizer)
Xtrainthis = Xthis[mask]
ytrainthis = ythis[mask]
clf = RandomForestClassifier()
parameters = {"n_estimators":[15, 20, 25, 30, 35], "max_features":['auto','sqrt','log2'], "min_samples_leaf":[1, 10, 20, 30, 40, 50]}
gridsearch = GridSearchCV(clf, param_grid=parameters, scoring="accuracy")
gridsearch.fit(Xtrainthis, ytrainthis)
if gridsearch.best_score_ > maxscore:
maxscore = gridsearch.best_score_
best_min_df = min_df
best_gridsearch = gridsearch
print("Best Parameters")
for key, value in best_gridsearch.best_params_.items():
print(key, ":", value)
print("\nBest min_df:", best_min_df)
print("\nBest score", best_gridsearch.best_score_)
clf = best_gridsearch.best_estimator_
In [30]:
vectorizer = CountVectorizer(min_df=best_min_df)
Xthis, ythis = make_xy(critics, vectorizer)
Xtrainthis = Xthis[mask]
ytrainthis = ythis[mask]
Xtestthis = Xthis[~mask]
ytestthis = ythis[~mask]
training_accuracy = clf.score(Xtrainthis, ytrainthis)
test_accuracy = clf.score(Xtestthis, ytestthis)
print("Accuracy on training data: {:2f}".format(training_accuracy))
print("Accuracy on test data: {:2f}".format(test_accuracy))