Data Mining · NLP · Python · Sentiment Analysis

Mining Twitter Data with Python (Part 6 – Sentiment Analysis Basics)

Sentiment Analysis is one of the interesting applications of text analytics. Although the term is often associated with sentiment classification of documents, broadly speaking it refers to the use of text analytics approaches applied to the set of problems related to identifying and extracting subjective material in text sources.

This article continues the series on mining Twitter data with Python, describing a simple approach for Sentiment Analysis and applying it to the rubgy data set (see Part 4).

Tutorial Table of Contents:

A Simple Approach for Sentiment Analysis

The technique we’re discussing in this post has been elaborated from the traditional approach proposed by Peter Turney in his paper Thumbs Up or Thumbs Down? Semantic Orientation Applied to Unsupervised Classification of Reviews. A lot of work has been done in Sentiment Analysis since then, but the approach has still an interesting educational value. In particular, it is intuitive, simple to understand and to test, and most of all unsupervised, so it doesn’t require any labelled data for training.

Firstly, we define the Semantic Orientation (SO) of a word as the difference between its associations with positive and negative words. In practice, we want to calculate “how close” a word is with terms like good and bad. The chosen measure of “closeness” is Pointwise Mutual Information (PMI), calculated as follows (t1 and t2 are terms):

\mbox{PMI}(t_1, t_2) = \log\Bigl(\frac{P(t_1 \wedge t_2)}{P(t_1) \cdot P(t_2)}\Bigr)

In Turney’s paper, the SO of a word was calculated against excellent and poor, but of course we can extend the vocabulary of positive and negative terms. Using V^{+} and a vocabulary of positive terms and V^{-} for the negative ones, the Semantic Orientation of a term t is hence defined as:

\mbox{SO}(t) = \sum_{t' \in V^{+}}\mbox{PMI}(t, t') - \sum_{t' \in V^{-}}\mbox{PMI}(t, t')

We can build our own list of positive and negative terms, or we can use one of the many resources available on-line, for example the opinion lexicon by Bing Liu.

Computing Term Probabilities

In order to compute P(t) (the probability of observing the term t) and P(t_1 \wedge t_2) (the probability of observing the terms t1 and t2 occurring together) we can re-use some previous code to calculate term frequencies and term co-occurrences. Given the set of documents (tweets) D, we define the Document Frequency (DF) of a term as the number of documents where the term occurs. The same definition can be applied to co-occurrent terms. Hence, we can define our probabilities as:

P(t) = \frac{\mbox{DF}(t)}{|D|}\\  P(t_1 \wedge t_2) = \frac{\mbox{DF}(t_1 \wedge t_2)}{|D|}

In the previous articles, the document frequency for single terms was stored in the dictionaries count_single and count_stop_single (the latter doesn’t store stop-words), while the document frequency for the co-occurrencies was stored in the co-occurrence matrix com

This is how we can compute the probabilities:

# n_docs is the total n. of tweets
p_t = {}
p_t_com = defaultdict(lambda : defaultdict(int))

for term, n in count_stop_single.items():
    p_t[term] = n / n_docs
    for t2 in com[term]:
        p_t_com[term][t2] = com[term][t2] / n_docs

Computing the Semantic Orientation

Given two vocabularies for positive and negative terms:

positive_vocab = [
    'good', 'nice', 'great', 'awesome', 'outstanding',
    'fantastic', 'terrific', ':)', ':-)', 'like', 'love',
    # shall we also include game-specific terms?
    # 'triumph', 'triumphal', 'triumphant', 'victory', etc.
]
negative_vocab = [
    'bad', 'terrible', 'crap', 'useless', 'hate', ':(', ':-(',
    # 'defeat', etc.
]

We can compute the PMI of each pair of terms, and then compute the
Semantic Orientation as described above:

pmi = defaultdict(lambda : defaultdict(int))
for t1 in p_t:
    for t2 in com[t1]:
        denom = p_t[t1] * p_t[t2]
        pmi[t1][t2] = math.log2(p_t_com[t1][t2] / denom)

semantic_orientation = {}
for term, n in p_t.items():
    positive_assoc = sum(pmi[term][tx] for tx in positive_vocab)
    negative_assoc = sum(pmi[term][tx] for tx in negative_vocab)
    semantic_orientation[term] = positive_assoc - negative_assoc

The Semantic Orientation of a term will have a positive (negative) value if the term is often associated with terms in the positive (negative) vocabulary. The value will be zero for neutral terms, e.g. the PMI’s for positive and negative balance out, or more likely a term is never observed together with other terms in the positive/negative vocabularies.

We can print out the semantic orientation for some terms:

semantic_sorted = sorted(semantic_orientation.items(), 
                         key=operator.itemgetter(1), 
                         reverse=True)
top_pos = semantic_sorted[:10]
top_neg = semantic_sorted[-10:]

print(top_pos)
print(top_neg)
print("ITA v WAL: %f" % semantic_orientation['#itavwal'])
print("SCO v IRE: %f" % semantic_orientation['#scovire'])
print("ENG v FRA: %f" % semantic_orientation['#engvfra'])
print("#ITA: %f" % semantic_orientation['#ita'])
print("#FRA: %f" % semantic_orientation['#fra'])
print("#SCO: %f" % semantic_orientation['#sco'])
print("#ENG: %f" % semantic_orientation['#eng'])
print("#WAL: %f" % semantic_orientation['#wal'])
print("#IRE: %f" % semantic_orientation['#ire'])

Different vocabularies will produce different scores. Using the opinion lexicon from Bing Liu, this is what we can observed on the Rugby data-set:

# the top positive terms
[('fantastic', 91.39950482011552), ('@dai_bach', 90.48767241244532), ('hoping', 80.50247748725415), ('#it', 71.28333427277785), ('days', 67.4394844955977), ('@nigelrefowens', 64.86112716005566), ('afternoon', 64.05064208341855), ('breathtaking', 62.86591435212975), ('#wal', 60.07283361352875), ('annual', 58.95378954406133)]
# the top negative terms
[('#england', -74.83306534609066), ('6', -76.0687215594536), ('#itavwal', -78.4558633116863), ('@rbs_6_nations', -80.89363516601993), ("can't", -81.75379628180468), ('like', -83.9319149443813), ('10', -85.93073078165587), ('italy', -86.94465165178258), ('#engvfra', -113.26188957010228), ('ball', -161.82146824640125)]
# Matches
ITA v WAL: -78.455863
SCO v IRE: -73.487661
ENG v FRA: -113.261890
# Individual team
#ITA: 53.033824
#FRA: 14.099372
#SCO: 4.426723
#ENG: -0.462845
#WAL: 60.072834
#IRE: 19.231722

Some Limitations

The PMI-based approach has been introduced as simple and intuitive, but of course it has some limitations. The semantic scores are calculated on terms, meaning that there is no notion of “entity” or “concept” or “event”. For example, it would be nice to aggregate and normalise all the references to the team names, e.g. #ita, Italy and Italia should all contribute to the semantic orientation of the same entity. Moreover, do the opinions on the individual teams also contribute to the overall opinion on a match?

Some aspects of natural language are also not captured by this approach, more notably modifiers and negation: how do we deal with phrases like not bad (this is the opposite of just bad) or very good (this is stronger than just good)?

Summary

This article has continued the tutorial on mining Twitter data with Python introducing a simple approach for Sentiment Analysis, based on the computation of a semantic orientation score which tells us whether a term is more closely related to a positive or negative vocabulary. The intuition behind this approach is fairly simple, and it can be implemented using Pointwise Mutual Information as a measure of association. The approach has of course some limitations, but it’s a good starting point to get familiar with Sentiment Analysis.

@MarcoBonzanini

26 thoughts on “Mining Twitter Data with Python (Part 6 – Sentiment Analysis Basics)

    1. Hey Marco,

      thanks a lot for this nice introduction!!
      I have streamed some data from twitter over a period of 2 Weeks.
      I kind of adjusted your code a bit for my matter.
      If I analyze the data now, my code clusters all the tweets into 1 day packages and gives me the PMI for a certain word for that day.
      The goal is to have a normalized set of pmi´s over a period of 14 days to see the difference in opinion between each day. If I dont normalize the data, the difference depends actually on the amount of tweets i have.

      Therefore i want to normalize the pmi, the line looks like that:

      pmi[t1][t2] = (math.log2(p_t_com[t1][t2] / denom)/(-(math.log2(p_t_com[t1][t2]))))

      is that formula correct?

      Now the value of all co-occurences is between -1 and 1, correct?

      If i add up all the different PMI´s of the co-occurence words which come together with that particular one i end up having a PMI still depending on the number of tweets i collect.

      positive_assoc = sum(pmi[term][tx] for tx in positive_vocab)
      negative_assoc = sum(pmi[term][tx] for tx in negative_vocab)
      semantic_orientation[term] = (positive_assoc – negative_assoc)/count2

      So i have to divide this PMI through the number of elements i added up?
      Is that right? Or do you have another suggestion? I am just a little bit lost…

      Hopefully you understand my problem, my english is not the best!

      Like

  1. Hi Marco,
    very nice articles.
    Just a question here: is there a possibility to print a hole Tweet with positiv words
    and not just the positiv / negative words alone?

    Like

  2. Hi pad, yes you can loop through the tweets again after building the vocabularies (e.g. top_pos and top_neg in the last example), and it’s a matter of checking:
    if any(word in tweet[‘text’] for word in top_pos):
    print(tweet[‘text’])

    Like

    1. hey marco, thank you i will try this.

      when you’ re printing out your PMI’s do you get the same results everytime?
      I once hat a PMI of -1.853720 and the next time 18.056733 for the same term :O

      Like

      1. Assuming the data set is the same, and the vocabularies V+ and V- are the same, you should end up with the same values for PMI. With different data (e.g. new tweets added to the data set) the term probabilities are most likely to be different

        Like

  3. Hello again!

    As usual, I have another question.
    The co-occurrence dict give the occurrence of two words that occur in the same tweet. I read the paper and it stated we use 2 words for somewhat deriving the context, and hence we look at 2 consecutive words. Doesn’t that mean we should use something like this in finding the co-occurances instead of using 2 for loops?

    #editing com
    for i in range(len(word_list)-1):
    word1, word2 = sorted([ word_list[i], word_list[ i+1 ]])
    if word1 != word2:
    com[word1][word2] +=1

    Like

    1. Hi Krishanu, the approach using a local context can be useful to extract phrases that follow a particular pattern (e.g. an adjective followed by a noun, for example “good movie”). If you follow this route, the downside is that you need to specify all these patterns manually. Experimenting with bigrams/trigrams is indeed worth exploring, but consider that you can fall short in many occasions. For example in the sentence “the movie was very good”, the connection movie-good is not captured by bigrams/trigrams because the two words are too distant. It makes sense to consider local context in reviews that can span multiple sentences, maybe less in a tweet that has just a few words. Long story short, try it :)
      It’s worth mentioning that there are also more advanced techniques, like word2vec, used to better represent words and to capture semantic relationships between words — I just haven’t had much time to write about it.
      Cheers,
      Marco

      Like

  4. does not work any help plzzzzzzzzzzzz
    —————————————-
    import sys
    import json
    from collections import Counter
    import re
    from nltk.corpus import stopwords
    import string
    from collections import defaultdict
    import operator

    p_t = {}
    p_t_com = defaultdict(lambda : defaultdict(int))
    com = defaultdict(lambda : defaultdict(int))
    pmi = defaultdict(lambda : defaultdict(int))
    punctuation = list(string.punctuation)
    stop = stopwords.words(‘english’) + punctuation + [‘rt’, ‘via’]

    emoticons_str = r”””
    (?:
    [:=;] # Eyes
    [oO\-]? # Nose (optional)
    [D\)\]\(\]/\\OpP] # Mouth
    )”””

    regex_str = [
    emoticons_str,
    r’]+>’, # HTML tags
    r'(?:@[\w_]+)’, # @-mentions
    r”(?:\#+[\w_]+[\w\’_\-]*[\w_]+)”, # hash-tags
    r’http[s]?://(?:[a-z]|[0-9]|[$-_@.&+]|[!*\(\),]|(?:%[0-9a-f][0-9a-f]))+’, # URLs

    r'(?:(?:\d+,?)+(?:\.?\d+)?)’, # numbers
    r”(?:[a-z][a-z’\-_]+[a-z])”, # words with – and ‘
    r'(?:[\w_]+)’, # other words
    r'(?:\S)’ # anything else
    ]

    tokens_re = re.compile(r'(‘+’|’.join(regex_str)+’)’, re.VERBOSE | re.IGNORECASE)
    emoticon_re = re.compile(r’^’+emoticons_str+’$’, re.VERBOSE | re.IGNORECASE)

    def tokenize(s):
    return tokens_re.findall(s)

    def preprocess(s, lowercase=True):
    tokens = tokenize(s)
    if lowercase:
    tokens = [token if emoticon_re.search(token) else token.lower() for token in tokens]
    return tokens

    if __name__ == ‘__main__’:
    fname = ‘news.json’

    with open(fname, ‘r’) as f:
    count_stop_single = Counter()
    for line in f:
    tweet = json.loads(line)
    terms_only = [term for term in preprocess(tweet[‘text’])
    if term not in stop and
    not term.startswith((‘#’, ‘@’, ‘http’)) and
    len(term) > 2]

    # Build co-occurrence matrix
    for i in range(len(terms_only)-1):
    for j in range(i+1, len(terms_only)):
    w1, w2 = sorted([terms_only[i], terms_only[j]])
    if w1 != w2:
    com[w1][w2] += 1
    com_max = []

    # For each term, look for the most common co-occurrent terms
    for t1 in com:
    t1_max_terms = sorted(com[t1].items(), key=operator.itemgetter(1), reverse=True)[:5]
    for t2, t2_count in t1_max_terms:
    com_max.append(((t1, t2), t2_count))
    terms_max = sorted(com_max, key=operator.itemgetter(1), reverse=True)
    print(terms_max[:5])
    #
    for term, n in count_stop_single.items():
    p_t[term] = n / n_docs
    for t2 in com[term]:
    p_t_com[term][t2] = com[term][t2] / n_docs
    print(p_t_com[term][t2]

    positive_vocab = [
    ‘good’, ‘nice’, ‘great’, ‘awesome’, ‘outstanding’,
    ‘fantastic’, ‘terrific’, ‘:)’, ‘:-)’, ‘like’, ‘love’,
    # shall we also include game-specific terms?
    # ‘triumph’, ‘triumphal’, ‘triumphant’, ‘victory’, etc.
    ]
    negative_vocab = [
    ‘bad’, ‘terrible’, ‘crap’, ‘useless’, ‘hate’, ‘:(‘, ‘:-(‘,
    # ‘defeat’, etc.
    ]
    for t1 in p_t:
    for t2 in com[t1]:
    denom = p_t[t1] * p_t[t2]
    pmi[t1][t2] = math.log2(p_t_com[t1][t2] / denom)

    semantic_orientation = {}
    for term, n in p_t.items():
    positive_assoc = sum(pmi[term][tx] for tx in positive_vocab)
    negative_assoc = sum(pmi[term][tx] for tx in negative_vocab)
    semantic_orientation[term] = positive_assoc – negative_assoc
    semantic_sorted = sorted(semantic_orientation.items(),
    key=operator.itemgetter(1),
    reverse=True)
    top_pos = semantic_sorted[:10]
    top_neg = semantic_sorted[-10:]

    print(top_pos)
    print(top_neg)
    print(“this: %f” % semantic_orientation[‘#new’])
    print(“SCO v IRE: %f” % semantic_orientation[‘#scovire’])
    print(“ENG v FRA: %f” % semantic_orientation[‘#engvfra’])
    print(“#ITA: %f” % semantic_orientation[‘#ita’])
    print(“#FRA: %f” % semantic_orientation[‘#fra’])
    print(“#SCO: %f” % semantic_orientation[‘#sco’])
    print(“#ENG: %f” % semantic_orientation[‘#eng’])
    print(“#WAL: %f” % semantic_orientation[‘#wal’])
    print(“#IRE: %f” % semantic_orientation[‘#ire’])

    Like

  5. my json is ok because is working with the other scripts from previous parts… the problem is can’t understand with which of yours previous scripts i must attach the semantic orientation to work correctly. I post my code which is basics yours from the article but i think is completely wrong the only change is the json file… i get this output

    []
    []

    Traceback (most recent call last):
    File “C:\Python27\Datamine\SemOr.py”, line 112, in
    print(“ITA v WAL: %f” % semantic_orientation[‘#itavwal’])
    KeyError: ‘#itavwal’

    Like

  6. Hi, I’m using Python 2.7 and it doesn’t allow log2:

    “pmi[t1][t2] = math.log2(p_t_com[t1][t2] / denom)”

    So I’m using “pmi[t1][t2] = math.log(p_t_com[t1][t2] / denom) / math.log(2)” instead… but I receive this error…

    pmi[t1][t2] = math.log(p_t_com[t1][t2] / denom) / math.log(2)
    ZeroDivisionError: integer division or modulo by zero

    any idea how to get around this?

    Thanks… I’ve been learning immensely from this blog. Most of this stuff is way over my head but I find it so interesting.

    Like

  7. Hey Marco, Mandla again…

    I would like to know what the measuring unit is for Semantic Orientation and/or how can it be represented as a percentage?

    Like

    1. Hi Mandla, the SO has no particular measuring unit. It’s defined through the Pointwise Mutual Information (PMI) formula as described above and in the paper by P.Turney linked in the article. If you check out the wikipedia page for PMI (https://en.wikipedia.org/wiki/Pointwise_mutual_information), there are option to normalise it, which I haven’t used. In summary, SO is a simple way to give you a hint about the polarity of words.
      Cheers,
      Marco

      Like

  8. Hi! Amazing tutorial!

    I did not understand this snipped though –

    # n_docs is the total n. of tweets
    p_t = {}
    p_t_com = defaultdict(lambda : defaultdict(int))

    for term, n in count_stop_single.items():
    p_t[term] = n / n_docs
    for t2 in com[term]:
    p_t_com[term][t2] = com[term][t2] / n_docs

    What is n_docs here?

    Thanks!!

    Like

    1. Hi Nachiket
      n_docs is the total number of tweets that you have collected. If you’re streaming into a JSON Lines file as described in the previous articles, it’s essentially the total number of lines in that file.

      Cheers
      Marco

      Like

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