Fisseha Berhane, PhD

Data Scientist

443-970-2353 fisseha@jhu.edu CV Resume Linkedin GitHub twitter twitter

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Building a word count application

The volume of unstructured text in existence is growing dramatically, and Spark is an excellent tool for analyzing this type of data. Here, I am calculating the most common words in the Complete Works of William Shakespeare retrieved from Project Gutenberg. This could also be scaled to find the most common words on the Internet.

During this lab we will cover:

Part 1: Creating a base RDD and pair RDDs

Part 2: Counting with pair RDDs

Part 3: Finding unique words and a mean value

Part 4: Apply word count to a file

Note that, for reference, you can look up the details of the relevant methods in Spark's Python API

Part 1: Creating a base RDD and pair RDDs

In this part of the lab, we will explore creating a base RDD with parallelize and using pair RDDs to count words.

(1a) Create a base RDD

We'll start by generating a base RDD by using a Python list and the sc.parallelize method. Then we'll print out the type of the base RDD.

In [38]:
wordsList = ['cat', 'elephant', 'rat', 'rat', 'cat']
wordsRDD = sc.parallelize(wordsList, 4)
# Print out the type of wordsRDD
print type(wordsRDD)
<class 'pyspark.rdd.RDD'>

(1b) Pluralize and test

Let's use a map() transformation to add the letter 's' to each string in the base RDD we just created. We'll define a Python function that returns the word with an 's' at the end of the word. Please replace <FILL IN> with your solution. If you have trouble, the next cell has the solution. After you have defined makePlural you can run the third cell which contains a test. If you implementation is correct it will print 1 test passed.

This is the general form that exercises will take, except that no example solution will be provided. Exercises will include an explanation of what is expected, followed by code cells where one cell will have one or more <FILL IN> sections. The cell that needs to be modified will have # TODO: Replace <FILL IN> with appropriate code on its first line. Once the <FILL IN> sections are updated and the code is run, the test cell can then be run to verify the correctness of your solution. The last code cell before the next markdown section will contain the tests.

In [39]:
# TODO: Replace <FILL IN> with appropriate code
def makePlural(word):
    """Adds an 's' to `word`.

    Note:
        This is a simple function that only adds an 's'.  No attempt is made to follow proper
        pluralization rules.

    Args:
        word (str): A string.

    Returns:
        str: A string with 's' added to it.
    """
    return (word+'s')

print makePlural('cat')
cats
In [40]:
# One way of completing the function
def makePlural(word):
    return word + 's'

print makePlural('cat')
cats
In [41]:
# Load in the testing code and check to see if your answer is correct
# If incorrect it will report back '1 test failed' for each failed test
# Make sure to rerun any cell you change before trying the test again
from test_helper import Test
# TEST Pluralize and test (1b)
Test.assertEquals(makePlural('rat'), 'rats', 'incorrect result: makePlural does not add an s')
1 test passed.

(1c) Apply makePlural to the base RDD

Now pass each item in the base RDD into a map() transformation that applies the makePlural() function to each element. And then call the collect() action to see the transformed RDD.

In [42]:
# TODO: Replace <FILL IN> with appropriate code
pluralRDD = wordsRDD.map(makePlural)
print pluralRDD.collect()
['cats', 'elephants', 'rats', 'rats', 'cats']
In [43]:
# TEST Apply makePlural to the base RDD(1c)
Test.assertEquals(pluralRDD.collect(), ['cats', 'elephants', 'rats', 'rats', 'cats'],
                  'incorrect values for pluralRDD')
1 test passed.

(1d) Pass a lambda function to map

Let's create the same RDD using a lambda function.

In [44]:
# TODO: Replace <FILL IN> with appropriate code
pluralLambdaRDD = wordsRDD.map(lambda x:x+'s')
print pluralLambdaRDD.collect()
['cats', 'elephants', 'rats', 'rats', 'cats']
In [45]:
# TEST Pass a lambda function to map (1d)
Test.assertEquals(pluralLambdaRDD.collect(), ['cats', 'elephants', 'rats', 'rats', 'cats'],
                  'incorrect values for pluralLambdaRDD (1d)')
1 test passed.

(1e) Length of each word

Now use map() and a lambda function to return the number of characters in each word. We'll collect this result directly into a variable.

In [46]:
# TODO: Replace <FILL IN> with appropriate code
pluralLengths = (pluralRDD
                 .map(lambda x: len(x))
                 .collect())
print pluralLengths
[4, 9, 4, 4, 4]
In [47]:
# TEST Length of each word (1e)
Test.assertEquals(pluralLengths, [4, 9, 4, 4, 4],
                  'incorrect values for pluralLengths')
1 test passed.

(1f) Pair RDDs

The next step in writing our word counting program is to create a new type of RDD, called a pair RDD. A pair RDD is an RDD where each element is a pair tuple (k, v) where k is the key and v is the value. In this example, we will create a pair consisting of ('<word>', 1) for each word element in the RDD.

We can create the pair RDD using the map() transformation with a lambda() function to create a new RDD.

In [48]:
# TODO: Replace <FILL IN> with appropriate code
wordPairs = wordsRDD.map(lambda x: (x,1))
print wordPairs.collect()
[('cat', 1), ('elephant', 1), ('rat', 1), ('rat', 1), ('cat', 1)]
In [49]:
# TEST Pair RDDs (1f)
Test.assertEquals(wordPairs.collect(),
                  [('cat', 1), ('elephant', 1), ('rat', 1), ('rat', 1), ('cat', 1)],
                  'incorrect value for wordPairs')
1 test passed.

Part 2: Counting with pair RDDs

Now, let's count the number of times a particular word appears in the RDD. There are multiple ways to perform the counting, but some are much less efficient than others.

A naive approach would be to collect() all of the elements and count them in the driver program. While this approach could work for small datasets, we want an approach that will work for any size dataset including terabyte- or petabyte-sized datasets. In addition, performing all of the work in the driver program is slower than performing it in parallel in the workers. For these reasons, we will use data parallel operations.

(2a) groupByKey() approach

An approach you might first consider (we'll see shortly that there are better ways) is based on using the groupByKey() transformation. As the name implies, the groupByKey() transformation groups all the elements of the RDD with the same key into a single list in one of the partitions. There are two problems with using groupByKey():

  • #### The operation requires a lot of data movement to move all the values into the appropriate partitions.
  • #### The lists can be very large. Consider a word count of English Wikipedia: the lists for common words (e.g., the, a, etc.) would be huge and could exhaust the available memory in a worker.

Use groupByKey() to generate a pair RDD of type ('word', iterator).

In [50]:
# TODO: Replace <FILL IN> with appropriate code
# Note that groupByKey requires no parameters
wordsGrouped = wordPairs.groupByKey().mapValues(lambda x: list(x))
for key, value in wordsGrouped.collect():
    print '{0}: {1}'.format(key, list(value))
rat: [1, 1]
elephant: [1]
cat: [1, 1]
In [51]:
# TEST groupByKey() approach (2a)
Test.assertEquals(sorted(wordsGrouped.mapValues(lambda x: list(x)).collect()),
                  [('cat', [1, 1]), ('elephant', [1]), ('rat', [1, 1])],
                  'incorrect value for wordsGrouped')
1 test passed.

(2b) Use groupByKey() to obtain the counts

Using the groupByKey() transformation creates an RDD containing 3 elements, each of which is a pair of a word and a Python iterator.

Now sum the iterator using a map() transformation. The result should be a pair RDD consisting of (word, count) pairs.

In [52]:
# TODO: Replace <FILL IN> with appropriate code
wordCountsGrouped = wordsGrouped.map(lambda (k, v): (k, sum(v)))
print wordCountsGrouped.collect()
[('rat', 2), ('elephant', 1), ('cat', 2)]
In [53]:
# TEST Use groupByKey() to obtain the counts (2b)
Test.assertEquals(sorted(wordCountsGrouped.collect()),
                  [('cat', 2), ('elephant', 1), ('rat', 2)],
                  'incorrect value for wordCountsGrouped')
1 test passed.

(2c) Counting using reduceByKey

A better approach is to start from the pair RDD and then use the reduceByKey() transformation to create a new pair RDD. The reduceByKey() transformation gathers together pairs that have the same key and applies the function provided to two values at a time, iteratively reducing all of the values to a single value. reduceByKey() operates by applying the function first within each partition on a per-key basis and then across the partitions, allowing it to scale efficiently to large datasets.

In [54]:
# TODO: Replace <FILL IN> with appropriate code
# Note that reduceByKey takes in a function that accepts two values and returns a single value
wordCounts = wordPairs.reduceByKey(lambda a,b: a+b)
print wordCounts.collect()
[('rat', 2), ('elephant', 1), ('cat', 2)]
In [55]:
# TEST Counting using reduceByKey (2c)
Test.assertEquals(sorted(wordCounts.collect()), [('cat', 2), ('elephant', 1), ('rat', 2)],
                  'incorrect value for wordCounts')
1 test passed.

(2d) All together

The expert version of the code performs the map() to pair RDD, reduceByKey() transformation, and collect in one statement.

In [56]:
# TODO: Replace <FILL IN> with appropriate code
wordCountsCollected =(wordsRDD.map(lambda x: (x,1)).reduceByKey(lambda a,b:a+b)
                       .collect())
print wordCountsCollected
[('rat', 2), ('elephant', 1), ('cat', 2)]
In [57]:
# TEST All together (2d)
Test.assertEquals(sorted(wordCountsCollected), [('cat', 2), ('elephant', 1), ('rat', 2)],
                  'incorrect value for wordCountsCollected')
1 test passed.

Part 3: Finding unique words and a mean value

(3a) Unique words

Calculate the number of unique words in wordsRDD. You can use other RDDs that you have already created to make this easier.

In [58]:
# TODO: Replace <FILL IN> with appropriate code
uniqueWords = len(wordCountsCollected)
print uniqueWords
3
In [59]:
# TEST Unique words (3a)
Test.assertEquals(uniqueWords, 3, 'incorrect count of uniqueWords')
1 test passed.

(3b) Mean using reduce

Find the mean number of words per unique word in wordCounts.

Use a reduce() action to sum the counts in wordCounts and then divide by the number of unique words. First map() the pair RDD wordCounts, which consists of (key, value) pairs, to an RDD of values.

In [60]:
# TODO: Replace <FILL IN> with appropriate code
from operator import add
totalCount = (wordCounts
              .map(lambda (x,v): v)
              .reduce(add))
average = totalCount / float(uniqueWords)
print totalCount
print round(average, 2)
5
1.67
In [61]:
# TEST Mean using reduce (3b)
Test.assertEquals(round(average, 2), 1.67, 'incorrect value of average')
1 test passed.

Part 4: Apply word count to a file

In this section we will finish developing our word count application. We'll have to build the wordCount function, deal with real world problems like capitalization and punctuation, load in our data source, and compute the word count on the new data.

(4a) wordCount function

First, define a function for word counting. You should reuse the techniques that have been covered in earlier parts of this lab. This function should take in an RDD that is a list of words like wordsRDD and return a pair RDD that has all of the words and their associated counts.

In [62]:
# TODO: Replace <FILL IN> with appropriate code
def wordCount(wordListRDD):
    """Creates a pair RDD with word counts from an RDD of words.

    Args:
        wordListRDD (RDD of str): An RDD consisting of words.

    Returns:
        RDD of (str, int): An RDD consisting of (word, count) tuples.
    """
    return (wordListRDD.map(lambda x: (x,1)).reduceByKey(lambda a,b:a+b))
print wordCount(wordsRDD).collect()
[('rat', 2), ('elephant', 1), ('cat', 2)]
In [63]:
# TEST wordCount function (4a)
Test.assertEquals(sorted(wordCount(wordsRDD).collect()),
                  [('cat', 2), ('elephant', 1), ('rat', 2)],
                  'incorrect definition for wordCount function')
1 test passed.

(4b) Capitalization and punctuation

Real world files are more complicated than the data we have been using in this lab. Some of the issues we have to address are:

  • #### Words should be counted independent of their capitialization (e.g., Spark and spark should be counted as the same word).
  • #### All punctuation should be removed.
  • #### Any leading or trailing spaces on a line should be removed.

Define the function removePunctuation that converts all text to lower case, removes any punctuation, and removes leading and trailing spaces. Use the Python re module to remove any text that is not a letter, number, or space. Reading help(re.sub) might be useful.

In [64]:
# TODO: Replace <FILL IN> with appropriate code
import re
def removePunctuation(text):
    """Removes punctuation, changes to lower case, and strips leading and trailing spaces.

    Note:
        Only spaces, letters, and numbers should be retained.  Other characters should should be
        eliminated (e.g. it's becomes its).  Leading and trailing spaces should be removed after
        punctuation is removed.

    Args:
        text (str): A string.

    Returns:
        str: The cleaned up string.
    """
    return re.sub(r'[^a-zA-Z0-9\s]+','',text).lower().strip()
print removePunctuation('Hi, you!')
print removePunctuation(' No under_score!')
hi you
no underscore
In [65]:
# TEST Capitalization and punctuation (4b)
Test.assertEquals(removePunctuation(" The Elephant's 4 cats. "),
                  'the elephants 4 cats',
                  'incorrect definition for removePunctuation function')
1 test passed.

(4c) Load a text file

For the next part of this lab, we will use the Complete Works of William Shakespeare from Project Gutenberg. To convert a text file into an RDD, we use the SparkContext.textFile() method. We also apply the recently defined removePunctuation() function using a map() transformation to strip out the punctuation and change all text to lowercase. Since the file is large we use take(15), so that we only print 15 lines.

In [66]:
# Just run this code
import os.path
baseDir = os.path.join('data')
inputPath = os.path.join('cs100', 'lab1', 'shakespeare.txt')
fileName = os.path.join(baseDir, inputPath)

shakespeareRDD = (sc
                  .textFile(fileName, 8)
                  .map(removePunctuation))
print '\n'.join(shakespeareRDD
                .zipWithIndex()  # to (line, lineNum)
                .map(lambda (l, num): '{0}: {1}'.format(num, l))  # to 'lineNum: line'
                .take(15))
0: 1609
1: 
2: the sonnets
3: 
4: by william shakespeare
5: 
6: 
7: 
8: 1
9: from fairest creatures we desire increase
10: that thereby beautys rose might never die
11: but as the riper should by time decease
12: his tender heir might bear his memory
13: but thou contracted to thine own bright eyes
14: feedst thy lights flame with selfsubstantial fuel

(4d) Words from lines

Before we can use the wordcount() function, we have to address two issues with the format of the RDD:

  • #### The first issue is that that we need to split each line by its spaces.
  • #### The second issue is we need to filter out empty lines.

Apply a transformation that will split each element of the RDD by its spaces. For each element of the RDD, you should apply Python's string split() function. You might think that a map() transformation is the way to do this, but think about what the result of the split() function will be.

In [67]:
# TODO: Replace <FILL IN> with appropriate code
shakespeareWordsRDD = shakespeareRDD.flatMap(lambda x:x.split(' ')) 
shakespeareWordCount = shakespeareWordsRDD.count()
print shakespeareWordsRDD.top(5)
print shakespeareWordCount
[u'zwaggerd', u'zounds', u'zounds', u'zounds', u'zounds']
927631
In [68]:
# TEST Words from lines (4d)
# This test allows for leading spaces to be removed either before or after
# punctuation is removed.
Test.assertTrue(shakespeareWordCount == 927631 or shakespeareWordCount == 928908,
                'incorrect value for shakespeareWordCount')
Test.assertEquals(shakespeareWordsRDD.top(5),
                  [u'zwaggerd', u'zounds', u'zounds', u'zounds', u'zounds'],
                  'incorrect value for shakespeareWordsRDD')
1 test passed.
1 test passed.

(4e) Remove empty elements

The next step is to filter out the empty elements. Remove all entries where the word is ''.

In [69]:
# TODO: Replace <FILL IN> with appropriate code
shakeWordsRDD = shakespeareWordsRDD.filter(lambda x: len(x)!=0)
shakeWordCount = shakeWordsRDD.count()
print shakeWordCount
882996
In [70]:
# TEST Remove empty elements (4e)
Test.assertEquals(shakeWordCount, 882996, 'incorrect value for shakeWordCount')
1 test passed.

(4f) Count the words

We now have an RDD that is only words. Next, let's apply the wordCount() function to produce a list of word counts. We can view the top 15 words by using the takeOrdered() action; however, since the elements of the RDD are pairs, we need a custom sort function that sorts using the value part of the pair.

You'll notice that many of the words are common English words. These are called stopwords. In a later lab, we will see how to eliminate them from the results.

Use the wordCount() function and takeOrdered() to obtain the fifteen most common words and their counts.

In [71]:
# TODO: Replace <FILL IN> with appropriate code
top15WordsAndCounts =wordCount(shakeWordsRDD).takeOrdered(15, lambda (x,v):-v)
print '\n'.join(map(lambda (w, c): '{0}: {1}'.format(w, c), top15WordsAndCounts))
the: 27361
and: 26028
i: 20681
to: 19150
of: 17463
a: 14593
you: 13615
my: 12481
in: 10956
that: 10890
is: 9134
not: 8497
with: 7771
me: 7769
it: 7678
In [73]:
# TEST Count the words (4f)
Test.assertEquals(top15WordsAndCounts,
                  [(u'the', 27361), (u'and', 26028), (u'i', 20681), (u'to', 19150), (u'of', 17463),
                   (u'a', 14593), (u'you', 13615), (u'my', 12481), (u'in', 10956), (u'that', 10890),
                   (u'is', 9134), (u'not', 8497), (u'with', 7771), (u'me', 7769), (u'it', 7678)],
                  'incorrect value for top15WordsAndCounts')
1 test passed.
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