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wordsList = ['cat', 'elephant', 'rat', 'rat', 'cat']
wordsRDD = sc.parallelize(wordsList, 4)
# Print out the type of wordsRDD
print type(wordsRDD)
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.<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.
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# 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')
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# One way of completing the function
def makePlural(word):
return word + 's'
print makePlural('cat')
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# 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')
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# TODO: Replace <FILL IN> with appropriate code
pluralRDD = wordsRDD.map(makePlural)
print pluralRDD.collect()
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# TEST Apply makePlural to the base RDD(1c)
Test.assertEquals(pluralRDD.collect(), ['cats', 'elephants', 'rats', 'rats', 'cats'],
'incorrect values for pluralRDD')
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# TODO: Replace <FILL IN> with appropriate code
pluralLambdaRDD = wordsRDD.map(lambda x : x + 's')
print pluralLambdaRDD.collect()
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# TEST Pass a lambda function to map (1d)
Test.assertEquals(pluralLambdaRDD.collect(), ['cats', 'elephants', 'rats', 'rats', 'cats'],
'incorrect values for pluralLambdaRDD (1d)')
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# TODO: Replace <FILL IN> with appropriate code
pluralLengths = (pluralRDD
.map(lambda x : len(x))
.collect())
print pluralLengths
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# TEST Length of each word (1e)
Test.assertEquals(pluralLengths, [4, 9, 4, 4, 4],
'incorrect values for pluralLengths')
(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.map() transformation with a lambda() function to create a new RDD.
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# TODO: Replace <FILL IN> with appropriate code
wordPairs = wordsRDD.map(lambda x : (x, 1))
print wordPairs.collect()
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# TEST Pair RDDs (1f)
Test.assertEquals(wordPairs.collect(),
[('cat', 1), ('elephant', 1), ('rat', 1), ('rat', 1), ('cat', 1)],
'incorrect value for wordPairs')
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.groupByKey() approach 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():groupByKey() to generate a pair RDD of type ('word', iterator).
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# TODO: Replace <FILL IN> with appropriate code
# Note that groupByKey requires no parameters
wordsGrouped = wordPairs.groupByKey()
for key, value in wordsGrouped.collect():
print '{0}: {1}'.format(key, list(value))
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# 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')
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# TODO: Replace <FILL IN> with appropriate code
wordCountsGrouped = wordsGrouped.map(lambda (k, v): (k, sum(v)))
print wordCountsGrouped.collect()
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# TEST Use groupByKey() to obtain the counts (2b)
Test.assertEquals(sorted(wordCountsGrouped.collect()),
[('cat', 2), ('elephant', 1), ('rat', 2)],
'incorrect value for wordCountsGrouped')
reduceByKey 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.
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# 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()
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# TEST Counting using reduceByKey (2c)
Test.assertEquals(sorted(wordCounts.collect()), [('cat', 2), ('elephant', 1), ('rat', 2)],
'incorrect value for wordCounts')
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# TODO: Replace <FILL IN> with appropriate code
wordCountsCollected = wordsRDD.map(lambda x : (x, 1)).reduceByKey(lambda a, b: a + b).collect()
print wordCountsCollected
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# TEST All together (2d)
Test.assertEquals(sorted(wordCountsCollected), [('cat', 2), ('elephant', 1), ('rat', 2)],
'incorrect value for wordCountsCollected')
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.
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# TODO: Replace <FILL IN> with appropriate code
uniqueWords = wordCounts.count()
print uniqueWords
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# TEST Unique words (3a)
Test.assertEquals(uniqueWords, 3, 'incorrect count of uniqueWords')
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# TODO: Replace <FILL IN> with appropriate code
from operator import add
totalCount = (wordCounts
.map(lambda (k, v) : v)
.reduce(add))
average = totalCount / float(uniqueWords)
print totalCount
print round(average, 2)
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# TEST Mean using reduce (3b)
Test.assertEquals(round(average, 2), 1.67, 'incorrect value of average')
wordCount function wordsRDD and return a pair RDD that has all of the words and their associated counts.
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# 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()
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# TEST wordCount function (4a)
Test.assertEquals(sorted(wordCount(wordsRDD).collect()),
[('cat', 2), ('elephant', 1), ('rat', 2)],
'incorrect definition for wordCount 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.
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# 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.
"""
import string
pattern = re.compile('[%s]' % string.punctuation)
return re.sub(pattern, '', text.lower().strip(" "))
print removePunctuation('Hi, you!')
print removePunctuation(' No under_score!')
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# TEST Capitalization and punctuation (4b)
Test.assertEquals(removePunctuation(" The Elephant's 4 cats. "),
'the elephants 4 cats',
'incorrect definition for removePunctuation function')
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.
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# 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))
wordcount() function, we have to address two issues with the format of the RDD:map() transformation is the way to do this, but think about what the result of the split() function will be.
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# TODO: Replace <FILL IN> with appropriate code
shakespeareWordsRDD = shakespeareRDD.flatMap(lambda x : x.split(" "))
shakespeareWordCount = shakespeareWordsRDD.count()
print shakespeareWordsRDD.top(5)
print shakespeareWordCount
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# 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')
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# TODO: Replace <FILL IN> with appropriate code
shakeWordsRDD = shakespeareWordsRDD.filter(lambda x : x!='')
shakeWordCount = shakeWordsRDD.count()
print shakeWordCount
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# TEST Remove empty elements (4e)
Test.assertEquals(shakeWordCount, 882996, 'incorrect value for shakeWordCount')
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.wordCount() function and takeOrdered() to obtain the fifteen most common words and their counts.
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# TODO: Replace <FILL IN> with appropriate code
top15WordsAndCounts = wordCount(shakeWordsRDD).takeOrdered(15,lambda (k, v) : -v)
print '\n'.join(map(lambda (w, c): '{0}: {1}'.format(w, c), top15WordsAndCounts))
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# 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')
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