CODEWITHSUNDEEP
apache-sparkpyspark
Krishnang K Dalal
Krishnang K Dalal

Reputation: 2564

PySpark Count Distinct By Group In A RDD

I have an RDD of date-time and hostname as tuple and I want to count the unique hostnames by date.

RDD:

X = [(datetime.datetime(1995, 8, 1, 0, 0, 1), u'in24.inetnebr.com'),
     (datetime.datetime(1995, 8, 1, 0, 0, 7), u'uplherc.upl.com'),
     (datetime.datetime(1995, 8, 1, 0, 0, 8), u'uplherc.upl.com'),
     (datetime.datetime(1995, 8, 2, 0, 0, 8), u'uplherc.upl.com'),
     (datetime.datetime(1995, 8, 2, 0, 0, 8), u'uplherc.upl.com'),
     (datetime.datetime(1995, 8, 2, 0, 0, 9), u'ix-esc-ca2-07.ix.netcom.com'),
     (datetime.datetime(1995, 8, 3, 0, 0, 10), u'uplherc.upl.com'),
     (datetime.datetime(1995, 8, 3, 0, 0, 10), u'slppp6.intermind.net'),
     (datetime.datetime(1995, 8, 4, 0, 0, 10), u'piweba4y.prodigy.com'),
     (datetime.datetime(1995, 8, 5, 0, 0, 11), u'slppp6.intermind.net')]

DESIRED OUTPUT:

[(datetime.datetime(1995, 8, 1, 0, 0, 1), 2),
 (datetime.datetime(1995, 8, 2, 0, 0, 8), 2),
 (datetime.datetime(1995, 8, 3, 0, 0, 10), 2),
 (datetime.datetime(1995, 8, 4, 0, 0, 10), 1),
 (datetime.datetime(1995, 8, 5, 0, 0, 11), 1)]

MY ATTEMPT:

dayGroupedHosts = X.groupBy(lambda x: x[0]).distinct()
dayHostCount = dayGroupedHosts.count()

I am getting an error while performing count operation. I am new to Spark and I would like to know the correct and efficient transformation to achieve such tasks.

Thanks a lot in advance.

Upvotes: 3

Views: 6511

Answers (2)

akuiper
akuiper

Reputation: 215117

Or convert to a DataFrame and use countDistinct method:

import pyspark.sql.functions as f

df = spark.createDataFrame(X, ["dt", "hostname"])
df.show()
+-------------------+--------------------+
|                 dt|            hostname|
+-------------------+--------------------+
|1995-08-01 00:00:01|   in24.inetnebr.com|
|1995-08-01 00:00:07|     uplherc.upl.com|
|1995-08-01 00:00:08|     uplherc.upl.com|
|1995-08-02 00:00:08|     uplherc.upl.com|
|1995-08-02 00:00:08|     uplherc.upl.com|
|1995-08-02 00:00:09|ix-esc-ca2-07.ix....|
|1995-08-03 00:00:10|     uplherc.upl.com|
|1995-08-03 00:00:10|slppp6.intermind.net|
|1995-08-04 00:00:10|piweba4y.prodigy.com|
|1995-08-05 00:00:11|slppp6.intermind.net|
+-------------------+--------------------+

df.groupBy(f.to_date('dt').alias('date')).agg(
  f.countDistinct('hostname').alias('hostname')
).show()
+----------+--------+
|      date|hostname|
+----------+--------+
|1995-08-02|       2|
|1995-08-03|       2|
|1995-08-01|       2|
|1995-08-04|       1|
|1995-08-05|       1|
+----------+--------+

Upvotes: 4

pault
pault

Reputation: 43544

You need to first convert the keys into dates. Then group by the key, and count the distinct values:

X.map(lambda x: (x[0].date(), x[1]))\
    .groupByKey()\
    .mapValues(lambda vals: len(set(vals)))\
    .sortByKey()\
    .collect()
#[(datetime.date(1995, 8, 1), 2),
# (datetime.date(1995, 8, 2), 2),
# (datetime.date(1995, 8, 3), 2),
# (datetime.date(1995, 8, 4), 1),
# (datetime.date(1995, 8, 5), 1)]

Upvotes: 4

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