University of Tampa Python Spark Project

Answer these questions using Spark code. Submit your code (in a py file) and the answers to the questions (in a text file). The answers should use the full dataset, not the small dataset. Start with the code shown below. (Hint: for any tasks that say max/largest, don’t use sortByKey, because that’s much slower than a better option.)

• Which day had the largest number of installed drives, and what was this number?
• How many distinct drives (by model+serial) are installed (i.e., that exist in the data) in each year?
• What’s the max drive capacity per year?

Full dataset: change the file path to: file:///ssd/data/backblaze.csv (146 million rows) – my solution took 17min

Run spark like this: spark-submit backblaze-spark.py –master=local[5]

Or to hide log messages: spark-submit backblaze-spark.py –master=local[5] 2> /dev/null

Look at /home/jeckroth/cinf201/2022-spring/spark/backblaze.py for some more example code.

Starting code with some examples that you can remove:

“”””

from pyspark.sql import SparkSession

spark = SparkSession.builder.appName("Backblaze").getOrCreate()

schema = "day DATE, serial STRING, model STRING, capacity LONG, failure INTEGER"
d = spark.read.schema(schema).load("file:///home/jeckroth/cinf201/spark/assignment/small-backblaze.csv", format="csv", sep=",", header="true")
d = d.rdd

# print first 10 rows
print(d.take(10))

## How many failures occurred each year?

# make key (year) & value (failure 0/1)
d2 = d.map(lambda row: (row.day.year, row.failure))
# add up failures per year
failureCounts = d2.reduceByKey(lambda cnt, rowcnt: cnt + rowcnt)
print(failureCounts.collect())

## Which model (not serial number) has the most failures overall?

# grab model & failure from data, model is the key
d3 = d.map(lambda row: (row.model, row.failure))
# count failures for that model; result so far: [(modelX, 55), (modelY, 2100)]
d3 = d3.reduceByKey(lambda cnt, rowcnt: cnt + rowcnt)
# flip keys and values; result so far: [(55, modelX), (2100, modelY)]
d3 = d3.map(lambda pair: (pair[1], pair[0]))
# sort by value (second in the pair)
d3 = d3.sortByKey(ascending=False) ### NOT EFFICIENT TECHNIQUE
print(d3.collect())