11 Aug Which day had the largest number of installed drives, and what was this number?

Hey there, all the requirements in here.

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

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())