Partitioning for query performance in Hadoop - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a partitioned table by the column 'year'.

Hadoop

CREATE TABLE sales_data (id INT, amount FLOAT) PARTITIONED BY ([1] STRING);

Drag options to blanks, or click blank then click option'

Amonth

Bregion

Cyear

Dcategory

Attempts:

3 left

💡 Hint

Common Mistakes

Choosing a non-time column like 'region' or 'category' for partitioning.

Forgetting to specify the partition column in the table definition.

The partition column is 'year' to organize data by year for better query performance.

2fill in blank

medium

Complete the code to insert data into the partition for year 2023.

Hadoop

INSERT INTO sales_data PARTITION (year=[1]) SELECT id, amount FROM raw_sales WHERE year = 2023;

Drag options to blanks, or click blank then click option'

A'2023'

B2023

C'2022'

Dyear

Attempts:

3 left

💡 Hint

Common Mistakes

Using the number 2023 without quotes.

Using a column name instead of a literal value.

The partition value must be a string literal, so '2023' is correct.

3fill in blank

hard

Fix the error in the query to select data only from the 2022 partition.

Hadoop

SELECT * FROM sales_data WHERE year [1] '2022';

Drag options to blanks, or click blank then click option'

A==

B=

CLIKE

DIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of '=' causes syntax errors.

Using 'LIKE' without wildcards is incorrect here.

In SQL, the equality operator is a single '=' sign, not '=='.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps each word to its length if the length is greater than 3.

Hadoop

{word: [1] for word in words if [2]

Drag options to blanks, or click blank then click option'

Alen(word)

Blen(word) > 3

Cword

Dword > 3

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'word' instead of 'len(word)' for the value.

Using 'word > 3' which compares string to number incorrectly.

The dictionary maps words to their lengths (len(word)) and filters words with length greater than 3 (len(word) > 3).

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps uppercase keys to values only if the value is greater than 0.

Hadoop

result = [1]: [2] for [3], [2] in data.items() if [2] > 0}

Drag options to blanks, or click blank then click option'

Ak.upper()

Bv

Ck

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'value' instead of 'v' as the loop variable.

Not converting keys to uppercase.

Keys are converted to uppercase with k.upper(), values are v, and the loop variables are k and v.