Why advanced patterns solve complex analytics in dbt - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select all columns from the 'orders' table.

dbt

SELECT [1] FROM orders;

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

Aeverything

Ball

C*

Dcolumns

Attempts:

3 left

💡 Hint

Common Mistakes

Using words like 'all' or 'columns' instead of the symbol *.

Leaving the blank empty.

The asterisk (*) selects all columns from the table.

2fill in blank

medium

Complete the code to filter rows where 'status' equals 'completed'.

dbt

SELECT * FROM orders WHERE status [1] 'completed';

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

A=

B!=

C<>

DLIKE

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' or '<>' which mean 'not equal'.

Using LIKE without wildcards.

The equals sign (=) is used to filter rows where the column matches a specific value.

3fill in blank

hard

Fix the error in the code to correctly join 'orders' and 'customers' on 'customer_id'.

dbt

SELECT orders.id, customers.name FROM orders JOIN customers ON orders.[1] = customers.customer_id;

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

Aname

Bid

Corder_id

Dcustomer_id

Attempts:

3 left

💡 Hint

Common Mistakes

Joining on 'id' or 'order_id' which do not link the two tables.

Using 'name' which is not an ID column.

The join should be on the 'customer_id' column in the 'orders' table to match 'customers.customer_id'.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps words to their lengths for words longer than 3 characters.

dbt

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

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

Alen(word)

Bword > 3

Clen(word) > 3

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

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

Not using len() to get word length.

The dictionary maps each word to its length using len(word). The condition filters words longer than 3 characters using len(word) > 3.

5fill in blank

hard

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

dbt

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

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

Ak.upper()

Bv

C>

Dk

Attempts:

3 left

💡 Hint

Common Mistakes

Using k instead of k.upper() for keys.

Using '<' or '=' instead of '>' in the condition.

The key is the uppercase version of k using k.upper(). The value is v. The condition filters items where v is greater than 0.