DataFrame API in Snowpark in Snowflake - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a Snowpark DataFrame from a table named 'employees'.

Snowflake

df = session.table([1])

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

Aemployees

B"employees"

C'employees'

Dtable_employees

Attempts:

3 left

💡 Hint

Common Mistakes

Not using quotes around the table name

Using a variable name instead of a string

The session.table() method requires the table name as a string. Using double or single quotes is the standard way to specify the table name.

2fill in blank

medium

Complete the code to select the 'name' and 'salary' columns from the DataFrame.

Snowflake

df_selected = df.select([1])

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

A"name", "salary"

B"name", salary

Dname, salary

Attempts:

3 left

💡 Hint

Common Mistakes

Not quoting column names

Using variables instead of strings

The select method takes column names as strings separated by commas. Both column names must be in quotes.

3fill in blank

hard

Fix the error in the code to filter rows where salary is greater than 50000.

Snowflake

df_filtered = df.filter(df[[1]] > 50000)

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

A"salary"

Bsalary

C'salary'

Ddf.salary

Attempts:

3 left

💡 Hint

Common Mistakes

Using the column name without quotes inside brackets

Using dot notation inside brackets

When accessing columns by name in a DataFrame, use the column name as a string inside square brackets.

4fill in blank

hard

Fill both blanks to create a new column 'bonus' which is 10% of 'salary' and select it along with 'name'.

Snowflake

df_new = df.select([1], (df[[2]] * 0.1).alias("bonus"))

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

A"name"

B"salary"

Csalary

Dname

Attempts:

3 left

💡 Hint

Common Mistakes

Using unquoted column names inside brackets

Mixing quoted and unquoted column names

Column names must be strings inside quotes when used with df[]. The select method takes column names as strings.

5fill in blank

hard

Fill all three blanks to group by 'department', calculate average 'salary' as 'avg_salary', and filter departments with avg_salary > 60000.

Snowflake

df_grouped = df.groupBy([1]).agg(df[[2]].avg().alias([3])).filter("avg_salary > 60000")

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

A"department"

B"salary"

C"avg_salary"

Ddepartment

Attempts:

3 left

💡 Hint

Common Mistakes

Not quoting column names

Using unquoted alias names

Use quoted column names for grouping and aggregation. The alias name must also be a string.