Why advanced grouping matters in Pandas - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to group the DataFrame by the 'Category' column.

Pandas

grouped = df.groupby([1])

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Adf['Category']

BCategory

C['Category']

D'Category'

Attempts:

3 left

💡 Hint

Common Mistakes

Passing the column without quotes causes an error.

Passing the whole column data instead of the column name.

To group by a column name, you pass the column name as a string to groupby.

2fill in blank

medium

Complete the code to calculate the mean of each group.

Pandas

mean_values = df.groupby('Category').[1]()

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Amean

Bmax

Ccount

Dsum

Attempts:

3 left

💡 Hint

Common Mistakes

Using sum() instead of mean() gives total, not average.

Using count() returns counts, not averages.

The mean() function calculates the average value for each group.

3fill in blank

hard

Fix the error in the code to group by 'Category' and 'Subcategory'.

Pandas

grouped = df.groupby([1])

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A['Category', 'Subcategory']

B'Category', 'Subcategory'

C'Category Subcategory'

Ddf[['Category', 'Subcategory']]

Attempts:

3 left

💡 Hint

Common Mistakes

Passing a single string with both column names combined.

Passing the DataFrame slice instead of column names.

To group by multiple columns, pass a list of column names as strings.

4fill in blank

hard

Fill both blanks to create a dictionary with word lengths greater than 3.

Pandas

lengths = {word: [1] for word in words if len(word) [2] 3}

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Alen(word)

B>

C<

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself as the value instead of its length.

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

The dictionary maps each word to its length, but only includes words longer than 3 characters.

5fill in blank

hard

Fill all three blanks to create a filtered dictionary with uppercase keys and values greater than 0.

Pandas

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

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Ak.upper()

Bv

C>

Dk.lower()

Attempts:

3 left

💡 Hint

Common Mistakes

Using k.lower() instead of k.upper().

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

This dictionary comprehension creates keys as uppercase versions of original keys, keeps values as is, and filters to only include values greater than zero.