Data aggregation reporting in Pandas - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate the average score from the DataFrame.

Pandas

average_score = df['score'].[1]()

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

Amean

Bsum

Ccount

Dmax

Attempts:

3 left

💡 Hint

Common Mistakes

Using sum() instead of mean() will give the total, not the average.

Using count() returns the number of entries, not the average.

The mean() function calculates the average value of the 'score' column.

2fill in blank

medium

Complete the code to group the DataFrame by 'category' and calculate the sum of 'sales'.

Pandas

grouped = df.groupby('[1]')['sales'].sum()

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

Aregion

Bscore

Cdate

Dcategory

Attempts:

3 left

💡 Hint

Common Mistakes

Grouping by 'score' or 'date' will not group sales by category.

Using 'region' groups by location, not category.

Grouping by 'category' allows us to sum sales within each category.

3fill in blank

hard

Fix the error in the code to calculate the maximum value in the 'price' column.

Pandas

max_price = df['price'].[1]()

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

Amax

Bmaximum

Cmax_value

DmaxPrice

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'maximum' or 'max_value' causes errors because they are not pandas functions.

Using camelCase like 'maxPrice' is incorrect in pandas.

The correct pandas function to get the maximum value is max().

4fill in blank

hard

Fill both blanks to create a dictionary with word lengths for words longer than 3 characters.

Pandas

lengths = {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' compares a string to a number, causing errors.

Using 'word' alone in the condition does not filter by length.

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

5fill in blank

hard

Fill all three blanks to create a dictionary of uppercase keys and values greater than zero.

Pandas

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.lower()

Attempts:

3 left

💡 Hint

Common Mistakes

Using k.lower() instead of uppercase keys.

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

Using keys or values incorrectly in the dictionary.

The dictionary comprehension uses k.upper() for keys, keeps values as v, and filters values greater than zero with v > 0.