Split-apply-combine mental model in Pandas - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

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

Pandas

grouped = df.[1]('team')

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

Amerge

Bsort

Cfilter

Dgroupby

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'sort' instead of 'groupby' will not create groups.

Using 'filter' or 'merge' does not split the data.

The groupby method splits the DataFrame into groups based on the values in the specified column.

2fill in blank

medium

Complete the code to calculate the mean score for each group.

Pandas

mean_scores = grouped['score'].[1]()

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

Acount

Bsum

Cmean

Dmax

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'sum' adds values instead of averaging.

Using 'count' counts items, not values.

Using 'max' finds the largest value, not the average.

The mean method calculates the average value for each group.

3fill in blank

hard

Fix the error in the code to apply a custom function to each group.

Pandas

result = grouped['score'].[1](lambda x: x.max() - x.min())

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

Aapply

Btransform

Caggregate

Dfilter

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'aggregate' expects named functions or dicts, not lambdas directly.

Using 'transform' returns a transformed version with the same shape.

Using 'filter' selects groups, not applies functions.

The apply method lets you run a custom function on each group.

4fill in blank

hard

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

Pandas

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

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Alen(word)

Blen(word) > 3

Cword.startswith('a')

Dword == 'data'

Attempts:

3 left

💡 Hint

Common Mistakes

Using a condition unrelated to length will filter wrong words.

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

The dictionary comprehension maps each word to its length if the word length is greater than 3.

5fill in blank

hard

Fill all three blanks to create a dictionary of uppercase keys and values filtered by positive values.

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 lowercase keys instead of uppercase.

Using '<' instead of '>' causing wrong filtering.

Using keys as values or vice versa.

This dictionary comprehension creates keys as uppercase strings and keeps values only if they are positive.

Practice

1. What is the main idea behind the split-apply-combine mental model in pandas?

easy

A. Create new columns, rename them, then delete old columns

B. Sort data, filter rows, then plot results

C. Split data into groups, apply a function to each group, then combine results

D. Load data, clean missing values, then save to file

Solution

1. Step 1: Understand the split step

   The data is divided into groups based on some criteria using groupby().

2. Step 2: Apply and combine steps

   A function is applied to each group separately, then the results are combined into a new DataFrame.

3. Final Answer:

   Split data into groups, apply a function to each group, then combine results -> Option C

4. Quick Check:

   Split-apply-combine = Group, apply, combine [OK]

Hint: Remember: split groups, apply function, combine results [OK]

Common Mistakes:

• Confusing split-apply-combine with sorting or filtering
• Thinking it only applies to plotting or saving files
• Mixing up the order of steps

2. Which of the following is the correct syntax to group a DataFrame df by the column 'Category' and calculate the mean of each group?

easy

A. df.groupby('Category').mean()

B. df.mean().groupby('Category')

C. df.groupby.mean('Category')

D. df.groupby('Category').apply(mean)

Solution

1. Step 1: Identify correct groupby syntax

   The correct way to group by a column is df.groupby('Category').

2. Step 2: Apply aggregation function

   To get the mean of each group, use .mean() after grouping.

3. Final Answer:

   df.groupby('Category').mean() -> Option A

4. Quick Check:

   Groupby then mean = df.groupby('Category').mean() [OK]

Hint: Use df.groupby('col').mean() to get group means [OK]

Common Mistakes:

• Calling mean before groupby
• Using apply(mean) without importing mean function
• Wrong method chaining order

3. Given the DataFrame df below:

  Category  Value
0        A      5
1        B      3
2        A      7
3        B      2
4        C      4

What is the output of df.groupby('Category').sum()?

medium

A. Category as column with sums: A=5, B=3, C=4

B. Error because sum() cannot be applied

C. Sum of all values ignoring groups: 21

D. Category as index with sums: A=12, B=5, C=4

Solution

1. Step 1: Group data by 'Category'

   Rows are grouped by 'Category': A, B, and C groups.

2. Step 2: Sum values in each group

   Sum values for A: 5+7=12, B: 3+2=5, C: 4.

3. Final Answer:

   Category as index with sums: A=12, B=5, C=4 -> Option D

4. Quick Check:

   Sum per group = A:12, B:5, C:4 [OK]

Hint: Sum values after groupby to get totals per group [OK]

Common Mistakes:

• Summing all values ignoring groups
• Expecting category as a column instead of index
• Thinking sum() causes error

4. Consider this code snippet:

result = df.groupby('Type').agg({'Score': 'mean', 'Age': 'max'})

What is wrong if the code raises a KeyError?

medium

A. The aggregation functions 'mean' and 'max' are invalid

B. The DataFrame does not have a column named 'Type'

C. The groupby method cannot take a string argument

D. agg() requires a list, not a dictionary

Solution

1. Step 1: Understand KeyError cause

   A KeyError usually means the column used for grouping does not exist in the DataFrame.

2. Step 2: Check groupby argument and agg usage

   Using a string in groupby and a dictionary in agg is correct syntax.

3. Final Answer:

   The DataFrame does not have a column named 'Type' -> Option B

4. Quick Check:

   KeyError means missing column [OK]

Hint: Check column names before groupby to avoid KeyError [OK]

Common Mistakes:

• Assuming aggregation functions are invalid
• Thinking groupby can't take string columns
• Misunderstanding agg() argument types

5. You have a DataFrame df with columns 'Team', 'Player', and 'Points'. You want to find the player with the highest points in each team. Which code correctly uses the split-apply-combine model to achieve this?

hard

A. df.groupby('Team').apply(lambda g: g[g['Points'] == g['Points'].max()])

B. df.groupby('Team')['Points'].max()

C. df.groupby('Player')['Points'].sum()

D. df.groupby('Team').mean()['Points']

Solution

1. Step 1: Understand the goal

   We want the player with the highest points per team, not just the max points value.

2. Step 2: Use groupby and apply to filter rows

   Using apply with a lambda filters each group to rows where 'Points' equals the group's max.

3. Step 3: Check other options

   df.groupby('Team')['Points'].max() returns max points per team but not the player. df.groupby('Player')['Points'].sum() groups by player, not team. df.groupby('Team').mean()['Points'] returns mean points, not max player.

4. Final Answer:

   df.groupby('Team').apply(lambda g: g[g['Points'] == g['Points'].max()]) -> Option A

5. Quick Check:

   Use apply with filter to get max player per group [OK]

Hint: Use groupby + apply with filter to get max per group row [OK]

Common Mistakes:

• Using max() alone returns only values, not rows
• Grouping by wrong column
• Using mean() instead of max()