Selecting data with MultiIndex in Pandas - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select the data for the first level index 'A' from the MultiIndex DataFrame.

Pandas

selected = df.loc[[1]]

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

A'D'

B'B'

C'C'

D'A'

Attempts:

3 left

💡 Hint

Common Mistakes

Using integer positions instead of labels.

Using .iloc instead of .loc for label-based selection.

Using df.loc['A'] selects all rows where the first level of the MultiIndex is 'A'.

2fill in blank

medium

Complete the code to select data for the MultiIndex where the first level is 'B' and the second level is 'two'.

Pandas

selected = df.loc[([1], 'two')]

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

A'D'

B'A'

C'B'

D'C'

Attempts:

3 left

💡 Hint

Common Mistakes

Not using a tuple for multi-level selection.

Swapping the order of levels in the tuple.

Using df.loc[('B', 'two')] selects the row where the first level is 'B' and the second level is 'two'.

3fill in blank

hard

Fix the error in the code to select all rows where the second level index is 'one'.

Pandas

selected = df.xs([1], level=1)

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

A'one'

B1

C'two'

D'zero'

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong level number.

Using an integer instead of the label string.

The xs method selects cross sections. Using xs('one', level=1) selects all rows where the second level index is 'one'.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps each word to its length only if the length is greater than 3.

Pandas

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

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Alen(word)

Blen(word) > 3

Cword

Dword > 3

Attempts:

3 left

💡 Hint

Common Mistakes

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

Using the word in the condition instead of its length.

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

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps uppercase words to their counts only if the count is greater than 1.

Pandas

result = [1]: [2] for [3], [2] in counts.items() if [2] > 1}}

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

Aword.upper()

Bcount

Cword

Ditem

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong variable names in the loop.

Not converting words to uppercase.

Using the count variable incorrectly.

The comprehension maps each uppercase word (word.upper()) to its count (count) only if the count is greater than 1. The loop variables are word and count.

Practice

1. What does df.loc[('A', 1)] do when df has a MultiIndex with two levels?

easy

A. Selects the row where the first level is 'A' and the second level is 1.

B. Selects all rows where the first level is 'A' regardless of the second level.

C. Selects all rows where the second level is 1 regardless of the first level.

D. Raises an error because tuples cannot be used with MultiIndex.

Solution

1. Step 1: Understand MultiIndex selection with tuples

   Using df.loc with a tuple selects rows matching each level in the MultiIndex exactly.

2. Step 2: Apply tuple to select specific row

   The tuple ('A', 1) matches the first level 'A' and second level 1, selecting that exact row.

3. Final Answer:

   Selects the row where the first level is 'A' and the second level is 1. -> Option A

4. Quick Check:

   Tuple in loc = exact MultiIndex row [OK]

Hint: Use full tuple in loc to select exact MultiIndex row [OK]

Common Mistakes:

• Thinking partial tuple selects exact row
• Confusing level order in tuple
• Believing tuple usage causes error

2. Which of the following is the correct tuple syntax to select all rows where the first level of a MultiIndex is 'B' using df.loc?

easy

A. df.loc[('B')]

B. df.loc[('B',)]

C. df.loc['B']

D. df.loc[['B']]

Solution

1. Step 1: Understand tuple syntax for single element

   To specify a single-element tuple in Python, a comma is needed: ('B',).

2. Step 2: Use tuple in df.loc to select first level

   Using df.loc[('B',)] selects all rows where the first level is 'B'.

3. Final Answer:

   df.loc[('B',)] -> Option B

4. Quick Check:

   Single-element tuple needs comma [OK]

Hint: Use comma for single-element tuple in loc [OK]

Common Mistakes:

• Using parentheses without comma for tuple
• Using list instead of tuple
• Omitting tuple and causing key error

3. Given the DataFrame df with MultiIndex levels 'Letter' and 'Number':

import pandas as pd
index = pd.MultiIndex.from_tuples([('A', 1), ('A', 2), ('B', 1), ('B', 2)], names=['Letter', 'Number'])
data = [10, 20, 30, 40]
df = pd.DataFrame(data, index=index, columns=['Value'])
print(df.loc['A'])

What is the output?

medium

A. Value\nLetter Number\nA 1 10\n 2 20

B. Value\nLetter Number\nB 1 30\n 2 40

C. KeyError because 'A' is not a valid index

D. Value\nNumber\n1 10\n2 20

Solution

1. Step 1: Understand partial selection with single label

   Using df.loc['A'] selects all rows where the first level is 'A'. The result is a DataFrame indexed by the second level only.

2. Step 2: Check output structure

   The output shows 'Value' column with index 'Number' 1 and 2, values 10 and 20 respectively.

3. Final Answer:

   Value Number 1 10 2 20 -> Option D

4. Quick Check:

   Partial index selection drops first level [OK]

Hint: Partial loc drops selected level from index [OK]

Common Mistakes:

• Expecting full MultiIndex in output
• Confusing which levels remain after selection
• Thinking KeyError occurs for valid label

4. Consider this code snippet:

import pandas as pd
index = pd.MultiIndex.from_tuples([('X', 1), ('X', 2), ('Y', 1)])
data = [100, 200, 300]
df = pd.DataFrame(data, index=index, columns=['Score'])

# Attempt to select all rows where first level is 'X'
result = df.loc[('X')]
print(result)

What is the issue and how to fix it?

medium

A. SyntaxError; fix by removing parentheses: df.loc['X'].

B. Raises KeyError; fix by using df.loc[('X',)] to specify single-element tuple.

C. No issue; code runs correctly and prints rows with first level 'X'.

D. Raises TypeError; fix by converting 'X' to list: df.loc[['X']].

Solution

1. Step 1: Analyze the tuple usage in df.loc

   Here, df.loc[('X')] is equivalent to df.loc['X'] because parentheses around a string do not create a tuple.

2. Step 2: Confirm code behavior

   This selects all rows where the first level is 'X' correctly, so no error occurs.

3. Final Answer:

   No issue; code runs correctly and prints rows with first level 'X'. -> Option C

4. Quick Check:

   Parentheses alone don't make tuple [OK]

Hint: Parentheses alone don't create tuple; string stays string [OK]

Common Mistakes:

• Assuming ('X') is a tuple
• Expecting KeyError for valid label
• Confusing tuple syntax with parentheses

5. You have a MultiIndex DataFrame df with levels 'City', 'Year', and 'Month'. How do you select all data for 'New York' in 2023 regardless of month using a tuple that specifies all three levels?

hard

A. df.loc[('New York', 2023, slice(None))]

B. df.loc[('New York', 2023)]

C. df.loc[('New York', 2023,)]

D. df.loc['New York', 2023]

Solution

1. Step 1: Understand MultiIndex with 3 levels

   The MultiIndex has three levels: 'City', 'Year', and 'Month'. To select all months for 'New York' and 2023, we must specify all three levels.

2. Step 2: Use slice(None) to select all in third level

   Using slice(None) in the third position selects all months. So df.loc[('New York', 2023, slice(None))] selects all rows matching city and year, any month.

3. Final Answer:

   df.loc[('New York', 2023, slice(None))] -> Option A

4. Quick Check:

   Use slice(None) for full level selection [OK]

Hint: Use slice(None) for full selection in MultiIndex level [OK]

Common Mistakes:

• Omitting third level causing KeyError
• Using incomplete tuple without slice
• Using comma-separated loc without tuple