Why indexing matters in Pandas - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select the first row of the DataFrame using .iloc.

Pandas

first_row = df.[1][0]

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

Ailoc

Bat

Cloc

Diat

Attempts:

3 left

💡 Hint

Common Mistakes

Using loc instead of iloc causes errors if index labels are not integers.

Using at or iat requires scalar access, not slicing.

The iloc method selects rows by integer position. Here, df.iloc[0] gets the first row.

2fill in blank

medium

Complete the code to set the 'id' column as the DataFrame index.

Pandas

df = df.[1]('id')

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

Aset_index

Brename

Creset_index

Ddrop

Attempts:

3 left

💡 Hint

Common Mistakes

Using reset_index removes the current index instead of setting a new one.

Using drop or rename does not change the index.

The set_index method sets a column as the DataFrame index.

3fill in blank

hard

Fix the error in the code to select rows where the index is greater than 5.

Pandas

filtered = df[df.index [1] 5]

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

A<=

B==

C>

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' selects only index equal to 5, not greater.

Using '<=' selects index less than or equal to 5.

To filter rows with index greater than 5, use the '>' operator.

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 len(word) [2] 3}

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

Alen(word)

B>

C<

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' filters words shorter than 3 characters.

Using 'word' instead of len(word) gives the word itself, not its length.

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 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() creates lowercase keys instead of uppercase.

Using '<' or '!=' in the condition filters wrong values.

This dictionary comprehension creates keys as uppercase strings and includes only items with values greater than zero.

Practice

1. Why is indexing important in pandas DataFrames?

easy

A. It deletes duplicate rows.

B. It changes the data types of columns.

C. It sorts the DataFrame automatically.

D. It labels rows to find data quickly.

Solution

1. Step 1: Understand the role of indexing

   Indexing assigns labels to rows, making it easier to locate and access data quickly.

2. Step 2: Compare with other options

   Changing data types, sorting, or deleting duplicates are different operations unrelated to indexing.

3. Final Answer:

   It labels rows to find data quickly. -> Option D

4. Quick Check:

   Indexing = labeling rows [OK]

Hint: Indexing helps find rows fast by labeling them [OK]

Common Mistakes:

• Confusing indexing with sorting
• Thinking indexing changes data types
• Assuming indexing deletes duplicates

2. Which of the following is the correct way to set the 'Name' column as the index of a DataFrame df?

easy

A. df.setIndex('Name')

B. df.index('Name')

C. df.set_index('Name')

D. df.index = 'Name'

Solution

1. Step 1: Recall the correct pandas method

   The method to set a column as index is set_index, called as df.set_index('column_name').

2. Step 2: Check other options for syntax errors

   df.index('Name') and df.setIndex('Name') are invalid methods. Assigning df.index = 'Name' sets index to a string, not a column.

3. Final Answer:

   df.set_index('Name') -> Option C

4. Quick Check:

   Use set_index('column') to set index [OK]

Hint: Use df.set_index('column') to set index [OK]

Common Mistakes:

• Using wrong method names like setIndex
• Trying to call df.index as a function
• Assigning a string directly to df.index

3. Given the DataFrame df below:

   Name  Age  Score
0  Alice   25     88
1    Bob   30     92
2  Carol   22     85

What is the output of df.loc[1, 'Score']?

medium

A. 30

B. 92

C. Bob

D. 85

Solution

1. Step 1: Understand loc usage

   loc selects data by label. Here, row label 1 and column 'Score' is requested.

2. Step 2: Find value at row 1, column 'Score'

   Row 1 corresponds to Bob, and his Score is 92.

3. Final Answer:

   92 -> Option B

4. Quick Check:

   df.loc[1, 'Score'] = 92 [OK]

Hint: loc uses row labels, so df.loc[1, 'Score'] = 92 [OK]

Common Mistakes:

• Confusing loc with iloc (position vs label)
• Selecting wrong column value
• Assuming row 1 is index 0

4. What is wrong with this code snippet?

df = pd.DataFrame({'A': [1,2], 'B': [3,4]})
df.set_index('C')

medium

A. Column 'C' does not exist, causing an error.

B. set_index cannot be used on DataFrames.

C. The code will run but not change the index.

D. The DataFrame is empty, so set_index fails.

Solution

1. Step 1: Check if column 'C' exists

   The DataFrame has columns 'A' and 'B' only; 'C' is missing.

2. Step 2: Understand set_index behavior

   Trying to set index to a non-existent column raises a KeyError.

3. Final Answer:

   Column 'C' does not exist, causing an error. -> Option A

4. Quick Check:

   Missing column in set_index causes error [OK]

Hint: Check if column exists before set_index [OK]

Common Mistakes:

• Assuming set_index works without error on missing columns
• Thinking set_index can't be used on DataFrames
• Ignoring error messages

5. You have a DataFrame df with columns 'City', 'Year', and 'Population'. You want to quickly select all rows for the city 'Paris' using indexing. Which steps correctly prepare and select the data?

hard

A. df.set_index('City', inplace=True)\ndf.loc['Paris']

B. df.set_index('Year', inplace=True)\ndf.loc['Paris']

C. df.set_index('Population', inplace=True)\ndf.loc['Paris']

D. df.loc[df['City'] == 'Paris']

Solution

1. Step 1: Set 'City' as the index

   Setting 'City' as index allows direct row selection by city name.

2. Step 2: Use loc to select rows labeled 'Paris'

   With 'City' as index, df.loc['Paris'] returns all rows for Paris efficiently.

3. Final Answer:

   df.set_index('City', inplace=True)\ndf.loc['Paris'] -> Option A

4. Quick Check:

   Index by City, then loc['Paris'] selects rows [OK]

Hint: Set index to City, then use loc['Paris'] to select [OK]

Common Mistakes:

• Setting wrong column as index
• Using loc with non-index column
• Not setting index before loc selection