Why indexing matters in Pandas - Challenge Your Understanding

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of selecting rows with index labels

What is the output of this code snippet?

import pandas as pd

df = pd.DataFrame({
    'A': [10, 20, 30],
    'B': [100, 200, 300]
}, index=['x', 'y', 'z'])

result = df.loc[['y', 'z'], 'A']
print(result)

Pandas

import pandas as pd

df = pd.DataFrame({
    'A': [10, 20, 30],
    'B': [100, 200, 300]
}, index=['x', 'y', 'z'])

result = df.loc[['y', 'z'], 'A']
print(result)

A KeyError: "['y', 'z'] not in index"

B

0    20
1    30
Name: A, dtype: int64

C

x    10
y    20
Name: A, dtype: int64

D

y    20
z    30
Name: A, dtype: int64

💡 Hint

Remember that loc uses the index labels, not integer positions.

The loc method selects rows by their index labels. Since the DataFrame has index labels 'x', 'y', 'z', selecting ['y', 'z'] returns the rows with those labels. The column 'A' values for 'y' and 'z' are 20 and 30 respectively.

❓ data_output

intermediate

Number of rows after resetting index

Given this DataFrame, what is the number of rows after resetting the index?

import pandas as pd

df = pd.DataFrame({
    'score': [88, 92, 95],
    'grade': ['B', 'A', 'A']
}, index=['s1', 's2', 's3'])

new_df = df.reset_index()
print(len(new_df))

Pandas

import pandas as pd

df = pd.DataFrame({
    'score': [88, 92, 95],
    'grade': ['B', 'A', 'A']
}, index=['s1', 's2', 's3'])

new_df = df.reset_index()
print(len(new_df))

A 4

B 3

C 2

D KeyError

💡 Hint

Resetting the index adds the old index as a column but does not change the number of rows.

The reset_index() method moves the index into a column and creates a new default integer index. The number of rows remains the same, which is 3.

🔧 Debug

advanced

Why does this code raise a KeyError?

Consider this code:

import pandas as pd

df = pd.DataFrame({
    'value': [5, 10, 15]
}, index=[1, 2, 3])

print(df.loc[0])

Why does it raise a KeyError?

Pandas

import pandas as pd

df = pd.DataFrame({
    'value': [5, 10, 15]
}, index=[1, 2, 3])

print(df.loc[0])

A Because the DataFrame has no columns named 0

B Because loc only accepts column names, not index labels

C Because 0 is not in the DataFrame's index labels

D Because loc requires integer positions, not labels

💡 Hint

Check the index labels of the DataFrame and what loc expects.

The DataFrame's index labels are [1, 2, 3]. Using loc[0] tries to access the row with label 0, which does not exist, causing a KeyError.

🚀 Application

advanced

Selecting rows efficiently with index

You have a large DataFrame with a unique index of user IDs. Which method is the fastest to select multiple users by their IDs?

A Use <code>df.loc[list_of_user_ids]</code> to select rows by index labels

B Use <code>df.iloc[list_of_user_ids]</code> to select rows by integer position

C Use <code>df[df['user_id'].isin(list_of_user_ids)]</code> to filter rows

D Use <code>df.query('user_id in @list_of_user_ids')</code> to filter rows

💡 Hint

Think about how pandas uses the index for fast lookups.

Using df.loc[list_of_user_ids] leverages the index for fast direct access. The other methods involve filtering or positional indexing, which are slower for large data.

🧠 Conceptual

expert

Why setting an index improves join performance

Why does setting a column as the index improve the performance of joining two DataFrames on that column?

A Because indexes are implemented as hash tables or trees, allowing faster lookups during joins

B Because setting an index compresses the data, reducing memory usage during joins

C Because setting an index sorts the data, and joins require sorted data

D Because joins only work on index columns, so setting the index is mandatory

💡 Hint

Think about how indexes help find matching rows quickly.

Indexes in pandas are often implemented using hash tables or balanced trees, which allow quick lookup of matching keys during joins. This speeds up the join operation compared to scanning all rows.