Inplace operations consideration in Pandas - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Effect of inplace=True on DataFrame drop

What is the output of the following code snippet?

Pandas

import pandas as pd

df = pd.DataFrame({'A': [1, 2, 3], 'B': [4, 5, 6]})
df.drop('A', axis=1, inplace=True)
print(df)

A

   B
0  4
1  5
2  6

B

   A  B
0  1  4
1  2  5
2  3  6

C

   B
0  4
1  5
2  6
   A  B
0  1  4
1  2  5
2  3  6

D KeyError: 'A'

💡 Hint

inplace=True modifies the original DataFrame directly.

Using inplace=True drops column 'A' from df directly, so printing df shows only column 'B'.

❓ Predict Output

intermediate

Return value of inplace operation

What does the variable 'result' contain after running this code?

Pandas

import pandas as pd

df = pd.DataFrame({'X': [10, 20, 30]})
result = df.drop('X', axis=1, inplace=True)
print(result)

A KeyError

B Empty DataFrame with no columns

C DataFrame with column 'X'

D None

💡 Hint

inplace operations return None.

When inplace=True is used, the method returns None instead of a new DataFrame.

❓ data_output

advanced

Effect of inplace=False on DataFrame sort

What is the output of the following code snippet?

Pandas

import pandas as pd

df = pd.DataFrame({'A': [3, 1, 2]})
sorted_df = df.sort_values('A', inplace=False)
print(df)
print(sorted_df)

A

   A
0  3
1  1
2  2
   A
1  1
2  2
0  3

B TypeError

C

   A
0  3
1  1
2  2
   A
0  3
1  1
2  2

D

   A
1  1
2  2
0  3
   A
1  1
2  2
0  3

💡 Hint

inplace=False returns a new sorted DataFrame, original stays same.

df remains unsorted, sorted_df is a new DataFrame sorted by column 'A'.

🔧 Debug

advanced

Why does this inplace operation not change the DataFrame?

Consider this code: import pandas as pd df = pd.DataFrame({'A': [1, 2, 3]}) df.drop('A', axis=1) print(df) Why does df still have column 'A' after drop?

Pandas

import pandas as pd

df = pd.DataFrame({'A': [1, 2, 3]})
df.drop('A', axis=1)
print(df)

A Because print(df) is called before drop.

B Because drop without inplace=True returns a new DataFrame and does not modify df.

C Because column 'A' does not exist in df.

D Because axis=1 is incorrect and drop did nothing.

💡 Hint

Check if drop modifies df or returns a new object.

drop returns a new DataFrame by default; df remains unchanged unless inplace=True is used.

🚀 Application

expert

Choosing inplace vs non-inplace for chained operations

You want to clean a DataFrame by dropping column 'B' and then sorting by column 'A'. Which code snippet correctly performs both operations and keeps the changes?

A

df.drop('B', axis=1)
df.sort_values('A')
print(df)

B

df = df.drop('B', axis=1, inplace=True)
df = df.sort_values('A', inplace=True)
print(df)

C

df.drop('B', axis=1, inplace=True)
df.sort_values('A', inplace=True)
print(df)

D

df = df.drop('B', axis=1)
df = df.sort_values('A')
print(df)

💡 Hint

inplace=True modifies df directly; assigning the result of inplace=True methods sets df to None.

Option C modifies df inplace for both operations, so df reflects both changes. Option C assigns None to df causing errors. Option C does not modify df. Option C correctly assigns new DataFrames but is not the inplace approach.