Creating new columns in Pandas - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to add a new column 'total' as the sum of columns 'a' and 'b'.

Pandas

import pandas as pd
df = pd.DataFrame({'a': [1, 2], 'b': [3, 4]})
df['total'] = df['a'] [1] df['b']
print(df)

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

A/

B-

C*

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' instead of '+' causes subtraction.

Using '*' or '/' will multiply or divide instead of adding.

Use the plus sign + to add the two columns element-wise.

2fill in blank

medium

Complete the code to create a new column 'is_high' that is True if 'score' is greater than 50.

Pandas

import pandas as pd
df = pd.DataFrame({'score': [45, 60, 30]})
df['is_high'] = df['score'] [1] 50
print(df)

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

A>

B<

C==

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' will check for less than, which is incorrect here.

Using '==' checks for equality, not greater than.

The operator > checks if values are greater than 50.

3fill in blank

hard

Fix the error in the code to create a new column 'double' that doubles the values in 'num'.

Pandas

import pandas as pd
df = pd.DataFrame({'num': [1, 2, 3]})
df['double'] = df['num'] [1] 2
print(df)

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

A*

B+

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' adds 2 instead of doubling.

Using '-' or '/' will not double the values.

Use the multiplication operator * to double the values.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps words to their lengths only if length is greater than 3.

Pandas

words = ['apple', 'bat', 'carrot', 'dog']
lengths = {word: [1] for word in words if [2] > 3}
print(lengths)

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

Alen(word)

Bword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself as value instead of its length.

Checking the word instead of its length in the condition.

Use len(word) to get the length and filter with len(word) > 3.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps uppercase words to their scores only if score is above 50.

Pandas

data = {'apple': 40, 'banana': 60, 'cherry': 70}
result = { [1]: [2] for [3], score in data.items() if score > 50 }
print(result)

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

Aword.upper()

Bscore

Cword

Dfruit

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names in the loop.

Not converting words to uppercase for keys.

Use word.upper() as key, score as value, and word as the loop variable.

Practice

1. What is the correct way to add a new column named age + 10 to a DataFrame df that adds 10 to the existing age column?

easy

A. df['age + 10'] == df['age'] + 10

B. df['age + 10'] = df['age'] + 10

C. df.add_column('age + 10', df['age'] + 10)

D. df.age + 10 = df.age + 10

Solution

1. Step 1: Understand how to create a new column in pandas

   In pandas, new columns are created by assigning a value to df['new_column_name'].

2. Step 2: Apply the addition operation to the existing column

   Adding 10 to the age column is done by df['age'] + 10. Assigning this to df['age + 10'] creates the new column.

3. Final Answer:

   df['age + 10'] = df['age'] + 10 -> Option B

4. Quick Check:

   New column assignment = df['age + 10'] = df['age'] + 10 [OK]

Hint: Use df['new_col'] = expression to add columns [OK]

Common Mistakes:

• Using double equals (==) instead of single equals (=) for assignment
• Trying to use a method like add_column which doesn't exist
• Using dot notation for new columns with spaces or special characters

2. Which of the following is the correct syntax to create a new column full_name by combining first_name and last_name columns with a space in between in DataFrame df?

easy

A. df['full_name'] = df['first_name'] + ' ' + df['last_name']

B. df['full_name'] = df['first_name'].append(' ').append(df['last_name'])

C. df['full_name'] = df['first_name'].join(' ', df['last_name'])

D. df['full_name'] = df['first_name'] & ' ' & df['last_name']

Solution

1. Step 1: Combine string columns with + operator

   In pandas, string columns can be combined using the + operator. To add a space, include ' ' between them.

2. Step 2: Assign the combined string to a new column

   Assign the result to df['full_name'] to create the new column.

3. Final Answer:

   df['full_name'] = df['first_name'] + ' ' + df['last_name'] -> Option A

4. Quick Check:

   String concat with + = df['full_name'] = df['first_name'] + ' ' + df['last_name'] [OK]

Hint: Use + to join string columns with spaces [OK]

Common Mistakes:

• Using append or join methods which don't work on pandas Series
• Using & operator which is for bitwise operations
• Forgetting to add space between names

3. Given the DataFrame df below:

   price  quantity
0    100         2
1    150         3
2    200         1

What will be the output of this code?

df['total'] = df['price'] * df['quantity']
df['total']

medium

A. [200, 450, 200]

B. [100, 150, 200]

C. [2, 3, 1]

D. [300, 450, 600]

Solution

1. Step 1: Multiply price and quantity element-wise

   Multiply each row's price by quantity: 100*2=200, 150*3=450, 200*1=200.

2. Step 2: Assign result to new column and output it

   The new column 'total' will have values [200, 450, 200]. Printing df['total'] shows this list.

3. Final Answer:

   [200, 450, 200] -> Option A

4. Quick Check:

   Element-wise multiply = [200, 450, 200] [OK]

Hint: Multiply columns element-wise for new column [OK]

Common Mistakes:

• Adding instead of multiplying columns
• Confusing row-wise with column-wise operations
• Expecting a scalar instead of a Series output

4. What is wrong with this code that tries to create a new column discounted_price by subtracting 10 from price in DataFrame df?

df.discounted_price = df['price'] - 10

medium

A. It will fail if the column name has spaces or special characters.

B. It will cause a syntax error because of dot notation.

C. It will create a new variable, not a DataFrame column.

D. It will create the column correctly.

Solution

1. Step 1: Understand dot notation for DataFrame columns

   Dot notation df.col = value works to create new columns only if 'col' is a valid Python identifier (no spaces, no special characters).

2. Step 2: Check the column name 'discounted_price'

   'discounted_price' consists of letters and underscores, which is valid. Therefore, the code creates the column correctly.

3. Final Answer:

   It will create the column correctly. -> Option D

4. Quick Check:

   Dot notation works for valid identifiers [OK]

Hint: Use df['col'] if column name has spaces or special chars [OK]

Common Mistakes:

• Thinking dot notation fails for valid column names
• Confusing DataFrame column assignment with local variable creation
• Believing underscores invalidate column names for dot notation

5. You have a DataFrame df with columns score and max_score. You want to create a new column percentage that shows the score as a percentage of max_score, but if max_score is zero, set percentage to None. Which code correctly does this?

hard

A. df['percentage'] = df['score'] / df['max_score'].replace(0, None) * 100

B. df.loc[df['max_score'] == 0, 'percentage'] = None
df['percentage'] = df['score'] / df['max_score'] * 100

C. df['percentage'] = df.apply(lambda row: row['score'] / row['max_score'] * 100 if row['max_score'] != 0 else None, axis=1)

D. df['percentage'] = df['score'] / df['max_score'] * 100
df['percentage'][df['max_score'] == 0] = None

Solution

1. Step 1: Use row-wise function to handle zero max_score

   Using df.apply with a lambda lets us check each row and apply the condition: if max_score is zero, return None; else calculate percentage.

2. Step 2: Assign the result to new column 'percentage'

   The lambda returns the correct value per row, and axis=1 ensures row-wise operation.

3. Final Answer:

   df['percentage'] = df.apply(lambda row: row['score'] / row['max_score'] * 100 if row['max_score'] != 0 else None, axis=1) -> Option C

4. Quick Check:

   Use apply with condition for safe division [OK]

Hint: Use apply with lambda for row-wise conditional columns [OK]

Common Mistakes:

• Replacing zero with None in Series causes errors
• Using chained indexing which raises warnings
• Not handling division by zero properly