DataFrame as labeled two-dimensional table in Pandas - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a DataFrame from a dictionary.

Pandas

import pandas as pd

data = {'Name': ['Alice', 'Bob'], 'Age': [25, 30]}
df = pd.[1](data)
print(df)

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

ASeries

BPanel

CDataFrame

Darray

Attempts:

3 left

💡 Hint

Common Mistakes

Using Series instead of DataFrame creates a one-dimensional structure.

Using Panel is deprecated and not suitable here.

The DataFrame constructor creates a labeled two-dimensional table from the dictionary.

2fill in blank

medium

Complete the code to select the 'Age' column from the DataFrame.

Pandas

import pandas as pd

data = {'Name': ['Alice', 'Bob'], 'Age': [25, 30]}
df = pd.DataFrame(data)
age_column = df[1]
print(age_column)

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

A[['Age']]

B['Age']

C.Age()

D.Age

Attempts:

3 left

💡 Hint

Common Mistakes

Using double brackets returns a DataFrame, not a Series.

Using dot notation with parentheses is invalid.

Using df['Age'] selects the 'Age' column as a Series.

3fill in blank

hard

Fix the error in the code to access the first row using .iloc.

Pandas

import pandas as pd

data = {'Name': ['Alice', 'Bob'], 'Age': [25, 30]}
df = pd.DataFrame(data)
first_row = df.iloc[1]
print(first_row)

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

A(0)

B0

C[[0]]

D[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of square brackets causes a TypeError.

Using double brackets returns a DataFrame instead of a Series.

The iloc indexer requires square brackets to select rows by position. Use df.iloc[0] to get the first row.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps words to their lengths for words longer than 3 characters.

Pandas

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

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

Alen(word)

Bword > 3

Clen(word) > 3

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself instead of its length.

Comparing the word string directly to a number.

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 lengths for words longer than 3 characters.

Pandas

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

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

Aword.upper()

Blen(word)

Clen(word) > 3

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the original word instead of uppercase.

Not filtering words by length.

The dictionary comprehension uses word.upper() as keys, maps to len(word), and filters words with length greater than 3.

Practice

1.

What is a pandas DataFrame best described as?

easy

A. A one-dimensional list of numbers

B. A type of database server

C. A function to plot graphs

D. A labeled two-dimensional table with rows and columns

Solution

1. Step 1: Understand the structure of a DataFrame

   A DataFrame organizes data in rows and columns with labels for both, like a spreadsheet.

2. Step 2: Compare options to definition

   Only "A labeled two-dimensional table with rows and columns" describes a labeled two-dimensional table, matching the DataFrame concept.

3. Final Answer:

   A labeled two-dimensional table with rows and columns -> Option D

4. Quick Check:

   DataFrame = labeled 2D table [OK]

Hint: Remember: DataFrame = table with row and column labels [OK]

Common Mistakes:

• Confusing DataFrame with a list or array
• Thinking DataFrame is a plotting tool
• Mixing DataFrame with database servers

2.

Which of the following is the correct way to create a pandas DataFrame from a dictionary?

import pandas as pd

# Which line correctly creates a DataFrame?

easy

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

B. df = pd.DataFrame['A': [1, 2], 'B': [3, 4]]

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

D. df = pd.DataFrame({'A': 1, 2, 'B': 3, 4})

Solution

1. Step 1: Check dictionary syntax for DataFrame creation

   The dictionary keys are column names and values are lists of column data. Correct syntax uses curly braces and colons.

2. Step 2: Identify correct pandas DataFrame constructor usage

   df = pd.DataFrame({'A': [1, 2], 'B': [3, 4]}) uses pd.DataFrame with a dictionary correctly. Others have syntax errors or wrong brackets.

3. Final Answer:

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

4. Quick Check:

   Use pd.DataFrame with dict and colons [OK]

Hint: Use curly braces and colons inside pd.DataFrame() [OK]

Common Mistakes:

• Using square brackets instead of parentheses
• Using '=' instead of ':' in dictionary
• Passing values without list brackets

3.

What will be the output of the following code?

import pandas as pd

data = {'Name': ['Anna', 'Bob'], 'Age': [25, 30]}
df = pd.DataFrame(data)
print(df.loc[1, 'Name'])

medium

A. Bob

B. Anna

C. 25

D. KeyError

Solution

1. Step 1: Understand DataFrame creation and indexing

   The DataFrame has two rows with index 0 and 1. 'Name' column has ['Anna', 'Bob'].

2. Step 2: Access value at row index 1 and column 'Name'

   df.loc[1, 'Name'] selects row 1, column 'Name', which is 'Bob'.

3. Final Answer:

   Bob -> Option A

4. Quick Check:

   df.loc[1, 'Name'] = Bob [OK]

Hint: loc uses row label and column name to get value [OK]

Common Mistakes:

• Confusing loc with iloc (positional index)
• Expecting row 1 to be 'Anna'
• Mistaking column name for index

4.

Identify the error in this code snippet that tries to create a DataFrame:

import pandas as pd

data = {'X': [1, 2], 'Y': [3]}
df = pd.DataFrame(data)
print(df)

medium

A. The value for key 'Y' has the wrong length (must match 'X')

B. The dictionary keys must be integers

C. pd.DataFrame cannot be created from a dictionary

D. Missing parentheses in print statement

Solution

1. Step 1: Check dictionary values for DataFrame columns

   Each column value must be a list or array of equal length. 'Y': [3] has length 1, 'X' has length 2, which is invalid.

2. Step 2: Identify correct error cause

   The value for key 'Y' has the wrong length (must match 'X') correctly points out the length mismatch.

3. Final Answer:

   The value for key 'Y' has the wrong length (must match 'X') -> Option A

4. Quick Check:

   All column values must have equal length [OK]

Hint: All DataFrame columns need list-like values of equal length [OK]

Common Mistakes:

• Using lists of unequal lengths for columns
• Thinking dictionary keys must be numbers
• Assuming print needs parentheses in Python 3

5.

You have this nested dictionary representing sales data:

sales = {
  'Store1': {'Jan': 100, 'Feb': 120},
  'Store2': {'Jan': 90, 'Feb': 110}
}

How can you create a DataFrame where rows are months and columns are stores?

hard

A. df = pd.DataFrame(sales).T

B. df = pd.DataFrame.from_dict(sales, orient='index').T

C. df = pd.DataFrame(sales, orient='index')

D. df = pd.DataFrame.from_dict(sales, orient='index')

Solution

1. Step 1: Understand the nested dictionary structure

   Keys are stores, values are dicts of months and sales. We want months as rows, stores as columns.

2. Step 2: Use from_dict with orient='index' and transpose

   from_dict with orient='index' treats outer keys as rows. Transpose (T) swaps rows and columns to get months as rows.

3. Final Answer:

   df = pd.DataFrame.from_dict(sales, orient='index').T -> Option B

4. Quick Check:

   Use from_dict with orient='index' + transpose [OK]

Hint: Use from_dict with orient='index' then transpose [OK]

Common Mistakes:

• Not transposing after from_dict
• Using wrong orient parameter
• Assuming pd.DataFrame(sales) arranges data correctly