Date feature extraction in Data Analysis Python - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to convert the 'date' column to datetime format.

Data Analysis Python

import pandas as pd

df = pd.DataFrame({'date': ['2023-01-01', '2023-02-15', '2023-03-20']})
df['date'] = pd.to_datetime(df['date'][1])

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

A()

B[]

C{}

D''

Attempts:

3 left

💡 Hint

Common Mistakes

Using square brackets instead of parentheses.

Forgetting to call the function.

To convert a column to datetime, use pd.to_datetime() with parentheses.

2fill in blank

medium

Complete the code to extract the year from the 'date' column into a new column 'year'.

Data Analysis Python

df['year'] = df['date'].dt.[1]

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

Amonth

Bhour

Cday

Dyear

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'month' or 'day' instead of 'year'.

Forgetting the dt accessor.

The dt.year attribute extracts the year from datetime values.

3fill in blank

hard

Fix the error in extracting the weekday name from the 'date' column.

Data Analysis Python

df['weekday'] = df['date'].dt.[1]()

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

Aweekday

Bweekday_name

Cday_name

Dday

Attempts:

3 left

💡 Hint

Common Mistakes

Using weekday_name() which does not exist.

Using weekday which returns an integer.

The correct method to get the weekday name is dt.day_name().

4fill in blank

hard

Fill both blanks to create a dictionary with words as keys and their lengths as values, but only for words longer than 4 letters.

Data Analysis Python

words = ['apple', 'bat', 'carrot', 'dog', 'elephant']
lengths = { [1] : [2] for word in words if len(word) > 4 }

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

Aword

Blen(word)

Cword.upper()

Dword.lower()

Attempts:

3 left

💡 Hint

Common Mistakes

Using the length as key instead of value.

Not filtering words longer than 4 letters.

The dictionary comprehension uses the word as key and its length as value.

5fill in blank

hard

Fill all three blanks to create a dictionary with uppercase words as keys and their lengths as values, but only for words with length greater than 3.

Data Analysis Python

words = ['sun', 'moon', 'star', 'sky', 'planet']
result = { [1] : [2] for word in words if len(word) [3] 3 }

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

Aword.upper()

Blen(word)

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using lowercase words as keys.

Using wrong comparison operator.

Not using length correctly.

The dictionary comprehension uses uppercase words as keys, their lengths as values, and filters words longer than 3 letters.

Practice

1. What does the dt.year attribute do when applied to a pandas datetime column?

easy

A. Extracts the year part from each date in the column

B. Extracts the day of the week as a number

C. Converts the date to a string format

D. Calculates the difference between dates

Solution

1. Step 1: Understand pandas datetime accessor

   The dt accessor allows extraction of date parts from datetime columns.

2. Step 2: Identify what dt.year extracts

   dt.year extracts the year component from each datetime value.

3. Final Answer:

   Extracts the year part from each date in the column -> Option A

4. Quick Check:

   dt.year = year extraction [OK]

Hint: Remember dt.year gives the year number from dates [OK]

Common Mistakes:

• Confusing dt.year with dt.day or dt.month
• Thinking dt.year converts dates to strings
• Assuming dt.year calculates date differences

2. Which of the following is the correct syntax to extract the month from a pandas datetime column named df['date']?

easy

A. df['date'].month

B. df['date'].dt.month

C. df['date'].get_month()

D. df['date'].extract('month')

Solution

1. Step 1: Recall pandas datetime accessor usage

   To extract date parts, pandas requires the dt accessor before the attribute.

2. Step 2: Identify correct syntax for month extraction

   The correct syntax is df['date'].dt.month. Other options are invalid methods or missing dt.

3. Final Answer:

   df['date'].dt.month -> Option B

4. Quick Check:

   Use dt.month to get month [OK]

Hint: Use .dt before date parts like month or day [OK]

Common Mistakes:

• Omitting .dt accessor
• Using non-existent methods like get_month()
• Trying to extract with extract('month')

3. Given the following code, what will be the output?

import pandas as pd
dates = pd.to_datetime(['2023-01-15', '2023-07-04', '2023-12-31'])
df = pd.DataFrame({'date': dates})
df['day_of_week'] = df['date'].dt.dayofweek
print(df['day_of_week'].tolist())

medium

A. [6, 1, 0]

B. [0, 1, 6]

C. [6, 2, 6]

D. [6, 1, 6]

Solution

1. Step 1: Convert dates to day of week numbers

   Using dt.dayofweek, Monday=0 ... Sunday=6. Dates: 2023-01-15 is Sunday (6), 2023-07-04 is Tuesday (1), 2023-12-31 is Sunday (6).

2. Step 2: List the dayofweek values

   The list is [6, 1, 6].

3. Final Answer:

   [6, 1, 6] -> Option D

4. Quick Check:

   Sunday=6, Tuesday=1, Sunday=6 [OK]

Hint: Remember Monday=0, Sunday=6 for dayofweek [OK]

Common Mistakes:

• Confusing dayofweek with weekday names
• Assuming Sunday=0 instead of 6
• Mixing up dayofweek and day_name()

4. What is wrong with this code snippet to extract the hour from a datetime column df['timestamp']?

df['hour'] = df['timestamp'].hour

medium

A. Missing .dt accessor before .hour

B. The column must be converted to string first

C. The .hour attribute does not exist in pandas

D. The code should use .time.hour instead

Solution

1. Step 1: Check pandas datetime attribute access

   Datetime properties require the dt accessor to access parts like hour.

2. Step 2: Identify missing .dt in code

   The code uses df['timestamp'].hour which is invalid; it should be df['timestamp'].dt.hour.

3. Final Answer:

   Missing .dt accessor before .hour -> Option A

4. Quick Check:

   Use .dt.hour to extract hour [OK]

Hint: Always use .dt before date parts like hour [OK]

Common Mistakes:

• Trying to access .hour directly on Series
• Converting datetime to string unnecessarily
• Using .time.hour which is invalid in pandas

5. You have a pandas DataFrame df with a datetime column 'order_date'. You want to create a new column 'is_weekend' that is True if the order was placed on Saturday or Sunday, and False otherwise. Which code snippet correctly achieves this?

hard

A. df['is_weekend'] = df['order_date'].dt.dayofweek == 6 or 5

B. df['is_weekend'] = df['order_date'].dt.weekday in [5,6]

C. df['is_weekend'] = df['order_date'].dt.dayofweek >= 5

D. df['is_weekend'] = df['order_date'].dt.day_name() in ['Saturday', 'Sunday']

Solution

1. Step 1: Understand dayofweek values for weekend

   In pandas, dt.dayofweek returns 0 for Monday through 6 for Sunday. Weekend days are 5 (Saturday) and 6 (Sunday).

2. Step 2: Check which code correctly marks weekend

   df['is_weekend'] = df['order_date'].dt.dayofweek >= 5 uses df['order_date'].dt.dayofweek >= 5, which is True for 5 and 6, correctly marking weekends. Other options use invalid syntax or incorrect logic.

3. Final Answer:

   df['is_weekend'] = df['order_date'].dt.dayofweek >= 5 -> Option C

4. Quick Check:

   Weekend days are dayofweek >= 5 [OK]

Hint: Weekend days have dayofweek 5 or 6, use >= 5 [OK]

Common Mistakes:

• Using 'in' without vectorized methods
• Using 'or' instead of bitwise operators
• Confusing dayofweek numbers or names