Date and time feature extraction in ML Python - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to extract the year from a datetime column in pandas.

ML Python

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

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

Aday

Byear

Chour

Dmonth

Attempts:

3 left

💡 Hint

Common Mistakes

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

Forgetting to use the .dt accessor before the attribute.

The dt.year attribute extracts the year part from a datetime column in pandas.

2fill in blank

medium

Complete the code to extract the weekday name from a datetime column in pandas.

ML Python

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

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

Aday_name

Bweekday_name

Cday

Dweekday

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'weekday' which returns an integer instead of the name.

Using 'weekday_name' which is deprecated.

The dt.day_name() method returns the weekday name (like Monday, Tuesday) from a datetime column.

3fill in blank

hard

Fix the error in the code to extract the hour from a datetime column.

ML Python

df['hour'] = df['date_column'].[1].hour

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

Atime

Bdatetime

Cdt

Ddate

Attempts:

3 left

💡 Hint

Common Mistakes

Trying to access .hour directly without .dt.

Using 'datetime' or 'time' which are not pandas accessors.

The .dt accessor is needed to access datetime properties like hour in pandas Series.

4fill in blank

hard

Fill both blanks to create a new column with the month and day extracted from a datetime column.

ML Python

df['month_day'] = df['date_column'].dt.[1].astype(str) + '-' + df['date_column'].dt.[2].astype(str)

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

Amonth

Bday

Chour

Dyear

Attempts:

3 left

💡 Hint

Common Mistakes

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

Not converting numbers to strings before concatenation.

Extracting month and day and converting them to strings allows combining them into a 'month-day' format.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps each date to its weekday number if the hour is greater than 12.

ML Python

result = {date: date.[1] for date in df['date_column'] if date.[2] [3] 12}

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

Aweekday

Bhour

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' in the comparison.

Using 'day' instead of 'weekday' for the first blank.

Using 'date' or 'time' instead of 'dt' accessor (not shown here but common).

The dictionary comprehension extracts the weekday number for dates where the hour is greater than 12.

Practice

1. Which of the following is a common feature extracted from a date to help machine learning models?

easy

A. Font size

B. Color

C. Month

D. Temperature

Solution

1. Step 1: Understand date features

   Date features include parts of a date like year, month, day, hour, and weekday.

2. Step 2: Identify relevant feature

   Among the options, only 'Month' is a part of a date and useful for models.

3. Final Answer:

   Month -> Option C

4. Quick Check:

   Date feature = Month [OK]

Hint: Pick the option that relates directly to date parts [OK]

Common Mistakes:

• Choosing unrelated features like color or font size
• Confusing date features with unrelated data

2. Which Python code correctly extracts the weekday from a pandas datetime column named 'date'?

easy

A. df['weekday'] = df['date'].dt.weekday

B. df['weekday'] = df['date'].weekday()

C. df['weekday'] = df['date'].weekday

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

Solution

1. Step 1: Recall pandas datetime accessor

   To extract weekday, use the .dt accessor followed by .weekday without parentheses.

2. Step 2: Check each option

   df['weekday'] = df['date'].dt.weekday uses .dt.weekday correctly. df['weekday'] = df['date'].weekday() calls weekday() directly on the series, which is invalid. df['weekday'] = df['date'].weekday misses .dt. df['weekday'] = df['date'].dt.weekday() incorrectly uses parentheses after .weekday.

3. Final Answer:

   df['weekday'] = df['date'].dt.weekday -> Option A

4. Quick Check:

   Use .dt.weekday without parentheses [OK]

Hint: Use .dt.weekday without parentheses for pandas datetime [OK]

Common Mistakes:

• Calling weekday() as a method on series
• Missing .dt accessor
• Adding parentheses after .weekday

3. Given the code:

import pandas as pd
df = pd.DataFrame({'date': pd.to_datetime(['2024-06-01 14:30', '2024-06-02 09:15'])})
df['hour'] = df['date'].dt.hour
df['is_weekend'] = df['date'].dt.weekday >= 5
print(df[['hour', 'is_weekend']].to_dict())

What is the printed output?

medium

A. {'hour': {0: 14, 1: 9}, 'is_weekend': {0: False, 1: False}}

B. {'hour': {0: 14, 1: 9}, 'is_weekend': {0: True, 1: True}}

C. {'hour': {0: 14, 1: 9}, 'is_weekend': {0: False, 1: True}}

D. SyntaxError

Solution

1. Step 1: Extract hour values

   The first date has hour 14, second has hour 9, so 'hour' column is {0:14, 1:9}.

2. Step 2: Determine weekend flags

   Weekday 5 and 6 are weekend. Dates are 2024-06-01 (Saturday=5) and 2024-06-02 (Sunday=6). Both are weekend, so 'is_weekend' should be True for both.

3. Step 3: Check code logic

   Code uses df['date'].dt.weekday >= 5, which is True for both dates. So 'is_weekend' is {0: True, 1: True}.

4. Final Answer:

   {'hour': {0: 14, 1: 9}, 'is_weekend': {0: True, 1: True}} -> Option B

5. Quick Check:

   Weekend days are 5 or 6, both dates match [OK]

Hint: Check weekday numbers: 5=Saturday, 6=Sunday for weekend [OK]

Common Mistakes:

• Assuming weekend is false for Saturday/Sunday
• Mixing hour extraction with weekend logic
• Misreading weekday numbers

4. The following code aims to add a 'month' feature from a datetime column but throws an error:

df['month'] = df['date'].month

What is the error and how to fix it?

medium

A. AttributeError because .month must be accessed via .dt; fix: df['date'].dt.month

B. SyntaxError due to missing parentheses; fix: df['date'].month()

C. TypeError because 'date' is not datetime; fix: convert to datetime first

D. No error; code is correct

Solution

1. Step 1: Understand pandas datetime access

   Datetime properties like month must be accessed with .dt when working on a pandas Series.

2. Step 2: Identify error cause

   Using df['date'].month tries to get 'month' attribute of the Series, causing AttributeError.

3. Step 3: Correct code

   Use df['date'].dt.month to extract month correctly.

4. Final Answer:

   AttributeError because .month must be accessed via .dt; fix: df['date'].dt.month -> Option A

5. Quick Check:

   Use .dt.month for pandas datetime columns [OK]

Hint: Always use .dt before datetime properties on pandas Series [OK]

Common Mistakes:

• Missing .dt accessor
• Trying to call .month() as a method
• Not converting column to datetime type

5. You have a dataset with a datetime column 'timestamp'. You want to create a feature that is 1 if the time is during business hours (9am to 5pm) on weekdays, else 0. Which code correctly creates this feature?

hard

A. df['business_hours'] = ((df['timestamp'].dt.hour > 9) & (df['timestamp'].dt.hour <= 17) & (df['timestamp'].dt.weekday <= 5)).astype(int)

B. df['business_hours'] = ((df['timestamp'].dt.hour > 9) & (df['timestamp'].dt.hour < 17) & (df['timestamp'].dt.weekday < 5)).astype(int)

C. df['business_hours'] = ((df['timestamp'].dt.hour >= 9) & (df['timestamp'].dt.hour <= 17) & (df['timestamp'].dt.weekday <= 5)).astype(int)

D. df['business_hours'] = ((df['timestamp'].dt.hour >= 9) & (df['timestamp'].dt.hour < 17) & (df['timestamp'].dt.weekday < 5)).astype(int)

Solution

1. Step 1: Define business hours range

   Business hours are from 9:00 (inclusive) to 17:00 (exclusive), so hour >= 9 and hour < 17.

2. Step 2: Define weekdays

   Weekdays are Monday (0) to Friday (4), so weekday < 5.

3. Step 3: Combine conditions and convert to int

   Use logical AND (&) to combine conditions and convert boolean to int with .astype(int).

4. Final Answer:

   df['business_hours'] = ((df['timestamp'].dt.hour >= 9) & (df['timestamp'].dt.hour < 17) & (df['timestamp'].dt.weekday < 5)).astype(int) -> Option D

5. Quick Check:

   Use inclusive start, exclusive end for hours and weekday < 5 [OK]

Hint: Use >=9 and <17 for hours, weekday <5 for Mon-Fri [OK]

Common Mistakes:

• Using >9 instead of >=9
• Including weekend days by using <=5
• Using <=17 instead of <17