Date feature extraction in Data Analysis Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Extracting Month and Day from Dates

What is the output of the following code that extracts the month and day from a pandas datetime column?

Data Analysis Python

import pandas as pd

df = pd.DataFrame({'date': pd.to_datetime(['2023-01-15', '2023-07-04', '2023-12-31'])})
df['month'] = df['date'].dt.month
df['day'] = df['date'].dt.day
print(df[['month', 'day']].to_dict(orient='list'))

A {"month": [1, 7, 12], "day": [15, 4, 31]}

B {"month": [15, 4, 31], "day": [1, 7, 12]}

C {"month": [2023, 2023, 2023], "day": [15, 4, 31]}

D {"month": [1, 7, 12], "day": [2023, 2023, 2023]}

💡 Hint

Look at how pandas datetime properties .dt.month and .dt.day extract parts of the date.

The .dt accessor allows extraction of datetime components. Here, month and day are extracted correctly from each date.

❓ data_output

intermediate

Counting Weekdays in a Date Range

Given a date range, how many Mondays are there in the DataFrame after extracting the weekday?

Data Analysis Python

import pandas as pd

dates = pd.date_range(start='2023-01-01', end='2023-01-31')
df = pd.DataFrame({'date': dates})
df['weekday'] = df['date'].dt.day_name()
monday_count = df[df['weekday'] == 'Monday'].shape[0]
print(monday_count)

A 3

B 4

C 6

D 5

💡 Hint

Count how many Mondays fall between Jan 1 and Jan 31, 2023.

January 2023 has Mondays on 2nd, 9th, 16th, 23rd, and 30th, totaling 5.

❓ visualization

advanced

Plotting Hourly Distribution of Events

Which option produces the correct bar plot showing counts of events per hour from a datetime column?

Data Analysis Python

import pandas as pd
import matplotlib.pyplot as plt

# Sample data
timestamps = pd.to_datetime([
    '2023-06-01 08:15', '2023-06-01 08:45', '2023-06-01 09:00',
    '2023-06-01 10:30', '2023-06-01 10:45', '2023-06-01 10:50'
])
df = pd.DataFrame({'timestamp': timestamps})
df['hour'] = df['timestamp'].dt.hour
counts = df['hour'].value_counts().sort_index()

plt.bar(counts.index, counts.values)
plt.xlabel('Hour of Day')
plt.ylabel('Event Count')
plt.title('Events per Hour')
plt.show()

A A line chart showing event counts over hours 8 to 10.

B A bar chart with bars at hours 8, 9, and 10 with heights 2, 1, and 3 respectively.

C A pie chart showing percentage of events per hour.

D A scatter plot with hours on x-axis and event counts on y-axis.

💡 Hint

Look for a bar chart with counts grouped by hour.

The code groups events by hour and plots a bar chart with counts 2 at 8, 1 at 9, and 3 at 10.

🔧 Debug

advanced

Identifying Error in Date Extraction Code

What error does the following code raise when trying to extract the week number from a datetime column?

Data Analysis Python

import pandas as pd

df = pd.DataFrame({'date': pd.to_datetime(['2023-01-01', '2023-06-15'])})
df['week_num'] = df['date'].dt.week
print(df)

A KeyError: 'week_num'

B TypeError: dt.week is not callable

C AttributeError: 'DatetimeProperties' object has no attribute 'week'

D No error, prints week numbers correctly

💡 Hint

Check if .dt.week is a valid attribute in pandas datetime accessor.

The .dt.week attribute was deprecated and removed; use .dt.isocalendar().week instead.

🚀 Application

expert

Calculating Age from Birthdate Column

Given a DataFrame with a 'birthdate' column, which option correctly calculates the age in years as of today?

Data Analysis Python

import pandas as pd
from datetime import datetime

df = pd.DataFrame({'birthdate': pd.to_datetime(['1990-05-15', '2000-12-01', '1985-07-30'])})

# Calculate age here

print(df['age'].tolist())

A df['age'] = (pd.to_datetime('today') - df['birthdate']).dt.days // 365

B df['age'] = datetime.now().year - df['birthdate'].dt.year

C df['age'] = (datetime.now() - df['birthdate']).days // 365

D df['age'] = (pd.Timestamp(datetime.now()) - df['birthdate']).dt.days // 365

💡 Hint

Use pandas datetime subtraction and .dt.days to get days difference.

Option A correctly uses pandas Timestamp subtraction and .dt.days to get days difference, then divides by 365 for age.

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