Date arithmetic (Timedelta) in Data Analysis Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Calculate the difference in days between two dates

What is the output of this code snippet that calculates the difference in days between two dates?

Data Analysis Python

import pandas as pd
start_date = pd.to_datetime('2024-01-01')
end_date = pd.to_datetime('2024-01-15')
diff = end_date - start_date
print(diff.days)

A 15

B 14

C 13

D Error

💡 Hint

Remember that the difference between dates counts full days between them.

The difference between January 15 and January 1 is 14 full days, so diff.days returns 14.

❓ data_output

intermediate

Add a timedelta to a date

What is the resulting date after adding 10 days to '2024-03-10'?

Data Analysis Python

import pandas as pd
initial_date = pd.to_datetime('2024-03-10')
new_date = initial_date + pd.Timedelta(days=10)
print(new_date.strftime('%Y-%m-%d'))

A 2024-03-20

B 2024-03-10

C 2024-03-21

D 2024-03-19

💡 Hint

Adding 10 days means moving forward 10 calendar days.

Adding 10 days to March 10 results in March 20, 2024.

❓ visualization

advanced

Plotting date ranges with timedelta

Which option correctly creates a plot showing dates from '2024-01-01' to '2024-01-10' with daily intervals?

Data Analysis Python

import pandas as pd
import matplotlib.pyplot as plt
dates = pd.date_range(start='2024-01-01', end='2024-01-10')
values = range(len(dates))
plt.plot(dates, values)
plt.xlabel('Date')
plt.ylabel('Value')
plt.title('Date Range Plot')
plt.show()

A A scatter plot with dates missing Jan 5

B A bar plot with dates on y-axis and values on x-axis

C A line plot with dates on x-axis from Jan 1 to Jan 10 and values 0 to 9 on y-axis

D An empty plot with no data

💡 Hint

Check if the dates cover the full range and if the plot type matches the code.

The code creates a line plot with dates from Jan 1 to Jan 10 on the x-axis and values 0 to 9 on the y-axis.

🧠 Conceptual

advanced

Understanding timedelta components

Given a timedelta of 3 days, 4 hours, and 30 minutes, what is the total number of seconds?

Data Analysis Python

from datetime import timedelta
td = timedelta(days=3, hours=4, minutes=30)
total_seconds = td.total_seconds()
print(int(total_seconds))

A 275400

B 259200

C 302400

D 273000

💡 Hint

Calculate seconds for days, hours, and minutes separately then add.

3 days = 259200 seconds, 4 hours = 14400 seconds, 30 minutes = 1800 seconds; total = 259200 + 14400 + 1800 = 275400 seconds.

🔧 Debug

expert

Identify the error in timedelta subtraction

What error does this code raise when subtracting a timedelta from a string date?

Data Analysis Python

from datetime import timedelta
start = '2024-05-01'
result = start - timedelta(days=5)
print(result)

A AttributeError: 'str' object has no attribute 'days'

B ValueError: invalid literal for int() with base 10

C No error, prints '2024-04-26'

D TypeError: unsupported operand type(s) for -: 'str' and 'datetime.timedelta'

💡 Hint

Check the data types involved in the subtraction.

You cannot subtract a timedelta directly from a string; the string must be converted to a date first.

Practice

1. What does the timedelta object represent in Python's date arithmetic?

easy

A. A method to convert strings to dates

B. A specific date and time value

C. A function to format dates

D. A duration or difference between two dates or times

Solution

1. Step 1: Understand the purpose of timedelta

   The timedelta object is used to represent a span of time, such as days, seconds, or microseconds.

2. Step 2: Differentiate timedelta from other date/time objects

   Unlike a specific date or time, timedelta measures the difference or duration between two dates or times.

3. Final Answer:

   A duration or difference between two dates or times -> Option D

4. Quick Check:

   timedelta = duration/difference [OK]

Hint: timedelta means time difference or duration [OK]

Common Mistakes:

• Confusing timedelta with a specific date
• Thinking timedelta formats dates
• Assuming timedelta converts strings

2. Which of the following is the correct way to create a timedelta of 5 days in Python?

easy

A. timedelta(5 days)

B. timedelta(seconds=5)

C. timedelta(days=5)

D. timedelta(day=5)

Solution

1. Step 1: Recall timedelta constructor syntax

   The timedelta constructor accepts keyword arguments like days, seconds, etc.

2. Step 2: Identify the correct keyword and format

   timedelta(5 days): syntax error ('days' undefined). timedelta(seconds=5): 5 seconds not days. timedelta(day=5): invalid keyword 'day'. timedelta(days=5): correct days=5.

3. Final Answer:

   timedelta(days=5) -> Option C

4. Quick Check:

   Use keyword days=5 for 5 days [OK]

Hint: Use keyword days=number to create timedelta days [OK]

Common Mistakes:

• Using 'days' without quotes causing NameError
• Using seconds=5 instead of days=5
• Misspelling 'days' as 'day'

3. What will be the output of this code?

from datetime import datetime, timedelta
start = datetime(2024, 6, 10)
new_date = start + timedelta(days=7)
print(new_date.strftime('%Y-%m-%d'))

medium

A. 2024-06-17

B. 2024-06-03

C. 2024-07-10

D. 2024-06-10

Solution

1. Step 1: Understand the initial date

   The variable start is June 10, 2024.

2. Step 2: Add 7 days using timedelta

   Adding 7 days to June 10 results in June 17, 2024.

3. Step 3: Format the date output

   The strftime formats the date as 'YYYY-MM-DD', so output is '2024-06-17'.

4. Final Answer:

   2024-06-17 -> Option A

5. Quick Check:

   June 10 + 7 days = June 17 [OK]

Hint: Add days to date, then format output string [OK]

Common Mistakes:

• Subtracting days instead of adding
• Misreading date format
• Confusing month and day positions

4. Identify the error in this code snippet:

from datetime import datetime, timedelta
start = datetime(2024, 6, 10)
new_date = start + timedelta(day=7)
print(new_date)

medium

A. timedelta uses invalid keyword 'day'

B. datetime object cannot be added to timedelta

C. print statement syntax is incorrect

D. datetime constructor missing time parameters

Solution

1. Step 1: Check timedelta constructor usage

   The timedelta constructor requires keyword arguments like days=7; day=7 is invalid keyword causing TypeError.

2. Step 2: Verify other parts of code

   Adding timedelta to datetime is valid, print syntax is correct, and time parameters are optional.

3. Final Answer:

   timedelta uses invalid keyword 'day' -> Option A

4. Quick Check:

   timedelta(days=7) needed, not day=7 [OK]

Hint: Always use keyword days= for timedelta days [OK]

Common Mistakes:

• Misspelling 'days' as 'day'
• Thinking datetime + timedelta is invalid
• Assuming print syntax error

5. You have a DataFrame with a column 'start_date' of type datetime. You want to create a new column 'deadline' that is 10 days after 'start_date'. Which code correctly does this using pandas and timedelta?

import pandas as pd
from datetime import timedelta

df = pd.DataFrame({'start_date': pd.to_datetime(['2024-06-01', '2024-06-05'])})
# Which line correctly adds 10 days?

hard

A. df['deadline'] = df['start_date'] + timedelta(seconds=10)

B. df['deadline'] = df['start_date'] + pd.Timedelta(days=10)

C. df['deadline'] = df['start_date'] + timedelta(day=10)

D. df['deadline'] = df['start_date'] + pd.timedelta(days=10)

Solution

1. Step 1: Understand pandas datetime addition

   When adding time to pandas datetime columns, use pd.Timedelta for vectorized operations.

2. Step 2: Check each option

   df['deadline'] = df['start_date'] + pd.Timedelta(days=10): correct pd.Timedelta(days=10). df['deadline'] = df['start_date'] + timedelta(seconds=10): timedelta(seconds=10) adds seconds not days. df['deadline'] = df['start_date'] + timedelta(day=10): invalid keyword day=10. df['deadline'] = df['start_date'] + pd.timedelta(days=10): pd.timedelta doesn't exist.

3. Final Answer:

   df['deadline'] = df['start_date'] + pd.Timedelta(days=10) -> Option B

4. Quick Check:

   Use pd.Timedelta for pandas datetime addition [OK]

Hint: Use pd.Timedelta(days=10) to add days to pandas datetime [OK]

Common Mistakes:

• Using seconds=10 instead of days=10
• Using invalid keyword 'day=10'
• Wrong capitalization pd.timedelta