Date arithmetic with timedelta in Pandas - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Adding timedelta to a pandas Series of dates

What is the output of this code snippet?

Pandas

import pandas as pd
from datetime import timedelta

dates = pd.Series(pd.to_datetime(['2024-01-01', '2024-01-02', '2024-01-03']))
result = dates + timedelta(days=3)
print(result)

A

0   2024-01-04
1   2024-01-05
2   2024-01-06
dtype: datetime64[ns]

B

0   2024-01-01
1   2024-01-02
2   2024-01-03
dtype: datetime64[ns]

C TypeError: unsupported operand type(s) for +: 'Series' and 'timedelta.timedelta'

D

0   2024-01-04 00:00:00
1   2024-01-05 00:00:00
2   2024-01-06 00:00:00
dtype: datetime64[ns]

💡 Hint

Try adding a timedelta object directly to a pandas Series of datetime values.

Adding a timedelta object to a pandas Series of datetime values adds the time delta to each element. The output shows the dates shifted by 3 days with time included as 00:00:00.

❓ data_output

intermediate

Subtracting timedelta from a DataFrame column

Given this DataFrame, what is the output of subtracting 5 hours from the 'timestamp' column?

Pandas

import pandas as pd
from datetime import timedelta

df = pd.DataFrame({'timestamp': pd.to_datetime(['2024-06-01 12:00', '2024-06-01 15:00'])})
df['new_time'] = df['timestamp'] - timedelta(hours=5)
print(df['new_time'])

A

0   2024-06-01 07:00:00
1   2024-06-01 10:00:00
Name: new_time, dtype: datetime64[ns]

B

0   2024-06-01 17:00:00
1   2024-06-01 20:00:00
Name: new_time, dtype: datetime64[ns]

C TypeError: unsupported operand type(s) for -: 'Series' and 'timedelta.timedelta'

D

0   2024-06-01 12:00:00
1   2024-06-01 15:00:00
Name: new_time, dtype: datetime64[ns]

💡 Hint

Subtracting timedelta decreases the datetime values by the specified duration.

Subtracting 5 hours from each timestamp shifts the time earlier by 5 hours.

🔧 Debug

advanced

Identify the error in timedelta addition to a DataFrame

What error does this code produce and why?

Pandas

import pandas as pd
from datetime import timedelta

df = pd.DataFrame({'date': ['2024-01-01', '2024-01-02']})
df['date'] = df['date'] + timedelta(days=1)
print(df)

A TypeError: unsupported operand type(s) for +: 'Series' and 'timedelta.timedelta'

B ValueError: could not convert string to datetime

C No error, dates incremented by 1 day

D AttributeError: 'list' object has no attribute 'dt'

💡 Hint

Check the data type of the 'date' column before adding timedelta.

The 'date' column contains strings, not datetime objects. Adding timedelta to strings causes a TypeError.

🚀 Application

advanced

Calculate the difference in days between two date columns

Given this DataFrame, which option correctly creates a new column 'diff_days' showing the number of days between 'end_date' and 'start_date'?

Pandas

import pandas as pd

df = pd.DataFrame({
    'start_date': pd.to_datetime(['2024-01-01', '2024-01-10']),
    'end_date': pd.to_datetime(['2024-01-05', '2024-01-15'])
})

A

df['diff_days'] = df['end_date'] - df['start_date']
print(df['diff_days'])

B

df['diff_days'] = (df['end_date'] - df['start_date']).dt.days
print(df['diff_days'])

C

df['diff_days'] = (df['start_date'] - df['end_date']).days
print(df['diff_days'])

D

df['diff_days'] = (df['end_date'] + df['start_date']).dt.days
print(df['diff_days'])

💡 Hint

Subtracting datetime columns gives a timedelta Series; use .dt.days to get integer days.

Subtracting two datetime columns returns a timedelta Series. Using .dt.days extracts the number of days as integers.

🧠 Conceptual

expert

Understanding timedelta behavior with mixed frequency data

You have a pandas Series with timestamps at irregular intervals. You add a timedelta of 1 day to this Series. What happens to the timestamps that fall on daylight saving time change days?

A A ValueError is raised because timedelta cannot handle daylight saving changes.

B The timestamps shift by 23 or 25 hours depending on the daylight saving change.

C The timestamps shift by exactly 24 hours regardless of daylight saving changes.

D The timestamps remain unchanged because timedelta ignores irregular intervals.

💡 Hint

Timedelta represents a fixed duration, not calendar-aware shifts.

Timedelta adds a fixed duration (e.g., 24 hours) to timestamps, ignoring daylight saving time shifts.

Practice

1. What does pd.Timedelta(days=3) represent in pandas?

easy

A. A string describing 3 days

B. A specific date 3 days from now

C. A time difference of 3 days

D. A timestamp at 3 AM

Solution

1. Step 1: Understand Timedelta purpose

   pd.Timedelta is used to represent a duration or difference in time, not a fixed date or time.

2. Step 2: Interpret the argument

   The argument days=3 means the duration is 3 days long.

3. Final Answer:

   A time difference of 3 days -> Option C

4. Quick Check:

   Timedelta(days=3) = 3 days [OK]

Hint: Timedelta means duration, not a fixed date [OK]

Common Mistakes:

• Thinking Timedelta is a fixed date
• Confusing Timedelta with Timestamp
• Assuming it returns a string

2. Which of the following is the correct way to add 5 days to a pandas Timestamp ts?

easy

A. ts.add(pd.Timedelta(5))

B. ts + pd.Timedelta('5')

C. ts + pd.Timedelta(5 days)

D. ts + pd.Timedelta(days=5)

Solution

1. Step 1: Check correct Timedelta syntax

   The correct way to specify 5 days is pd.Timedelta(days=5). The argument must be named or a valid string.

2. Step 2: Verify addition syntax

   Adding a Timedelta to a Timestamp uses the + operator, so ts + pd.Timedelta(days=5) is correct.

3. Final Answer:

   ts + pd.Timedelta(days=5) -> Option D

4. Quick Check:

   Correct syntax uses days=5 inside Timedelta [OK]

Hint: Use named argument days=5 inside Timedelta [OK]

Common Mistakes:

• Missing quotes or wrong argument format
• Using add() method incorrectly
• Passing string without units

3. What is the output of this code?

import pandas as pd
start = pd.Timestamp('2024-01-01')
new_date = start + pd.Timedelta(days=10) - pd.Timedelta(days=3)
print(new_date.strftime('%Y-%m-%d'))

medium

A. 2024-01-08

B. 2024-01-13

C. 2024-01-04

D. 2024-01-10

Solution

1. Step 1: Calculate addition of 10 days

   Starting from 2024-01-01, adding 10 days gives 2024-01-11.

2. Step 2: Subtract 3 days from the result

   Subtracting 3 days from 2024-01-11 results in 2024-01-08.

3. Final Answer:

   2024-01-08 -> Option A

4. Quick Check:

   2024-01-01 + 10 - 3 days = 2024-01-08 [OK]

Hint: Add then subtract days stepwise to find final date [OK]

Common Mistakes:

• Adding and subtracting days in wrong order
• Miscounting days
• Confusing string formatting output

4. The following code throws an error. What is the problem?

import pandas as pd
start = pd.Timestamp('2024-05-01')
new_date = start + pd.Timedelta(5)
print(new_date)

medium

A. Timedelta needs a unit like days=5

B. Timestamp cannot be added to Timedelta

C. pd.Timedelta requires string input only

D. Missing import for Timedelta

Solution

1. Step 1: Check Timedelta argument

   pd.Timedelta(5) is missing the unit. It needs a keyword like days=5 or a string like '5 days'.

2. Step 2: Understand error cause

   Without specifying units, pandas cannot interpret the number 5 alone, causing an error.

3. Final Answer:

   Timedelta needs a unit like days=5 -> Option A

4. Quick Check:

   Timedelta(5) missing units causes error [OK]

Hint: Always specify units when creating Timedelta [OK]

Common Mistakes:

• Passing integer without units
• Assuming default unit is days
• Confusing Timestamp and Timedelta usage

5. You have a pandas DataFrame df with a column 'date' of type datetime. You want to create a new column 'date_plus_7' that adds 7 days to each date. Which code correctly does this?

hard

A. df['date_plus_7'] = df['date'] + pd.Timedelta('7 days')

B. df['date_plus_7'] = df['date'] + pd.Timedelta(days=7)

C. df['date_plus_7'] = df['date'] + pd.Timedelta(7)

D. df['date_plus_7'] = df['date'] + pd.Timedelta(days=7) * len(df)

Solution

1. Step 1: Understand Timedelta broadcasting

   Adding a Timedelta to a pandas Series of dates works if the Timedelta is a scalar (single duration).

2. Step 2: Check each option

   df['date_plus_7'] = df['date'] + pd.Timedelta('7 days') uses a string '7 days' which works but is less common in vectorized operations. df['date_plus_7'] = df['date'] + pd.Timedelta(days=7) uses pd.Timedelta(days=7) which is the standard and recommended way. df['date_plus_7'] = df['date'] + pd.Timedelta(7) misses units and causes error. df['date_plus_7'] = df['date'] + pd.Timedelta(days=7) * len(df) multiplies timedelta by length, which incorrectly scales the timedelta.

3. Step 3: Choose the best practice

   df['date_plus_7'] = df['date'] + pd.Timedelta(days=7) is the correct and recommended way to add 7 days to each date in a column.

4. Final Answer:

   df['date_plus_7'] = df['date'] + pd.Timedelta(days=7) -> Option B

5. Quick Check:

   Use pd.Timedelta(days=7) for vectorized addition with Series [OK]

Hint: Use pd.Timedelta(days=7) for vectorized addition with Series [OK]

Common Mistakes:

• Passing integer without units
• Multiplying Timedelta by length of DataFrame
• Not using vectorized Timedelta correctly