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.