Date arithmetic with timedelta in Pandas - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to add 7 days to the date in the 'date' column.

Pandas

import pandas as pd
from datetime import timedelta

df = pd.DataFrame({'date': pd.to_datetime(['2024-01-01', '2024-01-02'])})
df['new_date'] = df['date'] + [1](days=7)
print(df)

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

Atimedelta

Bdatetime

Cdate

Dtime

Attempts:

3 left

💡 Hint

Common Mistakes

Using datetime instead of timedelta causes an error.

Trying to add an integer directly to a datetime column.

We use timedelta to add a specific number of days to a date.

2fill in blank

medium

Complete the code to subtract 3 days from the 'date' column.

Pandas

import pandas as pd
from datetime import timedelta

df = pd.DataFrame({'date': pd.to_datetime(['2024-03-10', '2024-03-15'])})
df['earlier_date'] = df['date'] - [1](days=3)
print(df)

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

Adatetime

Btime

Cdate

Dtimedelta

Attempts:

3 left

💡 Hint

Common Mistakes

Using datetime instead of timedelta.

Trying to subtract an integer directly from a datetime column.

Subtracting a timedelta of 3 days moves the date 3 days earlier.

3fill in blank

hard

Fix the error in the code to add 10 days to the 'date' column.

Pandas

import pandas as pd
from datetime import timedelta

df = pd.DataFrame({'date': pd.to_datetime(['2024-05-01', '2024-05-02'])})
df['future_date'] = df['date'] + [1](10)
print(df)

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

Atime

Btimedelta

Cdate

Ddatetime

Attempts:

3 left

💡 Hint

Common Mistakes

Using datetime instead of timedelta.

Not specifying days=10, which can cause confusion.

The timedelta constructor can use named arguments like days=10. Passing just 10 is allowed and sets days=10.

4fill in blank

hard

Fill both blanks to create a dictionary with words as keys and their lengths if length is greater than 4.

Pandas

words = ['apple', 'bat', 'grape', 'kiwi']
lengths = {word: [1] for word in words if [2]
print(lengths)

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

Alen(word)

Bword

Clen(word) > 4

Dword > 4

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself as value instead of its length.

Using the word variable in the condition instead of its length.

The dictionary comprehension uses len(word) as value and filters words with length greater than 4.

5fill in blank

hard

Fill all three blanks to create a dictionary with uppercase words as keys and their lengths if length is greater than 3.

Pandas

words = ['dog', 'cat', 'elephant', 'fox']
result = { [1]: [2] for word in words if [3] }
print(result)

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

Aword.upper()

Blen(word)

Clen(word) > 3

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the original word as key instead of uppercase.

Using the word itself as value instead of length.

Incorrect condition for filtering words.

The dictionary comprehension uses the uppercase word as key, length as value, and filters words longer than 3 characters.

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