Date arithmetic (Timedelta) in Data Analysis Python - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a timedelta of 5 days.

Data Analysis Python

from datetime import timedelta

five_days = timedelta([1]=5)
print(five_days)

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

Adays

Bhours

Cminutes

Dseconds

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'day' instead of 'days' as the parameter name.

Using 'hours' or 'minutes' when the question asks for days.

The timedelta function uses the days parameter to represent a duration of days.

2fill in blank

medium

Complete the code to calculate the date 10 days after January 1, 2024.

Data Analysis Python

from datetime import datetime, timedelta

start_date = datetime(2024, 1, 1)
new_date = start_date + timedelta([1]=10)
print(new_date.strftime('%Y-%m-%d'))

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

Adays

Bhours

Cminutes

Dseconds

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'day' instead of 'days'.

Using 'hours' or 'minutes' which changes the time, not the date.

Adding a timedelta with days=10 to the start date moves the date 10 days forward.

3fill in blank

hard

Fix the error in the code to subtract 3 hours from a datetime object.

Data Analysis Python

from datetime import datetime, timedelta

now = datetime(2024, 6, 15, 12, 0)
new_time = now - timedelta([1]=3)
print(new_time.strftime('%H:%M'))

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

Adays

Bhours

Chour

Dminute

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'hour' instead of 'hours'.

Using 'minute' when subtracting hours.

The correct parameter for hours in timedelta is hours (plural), not hour.

4fill in blank

hard

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

Data Analysis Python

words = ['apple', 'bat', 'grape', 'kiwi']
lengths = {word: [1] for word in words if len(word) [2] 4}
print(lengths)

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

Alen(word)

B>

C<

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself as value instead of its length.

Using '<' instead of '>' in the condition.

The dictionary comprehension uses len(word) to get the length and filters words with length greater than 4 using >.

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.

Data Analysis Python

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

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

Aword.upper()

Blen(word)

C>

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the original word as key instead of uppercase.

Using '<' instead of '>' in the condition.

The dictionary comprehension uses word.upper() as keys, len(word) as values, and filters words with length greater than 3 using >.