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 >.

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