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)

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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)

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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)

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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)

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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.