Extracting day of week and hour in Pandas - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to extract the day of the week from the 'date' column.

Pandas

df['day_of_week'] = df['date'].dt.[1]

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

Asecond

Bdayofweek

Cminute

Dhour

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'hour' instead of 'dayofweek' will extract the hour, not the day.

Trying to call a method like 'dayofweek()' instead of using the attribute.

The dayofweek attribute extracts the day of the week as an integer (Monday=0, Sunday=6) from a datetime column.

2fill in blank

medium

Complete the code to extract the hour from the 'timestamp' column.

Pandas

df['hour'] = df['timestamp'].dt.[1]

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

Aday

Bmonth

Chour

Dyear

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'day' or 'month' will extract wrong parts of the datetime.

Trying to call hour() as a method instead of using the attribute.

The hour attribute extracts the hour (0-23) from a datetime column.

3fill in blank

hard

Complete the code to correctly extract the day of week from 'date_time'.

Pandas

df['day_of_week'] = df['date_time'].dt.[1]

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

Aminute

Bhour

Csecond

Ddayofweek

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses after dayofweek causes an error.

Confusing attributes with methods.

The dayofweek is an attribute, not a method.

4fill in blank

hard

Fill both blanks to create a dictionary with day_of_week (0-6) as keys and record counts as values if count is greater than 3. Assume 'day_of_week' column exists.

Pandas

day_counts = {day: [1] for day in range(7) if [2]

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

A(df['day_of_week'] == day).sum()

B(df['day_of_week'] == day).sum() > 3

Cday

Dday > 3

Attempts:

3 left

💡 Hint

Common Mistakes

Using the day itself as the condition instead of its count.

Using parentheses incorrectly in the comprehension.

The dictionary comprehension maps each day to its count only if the count is greater than 3.

5fill in blank

hard

Fill all three blanks to create a dictionary with uppercase day names as keys, their record counts as values, only if count is greater than 4. Assume 'day_name' column exists.

Pandas

result = { [1]: [2] for day_name in df['day_name'].unique() if [3] }

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

Aday_name.upper()

B(df['day_name'] == day_name).sum()

C(df['day_name'] == day_name).sum() > 4

Dday_name.lower()

Attempts:

3 left

💡 Hint

Common Mistakes

Using lowercase instead of uppercase for keys.

Not filtering by count correctly.

The dictionary comprehension uses uppercase day names as keys, their counts as values, and filters those with more than 4 records.

Practice

1. What does the .dt.dayofweek attribute return when applied to a pandas datetime column?

easy

A. An integer from 0 (Monday) to 6 (Sunday) representing the day of the week

B. The name of the day as a string like 'Monday' or 'Tuesday'

C. The hour of the day from 0 to 23

D. The month number from 1 to 12

Solution

1. Step 1: Understand the .dt accessor in pandas

   The .dt accessor allows extraction of datetime properties from a pandas datetime column.

2. Step 2: Identify what .dt.dayofweek returns

   .dt.dayofweek returns an integer where Monday is 0 and Sunday is 6, representing the day of the week.

3. Final Answer:

   An integer from 0 (Monday) to 6 (Sunday) representing the day of the week -> Option A

4. Quick Check:

   .dt.dayofweek = 0 to 6 [OK]

Hint: Remember Monday starts at 0 for dayofweek [OK]

Common Mistakes:

• Thinking it returns day names instead of numbers
• Confusing dayofweek with day or dayofyear
• Expecting Sunday to be 0 instead of Monday

2. Which of the following is the correct way to extract the hour from a pandas datetime column named df['time']?

easy

A. df['time'].hour

B. df['time'].dt.hour()

C. df['time'].dt.hour

D. df['time'].dt['hour']

Solution

1. Step 1: Recall the pandas datetime accessor syntax

   To extract datetime components, use the .dt accessor followed by the attribute without parentheses.

2. Step 2: Identify the correct syntax for hour extraction

   The correct syntax is df['time'].dt.hour, which returns the hour as an integer from 0 to 23.

3. Final Answer:

   df['time'].dt.hour -> Option C

4. Quick Check:

   Use .dt.hour without parentheses [OK]

Hint: Use .dt.hour without parentheses to get hour [OK]

Common Mistakes:

• Using .hour directly without .dt
• Adding parentheses after .hour like a function
• Trying to access hour with dictionary syntax

3. Given the following code:

import pandas as pd

df = pd.DataFrame({
    'timestamp': ['2024-06-01 14:30:00', '2024-06-02 09:15:00']
})
df['timestamp'] = pd.to_datetime(df['timestamp'])
df['day'] = df['timestamp'].dt.dayofweek
df['hour'] = df['timestamp'].dt.hour
print(df[['day', 'hour']].to_dict())

What is the output?

medium

A. {'day': {0: 5, 1: 6}, 'hour': {0: 14, 1: 9}}

B. {'day': {0: 6, 1: 0}, 'hour': {0: 14, 1: 9}}

C. {'day': {0: 5, 1: 6}, 'hour': {0: 14, 1: 15}}

D. {'day': {0: 4, 1: 5}, 'hour': {0: 14, 1: 9}}

Solution

1. Step 1: Convert strings to datetime and extract dayofweek

   '2024-06-01' is a Saturday, which is day 5 (Monday=0). '2024-06-02' is Sunday, day 6.

2. Step 2: Extract hour from timestamps

   The hours are 14 and 9 respectively, matching the times given.

3. Final Answer:

   {'day': {0: 5, 1: 6}, 'hour': {0: 14, 1: 9}} -> Option A

4. Quick Check:

   Saturday=5, Sunday=6, hours 14 and 9 [OK]

Hint: Check weekday numbers: Monday=0, Saturday=5, Sunday=6 [OK]

Common Mistakes:

• Mixing up dayofweek numbering
• Incorrect hour extraction
• Not converting strings to datetime first

4. The following code throws an error. What is the main issue?

import pandas as pd

df = pd.DataFrame({'time': ['2024-06-01 14:30:00']})
df['hour'] = df['time'].hour

medium

A. The DataFrame is missing an index

B. Incorrect use of .hour instead of .dt.hour with parentheses

C. The 'time' column should be a list of integers, not strings

D. Missing conversion of 'time' column to datetime before accessing .hour

Solution

1. Step 1: Check the data type of 'time' column

   The 'time' column contains strings, not datetime objects, so .hour attribute is not available.

2. Step 2: Identify the fix

   Convert 'time' to datetime using pd.to_datetime() before accessing .dt.hour.

3. Final Answer:

   Missing conversion of 'time' column to datetime before accessing .hour -> Option D

4. Quick Check:

   Convert strings to datetime first [OK]

Hint: Always convert to datetime before .dt.hour [OK]

Common Mistakes:

• Trying to use .hour on string columns
• Using .hour without .dt accessor
• Ignoring data type conversion errors

5. You have a pandas DataFrame df with a column 'datetime' containing timestamps. You want to create a new column 'work_hour' that is True if the timestamp is on a weekday (Monday to Friday) and between 9 AM and 5 PM, otherwise False. Which code snippet correctly does this?

hard

A. df['work_hour'] = ((df['datetime'].dt.dayofweek >= 1) & (df['datetime'].dt.dayofweek <= 5) & (df['datetime'].dt.hour > 9) & (df['datetime'].dt.hour < 17))

B. df['work_hour'] = ((df['datetime'].dt.dayofweek >= 0) & (df['datetime'].dt.dayofweek <= 4) & (df['datetime'].dt.hour >= 9) & (df['datetime'].dt.hour < 17))

C. df['work_hour'] = (df['datetime'].dt.dayofweek > 0) & (df['datetime'].dt.dayofweek < 5) & (df['datetime'].dt.hour >= 9) & (df['datetime'].dt.hour <= 17)

D. df['work_hour'] = ((df['datetime'].dt.dayofweek >= 0) & (df['datetime'].dt.dayofweek <= 4) & (df['datetime'].dt.hour > 9) & (df['datetime'].dt.hour <= 17))

Solution

1. Step 1: Define weekdays correctly

   Weekdays are Monday (0) through Friday (4), so dayofweek should be between 0 and 4 inclusive.

2. Step 2: Define work hours correctly

   Work hours are from 9 AM (hour 9) up to but not including 5 PM (hour 17), so hour >= 9 and hour < 17.

3. Step 3: Combine conditions with & and parentheses

   Use & to combine all conditions and wrap each condition in parentheses for correct evaluation.

4. Final Answer:

   df['work_hour'] = ((df['datetime'].dt.dayofweek >= 0) & (df['datetime'].dt.dayofweek <= 4) & (df['datetime'].dt.hour >= 9) & (df['datetime'].dt.hour < 17)) -> Option B

5. Quick Check:

   Weekdays 0-4 and hours 9 to 16 inclusive [OK]

Hint: Use dayofweek 0-4 and hour >=9 and <17 for work hours [OK]

Common Mistakes:

• Using wrong dayofweek range (e.g., 1-5)
• Including 17 as work hour instead of less than 17
• Not using parentheses around conditions