Extracting day of week and hour in Pandas - Cheat Sheet & Quick Revision

Recall & Review

beginner

What pandas function helps convert a column to datetime format?

The function pd.to_datetime() converts a column or series to datetime format, enabling date and time operations.

beginner

How do you extract the day of the week from a pandas datetime column?

Use the .dt.day_name() accessor on a datetime column to get the day of the week as a string, like 'Monday' or 'Tuesday'.

beginner

How can you extract the hour from a pandas datetime column?

Use the .dt.hour accessor on a datetime column to get the hour as an integer from 0 to 23.

intermediate

Why is it useful to extract day of week and hour from datetime data?

Extracting day of week and hour helps find patterns like busiest hours or weekdays, which is useful for analysis and decision making.

intermediate

What is the difference between .dt.dayofweek and .dt.day_name() in pandas?

.dt.dayofweek returns an integer (0=Monday, 6=Sunday), while .dt.day_name() returns the full weekday name as a string.

Which pandas accessor extracts the hour from a datetime column?

A .dt.hour

B .dt.minute

C .dt.day_name()

D .dt.month

The .dt.hour accessor extracts the hour part from a datetime column.

What does .dt.day_name() return?

A Integer representing day of week

B Hour of the day

C Full name of the weekday as a string

D Month name

.dt.day_name() returns the weekday name like 'Monday' or 'Friday'.

How do you convert a string column to datetime in pandas?

A pd.to_datetime()

B pd.to_string()

C pd.to_numeric()

D pd.to_list()

Use pd.to_datetime() to convert strings to datetime format.

What integer does .dt.dayofweek return for Sunday?

A 7

B 5

C 0

D 6

.dt.dayofweek returns 6 for Sunday (0 is Monday).

Why extract day of week and hour from datetime data?

A To change data types

B To find patterns in time-based data

C To sort data alphabetically

D To remove missing values

Extracting day and hour helps find time patterns like peak hours or busy weekdays.

Explain how to extract the day of week and hour from a pandas datetime column.

Think about datetime conversion and using .dt accessor.

Describe why extracting day of week and hour can be useful in data analysis.

Consider how time affects behavior or events.

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