Outlier detection with IQR in Pandas - Mini Project: Build & Apply

Outlier detection with IQR

📖 Scenario: You work as a data analyst for a retail company. You have sales data for different stores. Sometimes, some sales numbers are unusually high or low. These are called outliers. Detecting outliers helps you understand if there are errors or special cases in the data.

🎯 Goal: You will learn how to detect outliers in sales data using the Interquartile Range (IQR) method with pandas. You will create the sales data, calculate IQR, find outliers, and display them.

📋 What You'll Learn

Use pandas to create and analyze data

Calculate quartiles and IQR

Detect outliers using IQR method

Print the detected outliers

💡 Why This Matters

🌍 Real World

Outlier detection helps businesses find unusual data points that may indicate errors, fraud, or special events.

💼 Career

Data analysts and data scientists often use IQR to clean data and improve the quality of their analysis.

1

Create sales data with pandas

Create a pandas DataFrame called sales_data with one column named sales and these exact values: 100, 150, 200, 250, 300, 350, 400, 1000.

Pandas

# Create the sales_data DataFrame with the given sales values
# Your code here

Hint

Use pd.DataFrame with a dictionary where the key is "sales" and the value is the list of numbers.

2

Calculate Q1, Q3 and IQR

Create three variables: Q1, Q3, and IQR. Calculate Q1 as the 25th percentile of sales_data["sales"], Q3 as the 75th percentile, and IQR as Q3 - Q1.

Pandas

import pandas as pd

sales_data = pd.DataFrame({"sales": [100, 150, 200, 250, 300, 350, 400, 1000]})

# Calculate Q1, Q3 and IQR below
# Your code here

Hint

Use the quantile() method on the sales column to get Q1 and Q3.

3

Detect outliers using IQR

Create a variable called outliers that contains the rows from sales_data where the sales value is less than Q1 - 1.5 * IQR or greater than Q3 + 1.5 * IQR.

Pandas

import pandas as pd

sales_data = pd.DataFrame({"sales": [100, 150, 200, 250, 300, 350, 400, 1000]})

Q1 = sales_data["sales"].quantile(0.25)
Q3 = sales_data["sales"].quantile(0.75)
IQR = Q3 - Q1

# Create outliers DataFrame using IQR method
# Your code here

Hint

Use boolean conditions combined with | inside the DataFrame filter brackets.

4

Print the detected outliers

Write a print statement to display the outliers DataFrame.

Pandas

import pandas as pd

sales_data = pd.DataFrame({"sales": [100, 150, 200, 250, 300, 350, 400, 1000]})

Q1 = sales_data["sales"].quantile(0.25)
Q3 = sales_data["sales"].quantile(0.75)
IQR = Q3 - Q1

outliers = sales_data[(sales_data["sales"]  (Q3 + 1.5 * IQR))]

# Print the outliers DataFrame
# Your code here

Hint

Use print(outliers) to show the rows with outliers.

Practice

1. What does the Interquartile Range (IQR) represent in a dataset?

easy

A. The range between the 25th percentile (Q1) and the 75th percentile (Q3)

B. The difference between the maximum and minimum values

C. The average value of the dataset

D. The total number of data points

Solution

1. Step 1: Understand quartiles

   Quartiles divide data into four equal parts. Q1 is the 25th percentile, Q3 is the 75th percentile.

2. Step 2: Define IQR

   IQR is the difference between Q3 and Q1, showing the middle 50% spread of data.

3. Final Answer:

   The range between the 25th percentile (Q1) and the 75th percentile (Q3) -> Option A

4. Quick Check:

   IQR = Q3 - Q1 [OK]

Hint: IQR = Q3 minus Q1, middle half range [OK]

Common Mistakes:

• Confusing IQR with total range
• Thinking IQR is the average
• Mixing up quartiles with percentiles

2. Which of the following pandas code snippets correctly calculates the IQR of a DataFrame column df['score']?

easy

A. iqr = df['score'].mean() - df['score'].median()

B. iqr = df['score'].max() - df['score'].min()

C. iqr = df['score'].quantile(0.75) - df['score'].quantile(0.25)

D. iqr = df['score'].std()

Solution

1. Step 1: Recall IQR formula

   IQR is Q3 minus Q1, where Q3 is 75th percentile and Q1 is 25th percentile.

2. Step 2: Check pandas quantile usage

   Using quantile(0.75) and quantile(0.25) correctly gets Q3 and Q1.

3. Final Answer:

   iqr = df['score'].quantile(0.75) - df['score'].quantile(0.25) -> Option C

4. Quick Check:

   Use quantile(0.75) - quantile(0.25) for IQR [OK]

Hint: Use quantile(0.75) minus quantile(0.25) for IQR [OK]

Common Mistakes:

• Using mean or median instead of quantiles
• Using max-min instead of IQR
• Using standard deviation instead of IQR

3. Given the following code, what will be the output of outliers?

import pandas as pd

data = {'values': [10, 12, 14, 15, 18, 100, 105]}
df = pd.DataFrame(data)
Q1 = df['values'].quantile(0.25)
Q3 = df['values'].quantile(0.75)
IQR = Q3 - Q1
lower_bound = Q1 - 1.5 * IQR
upper_bound = Q3 + 1.5 * IQR
outliers = df[(df['values'] < lower_bound) | (df['values'] > upper_bound)]
print(outliers)

medium

A. Rows with values 10 and 12

B. Rows with values 100 and 105

C. All rows because 1.5*IQR is too small

D. Empty DataFrame, no outliers detected

Solution

1. Step 1: Calculate Q1, Q3, and IQR

   Q1 = 12.0 (between 12 and 14), Q3 = 18.0 (between 18 and 100), IQR = 6.0.

2. Step 2: Calculate bounds and find outliers

   Lower bound = 12 - 1.5*6 = 3; Upper bound = 18 + 1.5*6 = 27. Values 100 and 105 are above upper bound, so they are outliers.

3. Final Answer:

   Rows with values 100 and 105 -> Option B

4. Quick Check:

   Outliers > Q3 + 1.5*IQR or < Q1 - 1.5*IQR: 100, 105 [OK]

Hint: Outliers lie beyond Q1-1.5*IQR or Q3+1.5*IQR [OK]

Common Mistakes:

• Miscomputing Q1 or Q3
• Using wrong inequality signs
• Ignoring both lower and upper bounds

4. Identify the error in this code snippet for detecting outliers using IQR:

Q1 = df['data'].quantile(0.25)
Q3 = df['data'].quantile(0.75)
IQR = Q3 - Q1
outliers = df[(df['data'] > Q3 + 1.5 * IQR) & (df['data'] < Q1 - 1.5 * IQR)]

medium

A. No error, code is correct

B. IQR calculation is incorrect

C. Quantile values are swapped

D. The condition uses '&' instead of '|' to combine bounds

Solution

1. Step 1: Understand outlier condition

   Outliers are values less than lower bound OR greater than upper bound, so conditions should be combined with OR (|).

2. Step 2: Check code condition

   The code uses AND (&), which means no value can be both greater than upper bound and less than lower bound simultaneously.

3. Final Answer:

   The condition uses '&' instead of '|' to combine bounds -> Option D

4. Quick Check:

   Use '|' for outlier condition, not '&' [OK]

Hint: Use OR (|) to combine outlier conditions, not AND (&) [OK]

Common Mistakes:

• Using AND instead of OR for outlier filtering
• Swapping Q1 and Q3 values
• Calculating IQR incorrectly

5. You have a DataFrame df with a column 'height'. You want to remove outliers using IQR but keep the original index. Which code correctly filters out outliers and keeps the index intact?

hard

A. Q1 = df['height'].quantile(0.25)
Q3 = df['height'].quantile(0.75)
IQR = Q3 - Q1
filtered_df = df[(df['height'] >= Q1 - 1.5*IQR) & (df['height'] <= Q3 + 1.5*IQR)]

B. Q1 = df['height'].quantile(0.25)
Q3 = df['height'].quantile(0.75)
IQR = Q3 - Q1
filtered_df = df.reset_index(drop=True)[(df['height'] >= Q1 - 1.5*IQR) & (df['height'] <= Q3 + 1.5*IQR)]

C. Q1 = df['height'].quantile(0.25)
Q3 = df['height'].quantile(0.75)
IQR = Q3 - Q1
filtered_df = df[(df['height'] > Q1 + 1.5*IQR) | (df['height'] < Q3 - 1.5*IQR)]

D. Q1 = df['height'].quantile(0.25)
Q3 = df['height'].quantile(0.75)
IQR = Q3 - Q1
filtered_df = df[(df['height'] < Q1 - 1.5*IQR) & (df['height'] > Q3 + 1.5*IQR)]

Solution

1. Step 1: Calculate IQR and bounds correctly

   Calculate Q1, Q3, and IQR. Define lower bound as Q1 - 1.5*IQR and upper bound as Q3 + 1.5*IQR.

2. Step 2: Filter rows within bounds and keep index

   Use boolean indexing with AND (&) to keep rows inside bounds. Do not reset index to keep original indices.

3. Final Answer:

   Code in option A correctly filters and keeps index -> Option A

4. Quick Check:

   Filter with AND inside bounds, no reset_index [OK]

Hint: Filter with AND inside bounds; avoid resetting index [OK]

Common Mistakes:

• Using OR instead of AND to filter
• Resetting index and losing original row labels
• Swapping bounds or using wrong inequalities