Outlier detection with IQR in Pandas - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate the first quartile (Q1) of the 'data' column using pandas.

Pandas

Q1 = df['data'].quantile([1])

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

A0.25

B0.5

C0.75

D1.0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0.5 which is the median instead of Q1.

Using 0.75 which is the third quartile.

The first quartile (Q1) is the 25th percentile, so we use 0.25 in the quantile function.

2fill in blank

medium

Complete the code to calculate the interquartile range (IQR) from Q1 and Q3.

Pandas

IQR = Q3 [1] Q1

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

A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Adding Q1 and Q3 instead of subtracting.

Multiplying or dividing Q3 and Q1.

IQR is the difference between the third quartile (Q3) and the first quartile (Q1), so we subtract Q1 from Q3.

3fill in blank

hard

Fix the error in the code to calculate the lower bound for outliers using IQR.

Pandas

lower_bound = Q1 [1] 1.5 * IQR

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

A-

B/

C+

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Adding 1.5 * IQR instead of subtracting.

Using multiplication or division incorrectly.

The lower bound is Q1 minus 1.5 times the IQR to detect outliers below the normal range.

4fill in blank

hard

Fill both blanks to create a dictionary of outliers below the lower bound and above the upper bound.

Pandas

outliers = {x: x for x in df['data'] if x [1] lower_bound or x [2] upper_bound}

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

A<

B>

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= or >= which includes boundary values.

Using wrong comparison operators that do not detect outliers.

Outliers are values less than the lower bound or greater than the upper bound.

5fill in blank

hard

Fill all three blanks to create a dictionary of non-outlier values within the IQR bounds.

Pandas

non_outliers = {x: x for x in df['data'] if x [1] lower_bound and x [2] upper_bound and x [3] None}

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

A>

B<

C!=

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operators that include outliers.

Not checking for None values.

Non-outliers are values greater than the lower bound and less than the upper bound, and not None.

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