Rolling standard deviation in Pandas - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of rolling standard deviation with window=3

What is the output of the following code snippet?

Pandas

import pandas as pd
s = pd.Series([1, 2, 3, 4, 5])
result = s.rolling(window=3).std()
print(result.to_list())

A [nan, nan, 0.816496580927726, 0.816496580927726, 0.816496580927726]

B [nan, nan, 0.5, 0.5, 0.5]

C [nan, nan, 1.224744871391589, 1.224744871391589, 1.224744871391589]

D [nan, nan, 1.0, 1.0, 1.0]

💡 Hint

Remember pandas uses sample standard deviation by default (ddof=1).

The rolling standard deviation with window=3 calculates the sample standard deviation (ddof=1) of each 3-element window. For the first two elements, there are not enough values, so the result is NaN. The standard deviation of [1,2,3] is approximately 0.8165 (sqrt(2/3)), and this repeats for subsequent windows.

❓ data_output

intermediate

Number of non-NaN values in rolling std result

Given a pandas Series of length 10, what is the number of non-NaN values in the result of rolling standard deviation with window=4?

Pandas

import pandas as pd
s = pd.Series(range(10))
result = s.rolling(window=4).std()
count_non_nan = result.count()
print(count_non_nan)

A 7

B 4

C 6

D 10

💡 Hint

Rolling window of size 4 produces NaN for the first 3 elements.

With window=4, the first 3 elements have insufficient data for std calculation, so they are NaN. The remaining 7 elements have valid std values, so count is 7.

🔧 Debug

advanced

Identify the error in rolling std calculation

What error does the following code raise?

Pandas

import pandas as pd
s = pd.Series([1, 2, 3, 4, 5])
result = s.rolling(window=3).std(ddof=0)
print(result)

A ValueError: window size must be positive

B TypeError: std() got an unexpected keyword argument 'ddof'

C No error, prints rolling std with population std (ddof=0)

D AttributeError: 'Rolling' object has no attribute 'std'

💡 Hint

Check if ddof is a valid argument for pandas rolling std.

The rolling std method accepts ddof parameter to specify degrees of freedom. ddof=0 calculates population standard deviation. No error occurs.

🚀 Application

advanced

Using rolling std to detect volatility spikes

You have a pandas Series of daily stock prices. Which code snippet correctly creates a boolean Series marking days where the 5-day rolling standard deviation exceeds 2.0?

Pandas

import pandas as pd
prices = pd.Series([10, 12, 11, 13, 15, 14, 16, 18, 17, 19])

A

volatility_spikes = prices.rolling(5).mean() > 2.0
print(volatility_spikes.tolist())

B

volatility_spikes = prices.rolling(window=5).var() > 2.0
print(volatility_spikes.tolist())

C

volatility_spikes = prices.rolling(window=5).std(ddof=0) < 2.0
print(volatility_spikes.tolist())

D

volatility_spikes = prices.rolling(window=5).std() > 2.0
print(volatility_spikes.tolist())

💡 Hint

Volatility is measured by standard deviation, not mean or variance threshold directly.

Option D correctly computes the 5-day rolling standard deviation and checks where it exceeds 2.0, producing a boolean Series marking volatility spikes.

🧠 Conceptual

expert

Effect of min_periods on rolling std output length

If you apply rolling standard deviation with window=4 and min_periods=2 on a Series of length 6, how many non-NaN values will the result contain?

Pandas

import pandas as pd
s = pd.Series([1, 2, 3, 4, 5, 6])
result = s.rolling(window=4, min_periods=2).std()
print(result.count())

A 6

B 5

C 3

D 4

💡 Hint

min_periods=2 means at least 2 values are needed to compute std, even if window is 4.

With min_periods=2, rolling std starts producing values from the 2nd element onward. For length 6, that means 5 non-NaN values.

Practice

1. What does the rolling(window).std() function calculate in pandas?

easy

A. The standard deviation of values within a moving window

B. The sum of values within a moving window

C. The average of values within a moving window

D. The maximum value within a moving window

Solution

1. Step 1: Understand rolling window concept

   The rolling function creates a moving window over data points.

2. Step 2: Apply std() to rolling window

   The std() function calculates the standard deviation of values inside each window.

3. Final Answer:

   The standard deviation of values within a moving window -> Option A

4. Quick Check:

   rolling(window).std() = standard deviation [OK]

Hint: Remember std() means standard deviation, not sum or mean [OK]

Common Mistakes:

• Confusing std() with sum() or mean()
• Thinking rolling() alone calculates std
• Assuming it returns a single value, not a series

2. Which of the following is the correct syntax to calculate a rolling standard deviation with a window size of 3 on a pandas Series named data?

easy

A. data.std().rolling(3)

B. data.rolling_std(3)

C. rolling(data, 3).std()

D. data.rolling(3).std()

Solution

1. Step 1: Identify correct method chaining

   The rolling() method is called on the Series, then std() is applied.

2. Step 2: Check syntax correctness

   data.rolling(3).std() uses correct pandas syntax: data.rolling(3).std()

3. Final Answer:

   data.rolling(3).std() -> Option D

4. Quick Check:

   Correct method chaining = data.rolling(3).std() [OK]

Hint: Call rolling() first, then std() to get rolling std dev [OK]

Common Mistakes:

• Calling std() before rolling()
• Using non-existent rolling_std() method
• Passing window size inside std()

3. Given the pandas Series data = pd.Series([2, 4, 6, 8, 10]), what is the output of data.rolling(3).std()?

medium

A. [NaN, 1.0, 2.0, 3.0, 4.0]

B. [NaN, NaN, 2.0, 2.0, 2.0]

C. [2.0, 2.0, 2.0, 2.0, 2.0]

D. [NaN, NaN, NaN, NaN, NaN]

Solution

1. Step 1: Understand rolling std with window=3

   For the first two values, window size 3 is not met, so result is NaN.

2. Step 2: Calculate std for windows

   Windows: [2,4,6], [4,6,8], [6,8,10] each have std deviation 2.0.

3. Final Answer:

   [NaN, NaN, 2.0, 2.0, 2.0] -> Option B

4. Quick Check:

   Rolling std with window 3 = [NaN, NaN, 2.0, 2.0, 2.0] [OK]

Hint: First (window-1) results are NaN, then std calculated [OK]

Common Mistakes:

• Expecting values for first two positions
• Confusing std with mean or sum
• Miscomputing standard deviation values

4. What is wrong with this code snippet to calculate rolling standard deviation on a DataFrame column df['values'] with window size 4?

df['rolling_std'] = df['values'].rolling.std(4)

medium

A. The code is correct and will run without error

B. The std() function should be called before rolling()

C. The rolling method is missing parentheses before std()

D. Window size should be passed to std(), not rolling()

Solution

1. Step 1: Check method chaining syntax

   rolling() requires parentheses with window size: rolling(4)

2. Step 2: Identify error in code

   Code uses rolling.std(4) which is invalid; should be rolling(4).std()

3. Final Answer:

   The rolling method is missing parentheses before std() -> Option C

4. Quick Check:

   rolling() needs parentheses before std() [OK]

Hint: Always use rolling(window).std(), not rolling.std(window) [OK]

Common Mistakes:

• Calling std() before rolling()
• Passing window size to std() instead of rolling()
• Omitting parentheses after rolling

5. You have daily sales data in a pandas DataFrame df with a column 'sales'. You want to detect days where sales variability over the last 5 days was unusually high. Which code correctly adds a column 'high_volatility' that is true when the rolling standard deviation exceeds 100?

hard

A. df['high_volatility'] = df['sales'].rolling(5).std() > 100

B. df['high_volatility'] = df['sales'].std().rolling(5) > 100

C. df['high_volatility'] = df['sales'].rolling(window=5).mean() > 100

D. df['high_volatility'] = df['sales'].rolling(5).sum() > 100

Solution

1. Step 1: Calculate rolling standard deviation over 5 days

   Use rolling(5).std() on 'sales' to get variability in last 5 days.

2. Step 2: Compare rolling std to threshold 100

   Create boolean column where rolling std is greater than 100.

3. Final Answer:

   df['high_volatility'] = df['sales'].rolling(5).std() > 100 -> Option A

4. Quick Check:

   rolling(5).std() > 100 detects high variability [OK]

Hint: Use rolling(5).std() then compare to threshold for volatility [OK]

Common Mistakes:

• Using std() before rolling()
• Using mean() or sum() instead of std()
• Comparing rolling sum or mean to threshold