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.