np.std() and np.var() for spread in NumPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of np.std() with axis parameter

What is the output of the following code snippet?

NumPy

import numpy as np
arr = np.array([[1, 2, 3], [4, 5, 6]])
result = np.std(arr, axis=0)
print(result)

A [1.11803399 1.11803399 1.11803399]

B [2.5 2.5 2.5]

C [2.0 2.0 2.0]

D [1.5 1.5 1.5]

💡 Hint

Remember that np.std calculates the standard deviation along the specified axis.

The standard deviation along axis 0 is calculated for each column. Each column has mean 2.5, sum of squared differences 4.5, divided by 2 (ddof=0) gives variance 2.25, std = 1.5.

❓ data_output

intermediate

Variance of a 1D array

What is the output of this code?

NumPy

import numpy as np
arr = np.array([2, 4, 4, 4, 5, 5, 7, 9])
print(np.var(arr))

A 3.0

B 2.0

C 4.0

D 1.5

💡 Hint

Variance is the average of squared differences from the mean.

The variance of the array is 3.0, calculated as the average squared distance from the mean.

🔧 Debug

advanced

Identify the error in variance calculation

What error will this code raise?

NumPy

import numpy as np
arr = np.array([1, 2, 3])
result = np.var(arr, ddof=2)
print(result)

A TypeError: ddof must be an integer

B RuntimeWarning: Degrees of freedom <= 0 for slice

C No error, output is 2.0

D ValueError: ddof must be less than or equal to the number of elements

💡 Hint

Check the degrees of freedom parameter and its effect on variance calculation.

Using ddof=2 with 3 elements gives degrees of freedom = 3-2=1 >0, so no warning or error. Mean=2, sum of squared differences=2, variance=2/1=2.0.

🚀 Application

advanced

Choosing between np.std() and np.var() for data spread

You want to compare the spread of two datasets visually. Which numpy function is better to use for direct comparison and why?

A np.var(), because variance is easier to interpret visually

B np.std(), because variance can be negative

C np.var(), because it is the square root of standard deviation

D np.std(), because it gives spread in the same units as the data

💡 Hint

Think about the units of measurement for variance and standard deviation.

Standard deviation is in the same units as the data, making it easier to interpret and compare visually. Variance is squared units, which can be confusing.

🧠 Conceptual

expert

Effect of axis and ddof on np.var() output shape

Given a 3D numpy array with shape (2,3,4), what is the shape of the output when calling np.var(arr, axis=(1,2), ddof=1)?

A (2,3,4)

B (2,)

C (1,)

D (3,4)

💡 Hint

Think about how variance reduces dimensions along specified axes.

Variance computed along axes 1 and 2 reduces those dimensions, leaving axis 0 intact, so output shape is (2,).

Practice

1. What does np.std() measure in a dataset?

easy

A. The maximum value in the dataset

B. The sum of all data points

C. The average distance of data points from the mean in original units

D. The total number of data points

Solution

1. Step 1: Understand the purpose of np.std()

   np.std() calculates the standard deviation, which shows how spread out the data is around the mean in the same units as the data.

2. Step 2: Compare options to definition

   Only The average distance of data points from the mean in original units correctly describes this as the average distance from the mean in original units.

3. Final Answer:

   The average distance of data points from the mean in original units -> Option C

4. Quick Check:

   Standard deviation = average spread in original units [OK]

Hint: Standard deviation = spread in original units [OK]

Common Mistakes:

• Confusing standard deviation with variance
• Thinking it counts data points
• Mixing it with sum or max values

2. Which of the following is the correct syntax to calculate variance of array arr using NumPy?

easy

A. np.std(arr)

B. np.variance(arr)

C. arr.var()

D. np.var(arr)

Solution

1. Step 1: Recall the NumPy function for variance

   The correct NumPy function to calculate variance is np.var().

2. Step 2: Check each option

   np.var(arr) uses np.var(arr) which is correct. np.variance(arr) uses a non-existent function np.variance(). arr.var() uses arr.var() which works only if arr is a NumPy array, but the question asks for NumPy function syntax. np.std(arr) calculates standard deviation, not variance.

3. Final Answer:

   np.var(arr) -> Option D

4. Quick Check:

   Variance function = np.var() [OK]

Hint: Variance uses np.var(), not np.variance() [OK]

Common Mistakes:

• Using np.variance() which does not exist
• Confusing variance with standard deviation function
• Using method on list instead of NumPy array

3. What is the output of the following code?

import numpy as np
arr = np.array([2, 4, 4, 4, 5, 5, 7, 9])
print(np.var(arr))

medium

A. 2.0

B. 4.0

C. 2.5

D. 3.0

Solution

1. Step 1: Calculate the mean of the array

   The array is [2,4,4,4,5,5,7,9]. Sum = 40, count = 8, mean = 40/8 = 5.

2. Step 2: Calculate variance

   Variance = average of squared differences from mean:
   (2-5)^2=9, (4-5)^2=1, (4-5)^2=1, (4-5)^2=1, (5-5)^2=0, (5-5)^2=0, (7-5)^2=4, (9-5)^2=16
   Sum = 9+1+1+1+0+0+4+16 = 32
   Variance = 32/8 = 4.0

3. Step 3: Check code output

   NumPy's np.var() by default calculates population variance (divides by N). So output is 4.0.

4. Final Answer:

   4.0 -> Option B

5. Quick Check:

   Variance = 4.0 [OK]

Hint: Variance = mean of squared differences [OK]

Common Mistakes:

• Calculating sample variance (dividing by N-1) instead of population
• Mixing variance with standard deviation
• Incorrect mean calculation

4. The code below is intended to print the standard deviation of data. What is wrong?

import numpy as np
data = [1, 2, 3, 4, 5]
print(np.std(data, ddof=1))

medium

A. Nothing is wrong; code runs correctly

B. ddof is not a valid argument for np.std()

C. The code will raise a TypeError because data is a list, not an array

D. The code calculates variance, not standard deviation

Solution

1. Step 1: Check if ddof is valid for np.std()

   The ddof parameter (delta degrees of freedom) is valid for np.std() and adjusts divisor for sample standard deviation.

2. Step 2: Check if input type is acceptable

   np.std() accepts lists and converts them internally to arrays, so no error occurs.

3. Final Answer:

   Nothing is wrong; code runs correctly -> Option A

4. Quick Check:

   ddof works with np.std() and lists accepted [OK]

Hint: ddof works with np.std() for sample std [OK]

Common Mistakes:

• Thinking ddof is invalid for standard deviation
• Assuming lists cause errors in NumPy functions
• Confusing variance and standard deviation

5. You have two datasets:
data1 = np.array([10, 12, 14, 16, 18])
data2 = np.array([10, 10, 10, 10, 10])
Which statement is true about their standard deviation and variance?

hard

A. data1 has higher variance and standard deviation than data2

B. data2 has zero variance and standard deviation

C. data2 has higher variance but lower standard deviation than data1

D. Both datasets have the same variance and standard deviation

Solution

1. Step 1: Analyze data2 values

   All values in data2 are the same (10), so variance and standard deviation must be zero because there is no spread.

2. Step 2: Analyze data1 values

   data1 values vary from 10 to 18, so variance and standard deviation are greater than zero, meaning data1 has higher spread.

3. Final Answer:

   data1 has higher variance and standard deviation than data2 -> Option A

4. Quick Check:

   Constant data variance = 0; varying data variance > 0 [OK]

Hint: Constant data has zero variance and std [OK]

Common Mistakes:

• Assuming equal variance for different spread data
• Confusing variance with mean
• Thinking zero variance means error