np.std() and np.var() for spread in NumPy

Introduction

We use np.std() and np.var() to understand how data points spread out from the average. This helps us see if data is close together or very different.

Checking how consistent students' test scores are in a class.

Measuring how much daily temperatures change in a week.

Understanding the variation in sales numbers for a product.

Comparing the spread of heights in two different groups of people.

Syntax

NumPy

np.std(array, axis=None, ddof=0)
np.var(array, axis=None, ddof=0)

array is your data, like a list or NumPy array.

axis decides if you calculate spread for the whole data or along rows/columns.

Examples

Calculates the standard deviation of all numbers in the list.

NumPy

np.std([1, 2, 3, 4, 5])

Calculates the variance of all numbers in the list.

NumPy

np.var([1, 2, 3, 4, 5])

Calculates standard deviation for each column.

NumPy

np.std([[1, 2], [3, 4]], axis=0)

Calculates variance for each row.

NumPy

np.var([[1, 2], [3, 4]], axis=1)

Sample Program

This program calculates how much daily sales numbers spread out from the average sales. Standard deviation and variance give us this spread in different ways.

NumPy

import numpy as np

# Sample data: daily sales for a week
sales = np.array([100, 120, 90, 110, 105, 115, 95])

# Calculate standard deviation
std_dev = np.std(sales)

# Calculate variance
variance = np.var(sales)

print(f"Standard Deviation: {std_dev:.2f}")
print(f"Variance: {variance:.2f}")

Important Notes

Standard deviation is the square root of variance.

Use ddof=1 if you want sample standard deviation or variance.

Lower spread means data points are close to the average; higher means more spread out.

Summary

np.std() finds how much data varies around the average in original units.

np.var() finds the average squared difference from the mean.

Both help understand data consistency and variation.

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