np.power() and np.square() in NumPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of np.power() with array and scalar

What is the output of this code snippet?

NumPy

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

A [1 6 9]

B [1 8 27]

C [3 6 9]

D [1 4 9]

💡 Hint

np.power raises each element to the given power.

np.power(arr, 3) raises each element of arr to the power 3, so 1^3=1, 2^3=8, 3^3=27.

❓ Predict Output

intermediate

Difference between np.square() and np.power()

What is the output of this code?

NumPy

import numpy as np
arr = np.array([2, 4, 6])
result = np.square(arr)
print(result)

A [1 16 36]

B [8 64 216]

C [2 4 6]

D [4 16 36]

💡 Hint

np.square is like np.power with exponent 2.

np.square(arr) squares each element: 2^2=4, 4^2=16, 6^2=36.

❓ data_output

advanced

Result shape after np.power with broadcasting

What is the shape of the result array after this operation?

NumPy

import numpy as np
arr1 = np.array([[1, 2], [3, 4]])
arr2 = np.array([2, 3])
result = np.power(arr1, arr2)
print(result.shape)

A (2, 2)

B (2,)

C (2, 3)

D (3, 2)

💡 Hint

Broadcasting matches shapes to perform element-wise operations.

arr2 shape (2,) broadcasts to arr1 shape (2,2), so result shape is (2,2).

🔧 Debug

advanced

Identify the error in np.power usage

What error does this code raise?

NumPy

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

A SyntaxError

B ValueError

C TypeError

D No error, outputs [1 4 9]

💡 Hint

The exponent must be a number, not a string.

np.power expects numeric types; passing a string causes TypeError.

🚀 Application

expert

Using np.square to compute Euclidean distances

Given two 2D points, which code correctly computes the squared Euclidean distance using np.square?

A np.sum(np.square(point1 - point2))

B np.square(np.sum(point1 - point2))

C np.square(point1) - np.square(point2)

D np.sum(point1 - point2)**2

💡 Hint

Squared distance is sum of squared differences of coordinates.

Option A squares element-wise differences then sums them, correctly computing squared distance.

Practice

1. What does the np.square() function do in NumPy?

easy

A. It raises each element of an array to the power of 2.

B. It calculates the square root of each element in an array.

C. It multiplies two arrays element-wise.

D. It returns the sum of squares of array elements.

Solution

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

   The function np.square() takes each element in an array and raises it to the power of 2.

2. Step 2: Compare with other options

   It does not calculate square roots, multiply arrays, or sum squares; it only squares each element.

3. Final Answer:

   It raises each element of an array to the power of 2. -> Option A

4. Quick Check:

   np.square(x) = x² [OK]

Hint: Remember: square means power of 2, not root or sum [OK]

Common Mistakes:

• Confusing square with square root
• Thinking it sums squares instead of element-wise operation
• Mixing with multiplication of arrays

2. Which of the following is the correct syntax to square all elements in a NumPy array arr using np.power()?

easy

A. np.power(arr)

B. np.power(2, arr)

C. np.power(arr, 2)

D. np.power(arr, arr)

Solution

1. Step 1: Understand np.power() syntax

   The function np.power(base, exponent) raises each element in base to the exponent.

2. Step 2: Apply to square elements

   To square elements of arr, use np.power(arr, 2). Other options misuse the order or miss the exponent.

3. Final Answer:

   np.power(arr, 2) -> Option C

4. Quick Check:

   np.power(arr, 2) = arr squared [OK]

Hint: np.power(base, exponent) order matters: base first [OK]

Common Mistakes:

• Swapping base and exponent
• Omitting the exponent argument
• Using the array as exponent incorrectly

3. What is the output of the following code?

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

medium

A. [1 6 9]

B. [1 8 27]

C. [3 6 9]

D. [1 4 9]

Solution

1. Step 1: Understand np.power(arr, 3)

   This raises each element of arr to the power of 3.

2. Step 2: Calculate each element

   1³=1, 2³=8, 3³=27, so the result is [1, 8, 27].

3. Final Answer:

   [1 8 27] -> Option B

4. Quick Check:

   np.power([1,2,3],3) = [1,8,27] [OK]

Hint: Power 3 means cube each element [OK]

Common Mistakes:

• Calculating square instead of cube
• Adding elements instead of powering
• Confusing element-wise with sum

4. Identify the error in the following code snippet:

import numpy as np
arr = np.array([2, 4, 6])
result = np.square(arr, 2)
print(result)

medium

A. np.square() takes only one argument, but two were given.

B. np.square() cannot be used on arrays.

C. The array must be converted to a list before squaring.

D. The code should use np.power(arr, 2) instead.

Solution

1. Step 1: Check np.square() function signature

   np.square() accepts only one argument: the array or number to square.

2. Step 2: Analyze the code error

   The code passes two arguments, which causes a TypeError.

3. Final Answer:

   np.square() takes only one argument, but two were given. -> Option A

4. Quick Check:

   np.square(x) needs 1 argument [OK]

Hint: np.square() needs exactly one argument [OK]

Common Mistakes:

• Passing extra arguments to np.square()
• Thinking np.square() needs base and exponent
• Confusing np.square() with np.power()

5. You have a NumPy array data = np.array([1, -2, 3, -4]). You want to create a new array where each element is squared, but negative values should be squared and then multiplied by -1 to keep their sign effect. Which code correctly achieves this?

hard

A. np.square(np.abs(data))

B. np.square(data)

C. np.power(data, 2)

D. np.where(data < 0, -np.square(data), np.square(data))

Solution

1. Step 1: Understand the requirement

   Negative values should be squared and then multiplied by -1; positive values just squared.

2. Step 2: Analyze each option

   np.where(data < 0, -np.square(data), np.square(data)) uses np.where to apply -np.square() for negatives and np.square() for positives, matching the requirement exactly.

3. Final Answer:

   np.where(data < 0, -np.square(data), np.square(data)) -> Option D

4. Quick Check:

   Conditional square with sign handled = np.where(data < 0, -np.square(data), np.square(data)) [OK]

Hint: Use np.where for conditionally applying functions [OK]

Common Mistakes:

• Multiplying sign before squaring
• Using np.sign(data) directly without condition
• Confusing absolute value with sign handling