outer() for outer operations in NumPy - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to compute the outer product of two arrays using numpy.

NumPy

import numpy as np

arr1 = np.array([1, 2, 3])
arr2 = np.array([4, 5, 6])
result = np.[1](arr1, arr2)
print(result)

Drag options to blanks, or click blank then click option'

Aouter

Bdot

Csum

Dmean

Attempts:

3 left

💡 Hint

Common Mistakes

Using np.dot instead of np.outer, which computes a dot product, not an outer product.

Using sum or mean which are aggregation functions, not for outer products.

The np.outer() function computes the outer product of two arrays, resulting in a matrix where each element is the product of elements from the two input arrays.

2fill in blank

medium

Complete the code to calculate the outer product of two 1D numpy arrays and store it in 'result'.

NumPy

import numpy as np

x = np.array([2, 3])
y = np.array([5, 7])
result = np.[1](x, y)
print(result)

Drag options to blanks, or click blank then click option'

Ainner

Bcross

Cdot

Douter

Attempts:

3 left

💡 Hint

Common Mistakes

Using np.dot or np.inner which compute dot and inner products, not outer.

Using np.cross which is for vector cross product, only for 3D vectors.

np.outer() calculates the outer product of two vectors, producing a matrix where each element is the product of elements from the two input arrays.

3fill in blank

hard

Fix the error in the code to correctly compute the outer product of arrays 'a' and 'b'.

NumPy

import numpy as np

a = np.array([1, 2])
b = np.array([3, 4])
result = np.[1](a, b)
print(result)

Drag options to blanks, or click blank then click option'

Amultiply

Bdot

Couter

Dsum

Attempts:

3 left

💡 Hint

Common Mistakes

Using np.dot which computes dot product, resulting in a scalar for 1D arrays.

Using np.multiply which does element-wise multiplication, not outer product.

The correct function to compute the outer product is np.outer(). Using np.dot() or others will not produce the outer product matrix.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps each word to its length only if the length is greater than 3.

NumPy

words = ['data', 'science', 'ai', 'ml']
lengths = {word: [1] for word in words if [2]
print(lengths)

Drag options to blanks, or click blank then click option'

Alen(word)

Bword > 3

Clen(word) > 3

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself instead of its length in the dictionary values.

Comparing the word string directly to a number instead of its length.

The dictionary comprehension uses len(word) to get the length of each word and filters words where the length is greater than 3 using len(word) > 3.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps each word in uppercase to its length only if the length is less than or equal to 4.

NumPy

words = ['data', 'science', 'ai', 'ml']
result = { [1]: [2] for word in words if [3] }
print(result)

Drag options to blanks, or click blank then click option'

Aword.upper()

Blen(word)

Clen(word) <= 4

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself as the key instead of uppercase.

Using the word instead of its length as the value.

Incorrect condition like comparing the word string directly to a number.

The dictionary comprehension maps each word in uppercase (word.upper()) to its length (len(word)) only if the length is less than or equal to 4 (len(word) <= 4).

Practice

1. What does the numpy.outer() function do with two input arrays?

easy

A. It concatenates the two arrays into one.

B. It adds the two arrays element-wise.

C. It returns the dot product of the two arrays.

D. It computes the outer product, combining every element of the first array with every element of the second array.

Solution

1. Step 1: Understand the purpose of numpy.outer()

   The function takes two arrays and pairs each element of the first array with each element of the second array.

2. Step 2: Identify the output type

   The result is a 2D array showing all combinations, which is the outer product.

3. Final Answer:

   It computes the outer product, combining every element of the first array with every element of the second array. -> Option D

4. Quick Check:

   Outer product = combine all pairs [OK]

Hint: Outer product pairs all elements from both arrays [OK]

Common Mistakes:

• Confusing outer product with dot product
• Thinking it adds arrays element-wise
• Assuming it concatenates arrays

2. Which of the following is the correct syntax to compute the outer product of arrays a and b using numpy?

easy

A. numpy.outer(a, b)

B. numpy.outerproduct(a, b)

C. numpy.outer(a * b)

D. numpy.outer(a + b)

Solution

1. Step 1: Recall the function name and parameters

   The correct function is numpy.outer() and it takes two arrays as separate arguments.

2. Step 2: Check the options for correct syntax

   Only numpy.outer(a, b) matches the correct function call with two arguments.

3. Final Answer:

   numpy.outer(a, b) -> Option A

4. Quick Check:

   Correct function call = numpy.outer(a, b) [OK]

Hint: Use numpy.outer(array1, array2) with two arguments [OK]

Common Mistakes:

• Using wrong function name like outerproduct
• Passing one combined argument instead of two
• Using addition or multiplication inside outer()

3. What is the output of the following code?

import numpy as np
x = np.array([1, 2])
y = np.array([3, 4])
result = np.outer(x, y)
print(result)

medium

A. [[3 6]
[4 8]]

B. [[3 4]
[6 8]]

C. [[4 6]
[8 12]]

D. [[1 2]
[3 4]]

Solution

1. Step 1: Calculate outer product manually

   Multiply each element of x by each element of y:
   1*3=3, 1*4=4
   2*3=6, 2*4=8

2. Step 2: Form the 2D array

   Arrange results as rows for each element of x:
   [[3, 4],
   [6, 8]]

3. Final Answer:

   [[3 4] [6 8]] -> Option B

4. Quick Check:

   Outer product matches [[3 4], [6 8]] [OK]

Hint: Multiply each element of first array by all in second [OK]

Common Mistakes:

• Mixing row and column order
• Confusing with dot product output
• Swapping elements of arrays

4. What is the output of the following code?

import numpy as np
x = np.array([1, 2, 3])
y = 5
result = np.outer(x, y)
print(result)

medium

A. [[5]
[10]
[15]]

B. [[5 10 15]]

C. [[1 2 3]
[5]]

D. Throws an error

Solution

1. Step 1: np.outer() handles scalar second argument

   It treats y=5 as a length-1 array, producing a (3,1) shaped array.

2. Step 2: Compute the products

   [[1*5], [2*5], [3*5]] = [[5], [10], [15]]

3. Final Answer:

   [[5] [10] [15]] -> Option A

4. Quick Check:

   Scalar second arg -> column vector [OK]

Hint: np.outer accepts scalars, treating as 1-element array [OK]

Common Mistakes:

• Thinking scalars cause errors
• Expecting row vector instead of column
• Confusing with concatenation

5. You have two arrays:

a = np.array([1, 0, 2])
b = np.array([3, np.nan])

You want to create a 2D array where each element is the product of elements from a and b, but you want to ignore zeros in a (replace their entire row with zeros). Which code correctly uses numpy.outer() and post-processing to achieve this?

hard

A. result = np.outer(a, b)
result[:, b == 0] = 0

B. result = np.outer(a, b)
result[a != 0, :] = 0

C. result = np.outer(a, b)
result[a == 0, :] = 0

D. result = np.outer(a, b)
result[b == 0, :] = 0

Solution

1. Step 1: Compute outer product

   Using np.outer(a, b) creates a 2D array with all products.

2. Step 2: Replace rows where a is zero with zeros

   Use boolean indexing result[a == 0, :] = 0 to zero out rows corresponding to zero elements in a.

3. Final Answer:

   result = np.outer(a, b) result[a == 0, :] = 0 -> Option C

4. Quick Check:

   Zero rows where a is zero [OK]

Hint: Zero rows where a is zero using boolean indexing [OK]

Common Mistakes:

• Zeroing columns instead of rows
• Using wrong condition for zeroing
• Confusing axis in indexing