np.setdiff1d() for difference in NumPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of np.setdiff1d() with simple arrays

What is the output of the following code snippet?

NumPy

import numpy as np

arr1 = np.array([1, 2, 3, 4, 5])
arr2 = np.array([3, 4, 6])
result = np.setdiff1d(arr1, arr2)
print(result)

A [1 2 3 4 5 6]

B [3 4 6]

C [1 2 5]

D [6]

💡 Hint

np.setdiff1d returns elements in the first array that are NOT in the second array.

np.setdiff1d(arr1, arr2) returns elements from arr1 that are not present in arr2. Here, 1, 2, and 5 are in arr1 but not in arr2.

❓ data_output

intermediate

Number of elements in difference array

How many elements are in the resulting array after running this code?

NumPy

import numpy as np

arr1 = np.array([10, 20, 30, 40, 50])
arr2 = np.array([20, 40, 60])
result = np.setdiff1d(arr1, arr2)
print(len(result))

A 3

B 2

C 4

D 5

💡 Hint

Count elements in arr1 that are not in arr2.

Elements 10, 30, and 50 are in arr1 but not in arr2, so the length is 3.

🔧 Debug

advanced

Identify the error in using np.setdiff1d with lists

What error will this code produce?

NumPy

import numpy as np

list1 = [1, 2, 3]
list2 = [2, 3, 4]
result = np.setdiff1d(list1, list2)
print(result)

A SyntaxError

B [1]

C TypeError: unhashable type: 'list'

D No error, output: [1]

💡 Hint

np.setdiff1d can accept lists as input because it converts them internally to arrays.

np.setdiff1d accepts lists and converts them to arrays internally, so no error occurs and output is [1].

❓ Predict Output

advanced

Output of np.setdiff1d with repeated elements

What is the output of this code?

NumPy

import numpy as np

arr1 = np.array([1, 2, 2, 3, 4, 4, 5])
arr2 = np.array([2, 4])
result = np.setdiff1d(arr1, arr2)
print(result)

A [1 3 5]

B [1 2 3 4 5]

C [1 3 4 5]

D [2 4]

💡 Hint

np.setdiff1d returns unique sorted values from arr1 not in arr2.

Repeated elements are ignored; only unique values from arr1 not in arr2 remain: 1, 3, 5.

🚀 Application

expert

Using np.setdiff1d to find missing data entries

Given two arrays representing IDs of users who completed two different surveys, which option correctly finds users who completed the first survey but NOT the second?

NumPy

import numpy as np

survey1 = np.array([101, 102, 103, 104, 105])
survey2 = np.array([102, 104, 106])
missing_users = ???
print(missing_users)

A np.union1d(survey1, survey2)

B np.setdiff1d(survey1, survey2)

C np.intersect1d(survey1, survey2)

D np.setdiff1d(survey2, survey1)

💡 Hint

We want users in survey1 but not in survey2.

np.setdiff1d(survey1, survey2) returns elements in survey1 missing from survey2, which is the correct way to find users who completed only the first survey.