np.in1d() for membership testing in NumPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of np.in1d() with integer arrays

What is the output of the following code?

import numpy as np
arr1 = np.array([1, 2, 3, 4, 5])
arr2 = np.array([2, 4, 6])
result = np.in1d(arr1, arr2)
print(result)

NumPy

import numpy as np
arr1 = np.array([1, 2, 3, 4, 5])
arr2 = np.array([2, 4, 6])
result = np.in1d(arr1, arr2)
print(result)

A [ True False True False True]

B [False True False True False]

C [False False False False False]

D [ True True True True True]

💡 Hint

np.in1d() checks if each element of the first array is in the second array.

The function returns a boolean array where each position corresponds to whether the element in arr1 is found in arr2. Only 2 and 4 are in arr2, so positions 1 and 3 are True.

❓ data_output

intermediate

Count of elements in arr1 found in arr2 using np.in1d()

Given the arrays below, how many elements of arr1 are present in arr2?

import numpy as np
arr1 = np.array([10, 20, 30, 40, 50])
arr2 = np.array([15, 20, 35, 40, 55])
mask = np.in1d(arr1, arr2)
count = np.sum(mask)
print(count)

NumPy

import numpy as np
arr1 = np.array([10, 20, 30, 40, 50])
arr2 = np.array([15, 20, 35, 40, 55])
mask = np.in1d(arr1, arr2)
count = np.sum(mask)
print(count)

A 1

B 3

C 2

D 0

💡 Hint

Count how many True values are in the boolean mask from np.in1d().

Only 20 and 40 from arr1 are found in arr2, so the count is 2.

🔧 Debug

advanced

Identify the error in np.in1d() usage

What error will this code raise?

import numpy as np
arr1 = [1, 2, 3]
arr2 = 2
result = np.in1d(arr1, arr2)
print(result)

NumPy

import numpy as np
arr1 = [1, 2, 3]
arr2 = 2
result = np.in1d(arr1, arr2)
print(result)

A TypeError: 'int' object is not iterable

B ValueError: operands could not be broadcast together

C No error, outputs [False False True]

D SyntaxError: invalid syntax

💡 Hint

np.in1d() expects the second argument to be iterable.

Passing a single integer instead of an iterable causes a TypeError because np.in1d tries to iterate over arr2.

🚀 Application

advanced

Filter DataFrame rows using np.in1d()

You have a DataFrame of students and their grades. You want to select only students whose names are in a given list.
Which code snippet correctly filters the DataFrame?

import pandas as pd
import numpy as np

data = {'Name': ['Alice', 'Bob', 'Charlie', 'David'], 'Grade': [85, 92, 78, 90]}
df = pd.DataFrame(data)
selected_names = ['Bob', 'David']

# Which line filters df to only rows with names in selected_names?

NumPy

import pandas as pd
import numpy as np

data = {'Name': ['Alice', 'Bob', 'Charlie', 'David'], 'Grade': [85, 92, 78, 90]}
df = pd.DataFrame(data)
selected_names = ['Bob', 'David']

A filtered_df = df[np.in1d(selected_names, df['Name'])]

B filtered_df = df[df['Name'] == selected_names]

C filtered_df = df[df['Name'].isin(selected_names)]

D filtered_df = df[np.in1d(df['Name'], selected_names)]

💡 Hint

np.in1d() returns a boolean array matching df['Name'] elements against selected_names.

Option D uses np.in1d() correctly to create a boolean mask for filtering rows. Option D uses pandas isin(), which is valid but not np.in1d(). Options A and B are incorrect.

🧠 Conceptual

expert

Understanding np.in1d() behavior with repeated elements

Consider the arrays:

import numpy as np
arr1 = np.array([1, 2, 2, 3, 4, 4, 4])
arr2 = np.array([2, 4])
result = np.in1d(arr1, arr2)
print(result)

What is the shape and content of the output array?

NumPy

import numpy as np
arr1 = np.array([1, 2, 2, 3, 4, 4, 4])
arr2 = np.array([2, 4])
result = np.in1d(arr1, arr2)
print(result)

A Shape (7,), [False, True, True, False, True, True, True]

B Shape (2,), [True, True]

C Shape (7,), [True, True, True, True, True, True, True]

D Shape (4,), [True, True, True, True]

💡 Hint

np.in1d() returns a boolean array matching the first array's shape.

The output array has the same length as arr1 (7). Each position is True if the element is in arr2 (2 or 4), otherwise False.