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.

Practice

1. What does the np.in1d() function do in NumPy?

easy

A. Finds the unique elements in an array.

B. Sorts the elements of an array in ascending order.

C. Calculates the sum of elements in an array.

D. Checks if elements of one array are present in another array and returns a boolean array.

Solution

1. Step 1: Understand the purpose of np.in1d()

   The function checks membership of each element in the first array against the second array.

2. Step 2: Identify the output type

   It returns a boolean array indicating True where elements are found and False otherwise.

3. Final Answer:

   Checks if elements of one array are present in another array and returns a boolean array. -> Option D

4. Quick Check:

   Membership test = Checks if elements of one array are present in another array and returns a boolean array. [OK]

Hint: Remember: np.in1d returns booleans for membership [OK]

Common Mistakes:

• Confusing np.in1d() with sorting or summing functions
• Expecting np.in1d() to return the matching elements instead of booleans
• Thinking np.in1d() modifies the original arrays

2. Which of the following is the correct syntax to check if elements of array a are in array b using np.in1d()?

easy

A. np.in1d(b, a)

B. np.in1d(a, b)

C. np.in1d(a == b)

D. np.in1d(a, b, axis=1)

Solution

1. Step 1: Recall np.in1d() parameter order

   The first argument is the array to test membership for, the second is the array to check against.

2. Step 2: Evaluate each option

   np.in1d(a, b) uses correct order: np.in1d(a, b). np.in1d(b, a) reverses arrays, np.in1d(a == b) uses invalid syntax, np.in1d(a, b, axis=1) uses unsupported axis parameter.

3. Final Answer:

   np.in1d(a, b) -> Option B

4. Quick Check:

   Correct syntax = np.in1d(a, b) [OK]

Hint: First array is tested, second array is reference [OK]

Common Mistakes:

• Swapping the order of arrays in np.in1d()
• Adding unsupported parameters like axis
• Using comparison operators inside np.in1d()

3. What is the output of the following code?

import numpy as np
x = np.array([1, 3, 5, 7])
y = np.array([3, 4, 5])
result = np.in1d(x, y)
print(result)

medium

A. [False True True False]

B. [True False True False]

C. [False True False False]

D. [True True True True]

Solution

1. Step 1: Check each element of x against y

   1 in y? No (False), 3 in y? Yes (True), 5 in y? Yes (True), 7 in y? No (False).

2. Step 2: Form the boolean array

   Result is [False, True, True, False].

3. Final Answer:

   [False True True False] -> Option A

4. Quick Check:

   Membership booleans = [False True True False] [OK]

Hint: Check each element one by one for membership [OK]

Common Mistakes:

• Mixing up True and False positions
• Assuming np.in1d returns matching elements instead of booleans
• Forgetting to import numpy

4. The following code throws an error. What is the mistake?

import numpy as np
x = [1, 2, 3]
y = np.array([2, 3, 4])
result = np.in1d(x, y, axis=0)
print(result)

medium

A. np.in1d() requires both inputs to be lists.

B. x should be converted to a NumPy array before using np.in1d().

C. np.in1d() does not accept the 'axis' parameter.

D. The arrays x and y must have the same shape.

Solution

1. Step 1: Check np.in1d() parameters

   np.in1d() accepts only two main parameters: the test array and the array to check against. It does not support an 'axis' parameter.

2. Step 2: Identify the error cause

   Passing axis=0 causes a TypeError because it's not a valid argument.

3. Final Answer:

   np.in1d() does not accept the 'axis' parameter. -> Option C

4. Quick Check:

   Invalid parameter = np.in1d() does not accept the 'axis' parameter. [OK]

Hint: np.in1d() only takes two main arguments [OK]

Common Mistakes:

• Trying to use axis parameter with np.in1d()
• Assuming input types must match exactly
• Thinking np.in1d() requires both inputs as arrays

5. You have two arrays:

data = np.array([10, 20, 30, 40, 50])
filter_vals = np.array([20, 40, 60])

You want to create a new array containing only elements from data that are present in filter_vals. Which code snippet correctly achieves this?

hard

A. filtered = data[np.in1d(data, filter_vals)]

B. filtered = filter_vals[np.in1d(filter_vals, data)]

C. filtered = np.in1d(data, filter_vals)

D. filtered = data[filter_vals]

Solution

1. Step 1: Use np.in1d() to get boolean mask

   np.in1d(data, filter_vals) returns a boolean array marking elements of data present in filter_vals.

2. Step 2: Use boolean mask to filter data

   Indexing data with this boolean mask selects only matching elements.

3. Final Answer:

   filtered = data[np.in1d(data, filter_vals)] -> Option A

4. Quick Check:

   Boolean mask indexing = filtered = data[np.in1d(data, filter_vals)] [OK]

Hint: Use np.in1d() mask to index original array [OK]

Common Mistakes:

• Indexing filter_vals instead of data
• Using np.in1d() without indexing
• Trying to index with filter_vals directly