Set operations on structured data in NumPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of numpy structured array intersection

What is the output of this code that finds the intersection of two structured numpy arrays based on all fields?

NumPy

import numpy as np

arr1 = np.array([(1, 'apple'), (2, 'banana'), (3, 'cherry')], dtype=[('id', 'i4'), ('fruit', 'U10')])
arr2 = np.array([(2, 'banana'), (3, 'cherry'), (4, 'date')], dtype=arr1.dtype)

result = np.intersect1d(arr1, arr2)
print(result)

A [(2, 'banana') (3, 'cherry')]

B [(1, 'apple') (2, 'banana') (3, 'cherry')]

C [(3, 'cherry') (4, 'date')]

D [(1, 'apple') (4, 'date')]

💡 Hint

Think about which records appear in both arrays exactly.

The intersection returns only the records present in both arrays. Here, (2, 'banana') and (3, 'cherry') are common.

❓ data_output

intermediate

Number of unique records after union

Given two structured numpy arrays, what is the number of unique records after performing a union operation?

NumPy

import numpy as np

arr1 = np.array([(10, 'red'), (20, 'blue')], dtype=[('code', 'i4'), ('color', 'U10')])
arr2 = np.array([(20, 'blue'), (30, 'green')], dtype=arr1.dtype)

union_result = np.union1d(arr1, arr2)
print(len(union_result))

A 2

B 3

C 4

D 1

💡 Hint

Count all unique records from both arrays combined.

The union combines all unique records from both arrays. Here, (10, 'red'), (20, 'blue'), and (30, 'green') are unique, so count is 3.

🔧 Debug

advanced

Error in set difference on structured arrays

What error does this code raise when trying to find the difference between two structured numpy arrays?

NumPy

import numpy as np

arr1 = np.array([(1, 2.0), (3, 4.0)], dtype=[('a', 'i4'), ('b', 'f4')])
arr2 = np.array([(3, 4.0)], dtype=[('a', 'i4'), ('b', 'f4')])

diff = np.setdiff1d(arr1, arr2, assume_unique=True)
print(diff)

A TypeError: unhashable type: 'numpy.void'

B ValueError: operands could not be broadcast together

C No error, outputs [(1, 2.0)]

D IndexError: index out of bounds

💡 Hint

Check if the operation is valid for structured arrays and what output is expected.

np.setdiff1d works on structured arrays and returns the records in arr1 not in arr2. No error occurs here.

❓ visualization

advanced

Visualizing set intersection of structured arrays

Which option correctly plots a Venn diagram showing the intersection size of two structured numpy arrays based on their 'id' field?

NumPy

import numpy as np
import matplotlib.pyplot as plt
from matplotlib_venn import venn2

arr1 = np.array([(1, 'x'), (2, 'y'), (3, 'z')], dtype=[('id', 'i4'), ('val', 'U1')])
arr2 = np.array([(2, 'y'), (3, 'z'), (4, 'w')], dtype=arr1.dtype)

ids1 = set(arr1['id'])
ids2 = set(arr2['id'])

plt.figure(figsize=(5,5))

A

venn2([list(arr1), list(arr2)], set_labels=('arr1', 'arr2'))
plt.show()

B

venn2([arr1, arr2], set_labels=('arr1', 'arr2'))
plt.show()

C

venn2([arr1['val'], arr2['val']], set_labels=('arr1', 'arr2'))
plt.show()

D

venn2([ids1, ids2], set_labels=('arr1', 'arr2'))
plt.show()

💡 Hint

Venn diagram requires sets of hashable elements, use the 'id' field.

Option D correctly converts the 'id' fields to sets and plots the Venn diagram showing intersection sizes.

🧠 Conceptual

expert

Why does np.intersect1d require structured arrays to have the same dtype?

Why must two structured numpy arrays have the same dtype to use np.intersect1d correctly?

A Because np.intersect1d compares raw bytes and different dtypes change memory layout, causing incorrect comparisons.

B Because np.intersect1d only works on 1D numeric arrays, not structured arrays.

C Because structured arrays with different dtypes are always empty when intersected, so dtype doesn't matter.

D Because numpy automatically converts dtypes internally, so different dtypes cause slowdowns but no errors.

💡 Hint

Think about how numpy compares structured array elements at the memory level.

np.intersect1d compares elements by their byte representation. Different dtypes mean different memory layouts, so comparisons fail or give wrong results.

Practice

1. What does the numpy.intersect1d function do when applied to two structured arrays?

easy

A. Finds rows present only in the second array

B. Combines all rows from both arrays without duplicates

C. Finds the common rows present in both arrays

D. Finds rows present only in the first array

Solution

1. Step 1: Understand intersect1d purpose

   numpy.intersect1d returns elements common to both input arrays.

2. Step 2: Apply to structured arrays

   For structured arrays, it compares rows and returns those present in both arrays.

3. Final Answer:

   Finds the common rows present in both arrays -> Option C

4. Quick Check:

   Intersection = common rows [OK]

Hint: Intersect means common elements only [OK]

Common Mistakes:

• Confusing intersect1d with union1d
• Thinking it returns unique rows from one array only
• Assuming it returns rows exclusive to one array

2. Which of the following is the correct syntax to find the union of two structured numpy arrays a and b?

easy

A. numpy.union(a | b)

B. numpy.union(a, b)

C. numpy.setunion(a, b)

D. numpy.union1d(a, b)

Solution

1. Step 1: Recall numpy union function

   The correct function to find union is numpy.union1d.

2. Step 2: Check syntax correctness

   The syntax is numpy.union1d(a, b) with two arguments.

3. Final Answer:

   numpy.union1d(a, b) -> Option D

4. Quick Check:

   Use union1d for union operation [OK]

Hint: Use union1d, not union or setunion [OK]

Common Mistakes:

• Using nonexistent functions like union or setunion
• Passing arguments incorrectly with bitwise operators
• Confusing union1d with intersect1d

3. Given two structured arrays:

a = np.array([(1, 'A'), (2, 'B'), (3, 'C')], dtype=[('id', int), ('val', 'U1')])
b = np.array([(2, 'B'), (4, 'D')], dtype=[('id', int), ('val', 'U1')])
print(np.setdiff1d(a, b))

What is the output?

medium

A. [(1, 'A') (3, 'C')]

B. [(2, 'B') (4, 'D')]

C. [(1, 'A') (2, 'B') (3, 'C')]

D. [(4, 'D')]

Solution

1. Step 1: Understand setdiff1d behavior

   np.setdiff1d(a, b) returns rows in a not in b.

2. Step 2: Compare rows of a and b

   Rows (2, 'B') is common, so excluded. Remaining are (1, 'A') and (3, 'C').

3. Final Answer:

   [(1, 'A') (3, 'C')] -> Option A

4. Quick Check:

   Difference = rows only in a [OK]

Hint: Setdiff1d returns items only in first array [OK]

Common Mistakes:

• Including common rows in output
• Confusing setdiff1d with union1d or intersect1d
• Expecting output from second array instead

4. Consider this code snippet:

a = np.array([(1, 'X'), (2, 'Y')], dtype=[('id', int), ('val', 'U1')])
b = np.array([(2, 'Y'), (3, 'Z')], dtype=[('id', int), ('val', 'U2')])
result = np.setxor1d(a, b)
print(result)

It raises an error. What is the likely cause?

medium

A. Arrays must be sorted before setxor1d

B. Structured arrays have different dtypes or field order

C. setxor1d does not support structured arrays

D. Missing import statement for numpy

Solution

1. Step 1: Check dtype compatibility

   For set operations on structured arrays, dtypes and field order must match exactly.

2. Step 2: Identify cause of error

   If dtypes differ or field order differs, setxor1d raises an error.

3. Final Answer:

   Structured arrays have different dtypes or field order -> Option B

4. Quick Check:

   Matching dtypes needed for set operations [OK]

Hint: Ensure structured arrays have identical dtypes [OK]

Common Mistakes:

• Assuming setxor1d can't handle structured arrays
• Forgetting to check dtype and field order
• Thinking arrays must be sorted first

5. You have two structured arrays representing employee records:

emp1 = np.array([(101, 'Alice'), (102, 'Bob'), (103, 'Carol')], dtype=[('id', int), ('name', 'U10')])
emp2 = np.array([(102, 'Bob'), (104, 'Dave')], dtype=[('id', int), ('name', 'U10')])

You want to find employees who are in either list but not both (exclusive employees). Which numpy function and code will give the correct result?

hard

A. np.setxor1d(emp1, emp2)

B. np.union1d(emp1, emp2)

C. np.intersect1d(emp1, emp2)

D. np.setdiff1d(emp1, emp2)

Solution

1. Step 1: Understand exclusive elements

   Exclusive employees are those in one array but not both, which is the symmetric difference.

2. Step 2: Identify correct numpy function

   np.setxor1d returns elements in either array but not in both.

3. Final Answer:

   np.setxor1d(emp1, emp2) -> Option A

4. Quick Check:

   Symmetric difference = setxor1d [OK]

Hint: Use setxor1d for exclusive elements [OK]

Common Mistakes:

• Using union1d which includes all elements
• Using intersect1d which finds common only
• Using setdiff1d which finds only one-sided difference