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.