Set operations on structured data in NumPy - Mini Project: Build & Apply

Set operations on structured data

📖 Scenario: You work in a store that tracks products by their name and price. You want to find which products are unique to each store's list and which products are common.

🎯 Goal: Build a program that uses numpy structured arrays to find the intersection, union, and difference of two product lists.

📋 What You'll Learn

Create two numpy structured arrays with product data

Define a configuration variable for the data type

Use numpy set operations on structured arrays

Print the results clearly

💡 Why This Matters

🌍 Real World

Stores and businesses often compare product lists from different branches or suppliers to manage inventory and pricing.

💼 Career

Data scientists use structured arrays and set operations to clean, compare, and analyze datasets with multiple fields efficiently.

1

DATA SETUP: Create two numpy structured arrays

Import numpy as np. Create a numpy structured array called store1 with these exact entries: ('apple', 0.50), ('banana', 0.30), ('orange', 0.80). Create another numpy structured array called store2 with these exact entries: ('banana', 0.30), ('kiwi', 1.00), ('apple', 0.50). Use a data type with fields 'name' as 'U10' and 'price' as 'f4'.

NumPy

# Import numpy as np
# Create store1 and store2 structured arrays with given entries and dtype

Hint

Use np.array with a list of tuples and specify dtype for structured arrays.

2

CONFIGURATION: Define the data type variable

Create a variable called dtype and set it to a list of tuples with fields ('name', 'U10') and ('price', 'f4').

NumPy

import numpy as np

# Define dtype variable for structured arrays
# Your code here

Hint

Set dtype exactly as a list of tuples with field names and types.

3

CORE LOGIC: Use numpy set operations on structured arrays

Use np.intersect1d with store1 and store2 to find common products and save to common. Use np.union1d to find all unique products and save to all_products. Use np.setdiff1d with store1 and store2 to find products only in store1 and save to only_store1.

NumPy

import numpy as np

dtype = [('name', 'U10'), ('price', 'f4')]
store1 = np.array([('apple', 0.50), ('banana', 0.30), ('orange', 0.80)], dtype=dtype)
store2 = np.array([('banana', 0.30), ('kiwi', 1.00), ('apple', 0.50)], dtype=dtype)

# Find common products
# Find all unique products
# Find products only in store1
# Your code here

Hint

Use np.intersect1d, np.union1d, and np.setdiff1d with the two arrays.

4

OUTPUT: Print the results clearly

Print the variables common, all_products, and only_store1 each on a separate line with labels: "Common products:", "All products:", and "Only in store1:" respectively.

NumPy

import numpy as np

dtype = [('name', 'U10'), ('price', 'f4')]
store1 = np.array([('apple', 0.50), ('banana', 0.30), ('orange', 0.80)], dtype=dtype)
store2 = np.array([('banana', 0.30), ('kiwi', 1.00), ('apple', 0.50)], dtype=dtype)

common = np.intersect1d(store1, store2)
all_products = np.union1d(store1, store2)
only_store1 = np.setdiff1d(store1, store2)

# Print common, all_products, and only_store1 with labels
# Your code here

Hint

Use print with labels and variables on separate lines.

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