Set operations on structured data in NumPy - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a structured numpy array with fields 'name' and 'age'.

NumPy

import numpy as np

data = np.array([('Alice', 25), ('Bob', 30)], dtype=[1])

Drag options to blanks, or click blank then click option'

A('name', 'U10'), ('age', 'i4')

B['name', 'age']

C{'name': 'U10', 'age': 'i4'}

D[('name', 'U10'), ('age', 'i4')]

Attempts:

3 left

💡 Hint

Common Mistakes

Using a tuple instead of a list for dtype

Using a dictionary instead of a list of tuples

Not specifying the dtype at all

The dtype for a structured array must be a list of tuples specifying field names and types.

2fill in blank

medium

Complete the code to find the union of two structured numpy arrays 'a' and 'b'.

NumPy

import numpy as np

a = np.array([(1, 'A'), (2, 'B')], dtype=[('id', 'i4'), ('val', 'U1')])
b = np.array([(2, 'B'), (3, 'C')], dtype=[('id', 'i4'), ('val', 'U1')])

union = np.[1](a, b)

Drag options to blanks, or click blank then click option'

Aunion1d

Bsetdiff1d

Cin1d

Dintersect1d

Attempts:

3 left

💡 Hint

Common Mistakes

Using intersect1d which finds common elements only

Using setdiff1d which finds difference

Using in1d which returns boolean mask

np.union1d returns the sorted unique values that are in either of the input arrays.

3fill in blank

hard

Fix the error in the code to find the intersection of two structured arrays 'x' and 'y'.

NumPy

import numpy as np

x = np.array([(1, 10), (2, 20)], dtype=[('id', 'i4'), ('score', 'i4')])
y = np.array([(2, 20), (3, 30)], dtype=[('id', 'i4'), ('score', 'i4')])

common = np.[1](x, y)

Drag options to blanks, or click blank then click option'

Aunion1d

Bintersect1d

Csetxor1d

Dsetdiff1d

Attempts:

3 left

💡 Hint

Common Mistakes

Using union1d which returns all unique elements

Using setxor1d which returns symmetric difference

Using setdiff1d which returns difference

np.intersect1d returns the sorted unique values that are in both arrays.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps each word to its length only if the length is greater than 3.

NumPy

words = ['data', 'science', 'ai', 'ml']
lengths = {word: [1] for word in words if [2]

Drag options to blanks, or click blank then click option'

Alen(word)

Bword

Clen(word) > 3

Dword > 3

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself as value instead of length

Checking if word > 3 which is invalid

Not using len(word) in condition

The dictionary comprehension maps each word to its length if the length is greater than 3.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps uppercase words to their values only if the value is greater than 0.

NumPy

data = {'a': 1, 'b': -1, 'c': 3}
result = [1]: [2] for k, v in data.items() if v [3] 0}

Drag options to blanks, or click blank then click option'

Ak.upper()

Bv

C>

Dk.lower()

Attempts:

3 left

💡 Hint

Common Mistakes

Using k.lower() instead of k.upper()

Using '<' instead of '>' in condition

Mapping keys instead of values

The comprehension maps uppercase keys to their values only if the value is positive.

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