Record arrays in NumPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of accessing fields in a record array

What is the output of the following code?

NumPy

import numpy as np

rec = np.rec.array([(1, 2.0, 'Hello'), (2, 3.0, 'World')],
                  dtype=[('foo', 'i4'), ('bar', 'f4'), ('baz', 'a5')])

print(rec.foo)
print(rec['baz'])

A

[1 2]
[b'Hello' b'World']

B

[1 2]
['Hello' 'World']

C

[1. 2.]
[b'Hello' b'World']

D

[1 2]
[b'Hello' b'World' b'!']

💡 Hint

Remember that record array fields keep their original data types, including byte strings.

The field 'foo' contains integers, so printing rec.foo outputs an integer array. The field 'baz' is a byte string type, so printing rec['baz'] outputs byte strings.

❓ data_output

intermediate

Number of elements in a record array after filtering

Given the record array below, how many elements remain after filtering where age > 30?

NumPy

import numpy as np

rec = np.rec.array([(25, 'Alice'), (35, 'Bob'), (40, 'Charlie')],
                  dtype=[('age', 'i4'), ('name', 'U10')])

filtered = rec[rec.age > 30]

print(len(filtered))

A 0

B 1

C 2

D 3

💡 Hint

Count how many ages are greater than 30.

Only 'Bob' (35) and 'Charlie' (40) have ages greater than 30, so 2 elements remain.

🔧 Debug

advanced

Identify the error in record array field assignment

What error does the following code raise?

NumPy

import numpy as np

rec = np.rec.array([(1, 2.0), (3, 4.0)], dtype=[('x', 'i4'), ('y', 'f4')])
rec.x = [10, 20, 30]

A ValueError: setting an array element with a sequence.

B IndexError: too many indices for array

C AttributeError: can't set attribute

D TypeError: invalid type assignment

💡 Hint

Check if the length of the assigned list matches the record array length.

The record array has length 2, but the list assigned has length 3, causing a ValueError due to shape mismatch.

🚀 Application

advanced

Create a record array from a structured array

You have a structured array arr with dtype [('a', 'i4'), ('b', 'f4')]. Which code correctly converts it to a record array?

NumPy

import numpy as np

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

A rec = np.recarray(arr.shape, dtype=arr.dtype)

B rec = np.rec.array(arr)

C rec = np.array(arr, dtype=np.recarray)

D rec = arr.view(np.recarray)

💡 Hint

Use the view method to change the array type without copying data.

Using arr.view(np.recarray) creates a record array view of the structured array without copying data. Option D creates a new record array but copies data. Option D is invalid dtype usage. Option D creates an empty record array.

🧠 Conceptual

expert

Why use record arrays instead of structured arrays?

Which statement best explains the advantage of using NumPy record arrays over structured arrays?

A Record arrays support multi-dimensional fields, unlike structured arrays.

B Record arrays allow accessing fields as attributes, making code more readable and concise.

C Record arrays automatically convert all string fields to Unicode.

D Record arrays use less memory than structured arrays for the same data.

💡 Hint

Think about how you access data fields in each array type.

Record arrays allow accessing fields using dot notation (e.g., rec.field), which can make code easier to write and read. Structured arrays require dictionary-like access (e.g., arr['field']). Memory usage and string handling are similar, and both support multi-dimensional fields.

Practice

1. What is the main advantage of using a record array in numpy?

easy

A. It speeds up numerical calculations on large arrays.

B. It automatically sorts data based on values.

C. It allows storing different data types in one array with named fields.

D. It compresses data to save memory.

Solution

1. Step 1: Understand record arrays

   Record arrays let you store mixed data types in one numpy array by using named fields.

2. Step 2: Compare options

   Only It allows storing different data types in one array with named fields. correctly describes this feature. Others describe unrelated features.

3. Final Answer:

   It allows storing different data types in one array with named fields. -> Option C

4. Quick Check:

   Record arrays = mixed types + named fields [OK]

Hint: Remember: record arrays hold mixed types with names [OK]

Common Mistakes:

• Confusing record arrays with regular numeric arrays
• Thinking record arrays sort data automatically
• Assuming record arrays compress data

2. Which of the following is the correct way to create a numpy record array with fields 'name' (string) and 'age' (integer)?

easy

A. np.rec.array([("Alice", 25), ("Bob", 30)], dtype=[('name', 'U10'), ('age', 'i4')])

B. np.array([("Alice", 25), ("Bob", 30)], dtype=[('name', 'i4'), ('age', 'U10')])

C. np.rec.array(["Alice", 25, "Bob", 30], dtype=[('name', 'U10'), ('age', 'i4')])

D. np.rec.array([(25, "Alice"), (30, "Bob")], dtype=[('name', 'U10'), ('age', 'i4')])

Solution

1. Step 1: Check data and dtype matching

   np.rec.array([("Alice", 25), ("Bob", 30)], dtype=[('name', 'U10'), ('age', 'i4')]) matches tuples of (string, int) with dtype [('name', 'U10'), ('age', 'i4')].

2. Step 2: Validate other options

   np.array([("Alice", 25), ("Bob", 30)], dtype=[('name', 'i4'), ('age', 'U10')]) swaps types incorrectly; C has wrong input format; A swaps field order.

3. Final Answer:

   np.rec.array([("Alice", 25), ("Bob", 30)], dtype=[('name', 'U10'), ('age', 'i4')]) -> Option A

4. Quick Check:

   Data matches dtype order and types [OK]

Hint: Match tuple order with dtype fields exactly [OK]

Common Mistakes:

• Swapping field order between data and dtype
• Using wrong data types in dtype
• Passing flat list instead of list of tuples

3. What will be the output of the following code?

import numpy as np
rec = np.rec.array([(1, 2.5), (3, 4.5)], dtype=[('x', 'i4'), ('y', 'f4')])
print(rec.x + rec.y)

medium

A. TypeError

B. [3 7]

C. [1 3]

D. [3.5 7.5]

Solution

1. Step 1: Understand data and fields

   rec.x is integer array [1, 3], rec.y is float array [2.5, 4.5].

2. Step 2: Add integer and float arrays element-wise

   Adding [1, 3] + [2.5, 4.5] results in [3.5, 7.5] as floats.

3. Final Answer:

   [3.5 7.5] -> Option D

4. Quick Check:

   1+2.5=3.5 and 3+4.5=7.5 [OK]

Hint: Adding int and float fields results in float array [OK]

Common Mistakes:

• Expecting integer output instead of float
• Confusing field names or types
• Thinking addition causes error

4. Identify the error in this code snippet:

import numpy as np
rec = np.rec.array([(1, 'a'), (2, 'b')], dtype=[('num', 'i4'), ('char', 'U1')])
print(rec.num + rec.char)

medium

A. You cannot add integer and string fields directly.

B. The dtype specification is incorrect.

C. The data tuples have wrong length.

D. The record array must be created with np.array, not np.rec.array.

Solution

1. Step 1: Analyze the operation

   rec.num is integer array, rec.char is string array.

2. Step 2: Check addition of int and string

   Adding int + string causes a TypeError in numpy.

3. Final Answer:

   You cannot add integer and string fields directly. -> Option A

4. Quick Check:

   int + string = TypeError [OK]

Hint: Cannot add numbers and strings directly in numpy [OK]

Common Mistakes:

• Assuming dtype is wrong instead of operation
• Thinking np.rec.array is incorrect here
• Ignoring type mismatch in addition

5. You have a numpy record array rec with fields 'id' (int), 'score' (float), and 'passed' (bool). How do you create a new record array containing only records where passed is True and score is above 80?

hard

A. rec[rec.passed or rec.score > 80]

B. rec[(rec.passed) & (rec.score > 80)]

C. rec[rec.passed and rec.score > 80]

D. rec[(rec.passed) | (rec.score > 80)]

Solution

1. Step 1: Understand filtering syntax

   Use boolean indexing with & for element-wise AND, parentheses needed.

2. Step 2: Evaluate options

   rec[(rec.passed) & (rec.score > 80)] correctly uses (rec.passed) & (rec.score > 80). Options B and C use Python 'or'/'and' which don't work element-wise. rec[(rec.passed) | (rec.score > 80)] uses | (OR) instead of AND.

3. Final Answer:

   rec[(rec.passed) & (rec.score > 80)] -> Option B

4. Quick Check:

   Use & with parentheses for element-wise AND [OK]

Hint: Use & with parentheses for element-wise conditions [OK]

Common Mistakes:

• Using 'and' or 'or' instead of '&' or '|' for arrays
• Forgetting parentheses around conditions
• Using | instead of & for AND condition