CSR format (Compressed Sparse Row) in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of CSR matrix multiplication

What is the output of the following code that multiplies two CSR matrices?

SciPy

import numpy as np
from scipy.sparse import csr_matrix

A = csr_matrix([[0, 0, 1], [1, 0, 0], [0, 2, 0]])
B = csr_matrix([[1, 0, 0], [0, 3, 0], [4, 0, 0]])

C = A.dot(B).toarray()
print(C)

A

[[0 3 0]
 [1 0 0]
 [0 0 0]]

B

[[0 0 0]
 [0 6 0]
 [4 0 0]]

C

[[4 0 0]
 [0 0 0]
 [0 6 0]]

D

[[4 0 0]
 [1 0 0]
 [0 6 0]]

💡 Hint

Remember that CSR matrix multiplication follows normal matrix multiplication rules.

Matrix A has non-zero elements at positions (0,2), (1,0), and (2,1). Matrix B has non-zero elements at (0,0), (1,1), and (2,0). Multiplying A and B, the element at (0,0) is A(0,2)*B(2,0) = 1*4=4, at (1,0) is A(1,0)*B(0,0)=1*1=1, and at (2,1) is A(2,1)*B(1,1)=2*3=6. All other elements are zero.

❓ data_output

intermediate

Number of non-zero elements in CSR matrix

Given the CSR matrix below, how many non-zero elements does it contain?

SciPy

from scipy.sparse import csr_matrix

matrix = csr_matrix([[0, 5, 0], [0, 0, 0], [7, 0, 8]])
print(matrix.nnz)

A 3

B 2

C 4

D 5

💡 Hint

Count all the non-zero values in the matrix.

The matrix has three non-zero values: 5 at (0,1), 7 at (2,0), and 8 at (2,2). So nnz (number of non-zero elements) is 3.

🔧 Debug

advanced

Identify the error in CSR matrix creation

What error will this code raise when trying to create a CSR matrix?

SciPy

from scipy.sparse import csr_matrix

# Data and indices lengths mismatch
data = [1, 2, 3]
indices = [0, 2]
indptr = [0, 2, 3]

csr = csr_matrix((data, indices, indptr), shape=(2, 3))

A No error, matrix created successfully

B ValueError: indices and data array must be the same length

C TypeError: unsupported operand type(s)

D IndexError: index out of bounds

💡 Hint

Check if the lengths of data and indices match.

The CSR format requires that the data and indices arrays have the same length because each data value corresponds to an index. Here, data has length 3 but indices has length 2, causing a ValueError.

🧠 Conceptual

advanced

Understanding CSR matrix components

Which of the following correctly describes the role of the 'indptr' array in a CSR matrix?

A 'indptr' stores the cumulative count of non-zero elements per row, marking start and end positions in 'data' and 'indices' arrays for each row.

B 'indptr' contains the column indices of non-zero elements.

C 'indptr' holds the actual non-zero data values of the matrix.

D 'indptr' is used to store the shape of the matrix.

💡 Hint

Think about how CSR stores rows efficiently.

'indptr' is an array of length (number_of_rows + 1). Each element points to the index in 'data' and 'indices' arrays where a row starts and ends. This allows quick access to each row's non-zero elements.

🚀 Application

expert

Efficient row slicing in CSR matrix

You want to extract the second row from a large CSR matrix efficiently. Which code snippet correctly extracts the second row as a CSR matrix?

SciPy

from scipy.sparse import csr_matrix

matrix = csr_matrix([[0, 1, 0], [2, 0, 3], [0, 0, 4]])

A

row = matrix[1, :]
print(type(row))

B

row = matrix[1]
print(type(row))

C

row = matrix.getrow(1)
print(type(row))

D

row = matrix.toarray()[1]
print(type(row))

💡 Hint

Look for a method designed for efficient row extraction in CSR.

matrix.getrow(1) returns the second row as a CSR matrix efficiently without converting the whole matrix to dense. matrix[1, :] returns a 1xN sparse matrix but can be less efficient. matrix[1] returns a 1D sparse matrix (not CSR). matrix.toarray()[1] converts entire matrix to dense, which is inefficient for large sparse matrices.