Sparse matrix operations in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of sparse matrix multiplication

What is the output of this code that multiplies two sparse 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)
print(C.toarray())

A

[[0 0 0]
 [1 0 0]
 [0 3 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 sparse matrix multiplication follows the same rules as dense matrix multiplication.

The multiplication of sparse matrices A and B results in matrix C where each element is the dot product of rows of A and columns of B. The output matches option D.

❓ data_output

intermediate

Number of non-zero elements after addition

Given two sparse matrices, what is the number of non-zero elements in their sum?

SciPy

from scipy.sparse import csc_matrix

X = csc_matrix([[0, 5, 0], [0, 0, 0], [7, 0, 0]])
Y = csc_matrix([[0, 0, 3], [0, 0, 0], [0, 0, 0]])

Z = X + Y
print(Z.nnz)

A 4

B 3

C 2

D 5

💡 Hint

Count unique non-zero positions after addition.

Matrix X has 2 non-zero elements, Y has 1 non-zero element at a different position. Their sum has 3 unique non-zero elements, but since addition does not overlap, total nnz is 3. However, the code prints 3, so option B seems correct. But option B says 4, which is incorrect. Rechecking: X has 2 non-zero elements, Y has 1, total 3 unique positions. So correct answer is 3.

🔧 Debug

advanced

Identify the error in sparse matrix slicing

What error does this code raise when slicing a sparse matrix?

SciPy

from scipy.sparse import lil_matrix

M = lil_matrix((3,3))
M[0, 0] = 1
M[1, 1] = 2

sub = M[0:2, 0:2]
print(sub)

A IndexError: index out of bounds

B TypeError: slicing with step is not supported

C No error, prints a sparse matrix

D ValueError: invalid slice

💡 Hint

Check if lil_matrix supports slicing.

The lil_matrix supports slicing and returns a sparse matrix of the sliced part. So no error occurs and the code prints the sliced sparse matrix.

❓ visualization

advanced

Visualizing sparsity pattern

Which code snippet correctly visualizes the sparsity pattern of a CSR sparse matrix using matplotlib?

SciPy

import matplotlib.pyplot as plt
from scipy.sparse import csr_matrix

A = csr_matrix([[0, 0, 1], [1, 0, 0], [0, 2, 0]])

# Visualization code here

A

plt.spy(A)
plt.show()

B

plt.scatter(A)
plt.show()

C

plt.plot(A)
plt.show()

D

plt.imshow(A)
plt.show()

💡 Hint

Use the function designed to show sparsity patterns.

plt.spy() is the correct function to visualize the sparsity pattern of a sparse matrix. Other options either cause errors or do not show sparsity correctly.

🚀 Application

expert

Efficient sparse matrix row normalization

Which code snippet correctly normalizes each row of a CSR sparse matrix so that each row sums to 1, without converting to dense format?

SciPy

from scipy.sparse import csr_matrix
import numpy as np

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

# Normalize rows here

A

row_sums = np.array(M.sum(axis=1)).flatten()
row_indices, _ = M.nonzero()
M.data /= row_sums[row_indices]

B M = M / M.sum(axis=1)

C

for i in range(M.shape[0]):
    M[i] = M[i] / M[i].sum()

D M = M.multiply(1 / M.sum(axis=0))

💡 Hint

Use the sparse matrix data and indices to normalize efficiently.

Option A correctly computes row sums and divides non-zero elements by their row sum using indexing. Option A and D cause errors or incorrect broadcasting. Option A tries to assign to sparse matrix slices which is not supported.