How to change elements in sparse matrix in Python's SciPy?

Question

I have built a small code that I want to use for solving eigenvalue problems involving large sparse matrices. It's working fine, all I want to do now is to set some elements in the sparse matrix to zero, i.e. the ones in the very top row (which corresponds to implementing boundary conditions). I can just adjust the column vectors (C0, C1, and C2) below to achieve that. However, I wondered if there is a more direct way. Evidently, NumPy indexing does not work with SciPy's sparse package.

import scipy.sparse as sp
import scipy.sparse.linalg  as la
import numpy as np
import matplotlib.pyplot as plt

#discretize x-axis
N = 11
x = np.linspace(-5,5,N)
print(x)
V = x * x / 2
h = len(x)/(N)
hi2 = 1./(h**2)
#discretize Schroedinger Equation, i.e. build 
#banded matrix from difference equation
C0 = np.ones(N)*30. + V
C1 = np.ones(N) * -16.
C2 = np.ones(N) * 1.
diagonals = np.array([-2,-1,0,1,2])
H = sp.spdiags([C2, C1, C0,C1,C2],[-2,-1,0,1,2], N, N)
H *= hi2 * (- 1./12.) * (- 1. / 2.)
#solve for eigenvalues
EV = la.eigsh(H,return_eigenvectors = False)

#check structure of H
plt.figure()
plt.spy(H)
plt.show()

This is a visualisation of the matrix that is build by the code above. I want so set the elements in the first row zero.

Inclinable

I found a work around. The format I am using (dia_matrix) is not good for want I want to achieve. I will use csr_matrix instead. Should I close this post then? – Inclinable

it's a well written question and may be useful to others in the future. How about posting what you found as an answer? – Eat

Answer 1

As suggested in the comments, I'll post the answer that I found to my own question. There are several matrix classes in in SciPy's sparse package, they are listed here. One can convert sparse matrices from one class to another. So for what I need to do, I choose to convert my sparse matrix to the class csr_matrix, simply by

H = sp.csr_matrix(H)

Then I can set the elements in the first row to 0 by using the regular NumPy notation:

H[0,0] = 0
H[0,1] = 0
H[0,2] = 0

For completeness, I post the full modified code snippet below.

#SciPy Sparse linear algebra takes care of sparse matrix computations
#http://docs.scipy.org/doc/scipy/reference/sparse.linalg.html
import scipy.sparse as sp
import scipy.sparse.linalg  as la

import numpy as np
import matplotlib.pyplot as plt

#discretize x-axis
N = 1100
x = np.linspace(-100,100,N)
V = x * x / 2.
h = len(x)/(N)
hi2 = 1./(h**2)

#discretize Schroedinger Equation, i.e. build 
#banded matrix from difference equation
C0 = np.ones(N)*30. + V
C1 = np.ones(N) * -16.
C2 = np.ones(N) * 1.

H = sp.spdiags([C2, C1, C0, C1, C2],[-2,-1,0,1,2], N, N)
H *= hi2 * (- 1./12.) * (- 1. / 2.)
H = sp.csr_matrix(H)
H[0,0] = 0
H[0,1] = 0
H[0,2] = 0

#check structure of H
plt.figure()
plt.spy(H)
plt.show()

EV = la.eigsh(H,return_eigenvectors = False)

Inclinable

If you have more rows than columns, csr is fast, but if you have more columns than rows, csc is faster. – Archetype

If you are going to modify the matrix a lot a Row-based linked list sparse matrix is faster - then use lil_matrix – Cambyses

Answer 2

Using lil_matrix is much more efficient in scipy to change elements than simple numpy method.

H = sp.csr_matrix(H)
HL = H.tolil()
HL[1,1] = 5  # same as the numpy indexing notation
print HL
print HL.todense() # if numpy style matrix is required
H = HL.tocsr()    # if csr is required

Dyanna