user1482714 user1482714 - 2 months ago 11
Python Question

Create sparse circulant matrix in python

I want to create a large (say 10^5 x 10^5) sparse circulant matrix in Python. It has 4 elements per row at positions

[i,i+1], [i,i+2], [i,i+N-2], [i,i+N-1]
, where I have assumed periodic boundary conditions for the indices (i.e.
[10^5,10^5]=[0,0], [10^5+1,10^5+1]=[1,1]
and so on). I looked at the scipy sparse matrices documentation but I am quite confused (I am new to Python).

I can create the matrix with numpy

import numpy as np

def Bc(i, boundary):
"""(int, int) -> int

Checks boundary conditions on index
"""
if i > boundary - 1:
return i - boundary
elif i < 0:
return boundary + i
else:
return i

N = 100
diffMat = np.zeros([N, N])
for i in np.arange(0, N, 1):
diffMat[i, [Bc(i+1, N), Bc(i+2, N), Bc(i+2+(N-5)+1, N), Bc(i+2+(N-5)+2, N)]] = [2.0/3, -1.0/12, 1.0/12, -2.0/3]


However, this is quite slow and for large
N
uses a lot of memory, so I want to avoid the creation with numpy and the converting to a sparse matrix and go directly to the latter.

I know how to do it in Mathematica, where one can use SparseArray and index patterns - is there something similar here?

Answer

To create a dense circulant matrix, you can use scipy.linalg.circulant. For example,

In [210]: from scipy.linalg import circulant

In [211]: N = 7

In [212]: vals = np.array([2.0/3, -1.0/12, 1.0/12, -2.0/3])

In [213]: offsets = np.array([1, 2, N-2, N-1])

In [214]: col0 = np.zeros(N)

In [215]: col0[offsets] = -vals

In [216]: c = circulant(col0)

In [217]: c
Out[217]: 
array([[ 0.    ,  0.6667, -0.0833,  0.    ,  0.    ,  0.0833, -0.6667],
       [-0.6667,  0.    ,  0.6667, -0.0833,  0.    ,  0.    ,  0.0833],
       [ 0.0833, -0.6667,  0.    ,  0.6667, -0.0833,  0.    ,  0.    ],
       [ 0.    ,  0.0833, -0.6667,  0.    ,  0.6667, -0.0833,  0.    ],
       [ 0.    ,  0.    ,  0.0833, -0.6667,  0.    ,  0.6667, -0.0833],
       [-0.0833,  0.    ,  0.    ,  0.0833, -0.6667,  0.    ,  0.6667],
       [ 0.6667, -0.0833,  0.    ,  0.    ,  0.0833, -0.6667,  0.    ]])

As you point out, for large N, that requires a lot of memory and most of the values are zero. To create a scipy sparse matrix, you can use scipy.sparse.diags. We have to create offsets (and corresponding values) for the diagonals above and below the main diagonal:

In [218]: from scipy import sparse

In [219]: N = 7

In [220]: vals = np.array([2.0/3, -1.0/12, 1.0/12, -2.0/3])

In [221]: offsets = np.array([1, 2, N-2, N-1])

In [222]: dupvals = np.concatenate((vals, vals[::-1]))

In [223]: dupoffsets = np.concatenate((offsets, -offsets))

In [224]: a = sparse.diags(dupvals, dupoffsets, shape=(N, N))

In [225]: a.toarray()
Out[225]: 
array([[ 0.    ,  0.6667, -0.0833,  0.    ,  0.    ,  0.0833, -0.6667],
       [-0.6667,  0.    ,  0.6667, -0.0833,  0.    ,  0.    ,  0.0833],
       [ 0.0833, -0.6667,  0.    ,  0.6667, -0.0833,  0.    ,  0.    ],
       [ 0.    ,  0.0833, -0.6667,  0.    ,  0.6667, -0.0833,  0.    ],
       [ 0.    ,  0.    ,  0.0833, -0.6667,  0.    ,  0.6667, -0.0833],
       [-0.0833,  0.    ,  0.    ,  0.0833, -0.6667,  0.    ,  0.6667],
       [ 0.6667, -0.0833,  0.    ,  0.    ,  0.0833, -0.6667,  0.    ]])

The matrix is stored in the "diagonal" format:

In [226]: a
Out[226]: 
<7x7 sparse matrix of type '<class 'numpy.float64'>'
    with 28 stored elements (8 diagonals) in DIAgonal format>

You can use the conversion methods of the sparse matrix to convert it to a different sparse format. For example, the following results in a matrix in CSR format:

In [227]: a.tocsr()
Out[227]: 
<7x7 sparse matrix of type '<class 'numpy.float64'>'
    with 28 stored elements in Compressed Sparse Row format>
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