Numpy ‘smart’ symmetric matrix

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最佳答案

If you can afford to symmetrize the matrix just before doing calculations, the following should be reasonably fast:

def symmetrize(a):
"""
Return a symmetrized version of NumPy array a.


Values 0 are replaced by the array value at the symmetric
position (with respect to the diagonal), i.e. if a_ij = 0,
then the returned array a' is such that a'_ij = a_ji.


Diagonal values are left untouched.


a -- square NumPy array, such that a_ij = 0 or a_ji = 0,
for i != j.
"""
return a + a.T - numpy.diag(a.diagonal())

This works under reasonable assumptions (such as not doing both a[0, 1] = 42 and the contradictory a[1, 0] = 123 before running symmetrize).

If you really need a transparent symmetrization, you might consider subclassing numpy.ndarray and simply redefining __setitem__:

class SymNDArray(numpy.ndarray):
"""
NumPy array subclass for symmetric matrices.


A SymNDArray arr is such that doing arr[i,j] = value
automatically does arr[j,i] = value, so that array
updates remain symmetrical.
"""


def __setitem__(self, (i, j), value):
super(SymNDArray, self).__setitem__((i, j), value)
super(SymNDArray, self).__setitem__((j, i), value)


def symarray(input_array):
"""
Return a symmetrized version of the array-like input_array.


The returned array has class SymNDArray. Further assignments to the array
are thus automatically symmetrized.
"""
return symmetrize(numpy.asarray(input_array)).view(SymNDArray)


# Example:
a = symarray(numpy.zeros((3, 3)))
a[0, 1] = 42
print a  # a[1, 0] == 42 too!

(or the equivalent with matrices instead of arrays, depending on your needs). This approach even handles more complicated assignments, like a[:, 1] = -1, which correctly sets a[1, :] elements.

Note that Python 3 removed the possibility of writing def …(…, (i, j),…), so the code has to be slightly adapted before running with Python 3: def __setitem__(self, indexes, value): (i, j) = indexes…

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The more general issue of optimal treatment of symmetric matrices in numpy bugged me too.

After looking into it, I think the answer is probably that numpy is somewhat constrained by the memory layout supportd by the underlying BLAS routines for symmetric matrices.

While some BLAS routines do exploit symmetry to speed up computations on symmetric matrices, they still use the same memory structure as a full matrix, that is, n^2 space rather than n(n+1)/2. Just they get told that the matrix is symmetric and to use only the values in either the upper or the lower triangle.

Some of the scipy.linalg routines do accept flags (like sym_pos=True on linalg.solve) which get passed on to BLAS routines, although more support for this in numpy would be nice, in particular wrappers for routines like DSYRK (symmetric rank k update), which would allow a Gram matrix to be computed a fair bit quicker than dot(M.T, M).

(Might seem nitpicky to worry about optimising for a 2x constant factor on time and/or space, but it can make a difference to that threshold of how big a problem you can manage on a single machine...)

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This is plain python and not numpy, but I just threw together a routine to fill a symmetric matrix (and a test program to make sure it is correct):

import random


# fill a symmetric matrix with costs (i.e. m[x][y] == m[y][x]
# For demonstration purposes, this routine connect each node to all the others
# Since a matrix stores the costs, numbers are used to represent the nodes
# so the row and column indices can represent nodes


def fillCostMatrix(dim):        # square array of arrays
# Create zero matrix
new_square = [[0 for row in range(dim)] for col in range(dim)]
# fill in main diagonal
for v in range(0,dim):
new_square[v][v] = random.randrange(1,10)


# fill upper and lower triangles symmetrically by replicating diagonally
for v in range(1,dim):
iterations = dim - v
x = v
y = 0
while iterations > 0:
new_square[x][y] = new_square[y][x] = random.randrange(1,10)
x += 1
y += 1
iterations -= 1
return new_square


# sanity test
def test_symmetry(square):
dim = len(square[0])
isSymmetric = ''
for x in range(0, dim):
for y in range(0, dim):
if square[x][y] != square[y][x]:
isSymmetric = 'NOT'
print "Matrix is", isSymmetric, "symmetric"


def showSquare(square):
# Print out square matrix
columnHeader = ' '
for i in range(len(square)):
columnHeader += '  ' + str(i)
print columnHeader


i = 0;
for col in square:
print i, col    # print row number and data
i += 1


def myMain(argv):
if len(argv) == 1:
nodeCount = 6
else:
try:
nodeCount = int(argv[1])
except:
print  "argument must be numeric"
quit()


# keep nodeCount <= 9 to keep the cost matrix pretty
costMatrix = fillCostMatrix(nodeCount)
print  "Cost Matrix"
showSquare(costMatrix)
test_symmetry(costMatrix)   # sanity test
if __name__ == "__main__":
import sys
myMain(sys.argv)


# vim:tabstop=8:shiftwidth=4:expandtab

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There are a number of well-known ways of storing symmetric matrices so they don't need to occupy n^2 storage elements. Moreover, it is feasible to rewrite common operations to access these revised means of storage. The definitive work is Golub and Van Loan, Matrix Computations, 3rd edition 1996, Johns Hopkins University Press, sections 1.27-1.2.9. For example, quoting them from form (1.2.2), in a symmetric matrix only need to store A = [a_{i,j} ] fori >= j. Then, assuming the vector holding the matrix is denoted V, and that A is n-by-n, put a_{i,j} in

V[(j-1)n - j(j-1)/2 + i]

This assumes 1-indexing.

Golub and Van Loan offer an Algorithm 1.2.3 which shows how to access such a stored V to calculate y = V x + y.

Golub and Van Loan also provide a way of storing a matrix in diagonal dominant form. This does not save storage, but supports ready access for certain other kinds of operations.

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It is trivial to Pythonically fill in [i][j] if [j][i] is filled in. The storage question is a little more interesting. One can augment the numpy array class with a packed attribute that is useful both to save storage and to later read the data.

class Sym(np.ndarray):


# wrapper class for numpy array for symmetric matrices. New attribute can pack matrix to optimize storage.
# Usage:
# If you have a symmetric matrix A as a shape (n,n) numpy ndarray, Sym(A).packed is a shape (n(n+1)/2,) numpy array
# that is a packed version of A.  To convert it back, just wrap the flat list in Sym().  Note that Sym(Sym(A).packed)




def __new__(cls, input_array):
obj = np.asarray(input_array).view(cls)


if len(obj.shape) == 1:
l = obj.copy()
p = obj.copy()
m = int((np.sqrt(8 * len(obj) + 1) - 1) / 2)
sqrt_m = np.sqrt(m)


if np.isclose(sqrt_m, np.round(sqrt_m)):
A = np.zeros((m, m))
for i in range(m):
A[i, i:] = l[:(m-i)]
A[i:, i] = l[:(m-i)]
l = l[(m-i):]
obj = np.asarray(A).view(cls)
obj.packed = p


else:
raise ValueError('One dimensional input length must be a triangular number.')


elif len(obj.shape) == 2:
if obj.shape[0] != obj.shape[1]:
raise ValueError('Two dimensional input must be a square matrix.')
packed_out = []
for i in range(obj.shape[0]):
packed_out.append(obj[i, i:])
obj.packed = np.concatenate(packed_out)


else:
raise ValueError('Input array must be 1 or 2 dimensional.')


return obj


def __array_finalize__(self, obj):
if obj is None: return
self.packed = getattr(obj, 'packed', None)

```

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To construct a NxN matrix that is symmetric along the main diagonal, and with 0's on the main diagonal you can do :

a = np.array([1, 2, 3, 4, 5])
b = np.zeros(shape=(a.shape[0], a.shape[0]))
upper = np.triu(b + a)
lower = np.tril(np.transpose(b + a))
D = (upper + lower) * (np.full(a.shape[0], fill_value=1) - np.eye(a.shape[0]))

This is kind of a special case, but recently I've used this kind of matrix for network adjacency representation.

Hope that helps. Cheers.