如果您有能力在进行计算之前就对称矩阵，则以下操作应相当快：
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())
这在合理的假设下有效（例如，在运行之前不做任何事情
a[0, 1] = 42并且矛盾）。
a[1, 0] = 123``symmetrize
如果您确实需要透明的对称化，则可以考虑子类化numpy.ndarray并简单地重新定义
__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] = 42print a # a[1, 0] == 42 too!
（或等价于矩阵而不是数组，具体取决于您的需求）。这种方法甚至可以处理更复杂的分配，例如
a[:, 1] = -1，可以正确设置
a[1, :]元素。
请注意，Python 3消除了编写的可能性
def …(…, (i, j),…)，因此在使用Python 3进行运行之前，必须对代码进行些微调整
def__setitem__(self, indexes, value): (i, j) = indexes。
