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在scipy中删除/设置稀疏矩阵的非零对角元素

[英]Remove/set the non-zero diagonal elements of a sparse matrix in scipy

Say I would like to remove the diagonal from a scipy.sparse.csr_matrix . 假设我想从scipy.sparse.csr_matrix删除对角线。 Is there an efficient way of doing so? 这样做有效吗? I saw that in the sparsetools module there are C functions to return the diagonal. 我在sparsetools模块中看到有C函数返回对角线。

Based on other SO answers here and here my current approach is the following: 基于其他SO答案在这里这里我目前的方法如下:

def csr_setdiag_val(csr, value=0):
    """Set all diagonal nonzero elements
    (elements currently in the sparsity pattern)
    to the given value. Useful to set to 0 mostly.
    """
    if csr.format != "csr":
        raise ValueError('Matrix given must be of CSR format.')
    csr.sort_indices()
    pointer = csr.indptr
    indices = csr.indices
    data = csr.data
    for i in range(min(csr.shape)):
        ind = indices[pointer[i]: pointer[i + 1]]
        j =  ind.searchsorted(i)
        # matrix has only elements up until diagonal (in row i)
        if j == len(ind):
            continue
        j += pointer[i]
        # in case matrix has only elements after diagonal (in row i)
        if indices[j] == i:
            data[j] = value

which I then follow with 然后我跟着它

csr.eliminate_zeros()

Is that the best I can do without writing my own Cython code? 如果不编写我自己的Cython代码,这是我能做的最好的吗?

Based on @hpaulj's comment, I created an IPython Notebook which can be seen on nbviewer . 根据@ hpaulj的评论,我创建了一个可以在nbviewer上看到的IPython Notebook。 This shows that out of all methods mentioned the following is the fastest (assume that mat is a sparse CSR matrix): 这表明在提到的所有方法中,以下是最快的(假设mat是一个稀疏的CSR矩阵):

mat - scipy.sparse.dia_matrix((mat.diagonal()[scipy.newaxis, :], [0]), shape=(one_dim, one_dim))

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