删除/设置scipy中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

问题描述

假设我想从 scipy.sparse.csr_matrix 中删除对角线.有没有一种有效的方法呢?我看到在 sparsetools 模块中有 C 函数可以返回对角线.

Say I would like to remove the diagonal from a scipy.sparse.csr_matrix. Is there an efficient way of doing so? I saw that in the sparsetools module there are C functions to return the diagonal.

基于其他 SO 答案此处和这里 我目前的方法如下:

Based on other SO answers here and here my current approach is the following:

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

然后我跟着

csr.eliminate_zeros()

如果不编写自己的 Cython 代码，这是我能做的最好的事情吗?

Is that the best I can do without writing my own Cython code?

推荐答案

基于@hpaulj 的评论，我创建了一个 IPython Notebook，它 可以在 nbviewer 上看到.这表明在所有提到的方法中，以下是最快的(假设 mat 是一个稀疏的 CSR 矩阵):

Based on @hpaulj's comment, I created an IPython Notebook which can be seen on nbviewer. This shows that out of all methods mentioned the following is the fastest (assume that mat is a sparse CSR matrix):

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

这篇关于删除/设置scipy中稀疏矩阵的非零对角元素的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！