方法1:我们可以将稀疏元素的行索引用作ID，并对这些元素的对应值进行乘法运算与 np.multiply.reduceat 输出.

Approach #1: We can use the row indices of the sparse elements as IDs and perform multiplication of the corresponding values of those elements with np.multiply.reduceat to get the desired output.

因此，一个实现将是-

from scipy import sparse
from scipy.sparse import csc_matrix

r,c,v = sparse.find(a) # a is input sparse matrix
out = np.zeros(a.shape[0],dtype=a.dtype)
unqr, shift_idx = np.unique(r,return_index=1)
out[unqr] = np.multiply.reduceat(v, shift_idx)

样品运行-

In [89]: # Let's create a sample csc_matrix
    ...: A = np.array([[-1,2,0,0],[0,0,0,0],[2,0,3,0],[4,5,6,0],[1,9,0,2]])
    ...: a = csc_matrix(A)
    ...: 

In [90]: a
Out[90]: 
<5x4 sparse matrix of type '<type 'numpy.int64'>'
    with 10 stored elements in Compressed Sparse Column format>

In [91]: a.toarray()
Out[91]: 
array([[-1,  2,  0,  0],
       [ 0,  0,  0,  0],
       [ 2,  0,  3,  0],
       [ 4,  5,  6,  0],
       [ 1,  9,  0,  2]])

In [92]: out
Out[92]: array([ -2,   0,   6, 120,   0,  18])

方法2::我们正在执行基于bin的乘法.我们有一个基于bin的求和解决方案，其中 np.bincount .因此，可以在此处使用的技巧是将数字转换为对数，执行基于bin的求和，然后使用exponential(对数反转)转换回原始格式，仅此而已！对于负数，我们可能要增加一个步骤或更多，但让我们看看非负数的实现是什么-

Approach #2: We are performing bin-based multiplication. We have bin-based summing solution with np.bincount. So, a trick that could be use here would be converting the numbers to logarithmic numbers, perform bin-based summing and then convert back to original format with exponential (reverse of log) and that's it! For negative numbers, we might to add a step or more, but let's see what the implementation be like for non-negative numbers -

r,c,v = sparse.find(a)
out = np.exp(np.bincount(r,np.log(v),minlength = a.shape[0]))
out[np.setdiff1d(np.arange(a.shape[0]),r)] = 0

带有非负数的样本运行-

A sample run with non-negative numbers -

In [118]: a.toarray()
Out[118]: 
array([[1, 2, 0, 0],
       [0, 0, 0, 0],
       [2, 0, 3, 0],
       [4, 5, 6, 0],
       [1, 9, 0, 2]])

In [120]: out  # Using listed code
Out[120]: array([   2.,    0.,    6.,  120.,   18.])