在csc_matrix中查找零元素的行索引

Question

I have a csc_matrix like this:

>>> arr_csc = arr.tocsc()
>>> arr_csc
<2x3 sparse matrix of type '<type 'numpy.int64'>'
    with 5 stored elements in Compressed Sparse Column format>
>>> arr_csc.todense()
matrix([[0, 1, 0],
        [3, 4, 0]])

Now, what I want is the row indices of all zero elements in each column. For example:

For column 0, I want "[0]"
For column 1, I want "[]"
For column 2. I want "[0, 1]"

What is the fastest way to do this?

Thanks!

Answer 1

How about something like this:

The main idea is to use .indptr and .indices , the rest of my solution can probably be improved on.

from scipy import sparse
import numpy as np 

arr_csc = sparse.csc_matrix([[0, 1, 0],
                            [3, 4, 0]])

result = []

all_rows = np.arange(arr_csc.shape[0])

for i in xrange(len(arr_csc.indptr) - 1):
    start = arr_csc.indptr[i]
    end = arr_csc.indptr[i+1]

    result.append(np.setdiff1d(all_rows, arr_csc.indices[start : end]))

print result 

Result:

[array([0]), array([], dtype=int64), array([0, 1])]

Answer 2

With your sample, this works:

In [808]: arr=sparse.csc_matrix([[0,1,0],[3,4,0]])\    
In [809]: arr1=arr==0
In [810]: arr1.T.tolil().rows
Out[810]: array([[0], [], [0, 1]], dtype=object)

Beware that when you do arr==0 you will get a warning:

/usr/lib/python3/dist-packages/scipy/sparse/compressed.py:220: SparseEfficiencyWarning: Comparing a sparse matrix with 0 using == is inefficient, try using != instead. ", try using != instead.", SparseEfficiencyWarning)

In your sample there are equal number of 0s and nonzeros. But in a typical sparse matrix, there are many more 0s. Lots of 0s means long lists of rows.