For example I have a coo_matrix A :
import scipy.sparse as sp
A = sp.coo_matrix([3,0,3,0],
[0,0,2,0],
[2,5,1,0],
[0,0,0,0])
np.nonzeros
Approach #1 We could do something like this 
# Get the columns indices of the input sparse matrix
C = sp.find(A)[1]
# Use np.in1d to create a mask of nonzero columns.
# So, we invert it and convert to int dtype for desired output.
out = (~np.in1d(np.arange(A.shape[1]),C)).astype(int)
Alternatively, to make the code shorter, we can use subtraction 
out = 1np.in1d(np.arange(A.shape[1]),C)
Stepbystep run 
1) Input array and sparse matrix from it :
In [137]: arr # Regular dense array
Out[137]:
array([[3, 0, 3, 0],
[0, 0, 2, 0],
[2, 5, 1, 0],
[0, 0, 0, 0]])
In [138]: A = sp.coo_matrix(arr) # Convert to sparse matrix as input here on
2) Get nonzero column indices :
In [139]: C = sp.find(A)[1]
In [140]: C
Out[140]: array([0, 2, 2, 0, 1, 2], dtype=int32)
3) Use np.in1d
to get mask of nonzero columns :
In [141]: np.in1d(np.arange(A.shape[1]),C)
Out[141]: array([ True, True, True, False], dtype=bool)
4) Invert it :
In [142]: ~np.in1d(np.arange(A.shape[1]),C)
Out[142]: array([False, False, False, True], dtype=bool)
5) Finally convert to int dtype :
In [143]: (~np.in1d(np.arange(A.shape[1]),C)).astype(int)
Out[143]: array([0, 0, 0, 1])
Alternative subtraction approach :
In [145]: 1np.in1d(np.arange(A.shape[1]),C)
Out[145]: array([0, 0, 0, 1])
Approach #2 Here's another way and possibly a faster one using matrixmultiplication

out = 1np.ones(A.shape[0],dtype=bool)*A.astype(bool)
Runtime test
In [166]: arr = np.random.randint(0,3,(1000,1000))
In [167]: A = sp.coo_matrix(arr)
In [168]: %timeit 1np.in1d(np.arange(A.shape[1]),sp.find(A)[1])
10 loops, best of 3: 47.3 ms per loop
In [169]: %timeit 1np.ones(A.shape[0],dtype=bool)*A.astype(bool)
100 loops, best of 3: 12.9 ms per loop