What is the most efficient way to compute a sparse boolean matrix I from one or two arrays a,b, with I[i,j]==True where a[i]==b[j]? The following is fast but memory-inefficient:
I = a[:,None]==b
The following is slow and still memory-inefficient during creation:
I = csr((a[:,None]==b),shape=(len(a),len(b)))
The following gives at least the rows,cols for better csr_matrix initialization, but it still creates the full dense matrix and is equally slow:
z = np.argwhere((a[:,None]==b))
Any ideas?
One way to do it would be to first identify all different elements that a and b have in common using sets. This should work well if there are not very many different possibilities for the values in a and b. One then would only have to loop over the different values (below in variable values) and use np.argwhere to identify the indices in a and b where these values occur. The 2D indices of the sparse matrix can then be constructed using np.repeat and np.tile:
import numpy as np
from scipy import sparse
a = np.random.randint(0, 10, size=(400,))
b = np.random.randint(0, 10, size=(300,))
## matrix generation after OP
I1 = sparse.csr_matrix((a[:,None]==b),shape=(len(a),len(b)))
##identifying all values that occur both in a and b:
values = set(np.unique(a)) & set(np.unique(b))
##here we collect the indices in a and b where the respective values are the same:
rows, cols = [], []
##looping over the common values, finding their indices in a and b, and
##generating the 2D indices of the sparse matrix with np.repeat and np.tile
for value in values:
x = np.argwhere(a==value).ravel()
y = np.argwhere(b==value).ravel()
rows.append(np.repeat(x, len(x)))
cols.append(np.tile(y, len(y)))
##concatenating the indices for different values and generating a 1D vector
##of True values for final matrix generation
rows = np.hstack(rows)
cols = np.hstack(cols)
data = np.ones(len(rows),dtype=bool)
##generating sparse matrix
I3 = sparse.csr_matrix( (data,(rows,cols)), shape=(len(a),len(b)) )
##checking that the matrix was generated correctly:
print((I1 != I3).nnz==0)
The syntax for generating the csr matrix is taken from the documentation. The test for sparse matrix equality is taken from this post.
Old Answer:
I don't know about performance, but at least you can avoid constructing the full dense matrix by using a simple generator expression. Here some code that uses two 1d arras of random integers to first generate the sparse matrix the way that the OP posted and then uses a generator expression to test all elements for equality:
import numpy as np
from scipy import sparse
a = np.random.randint(0, 10, size=(400,))
b = np.random.randint(0, 10, size=(300,))
## matrix generation after OP
I1 = sparse.csr_matrix((a[:,None]==b),shape=(len(a),len(b)))
## matrix generation using generator
data, rows, cols = zip(
*((True, i, j) for i,A in enumerate(a) for j,B in enumerate(b) if A==B)
)
I2 = sparse.csr_matrix((data, (rows, cols)), shape=(len(a), len(b)))
##testing that matrices are equal
## from https://stackoverflow.com/a/30685839/2454357
print((I1 != I2).nnz==0) ## --> True
I think there is no way around the double loop and ideally this would be pushed into numpy, but at least with the generator the loops are somewhat optimised ...