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Is there a numpy/scipy dot product for sparse matrix, calculating only the diagonal entries of the result?

Imagine having 2 sparse matrix:
> A, A.shape = (n,m)
> B, B.shape = (m,n)
I would like to compute the dot product A*B, but then only keep the diagonal. The matrices being big, I actually don't want to compute other values than the ones in the diagonal.
This is a variant of the question Is there a numpy/scipy dot product, calculating only the diagonal entries of the result?
Where the most relevant answer seems to be to use np.einsum:
np.einsum('ij,ji->i', A, B)
However this does not work:
ValueError: einstein sum subscripts string contains too many subscripts for operand 0
The solution is to use todense(), but it increases a lot the memory usage: np.einsum('ij,ji->i', A.todense(), B.todense())
The other solution, that I currently use, is to iterate over all the rows of A and compute each product in the loop :
for i in range(len_A):
result = np.float32(A[i].dot(B[:, i])[0, 0])
...
None of these solutions seems perfect. Is there an equivalent to np.einsum that could work with sparse matrices ?

[sum(A[i]*B.T[i]) for i in range(min(A.shape[0], B.shape[1]))]
otherwise this is faster:
l = min(A.shape[0], B.shape[1])
(A[np.arange(l)]*B.T[np.arange(l)]).sum(axis=1)

In general you shouldn't try to use numpy functions on the scipy.sparse arrays. In your case I'd first make sure both arrays actually have a compatible shape, that is
A, A.shape = (r,m)
B, B.shape = (m,r)
where r = min(n, p). Then we can compute the diagonal of the matrix product using
d = (A.multiply(B.T)).sum(axis=1)
Here we compute the entry wise row-column products, and manually sum them up. This avoids all the unnecessary computations you'd get using dot/#/*. (Note that unlike in numpy, both * and # perform matrix multiplication.)

incrementing a multidimensional numpy array (python) with products generated from a set of vectors corresponding to axes of the array

A is a k dimensional numpy array of floats (k could be pretty big, e.g. up to 10)
I need to implement an update to A by incrementing each of the values (as described below). I'm wondering if there is a numpy-style way that would be fast.
Let L_i be the length of axis i
An update to this array is generated in two steps follows:
For each axis of A a corresponding vector G is generated.
For example, corresponding to axis i a vector G_i of length L_i is generated (from data).
Update A at all positions by calculating an increment from the G vectors for each position in A
To do this at any particular position, let p be an array of k indices, corresponding to a position in A. Then A at p is incremented by a value calculated as the product:
Product(G_i[p[i]], for i from 0 to k-1)
A full update to A involves doing this operation for all locations in A (i.e. all possible values of p)
This operation would be very slow doing positions one by one via loops.
Is there a numpy style way to do this that would be fast?
edit
## this for three dimensions, final matrix at pos i,j,k has the
## product of c[i]*b[j]*a[k]
## but for arbitrary # of dimensions it will have a loop in a loop
## and will be slow
import numpy as np
a = np.array([1,2])
b = np.array([3,4,5])
c = np.array([6,7,8,9])
ab = []
for bval in b:
ab.append(bval*a)
ab = np.stack(ab)
abc = []
for cval in c:
abc.append(cval*ab)
abc = np.stack(abc)
as a function
def loopfunc(arraylist):
ndim = len(arraylist)
m = arraylist[0]
for i in range(1,ndim):
ml = []
for val in arraylist[i]:
ml.append(val*m)
m = np.stack(ml)
return m

This is a wacky problem, but I like it.
If I understand what you need from your example, you can accomplish this with some reshaping trickery and NumPy's usual broadcasting rules. The idea is to reshape each array so it has the right number of dimensions, then just directly multiply.
Here's a function that implements this.
from functools import reduce
import operator
import numpy as np
import scipy.linalg
def wacky_outer_product(*arrays):
assert len(arrays) >= 2
assert all(arr.ndim == 1 for arr in arrays)
ndim = len(arrays)
shapes = scipy.linalg.toeplitz((-1,) + (1,) * (ndim - 1))
reshaped = (arr.reshape(new_shape) for arr, new_shape in zip(arrays, shapes))
return reduce(operator.mul, reshaped).T
Testing this on your example arrays, we have:
>>> foo = wacky_outer_product(a, b, c)
>>> np.all(foo, abc)
True
Edit
Ok, the above function is fun, but the below is probably much better. No transposing, clearer, and much smaller:
from functools import reduce
import operator
import numpy as np
def wacky_outer_product(*arrays):
return reduce(operator.mul, np.ix_(*reversed(arrays)))

Scipy: Sparse indicator matrix from array(s)

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 ...

You could use numpy.isclose with small tolerance:
np.isclose(a,b)
Or pandas.DataFrame.eq:
a.eq(b)
Note this returns an array of True False.

Vectorise Python code

I have coded a kriging algorithm but I find it quite slow. Especially, do you have an idea on how I could vectorise the piece of code in the cons function below:
import time
import numpy as np
B = np.zeros((200, 6))
P = np.zeros((len(B), len(B)))
def cons():
time1=time.time()
for i in range(len(B)):
for j in range(len(B)):
P[i,j] = corr(B[i], B[j])
time2=time.time()
return time2-time1
def corr(x,x_i):
return np.exp(-np.sum(np.abs(np.array(x) - np.array(x_i))))
time_av = 0.
for i in range(30):
time_av+=cons()
print "Average=", time_av/100.
Edit: Bonus questions
What happens to the broadcasting solution if I want corr(B[i], C[j]) with C the same dimension than B
What happens to the scipy solution if my p-norm orders are an array:
p=np.array([1.,2.,1.,2.,1.,2.])
def corr(x, x_i):
return np.exp(-np.sum(np.abs(np.array(x) - np.array(x_i))**p))
For 2., I tried P = np.exp(-cdist(B, C,'minkowski', p)) but scipy is expecting a scalar.

Your problem seems very simple to vectorize. For each pair of rows of B you want to compute
P[i,j] = np.exp(-np.sum(np.abs(B[i,:] - B[j,:])))
You can make use of array broadcasting and introduce a third dimension, summing along the last one:
P2 = np.exp(-np.sum(np.abs(B[:,None,:] - B),axis=-1))
The idea is to reshape the first occurence of B to shape (N,1,M) while the second B is left with shape (N,M). With array broadcasting, the latter is equivalent to (1,N,M), so
B[:,None,:] - B
is of shape (N,N,M). Summing along the last index will then result in the (N,N)-shape correlation array you're looking for.
Note that if you were using scipy, you would be able to do this using scipy.spatial.distance.cdist (or, equivalently, a combination of scipy.spatial.distance.pdist and scipy.spatial.distance.squareform), without unnecessarily computing the lower triangular half of this symmetrix matrix. Using #Divakar's suggestion in comments for the simplest solution this way:
from scipy.spatial.distance import cdist
P3 = 1/np.exp(cdist(B, B, 'minkowski',1))
cdist will compute the Minkowski distance in 1-norm, which is exactly the sum of the absolute values of coordinate differences.

numpy - Computing "element-wise" difference between two arrays along first axis

Suppose I have two arrays A and B with dimensions (n1,m1,m2) and (n2,m1,m2), respectively. I want to compute the matrix C with dimensions (n1,n2) such that C[i,j] = sum((A[i,:,:] - B[j,:,:])^2). Here is what I have so far:
import numpy as np
A = np.array(range(1,13)).reshape(3,2,2)
B = np.array(range(1,9)).reshape(2,2,2)
C = np.zeros(shape=(A.shape[0], B.shape[0]) )
for i in range(A.shape[0]):
for j in range(B.shape[0]):
C[i,j] = np.sum(np.square(A[i,:,:] - B[j,:,:]))
C
What is the most efficient way to do this? In R I would use a vectorized approach, such as outer. Is there a similar method for Python?
Thanks.

You can use scipy's cdist, which is pretty efficient for such calculations after reshaping the input arrays to 2D, like so -
from scipy.spatial.distance import cdist
C = cdist(A.reshape(A.shape[0],-1),B.reshape(B.shape[0],-1),'sqeuclidean')
Now, the above approach must be memory efficient and thus a better one when working with large datasizes. For small input arrays, one can also use np.einsum and leverage NumPy broadcasting, like so -
diffs = A[:,None]-B
C = np.einsum('ijkl,ijkl->ij',diffs,diffs)