Numpy “vectorized” row-wise dot product running slower than a for loop

Question

Given a matrix A with shape (n,k) and a vector s of size n , I want to compute a matrix G with shape (k,k) as follows:

G += s[i] * A[i].T * A[i] , for all i in {0,...,n-1}

I tried to implement that using a for loop ( Method 1 ) and in a vectorized manner ( Method 2 ), but the for loop implementation is faster for large values of k (specially when k > 500 ).

The code was written as follows:

import numpy as np
k = 200
n = 50000
A = np.random.randint(0, 1000, (n,k)) # generates random data for the matrix A (n,k)
G1 = np.zeros((k,k)) # initialize G1 as a (k,k) matrix
s = np.random.randint(0, 1000, n) * 1.0 # initialize a random vector of size n

# METHOD 1
for i in xrange(n):
    G1 += s[i] * np.dot(np.array([A[i]]).T, np.array([A[i]]))

# METHOD 2
G2 = np.dot(A[:,np.newaxis].T, s[:,np.newaxis]*A)
G2 = np.squeeze(G2) # reduces dimension from (k,1,k) to (k,k)

The matrices G1 and G2 are the same (they are the matrix G ), and the only difference is how they were computed. Is there a more clever and efficient way to compute this?

Finally, these are the times I got with random sizes for k and n :

Test #: 1
k,n: (866, 45761)
Method1: 337.457569838s
Method2: 386.290487051s
--------------------
Test #: 2
k,n: (690, 48011)
Method1: 152.999140978s
Method2: 226.080267191s
--------------------
Test #: 3
k,n: (390, 5317)
Method1: 5.28722500801s
Method2: 4.86999702454s
--------------------
Test #: 4
k,n: (222, 5009)
Method1: 1.73456382751s
Method2: 0.929286956787s
--------------------
Test #: 5
k,n: (915, 16561)
Method1: 101.782826185s
Method2: 159.167108059s
--------------------
Test #: 6
k,n: (130, 11283)
Method1: 1.53138184547s
Method2: 0.64450097084s
--------------------
Test #: 7
k,n: (57, 37863)
Method1: 1.44776391983s
Method2: 0.494270086288s
--------------------
Test #: 8
k,n: (110, 34599)
Method1: 3.51851701736s
Method2: 1.61688089371s

Answer 1

Two much more improved versions would be -

(A.T*s).dot(A)
(A.T).dot(A*s[:,None])

Issue(s) with method2 :

With method2 , we are creating A[:,np.newaxis].T , which would be of shape (k,1,n) , that's a 3D array. I think with a 3D array, np.dot goes into some kind of loop and isn't truly vectorized (source code could reveal more info here).

For such 3D tensor multiplications, it's better to use the tensor equivalent : np.tensordot . Thus, an improved version of method2 becomes :

G2 = np.tensordot(A[:,np.newaxis].T, s[:,np.newaxis]*A, axes=((2),(0)))
G2 = np.squeeze(G2)

Since, we are sum-reducing just one axis from each of those inputs with np.tensordot , we don't really need tensordot here and simply np.dot on the squeezed-in version would suffice. This will lead us back to method4 .

Runtime test

Approaches -

def method1(A, s):
    G1 = np.zeros((k,k)) # initialize G1 as a (k,k) matrix
    for i in xrange(n):
        G1 += s[i] * np.dot(np.array([A[i]]).T, np.array([A[i]]))
    return G1

def method2(A, s):
    G2 = np.dot(A[:,np.newaxis].T, s[:,np.newaxis]*A)
    G2 = np.squeeze(G2) # reduces dimension from (k,1,k) to (k,k)
    return G2

def method3(A, s):
    return (A.T*s).dot(A)

def method4(A, s):
    return (A.T).dot(A*s[:,None])

def method2_improved(A, s):
    G2 = np.tensordot(A[:,np.newaxis].T, s[:,np.newaxis]*A, axes=((2),(0)))
    G2 = np.squeeze(G2)
    return G2

Timings and verification -

In [56]: k = 200
    ...: n = 5000
    ...: A = np.random.randint(0, 1000, (n,k))
    ...: s = np.random.randint(0, 1000, n) * 1.0
    ...: 

In [72]: print np.allclose(method1(A, s), method2(A, s))
    ...: print np.allclose(method1(A, s), method3(A, s))
    ...: print np.allclose(method1(A, s), method4(A, s))
    ...: print np.allclose(method1(A, s), method2_improved(A, s))
    ...: 
True
True
True
True

In [73]: %timeit method1(A, s)
    ...: %timeit method2(A, s)
    ...: %timeit method3(A, s)
    ...: %timeit method4(A, s)
    ...: %timeit method2_improved(A, s)
    ...: 
1 loops, best of 3: 1.12 s per loop
1 loops, best of 3: 693 ms per loop
100 loops, best of 3: 8.12 ms per loop
100 loops, best of 3: 8.17 ms per loop
100 loops, best of 3: 8.28 ms per loop