CODEWITHSUNDEEP
numpy
ejang
ejang

Reputation: 4062

Dot product of ith row with ith column

In NumPy:

A = np.array([[1,2,3],[4,5,6]])
array([[1, 3, 5],
       [2, 4, 6]])

B = np.array([[1,2],[3,4],[5,6]])
array([[1, 2],
       [3, 4],
       [5, 6]])

A.dot(B)
array([[35, 44],
       [44, 56]])

I only care about getting A.dot(B).diagonal() = array([35, 56])

Is there a way I can get array([35, 56]) without having to compute the inner products of all the rows and columns? I.e. the inner product of the ith row with ith column?

I ask because the performance difference becomes more significant for larger matrices.

Upvotes: 3

Views: 1638

Answers (2)

mathematical.coffee
mathematical.coffee

Reputation: 56905

This is just matrix multiplication for 2D arrays:

C[i, j] = sum(A[i, ] * B[, j])

So since you just want the diagonal elements, looks like you're after

sum(A[i, ] * B[, i]) # for each i

So you could just use list comprehension:

[np.dot(A[i,:], B[:, i]) for i in xrange(A.shape[0])]
# [22, 64]

OR, (and this only works because you want a diagonal so this assumes that if A's dimensions are n x m, B's dimensions will be m x n):

np.sum(A * B.T, axis=1)
# array([22, 64])

(no fancy numpy tricks going on here, just playing around with the maths).

Upvotes: 6

nair.ashvin
nair.ashvin

Reputation: 801

Can you simply leave out the row in the parameter you don't care about?

The 2x3 x 3x2 gives you a 2x2 result.

A 1x3 x 3x2 matrices will give you only the top row of [A][B], a 1x2 matrix.

EDIT: misread the question. Still, each value in the matrix is produced by the product of the transpose of a column and a row.

Upvotes: 0

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