Warning

This page was created from a pull request.

jax.numpy.dot¶

jax.numpy.dot(a, b, *, precision=None)[source]¶

Dot product of two arrays. Specifically,

LAX-backend implementation of dot(). In addition to the original NumPy arguments listed below, also supports precision for extra control over matrix-multiplication precision on supported devices. precision may be set to None, which means default precision for the backend, a lax.Precision enum value (Precision.DEFAULT, Precision.HIGH or Precision.HIGHEST) or a tuple of two lax.Precision enums indicating separate precision for each argument.

Original docstring below.

dot(a, b, out=None)

  • If both a and b are 1-D arrays, it is inner product of vectors (without complex conjugation).

  • If both a and b are 2-D arrays, it is matrix multiplication, but using matmul() or a @ b is preferred.

  • If either a or b is 0-D (scalar), it is equivalent to multiply() and using numpy.multiply(a, b) or a * b is preferred.

  • If a is an N-D array and b is a 1-D array, it is a sum product over the last axis of a and b.

  • If a is an N-D array and b is an M-D array (where M>=2), it is a sum product over the last axis of a and the second-to-last axis of b:

    dot(a, b)[i,j,k,m] = sum(a[i,j,:] * b[k,:,m])
Returns
outputndarray

Returns the dot product of a and b. If a and b are both scalars or both 1-D arrays then a scalar is returned; otherwise an array is returned. If out is given, then it is returned.

ValueError

If the last dimension of a is not the same size as the second-to-last dimension of b.

vdot : Complex-conjugating dot product. tensordot : Sum products over arbitrary axes. einsum : Einstein summation convention. matmul : ‘@’ operator as method with out parameter.

>>> np.dot(3, 4)
12

Neither argument is complex-conjugated:

>>> np.dot([2j, 3j], [2j, 3j])
(-13+0j)

For 2-D arrays it is the matrix product:

>>> a = [[1, 0], [0, 1]]
>>> b = [[4, 1], [2, 2]]
>>> np.dot(a, b)
array([[4, 1],
       [2, 2]])

>>> a = np.arange(3*4*5*6).reshape((3,4,5,6))
>>> b = np.arange(3*4*5*6)[::-1].reshape((5,4,6,3))
>>> np.dot(a, b)[2,3,2,1,2,2]
499128
>>> sum(a[2,3,2,:] * b[1,2,:,2])
499128