numpy.dot#

numpy.dot(a, b, out=None)#

Dot product of two arrays. Specifically,

  • 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])
    

It uses an optimized BLAS library when possible (see numpy.linalg).

Parameters:
aarray_like

First argument.

barray_like

Second argument.

outndarray, optional

Output argument. This must have the exact kind that would be returned if it was not used. In particular, it must have the right type, must be C-contiguous, and its dtype must be the dtype that would be returned for dot(a,b). This is a performance feature. Therefore, if these conditions are not met, an exception is raised, instead of attempting to be flexible.

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.

Raises:
ValueError

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

See also

vdot

Complex-conjugating dot product.

tensordot

Sum products over arbitrary axes.

einsum

Einstein summation convention.

matmul

‘@’ operator as method with out parameter.

linalg.multi_dot

Chained dot product.

Examples

>>> 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