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

>>> import numpy as np
>>> 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