/NumPy 1.17

# 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])
```
Parameters: `a : array_like` First argument. `b : array_like` Second argument. `out : ndarray, 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. `output : ndarray` 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`.

See also

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

#### 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
```

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Licensed under the 3-clause BSD License.
https://docs.scipy.org/doc/numpy-1.17.0/reference/generated/numpy.dot.html