# xorbits.numpy.linalg.multi_dot#

xorbits.numpy.linalg.multi_dot(arrays, *, out=None)#

Compute the dot product of two or more arrays in a single function call, while automatically selecting the fastest evaluation order.

multi_dot chains numpy.dot and uses optimal parenthesization of the matrices 1 2. Depending on the shapes of the matrices, this can speed up the multiplication a lot.

If the first argument is 1-D it is treated as a row vector. If the last argument is 1-D it is treated as a column vector. The other arguments must be 2-D.

Think of multi_dot as:

def multi_dot(arrays): return functools.reduce(np.dot, arrays)

Parameters
• arrays (sequence of array_like) – If the first argument is 1-D it is treated as row vector. If the last argument is 1-D it is treated as column vector. The other arguments must be 2-D.

• 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.

New in version 1.19.0(numpy.linalg).

Returns

output – Returns the dot product of the supplied arrays.

Return type

ndarray

numpy.dot

dot multiplication with two arguments.

References

1

Cormen, “Introduction to Algorithms”, Chapter 15.2, p. 370-378

2

https://en.wikipedia.org/wiki/Matrix_chain_multiplication

Examples

multi_dot allows you to write:

>>> from numpy.linalg import multi_dot
>>> # Prepare some data
>>> A = np.random.random((10000, 100))
>>> B = np.random.random((100, 1000))
>>> C = np.random.random((1000, 5))
>>> D = np.random.random((5, 333))
>>> # the actual dot multiplication
>>> _ = multi_dot([A, B, C, D])


>>> _ = np.dot(np.dot(np.dot(A, B), C), D)
>>> # or
>>> _ = A.dot(B).dot(C).dot(D)


Notes

The cost for a matrix multiplication can be calculated with the following function:

def cost(A, B):
return A.shape * A.shape * B.shape


Assume we have three matrices $$A_{10x100}, B_{100x5}, C_{5x50}$$.

The costs for the two different parenthesizations are as follows:

cost((AB)C) = 10*100*5 + 10*5*50   = 5000 + 2500   = 7500
cost(A(BC)) = 10*100*50 + 100*5*50 = 50000 + 25000 = 75000


Warning

This method has not been implemented yet. Xorbits will try to execute it with numpy.linalg.

This docstring was copied from numpy.linalg.