numpy.linalg.multi_dot¶

linalg.
multi_dot
(arrays, *, out=None)[source]¶ Compute the dot product of two or more arrays in a single function call, while automatically selecting the fastest evaluation order.
multi_dot
chainsnumpy.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 1D it is treated as a row vector. If the last argument is 1D it is treated as a column vector. The other arguments must be 2D.
Think of
multi_dot
as:def multi_dot(arrays): return functools.reduce(np.dot, arrays)
 Parameters
 arrayssequence of array_like
If the first argument is 1D it is treated as row vector. If the last argument is 1D it is treated as column vector. The other arguments must be 2D.
 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 Ccontiguous, 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.
 Returns
 outputndarray
Returns the dot product of the supplied arrays.
See also
numpy.dot
dot multiplication with two arguments.
Notes
The cost for a matrix multiplication can be calculated with the following function:
def cost(A, B): return A.shape[0] * A.shape[1] * B.shape[1]
Assume we have three matrices .
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
References
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])
instead of:
>>> _ = np.dot(np.dot(np.dot(A, B), C), D) >>> # or >>> _ = A.dot(B).dot(C).dot(D)