numpy.linalg.multi_dot¶
-
numpy.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 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
- arrayssequence 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.
- 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.
New in version 1.19.0.
- Returns
- outputndarray
Returns the dot product of the supplied arrays.
See also
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)