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This is documentation for an old release of NumPy (version 1.14). Read this page in the documentation of the latest stable release (version 2.2).

numpy.linalg.multi_dot

numpy.linalg.multi_dot(arrays)[source]

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 [R46] [R47]. 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.

Returns:

output : ndarray

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]

Let’s 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

References

[R46](1, 2) Cormen, “Introduction to Algorithms”, Chapter 15.2, p. 370-378
[R47](1, 2) http://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])

instead of:

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