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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_dotchainsnumpy.dotand 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_dotas: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. 
 NotesThe 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 = 75000References- 1
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      Cormen, “Introduction to Algorithms”, Chapter 15.2, p. 370-378 
- 2
 Examplesmulti_dotallows 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)
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 https://numpy.org/doc/1.19/reference/generated/numpy.linalg.multi_dot.html