xorbits.numpy.dot(a, b, out=None)[source]#

Dot product of two arrays. Specifically,

  • If both a and b are 1-D arrays, it is inner product of vectors (without complex conjugation).

  • If both a and b are 2-D arrays, it is matrix multiplication, but using matmul() or a @ b is preferred.

  • If either a or b is 0-D (scalar), it is equivalent to multiply() and using numpy.multiply(a, b) or a * b is preferred.

  • If a is an N-D array and b is a 1-D array, it is a sum product over the last axis of a and b.

  • If a is an N-D array and b is an M-D array (where M>=2), it is a sum product over the last axis of a and the second-to-last axis of b:

    dot(a, b)[i,j,k,m] = sum(a[i,j,:] * b[k,:,m])

It uses an optimized BLAS library when possible (see numpy.linalg).

  • a (array_like) – First argument.

  • b (array_like) – Second argument.

  • 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.


output – Returns the dot product of a and b. If a and b are both scalars or both 1-D arrays then a scalar is returned; otherwise an array is returned. If out is given, then it is returned.

Return type



ValueError – If the last dimension of a is not the same size as the second-to-last dimension of b.

See also


Complex-conjugating dot product.


Sum products over arbitrary axes.


Einstein summation convention.


‘@’ operator as method with out parameter.


Chained dot product.


>>> np.dot(3, 4)  

Neither argument is complex-conjugated:

>>> np.dot([2j, 3j], [2j, 3j])  

For 2-D arrays it is the matrix product:

>>> a = [[1, 0], [0, 1]]  
>>> b = [[4, 1], [2, 2]]  
>>> np.dot(a, b)  
array([[4, 1],
       [2, 2]])
>>> a = np.arange(3*4*5*6).reshape((3,4,5,6))  
>>> b = np.arange(3*4*5*6)[::-1].reshape((5,4,6,3))  
>>> np.dot(a, b)[2,3,2,1,2,2]  
>>> sum(a[2,3,2,:] * b[1,2,:,2])  

This docstring was copied from numpy.