Numpy基础 dot 矩阵乘积

Python : 3.8.11

numpy : 1.20.1

OS : Ubuntu Kylin 20.04

Conda : 4.10.1

jupyter lab : 3.1.4

代码示例

import numpy as np
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a = np.array([[1,2,3,4],[5,6,7,8]])
b = np.array([[1,2,3],[4,5,6],[7,8,9],[10,11,12]])
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a

array([[1, 2, 3, 4],
       [5, 6, 7, 8]])
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b

array([[ 1,  2,  3],
       [ 4,  5,  6],
       [ 7,  8,  9],
       [10, 11, 12]])
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a.shape,b.shape

((2, 4), (4, 3))
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# 第一个矩阵的列数（column）和第二个矩阵的行数（row）相同
np.dot(a,b)

array([[ 70,  80,  90],
       [158, 184, 210]])
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np.dot(b,a)

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-69-e6bd3a7b39a0> in <module>
----> 1 np.dot(b,a)

<__array_function__ internals> in dot(*args, **kwargs)

ValueError: shapes (4,3) and (2,4) not aligned: 3 (dim 1) != 2 (dim 0)
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源码学习

help(np.dot)

Help on function dot in module numpy:

dot(...)
    dot(a, b, out=None)
    
    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 :func:`matmul` or ``a @ b`` is preferred.
    
    - If either `a` or `b` is 0-D (scalar), it is equivalent to :func:`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])

......
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Numpy基础 dot 矩阵乘积

代码示例

源码学习

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