基于Numpy的矩阵相乘

2024-07-17 11:42| 来源: 网络整理| 查看: 265

基于Numpy的矩阵相乘

@author: Heisenberg

主要介绍了矩阵乘法、哈达玛积、克罗内克积的概念及基于numpy的实现。

1. 矩阵乘法 1.1 概念

即线性代数中不满足交换律的矩阵乘法，又称为矩阵内积、点积（dot-product）。 i . e . A B ≠ B A i.e. AB\neq BA i.e.AB=BA。一般记为 A ⋅ B A\cdot B A⋅B。

A ( m × n ) ⋅ B ( n × k ) = C ( m × k ) A_{(m\times n)}\cdot B_{(n\times k)}=C_{(m\times k)} A(m×n)⋅B(n×k)=C(m×k)基本算法是 A A A的第一行乘以 B B B的第一列求和作为新矩阵第一个元素 C 11 C_{11} C11， A A A的第一行乘以 B B B的第二列求和作为新矩阵第一个元素 C 12 C_{12} C12，以此类推。

举个栗子： [ 1 0 2 − 1 3 1 ] ⋅ [ 3 1 2 1 1 0 ] = [ 5 1 4 2 ] \left[\begin{array}{ccc} 1 & 0 & 2 \\ -1 & 3 & 1 \end{array}\right] \cdot\left[\begin{array}{ll} 3 & 1 \\ 2 & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{ll} 5 & 1 \\ 4 & 2 \end{array}\right] [1−10321]⋅⎣⎡321110⎦⎤=[5412]

1.2 实现方法 1.2.1 @方法 import numpy as np A = np.array([[1,0,2],[-1,3,1]]) B = np.array([[3,1],[2,1],[1,0]]) A@B #Out[1]: #array([[5, 1], # [4, 2]]) 1.2.2 dot方法 A.dot(B) #or np.dot(A,B) #Out[2]: #array([[5, 1], # [4, 2]]) 1.2.3 matmul方法 np.matmul(A,B) #Out[3]: #array([[5, 1], # [4, 2]]) 1.2.4 dot() vs matmul()

在两个array的维度均为两维的情况下结果相同。

但在3个维度时，结合下例来看：

matmul()方法将最后两个维度作为矩阵的两维，相当于有2个 2 × 2 2\times 2 2×2的矩阵，在这个维度上进行对应位置的矩阵乘法，即可以视为 [ [ 2 × 3 ] [ 2 × 3 ] ] ⋅ [ [ 3 × 2 ] [ 3 × 2 ] ] = [ [ 2 × 3 ] ⋅ [ 3 × 2 ] [ 2 × 3 ] ⋅ [ 3 × 2 ] ] = [ [ 2 × 2 ] [ 2 × 2 ] ] \left[\begin{array}{c}[2\times 3] \\ [2\times 3] \end{array}\right]\cdot\left[\begin{array}{c}[3\times 2] \\ [3\times 2] \end{array}\right] = \left[\begin{array}{c}[2\times 3]\cdot [3\times 2] \\ [2\times 3]\cdot [3\times 2] \end{array}\right]=\left[\begin{array}{c}[2\times 2] \\ [2\times 2] \end{array}\right] [[2×3][2×3]]⋅[[3×2][3×2]]=[[2×3]⋅[3×2][2×3]⋅[3×2]]=[[2×2][2×2]]

np.matmul中，多维的矩阵，将前n-2维视为后2维的元素后，进行乘法运算；且禁止矩阵与标量的乘法。

dot()方法将 A A A的最后一个维度作为向量，并将 B B B的倒数第二维看作另一个向量，因此 A A A可以看成有 2 × 2 2\times2 2×2个向量， B B B中有 2 × 2 2\times2 2×2个向量，dot会将 A A A的向量与 B B B的向量全部组合在一起，因此会有种 ( 2 × 2 ) × ( 2 × 2 ) (2\times2)\times(2\times2) (2×2)×(2×2)个结果，即可以视为

[ [ 1 × 3 ] [ 1 × 3 ] [ 1 × 3 ] [ 1 × 3 ] ] ⋅ [ [ 3 × 1 ] [ 3 × 1 ] [ 3 × 1 ] [ 3 × 1 ] ] = [ 1 × 3 ] ⋅ [ [ 3 × 1 ] [ 3 × 1 ] [ 3 × 1 ] [ 3 × 1 ] ] + [ 1 × 3 ] ⋅ [ [ 3 × 1 ] [ 3 × 1 ] [ 3 × 1 ] [ 3 × 1 ] ] + [ 1 × 3 ] ⋅ [ [ 3 × 1 ] [ 3 × 1 ] [ 3 × 1 ] [ 3 × 1 ] ] + [ 1 × 3 ] ⋅ [ [ 3 × 1 ] [ 3 × 1 ] [ 3 × 1 ] [ 3 × 1 ] ] = [ [ 2 × 2 ] [ 2 × 2 ] [ 2 × 2 ] [ 2 × 2 ] ] \left[\begin{array}{llll}[1\times 3] \\ [1\times 3] \\ [1\times 3] \\ [1\times 3] \end{array}\right]\cdot\left[\begin{array}{cc}[3\times 1]&[3\times 1] \\ [3\times 1]&[3\times 1] \end{array}\right] \\= [1\times 3]\cdot\left[\begin{array}{cc}[3\times 1]&[3\times 1] \\ [3\times 1]&[3\times 1] \end{array}\right]+[1\times 3]\cdot\left[\begin{array}{cc}[3\times 1]&[3\times 1] \\ [3\times 1]&[3\times 1] \end{array}\right]+[1\times 3]\cdot\left[\begin{array}{cc}[3\times 1]&[3\times 1] \\ [3\times 1]&[3\times 1] \end{array}\right]+[1\times 3]\cdot\left[\begin{array}{cc}[3\times 1]&[3\times 1] \\ [3\times 1]&[3\times 1] \end{array}\right]\\ =\left[\begin{array}{c}[2\times 2] \\ [2\times 2]\\ [2\times 2]\\ [2\times 2] \end{array}\right] ⎣⎢⎢⎡[1×3][1×3][1×3][1×3]⎦⎥⎥⎤⋅[[3×1][3×1][3×1][3×1]]=[1×3]⋅[[3×1][3×1][3×1][3×1]]+[1×3]⋅[[3×1][3×1][3×1][3×1]]+[1×3]⋅[[3×1][3×1][3×1][3×1]]+[1×3]⋅[[3×1][3×1][3×1][3×1]]=⎣⎢⎢⎡[2×2][2×2][2×2][2×2]⎦⎥⎥⎤

>>>A = np.array([i for i in range(12)]).reshape([2,2,3]) >>>B = np.array([i for i in range(12)]).reshape([2,3,2]) """ A [[[ 0 1 2] [ 3 4 5]] [[ 6 7 8] [ 9 10 11]]] B [[[ 0 1] [ 2 3] [ 4 5]] [[ 6 7] [ 8 9] [10 11]]] """ >>>np.matmul(A,B) """ array([[[ 10, 13], [ 28, 40]], [[172, 193], [244, 274]]]) """ >>>np.dot(A,B) ''' array([[[[ 10, 13], [ 28, 31]], [[ 28, 40], [100, 112]]], [[[ 46, 67], [172, 193]], [[ 64, 94], [244, 274]]]]) ''' 2. 哈达玛积（按位相乘） 2.1 概念

哈达马积，将矩阵中相同位置的元素相乘，又称element-wise product。一般记为 A ∗ B A*B A∗B。

举个栗子： ( 1 3 2 1 0 0 1 2 2 ) ∗ ( 0 0 2 7 5 0 2 1 1 ) = ( 1 ⋅ 0 3 ⋅ 0 2 ⋅ 2 1 ⋅ 7 0 ⋅ 5 0 ⋅ 0 1 ⋅ 2 2 ⋅ 1 2 ⋅ 1 ) = ( 0 0 4 7 0 0 2 2 2 ) \left(\begin{array}{lll} 1 & 3 & 2 \\ 1 & 0 & 0 \\ 1 & 2 & 2 \end{array}\right) *\left(\begin{array}{lll} 0 & 0 & 2 \\ 7 & 5 & 0 \\ 2 & 1 & 1 \end{array}\right)=\left(\begin{array}{lll} 1 \cdot 0 & 3 \cdot 0 & 2 \cdot 2 \\ 1 \cdot 7 & 0 \cdot 5 & 0 \cdot 0 \\ 1 \cdot 2 & 2 \cdot 1 & 2 \cdot 1 \end{array}\right)=\left(\begin{array}{lll} 0 & 0 & 4 \\ 7 & 0 & 0 \\ 2 & 2 & 2 \end{array}\right) ⎝⎛111302202⎠⎞∗⎝⎛072051201⎠⎞=⎝⎛1⋅01⋅71⋅23⋅00⋅52⋅12⋅20⋅02⋅1⎠⎞=⎝⎛072002402⎠⎞

2.2 实现方法 2.2.1 *方法 A = np.array([[1,2],[3,4]]) B = np.array([[5,6],[7,8]]) A*B #array([[ 5, 12], # [21, 32]]) 2.2.2 multiply()方法 np.multiply(A,B) #array([[ 5, 12], # [21, 32]]) 3. 克罗内克积 3.1 概念

克罗内克积也被称为直积或张量积，适用于任意矩阵大小，一般记作 A ⊗ B A\otimes B A⊗B。

举个栗子： ( 1 2 3 1 ) ⊗ ( 0 3 2 1 ) = ( 1 ⋅ 0 1 ⋅ 3 2 ⋅ 0 2 ⋅ 3 1 ⋅ 2 1 ⋅ 1 2 ⋅ 2 2 ⋅ 1 3 ⋅ 0 3 ⋅ 3 1 ⋅ 0 1 ⋅ 3 3 ⋅ 2 3 ⋅ 1 1 ⋅ 2 1 ⋅ 1 ) = ( 0 3 0 6 2 1 4 2 0 9 0 3 6 3 2 1 ) \left(\begin{array}{ll} 1 & 2 \\ 3 & 1 \end{array}\right) \otimes\left(\begin{array}{ll} 0 & 3 \\ 2 & 1 \end{array}\right)=\left(\begin{array}{llll} 1 \cdot 0 & 1 \cdot 3 & 2 \cdot 0 & 2 \cdot 3 \\ 1 \cdot 2 & 1 \cdot 1 & 2 \cdot 2 & 2 \cdot 1 \\ 3 \cdot 0 & 3 \cdot 3 & 1 \cdot 0 & 1 \cdot 3 \\ 3 \cdot 2 & 3 \cdot 1 & 1 \cdot 2 & 1 \cdot 1 \end{array}\right)=\left(\begin{array}{llll} 0 & 3 & 0 & 6 \\ 2 & 1 & 4 & 2 \\ 0 & 9 & 0 & 3 \\ 6 & 3 & 2 & 1 \end{array}\right) (1321)⊗(0231)=⎝⎜⎜⎛1⋅01⋅23⋅03⋅21⋅31⋅13⋅33⋅12⋅02⋅21⋅01⋅22⋅32⋅11⋅31⋅1⎠⎟⎟⎞=⎝⎜⎜⎛0206319304026231⎠⎟⎟⎞

3.2 np.kron()方法 A = np.array([[1,2],[3,4]]) B = np.array([[5,6],[7,8]]) np.kron(A,B) #array([[ 5, 6, 10, 12], # [ 7, 8, 14, 16], # [15, 18, 20, 24], # [21, 24, 28, 32]])

【本文地址】

公司简介

联系我们

今日新闻

推荐新闻

专题文章