问题描述
分别实现矩阵相乘的3种算法，比较三种算法在矩阵大小分别为22∗2222∗22, 23∗2323∗23, 24∗2424∗24, 25∗2525∗25, 26∗2626∗26, 27∗2727∗27, 28∗2828∗28, 29∗2929∗29时的运行时间与MATLAB自带的矩阵相乘的运行时间，绘制时间对比图。
解题方法
本文采用了以下方法进行求值：矩阵计算法、定义法、分治法和Strassen方法。这里我们使用Matlab以及Python对这个问题进行处理，比较两种语言在一样的条件下，运算速度的差别。
编程语言
Python
具体代码
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
|
#-*- coding: utf-8 -*-from matplotlib.font_manager import FontPropertiesimport numpy as npimport timeimport randomimport mathimport copyimport matplotlib.pyplot as plt#n = [2**2, 2**3, 2**4, 2**5, 2**6, 2**7, 2**8, 2**9, 2**10, 2**11, 2**12]n = [2**2, 2**3, 2**4, 2**5, 2**6, 2**7, 2**8, 2**9, 2**10, 2**11]Sum_time1 = []Sum_time2 = []Sum_time3 = []Sum_time4 = []for m in n: A = np.random.randint(0, 2, [m, m]) B = np.random.randint(0, 2, [m, m]) A1 = np.mat(A) B1 = np.mat(B) time_start = time.time() C1 = A1*B1 time_end = time.time() Sum_time1.append(time_end - time_start) C2 = np.zeros([m, m], dtype = np.int) time_start = time.time() for i in range(m): for k in range(m): for j in range(m): C2[i, j] = C2[i, j] + A[i, k] * B[k, j] time_end = time.time() Sum_time2.append(time_end - time_start) A11 = np.mat(A[0:m//2, 0:m//2]) A12 = np.mat(A[0:m//2, m//2:m]) A21 = np.mat(A[m//2:m, 0:m//2]) A22 = np.mat(A[m//2:m, m//2:m]) B11 = np.mat(B[0:m//2, 0:m//2]) B12 = np.mat(B[0:m//2, m//2:m]) B21 = np.mat(B[m//2:m, 0:m//2]) B22 = np.mat(B[m//2:m, m//2:m]) time_start = time.time() C11 = A11 * B11 + A12 * B21 C12 = A11 * B12 + A12 * B22 C21 = A21 * B11 + A22 * B21 C22 = A21 * B12 + A22 * B22 C3 = np.vstack((np.hstack((C11, C12)), np.hstack((C21, C22)))) time_end = time.time() Sum_time3.append(time_end - time_start) time_start = time.time() M1 = A11 * (B12 - B22) M2 = (A11 + A12) * B22 M3 = (A21 + A22) * B11 M4 = A22 * (B21 - B11) M5 = (A11 + A22) * (B11 + B22) M6 = (A12 - A22) * (B21 + B22) M7 = (A11 - A21) * (B11 + B12) C11 = M5 + M4 - M2 + M6 C12 = M1 + M2 C21 = M3 + M4 C22 = M5 + M1 - M3 - M7 C4 = np.vstack((np.hstack((C11, C12)), np.hstack((C21, C22)))) time_end = time.time() Sum_time4.append(time_end - time_start)f1 = open('python_time1.txt', 'w')for ele in Sum_time1: f1.writelines(str(ele) + '\n')f1.close()f2 = open('python_time2.txt', 'w')for ele in Sum_time2: f2.writelines(str(ele) + '\n')f2.close()f3 = open('python_time3.txt', 'w')for ele in Sum_time3: f3.writelines(str(ele) + '\n')f3.close()f4 = open('python_time4.txt', 'w')for ele in Sum_time4: f4.writelines(str(ele) + '\n')f4.close()font = FontProperties(fname=r"c:\windows\fonts\simsun.ttc", size=8)plt.figure(1)plt.subplot(221)plt.semilogx(n, Sum_time1, 'r-*')plt.ylabel(u"时间(s)", fontproperties=font)plt.xlabel(u"矩阵的维度n", fontproperties=font)plt.title(u'python自带的方法', fontproperties=font)plt.subplot(222)plt.semilogx(n, Sum_time2, 'b-*')plt.ylabel(u"时间(s)", fontproperties=font)plt.xlabel(u"矩阵的维度n", fontproperties=font)plt.title(u'定义法', fontproperties=font)plt.subplot(223)plt.semilogx(n, Sum_time3, 'y-*')plt.ylabel(u"时间(s)", fontproperties=font)plt.xlabel(u"矩阵的维度n", fontproperties=font)plt.title( u'分治法', fontproperties=font)plt.subplot(224)plt.semilogx(n, Sum_time4, 'g-*')plt.ylabel(u"时间(s)", fontproperties=font)plt.xlabel(u"矩阵的维度n", fontproperties=font)plt.title( u'Strasses法', fontproperties=font)plt.figure(2)plt.semilogx(n, Sum_time1, 'r-*', n, Sum_time2, 'b-+', n, Sum_time3, 'y-o', n, Sum_time4, 'g-^')#plt.legend(u'python自带的方法', u'定义法', u'分治法', u'Strasses法', fontproperties=font)plt.show() |
以上这篇Python实现矩阵相乘的三种方法小结就是小编分享给大家的全部内容了，希望能给大家一个参考，也希望大家多多支持服务器之家。
原文链接：https://blog.csdn.net/zhenguipa8450/article/details/78986540
