Posted in Python onJuly 26, 2018
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
#-*- coding: utf-8 -*- from matplotlib.font_manager import FontProperties import numpy as np import time import random import math import copy import 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三水点靠木。
Python实现矩阵相乘的三种方法小结
- Author -
Mr_Leeeee声明:登载此文出于传递更多信息之目的,并不意味着赞同其观点或证实其描述。
Reply on: @reply_date@
@reply_contents@