Python

来源：互联网发布：程序员一天工作几小时编辑：程序博客网时间：2024/05/21 03:55
操作系统课程作业代码
from math import *from random import *from time import clockimport systry:    #生成随机序列    def stream(n,m):        stm = []        for i in range(n):            stm.append(randint(0,m))        print("输入流为："+str(stm))        return stm    #返回去重序列    def uniquelist(lst):        newlist = []        for i in lst:            if i not in newlist:                newlist.append(i)        return newlist    #LRU算法    def LRU(b,stm):        lst = []        count = 0        index = 0        for s in stm:            if s in lst:                index +=1                print(s , ' ', lst)                continue          #如果在物理块中，不缺页            else:                if len(lst)<b:                    lst.append(s)                else:                    lstb = stm[:index]    #反之，将所求数据流中前第b个非重复值置换出来                    lstb = lstb[::-1]                    lstb = uniquelist(lstb)                    lstb[b-1]= s                    lst = lstb[:b]                print(s , ' ', lst , '缺页')                count += 1             index += 1        print("LRU算法缺页率为:{:.5f}".format(count/len(stm)))    #time+1    def timeadd1(lst):        tranlist = []        for i in lst[1::2]:            i += 1            tranlist.append(i)        lst[1::2] = tranlist        return lst    #FIFO算法    def FIFO(b,stm):        lst = []        count = 0                      for s in stm:            if s in lst[::2]:                lsta = lst[::2]                lst[lsta.index(s)*2+1] += 1                print(s , ' ', lst[::2])                continue          #如果在物理块中，不缺页            else:                lstb = lst[1::2]                                if len(lst)< 2*b:                    lst.append(s)                    timeadd1(lst)                    lst.append(1)                else:                    c = lstb.index(max(lstb))                    lst[2*c] = s                    timeadd1(lst)                    lst[2*c+1] = 1                print(s , ' ', lst[::2] , '缺页')                count += 1        print("FIFO算法缺页率为:{:.5f}".format(count/len(stm)))    #主函数    def main():        n = eval(input("Enter the length of input stream:"))        #n = 输入流的长度，由用户输入        m = eval(input("Enter the range of the input stream(0-m):"))#m = 输入流的范围，由用户输入，即范围为0-m之间        b = eval(input("Enter the block size:"))                    #b = 物理块的大小，由用户输入        print("物理块大小为：%d"%b)                                 # 打印物理块的大小        stm = stream(n,m)                                           # 调用stream()函数生成输入流        t1 = clock()        FIFO(b,stm)                                                 #调用FIFO()函数计算FIFO算法的缺页率        print("FIFO算法用时为:{:.5f}".format(clock()-t1))           #打印FIFO算法大概用时        t2 = clock()        LRU(b,stm)                                                  #调用LRU()函数计算LRU算法的缺页率        print("LRU算法用时为:{:.5f}".format(clock()-t2))    #测试函数    def test():        n = 30        m = 9        b = 3        stm = [1,2,5,6,8,3,6,5,3,6,5,6,8,9,2,7,0,4,9,5,3,6,7,4,5,8,7,3,4,5]        print("物理块大小为：%d"%b)        print(stm)        FIFO(b,stm)        LRU(b,stm)    #调用主函数    if __name__ == '__main__':        main()    except:    print ("Unexpected error:", sys.exc_info() )# sys.exc_info()返回出错信息
效果为：
当时还用py2exe模块封装成exe格式，py2exe仅支持Python2.X，具体使用自行摸索
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