• 个人项目词频统计2


    SSH:git@git.coding.net:shenbaishan/cipintongji_.git

    再修正SSH:git@git.coding.net:shenbaishan/cipintongji_.git

    需求分析: 

          1. 小文件输入——从控制台由用户输入到文件中,再对文件进行统计;

      2.支持命令行输入英文作品的文件名;

      3.支持命令行输入存储有英文作品文件的目录名,批量统计;

      4.从控制台读入英文单篇作品,重定向输入流。

     部分代码分析:

    See if the transformed @string repeat,than add the frequency,otherwise create class @String to contain the new @string

     1 void If_Repet(char *WordP)
     2 {
     3     vector<String>::iterator iter;
     4     for (iter = Sort_Word.begin(); iter != Sort_Word.end(); iter++)
     5     if (!strcmp(iter->Word_Point, WordP))
     6     {
     7         Total++;
     8         iter->Frequency++;
     9         return;
    10     }
    11     Sort_Word.push_back(String(WordP));
    12 }

    read each letter and transform them into @string

     1 void Catch_Letter(char *File_P)
     2 {
     3     char *Wor_Po = (char *)malloc(4);
     4     FILE *Article = fopen(File_P, "r");
     5     char *Letter_Point;
     6     while (!feof(Article))
     7     {
     8         fscanf(Article, "%s", Wor_Po);
     9         Letter_Point = Wor_Po;
    10         while (*Letter_Point++)
    11         {
    12             if (*Letter_Point == ',' || *Letter_Point == '.')
    13                 *Letter_Point = 0;
    14         }
    15         If_Repet(Wor_Po);
    16     }
    17 }

    after this function,all unique word should belong to an unique class @String     

     1 void Stat_Alone(char *Master_P)
     2 {
     3         Catch_Letter(Master_P);
     4         sort(Sort_Word.begin(), Sort_Word.end());
     5         reverse(Sort_Word.begin() ,Sort_Word.end());
     6         cout << "total :" << Total << endl << endl;
     7 
     8         int count = 20;
     9         for (vector<String>::iterator iter = Sort_Word.begin(); iter != Sort_Word.end(); iter++)/
    10         {
    11             printf("Word : %-15s Frequency :%d
    ", iter->Word_Point, iter->Frequency);
    12             if (!count--) break;
    13         }
    14         printf(".......");
    15         Total = 0;
    16         Sort_Word.clear();
    17 }

    distinguish which funtion according to the @cin from @console  and   read each file form the folder, and statist that file alone

     1 void main(int argv ,char **argc)
     2 {
     3     time_t start = clock();
     4 
     5     if (strcmp(argc[1], "folder"))    Stat_Alone(argc[1]);
     6     else
     7     {
     8         char *Folder = (char *)malloc(40*sizeof(int));
     9         FILE *Folder_P = fopen("folder.txt" ,"r");
    10         while (!feof(Folder_P))
    11         {
    12             if (feof(Folder_P)) break;
    13             
    14             fscanf(Folder_P, "%s", Folder);
    15             Stat_Alone(strcat(Folder, "\article.txt"));
    16         }
    17     }
    18 
    19     printf("Use time: %.0f ms" ,double(clock() - start)*1000/CLOCKS_PER_SEC);
    20     getchar();
    21 }

    单元测试:

    功能1

     

    功能2

    功能3

    功能4

    实现了功能,但是巨慢,但还好没崩溃

  • 相关阅读:
    wmware虚拟机的克隆
    解决SecureCRT无法用非root账号登录ssh
    Docker容器操作
    Docker镜像操作
    Docker的安装和启动
    linux安装tomcat
    POJ 2456 Aggressive cows ( 二分搜索)
    POJ 1064 Cable master (二分查找)
    2008 APAC local onsites C Millionaire (动态规划,离散化思想)
    贿赂囚犯 Bribe the prisoners ( 动态规划+剪枝)
  • 原文地址:https://www.cnblogs.com/shenbaishan/p/5871105.html
Copyright © 2020-2023  润新知