计算文本C中的单词频率 [英] Count word frequency in text C

问题描述

我找到了一个用于计算文本文件中单词频率的C代码，但它仅适用于> 1000个单词，我需要将其用于具有+40000个单词的文件。

如何修复它以处理大文件？



代码：

I've found a C code to count word frequency in a text file but it works only with >1000 words and I need to use it with files having +40000 words.
How can I fix it to work with big files?

Code:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main(int argc, char* argv[])
{
   if (argc == 1) {
   printf("The input file name has not been provided\n");
   }
   else if (argc == 2) {
   FILE *f = fopen(argv[1], "rb");
   fseek(f, 0, SEEK_END);
   long fsize = ftell(f);
   fseek(f, 0, SEEK_SET);

   char *str = malloc(fsize + 1);
   fread(str, fsize, 1, f);
   fclose(f);

   str[fsize] = 0;
   int count = 0, c = 0, i, j = 0, k, space = 0;
   char p[1000][512], str1[512], ptr1[1000][512];
   char *ptr;
   for (i = 0;i<strlen(str);i++)
   {
   if ((str[i] == ' ')||(str[i] == ',')||(str[i] == '.'))
   {
   space++;
   }
   }
   for (i = 0, j = 0, k = 0;j < strlen(str);j++)
   {
   if ((str[j] == ' ')||(str[j] == 44)||(str[j] == 46))
   {
   p[i][k] = '\0';
   i++;
   k = 0;
   }
   else
   p[i][k++] = str[j];
   }
   k = 0;
   for (i = 0;i <= space;i++)
   {
   for (j = 0;j <= space;j++)
   {
   if (i == j)
   {
   strcpy(ptr1[k], p[i]);
   k++;
   count++;
   break;
   }
   else
   {
   if (strcmp(ptr1[j], p[i]) != 0)
   continue;
   else
   break;
   }
   }
   }
   for (i = 0;i < count;i++)
   {
   for (j = 0;j <= space;j++)
   {
   if (strcmp(ptr1[i], p[j]) == 0)
   c++;
   }
   printf("%s %d \n", ptr1[i], c);
   c = 0;
   }
   }
   return 0;

}

我的尝试：



我认为这个问题与：p [1000] [512]，str1 [512]，ptr1 [1000] [512]

What I have tried:

I think the problem is something related to: p[1000][512], str1[512], ptr1[1000][512]

推荐答案

学会正确缩进代码，显示其结构，有助于阅读和理解。它还有助于发现结构错误。

Learn to indent properly your code, it show its structure and it helps reading and understanding. It also helps spotting structures mistakes.

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main(int argc, char* argv[])
{
  if (argc == 1) {
    printf("The input file name has not been provided\n");
  }
  else if (argc == 2) {
    FILE *f = fopen(argv[1], "rb");
    fseek(f, 0, SEEK_END);
    long fsize = ftell(f);
    fseek(f, 0, SEEK_SET);

    char *str = malloc(fsize + 1);
    fread(str, fsize, 1, f);
    fclose(f);

    str[fsize] = 0;
    int count = 0, c = 0, i, j = 0, k, space = 0;
    char p[1000][512], str1[512], ptr1[1000][512];
    char *ptr;
    for (i = 0;i<strlen(str);i++)
    {
      if ((str[i] == ' ')||(str[i] == ',')||(str[i] == '.'))
      {
        space++;
      }
    }
    for (i = 0, j = 0, k = 0;j < strlen(str);j++)
    {
      if ((str[j] == ' ')||(str[j] == 44)||(str[j] == 46))
      {
        p[i][k] = '\0';
        i++;
        k = 0;
      }
      else
        p[i][k++] = str[j];
    }
    k = 0;
    for (i = 0;i <= space;i++)
    {
      for (j = 0;j <= space;j++)
      {
        if (i == j)
        {
          strcpy(ptr1[k], p[i]);
          k++;
          count++;
          break;
        }
        else
        {
          if (strcmp(ptr1[j], p[i]) != 0)
            continue;
          else
            break;
        }
      }
    }
    for (i = 0;i < count;i++)
    {
      for (j = 0;j <= space;j++)
      {
        if (strcmp(ptr1[i], p[j]) == 0)
          c++;
      }
      printf("%s %d \n", ptr1[i], c);
      c = 0;
    }
  }
  return 0;

}

专业程序员的编辑器具有此功能，其他功能包括括号匹配和语法高亮。

< a href =https://notepad-plus-plus.org/> Notepad ++ Home [ ^ ]

ultraedit [ ^ ]



代码中的注释也是一个好主意。

Professional programmer's editors have this feature and others ones such as parenthesis matching and syntax highlighting.
Notepad++ Home[^]
ultraedit[^]

Comments in code are also a good idea.

引用：

我认为这个问题与以下内容有关：p [ 1000] [512]，str1 [512]，ptr1 [1000] [512]

I think the problem is something related to: p[1000][512], str1[512], ptr1[1000][512]

有一种简单的方法可以知道，尝试，你会看到。



据我所知，这个代码非常低效。它是运行时和内存中的强制力。

There is an easy way to know, try and you will see.

As far as I understand this code, it is highly inefficient. It is brut force, both runtime and in memory.

这篇关于计算文本C中的单词频率的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！