计算文本C中的单词频率

本文介绍了计算文本C中的单词频率的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 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。 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.