计算文本中某个单词的频率数

Question

我写了一个用于计算文本中特定单词频率的函数。该程序每次都返回零。我该如何改进它？

while (fgets(sentence, sizeof sentence, cfPtr)) 
{ 
for(j=0;j<total4;j++) 
     { 
      frequency[j] = comparision(sentence,&w); 
      all_frequency+=frequency[j]; 
}} 
. 
. 
. 
int comparision(const char sentence[ ],char *w) 
{ 
    int length=0,count=0,l=0,i; 
    length= strlen(sentence); 
    l= strlen(w); 
    while(sentence[i]!= '\n') 
    if(strncmp(sentence,w,l)) 
     count++; 
    i++; 
    return count; 
    } 

Answer 1

我校对过你的代码，并对编码风格和变量名称进行了评论。 仍然是我留下的有条件的缺陷，这是因为没有遍历 句子。

这里是你的代码标记起来：

while(fgets(sentence, sizeof sentence, cfPtr)) { 
    for(j=0;j<total4;j++){ 
     frequency[j] = comparision(sentence,&w); 
     all_frequency+=frequency[j]; 
    } 

} 

// int comparision(const char sentence[ ],char *w) w is a poor variable name in this case. 

int comparison(const char sentence[ ], char *word) //word is a better name. 
{ 

    //int length=0,count=0,l=0,i; 

    //Each variable should get its own line. 
    //Also, i should be initialized and l is redundant. 
    //Here are properly initialized variables: 

    int length = 0; 
    int count = 0; 
    int i = 0; 

    //length= strlen(sentence); This is redundant, as you know that the line ends at '\n' 

    length = strlen(word); //l is replaced with length. 

    //while(sentence[i]!= '\n') 

    //The incrementor and the if statement should be stored inside of a block 
    //(Formal name for curley braces). 

    while(sentence[i] != '\n'){ 
     if(strncmp(sentence, word, length) == 0) //strncmp returns 0 if equal, so you  
      count++;        //should compare to 0 for equality 
     i++; 
    } 
    return count; 
}