Question

    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>
    #define FILE_NAME 20
    #define LIST_SIZE 50

    typedef struct
    {
    char *name;
    int score;
    }RECORD;

    float calMean(RECORD list[], int count);
    void sortData(RECORD list[], int count);
    float calMedian(RECORD list[], int size);
    int calMode(RECORD list[], int count);
    int main (void)
{
       // Declarations
         float mean;
         float median;
         int mode;
         FILE *fp;
           char fileName[FILE_NAME];
           RECORD list[LIST_SIZE];
           char buffer[100];
           int count = 0;
           int i;
       // Statements
           printf("Enter the file name: ");
           gets(fileName);
           fp = fopen(fileName, "r");
           if(fp == NULL)
           printf("Error cannot open the file!\n");
           while(fgets(buffer, 100, fp) != NULL)
           {
             if( count >= LIST_SIZE)
             {
                printf("Only the first 50 data will be read!\n");
                 break;
             }
             if( count < LIST_SIZE)
            {
                list[count].name = (char*) malloc(strlen(buffer)*sizeof(char));
                   sscanf(buffer,"%[^,], %d", list[count].name, &list[count].score);
                  printf("name is %s and score is %d\n", list[count].name, list[count].score);
                  count++;
            }
                for( i =0; i < (LIST_SIZE - count); i++)
            {
                list[count + i].name = 0;
                list[count + i].score = 0;
            }
         }
           printf("Read in %d data records\n", count);
           mean = calMean(list, count);
         sortData(list, count);
         mode = calMode(list, count);
           printf("%2.2f\n", mean);
           fclose(fp);
           return 0;
}

float calMean(RECORD list[], int count)
{
       float tempMean;
        int sum = 0;
        int i;
        for(i = 0; i < count; i++)
        sum += list[i].score;
        tempMean = (float) sum/count;
        return tempMean;
}

void sortData(RECORD list[], int count)
{
    int temp;
    int current;
    int walker;
    float median;
    int size = count;
    for(current = 0; current  < count; current++)
    {
       for( walker = count-1; walker > current; walker--)
          if(list[walker].score < list[walker -1].score)
          {
             temp = list[walker].score;
             list[walker].score = list[walker -1].score;
             list[walker -1].score = temp;
          }
          printf("%d\n", list[current].score);
    }
    median = calMedian(list, size);
    printf("%2.2f\n", median);
    return;
}

float calMedian(RECORD list[], int size)
{
    float tempMedian;
    printf("size is: %d\n", size);
    if ( size % 2 == 0) 
       tempMedian = (float) ((list[size/2].score + list[size/2-1].score)/2.0) ;                   
       else                                                    
         tempMedian = (float) list[size/2 - 1].score; 
return tempMedian;
}

int calMode(RECORD list[], int count)
{
    int tempMode = 1;
    int i, j;

    for(i=0;i<10;i++)
    {
       list[list[i].score].score++;
    }
    for(i = 0; i < count; i++)
    {
       printf("\n%d:",list[i].score);
       for(j = 0; j < list[i].score; j++)
       {
          printf("*");
       }
    }
    printf("\n\n");
    return tempMode;
}

您好我正在尝试编写一个直方图，列出所有得分和出现的频率，然后使用直方图找到所有得分的模式，它们都在calMode函数中。

上面的代码是我尝试编写这个直方图，但它不正确

我得到了这个输出：


    Enter the file name: in.txt
    name is Ada Lovelace and score is 66
    name is Linus Torvalds and score is 75
    name is Peter Norton and score is 82
    name is Ken Thompson and score is 82
    name is Steve Wozniak and score is 79
    name is Marc Andreessen and score is 60
    name is Donald Knuth and score is 60
    name is Adele Goldberg and score is 71
    name is Grace Hopper and score is 82
    name is Bill Joy and score is 91
    name is Andrew Tanenbaum and score is 71
    name is Brian Kernighan and score is 72
    Read in 12 data records
    60
    60
    66
    71
    71
    72
    75
    79
    82
    82
    82
    91
    size is: 12
    73.50

    60:************************************************************
    60:************************************************************
    66:******************************************************************
    71:***********************************************************************
    71:***********************************************************************
    72:************************************************************************
    75:***************************************************************************
    79:*****************************************************************************
    **
     82:*****************************************************************************
*****
    82:*****************************************************************************
*****
    82:*****************************************************************************
*****
    91:*****************************************************************************
**************

74.25

这个直方图函数的算法有什么建议吗？

Answer 1

您的代码正在写入正确数量的星号，但是您运行此代码的控制台窗口不够宽，无法显示该数量的星号而没有换行。您可以更改控制台窗口的宽度。我不确定您正在运行哪个操作系统，因此我无法指导您完成更改控制台窗口宽度的过程。

尽管如此，改变比例是一个简单的选择。只需改变你的循环，使其以分数除以任何比例结束。例如，如果您想要一个星号代表两个：

   for(j = 0; j < list[i].score / 2; j++)
   {
      putchar('*');
   }

现在您遇到的问题是直方图中没有显示两个剩余部分。也许您可以将缩放的输出':'更改为代表两个，并为任何余数更改'.'：

   for(j = 0; j < list[i].score / 2; j++)
   {
      putchar(':');
   }

   if (list[i].score % 2 == 1) {
      putchar('.');
   }

写函数输出分数的直方图

1 个答案: