Question

美好的一天。

我有一个包含数据的文件，其中部分内容如下所示。它包含每3秒记录一次，持续30天。我想对数据执行以下操作。

2010-03-03 16:00:31;    66.89;  24.0;   14.89;  0.08;   

2010-03-03 16:00:35;    66.15;  24.1;   14.85;  0.08;   

2010-03-03 16:00:38;    67.10;  24.2;   14.81;  0.08;   

2010-03-03 16:00:42;    66.36;  24.3;   14.78;  0.08;

2010-03-03 16:00:46;    65.83;  24.4;   14.75;  0.09;   
.
.
.
2010-03-03 17:00:31;    62.78;  25.2;   13.96;  0.12;   

2010-03-03 17:00:35;    63.94;  25.3;   13.92;  0.11;   

2010-03-03 17:00:39;    61.94;  25.3;   13.89;  0.11;   

2010-03-03 17:00:43;    60.99;  25.4;   13.88;  0.12;   

2010-03-03 17:00:46;    62.67;  25.4;   13.89;  0.13;   

2010-03-03 17:00:50;    62.57;  25.3;   13.91;  0.13;   

2010-03-03 17:00:54;    61.51;  25.3;   13.91;  0.14;
.
.
.   
2010-03-03 18:00:29;    66.04;  20.7;   13.63;  0.56;   

2010-03-03 18:00:33;    66.04;  20.7;   13.63;  0.56;   

2010-03-03 18:00:37;    65.52;  20.7;   13.59;  0.56;   

2010-03-03 18:00:40;    64.46;  20.7;   13.56;  0.56;   

2010-03-03 18:00:44;    64.88;  20.8;   13.56;  0.56;
.
.
.

从第一行开始读取文件中的所有行。

2.对于每小时，我只想计算第二列中数据的总和以及产生此总和的记录数。

3.然后将以下信息打印到文件中：

i.Date;sum(1st hour),number of records that gave rise to this sum;sum(2nd hour),num_records;sum(3rd hour),num_records;...; sum total(24 hours),totol_records;mean

这是我想要打印到文件的例子，如上所述;

03\03\2010; 15093.47; 379; 16025.46; 380; 14800.58; 379; 14605.34; 380; 21754.27,379;...;82279.12,1897;43,37

04\03\2010; 6842.051; 379; 7137.491; 380; 7215.16; 380; 7159.189; 379; 6594.672; 380;...;34948.56,1898;18,41

05\03\2010; 9938.37; 379; 9670.438; 380; 8232.032; 380; 9198.899; 379; 7083.687; 380;...;44123.426,1898;23,25

我已经开始使用此代码

int file_readline(char *file_in,char *outfile,char *strline) {
    FILE *fd=NULL;
    FILE *fo= NULL
    char *date, *tmp,*time;
    double sum=0;
    double mean = 0;
    strline=calloc(MAX_BUFFER_SIZE,sizeof(strline));
    if (strline==NULL) {
        printf ("Error calloc strline.................");
        exit(EXIT_FAILURE);
    }
    file_in = calloc((strlen(strline)+strlen(file_in)),sizeof(file_in));
    if (file_in==NULL) 
    {
        printf ("Error calloc strline.................");
        exit(EXIT_FAILURE);
    }
    fd=fopen(file_in,"r");
    int i = 0;
    int j = 0;
    while ((fgets (strline, BUFSIZ, fd))>0 && !feof(fd)){
        date = strtok(strline, " ");
        time=strtok(NULL, " "); 
        tmp = strtok(NULL, ";");
        if (i == 3) { // get only the 3rd value
            sum += strtod(tmp, NULL);
            ++i;
            // don't know how to proceed from here

Answer 1

我会通过使每一行成为一个数组来解决这个问题，然后解析数组并按照你正在做的计算进行计算。看起来您的数据文件格式一致，所以不妨利用。将数组单元存储在特定变量中然后将它们转换为浮动数据可能是最简单的方法。

Answer 2

enum {OUT,ADD};
void processline(int action,int hour,float val)
{
    static int lasthour,z;
    static float vals[25];
    static int nums[25];
    if( OUT==action )
    {
        if(lasthour)
        {
            int i;
            for(i=1;i<=z;++i)
                printf("%.2f; %d; ",vals[i],nums[i]);
            printf("%d; %.2f\n",*nums,*vals);
            lasthour=z=0;memset(vals,0,sizeof vals);memset(nums,0,sizeof nums);
        }
        return;
    }
    if( hour!=lasthour )
        ++z,lasthour=hour;
    vals[z]+=val;
    nums[z]++;
    *vals+=val;
    *nums+=1;
}
main()
{
    char z[100],lastday[20]="";
    FILE *f=fopen("test.txt","rt");
    while( fgets(z,sizeof z,f) )
    {
        char a[100],b[100];
        float fl;
        if( 3==sscanf(z,"%[^ ]%[^;];%f",a,b,&fl) )
        {
            if( strcmp(lastday,a) )
            {
                if(*lastday)
                    printf("%s; ",lastday),processline(OUT,0,0);
                strcpy(lastday,a);              
            }
            processline(ADD,atoi(b),fl);
        }
    }
    if(*lastday)
        printf("%s; ",lastday),processline(OUT,0,0);
    fclose(f);
    return 0;
}

应该有效。需要多少个Perl-LoC？！

如何从文件中读取数据并清理并存储到文件中

2 个答案: