Question

执行从struct CSV D获取2个数组（column1和column2）的函数，并从中绘制图形。

想法是找到每个数组的最大值，最小值，然后在 min-EPSILON 和 max + EPSILON 之间中断范围到600个相等的区域，其中 EPSILON = 10 ^（ - 6）

问题是函数没有正确绘制最低行，我认为问题是将数组中的值与 min-EPSILON 进行比较时，不确定。请指教。

这是我的代码。

void   
do_plot(CSV *D, int column1, int column2) {
#define Y_REGIONS 600
#define X_REGIONS 600
#define EPSILON 0.000001
int col1=column1-1;                     //since indexing in C language starts from 0, to be more user friendly values increased by 1
int col2=column2-1;
double  min_y = D->values[0][col1]; //min val of column
double  max_y = D->values[0][col1]; //max val of column
double  min_x = D->values[0][col2]; //min val of column
double  max_x = D->values[0][col2]; //max val of column
int     i=0,j=0,k=0;                //iteration variables
double  interval_x, interval_y;     //region
int     counter;                    //counts how many elements of "col1" and "column2" are in bucket
int     plotval;                    //plotted value
double  upper_bound_y[Y_REGIONS+1],lower_bound_y[Y_REGIONS+1];      //arrays for lower and upper bounds of regions in y (added extra 1 not to run out of regions)
double  upper_bound_x[X_REGIONS+1],lower_bound_x[X_REGIONS+1];      //arrays for lower and upper bounds of regions in x
while (i < D->number_of_rows){      
    if (D->values[i][col1] > max_y){
        max_y = D->values[i][col1];
    }
    if (D->values[i][col1] < min_y){
        min_y = D->values[i][col1];
    }
    if (D->values[i][col2] > max_x){
        max_x = D->values[i][col2];
    }
    if (D->values[i][col2] < min_x){
        min_x = D->values[i][col2];
    }
    i++;
}
/* adding EPSILON val to max and min */
max_x=max_x+EPSILON;
max_y=max_y+EPSILON;
min_x=min_x-EPSILON;
min_y=min_y-EPSILON;
interval_y=(max_y-min_y)/Y_REGIONS; //breaking y axis into Y_REGIONS equal regions
interval_x=(max_x-min_x)/X_REGIONS; //breaking x axis into Y_REGIONS equal regions
/* calculating regions of y*/
upper_bound_y[0]=max_y;             //upper bound of the first region in y
lower_bound_y[0]=max_y-interval_y;  //lower bound of the first region in y
for (j=0; j<Y_REGIONS; j++){
    upper_bound_y[j+1]=upper_bound_y[j]-interval_y;
    lower_bound_y[j+1]=lower_bound_y[j]-interval_y;
}
/* calculating regions of x */
upper_bound_x[0]=min_x+interval_x;  //upper bound of the first region in y
lower_bound_x[0]=min_x;             //lower bound of the first region in y
for (j=0; j<X_REGIONS; j++){
    upper_bound_x[j+1]=upper_bound_x[j]+interval_x;
    lower_bound_x[j+1]=lower_bound_x[j]+interval_x;
}
/* plotting the graph */
for (i=0; i<Y_REGIONS; i++){
    printf("\n%6.20lf--%6.20lf: ", lower_bound_y[i], upper_bound_y[i]); //plotting y axis
    for (j=0; j<X_REGIONS; j++){    //x axis
        counter=0;          //resetting counter
        while (k <= D->number_of_rows){
            k++;
            /* checking whether element of input lies within region and counting number of elements */
            if (D->values[k][col1] < upper_bound_y[i] && D->values[k][col1] > lower_bound_y[i]){
                if (D->values[k][col2] < upper_bound_x[j] && D->values[k][col2] > lower_bound_x[j] ){
                    counter++;
                }
            }               
        }
        k=0; //resetting counter
        plotval=floor(log(counter+1)/log(2)); //formula to show number of values in bucket
        /* plotting x lines */
        if (plotval==0){
            printf(".");
        }
        else{
            printf("%d",plotval);
        }
    }
}
printf("\n");
return;
}

Answer 1

Bounds计算是错综复杂的并且有漏洞。

见upper_bound_x[n] == lower_bound_x[n+1]。然后，当与(D->values[k][col2] == upper_bound_x[n]进行比较时，它既不适合区域n也不适合区域n+1。

// Existing code
upper_bound_x[0]=min_x+interval_x;  //upper bound of the first region in y
lower_bound_x[0]=min_x;             //lower bound of the first region in y
for (j=0; j<X_REGIONS; j++){
    upper_bound_x[j+1]=upper_bound_x[j]+interval_x;
    lower_bound_x[j+1]=lower_bound_x[j]+interval_x;
}
....
if (D->values[k][col2] < upper_bound_x[j] && D->values[k][col2] > lower_bound_x[j] ){

建议重新编写并使用bound_x[X_REGIONS+1]数组，然后使用compare：

if (D->values[k][col2] >= bound_x[j] && D->values[k][col2] < bound_x[j] ){

或者，代码可以跳过bound[]数组（x＆amp; y）并动态计算边界。

次要：

重复代码：使辅助函数计算最小值和最大值，然后每次计算一次以计算x和y。

代码应发布CSV的定义。将x放在一列中，y放在另一列中会很困惑。最好有一个point数组（自己的结构包含x和y），而不是double对的数组。

请务必#include <math.h>

将双重值与EPS中的EPSILON进行比较的精度

1 个答案: