在C双键值与EPSILON比较精度(precision of comparing double va

2019-10-23 04:33发布

做功能,从结构CSV d需要2门阵列(列1列2和),并从它生成的图表。

思想是找到最大值,每个阵列的分钟值,然后分解最小EPSILONmax + EPSILON之间范围在600个相等的区域,其中,EPSILON = 10 ^( - 6)

问题是,功能无法正常绘制的最低线,我认为这个问题是从阵列分钟,EPSILON,不知道比较值当。 请指教。

这里是我的代码。

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:

边界的计算是令人费解的和有洞。

看到upper_bound_x[n] == lower_bound_x[n+1] 。 然后,当一个比较发生(D->values[k][col2] == upper_bound_x[n] ,它既不适合在区域n也不区域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,然后使用比较:

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

可替换地,代码可以跳过bound[]阵列(X,Y),并在运行中计算边界。

次要:

重复码:请帮助函数来计算最小值和最大值,然后CAL各一次来计算xy

代码应该张贴的定义CSV 。 它是混乱有x在一列和y在另一个。 最好是有一个阵列point (制作自己的结构拿着xy ),而不是数组double对。

时一定要#include <math.h>



文章来源: precision of comparing double values with EPSILON in C