在 C 中将双精度值与 EPSILON 进行比较的精度
precision of comparing double values with EPSILON in C
执行从 struct CSV D 获取 2 个数组（column1 和 column2）并从中绘制图形的函数。
想法是找到每个数组的最大，最小值，然后打破min−EPSILON和之间的范围max+EPSILON 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线，我认为问题是将数组中的值与 min-EPSILON[=28 进行比较时的问题=]，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;
}
边界计算复杂且有漏洞。
看到 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] ){
或者，代码可以跳过 bound[] 数组 (x&y) 并即时计算边界。
未成年人：
重复代码：使辅助函数计算最小值和最大值，然后分别调用一次以计算 x 和 y。
代码应该post定义CSV。在一列中有 x 而在另一列中有 y 是混乱的。最好有一个 point 的数组（制作自己的包含 x 和 y 的结构），而不是 double 对的数组。
一定要#include <math.h>
执行从 struct CSV D 获取 2 个数组（column1 和 column2）并从中绘制图形的函数。
想法是找到每个数组的最大，最小值，然后打破min−EPSILON和之间的范围max+EPSILON 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线，我认为问题是将数组中的值与 min-EPSILON[=28 进行比较时的问题=]，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;
}
边界计算复杂且有漏洞。
看到 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] ){
或者，代码可以跳过 bound[] 数组 (x&y) 并即时计算边界。
未成年人：
重复代码：使辅助函数计算最小值和最大值，然后分别调用一次以计算 x 和 y。
代码应该post定义CSV。在一列中有 x 而在另一列中有 y 是混乱的。最好有一个 point 的数组（制作自己的包含 x 和 y 的结构），而不是 double 对的数组。
一定要#include <math.h>