标签:24bitRGB 为例 int double height ++ 256 width
思路:开辟3个width*height的unsigned char型数组;打开要读出的RGB文件(以“rb”方式打开),打开3个要输出的数据统计文件(以“w”方式打开,可命名为R_sat.txt等);将RGB数据从RGB文件中读出,并分别保存到3个数组中,期间计算数据的概率分布和熵,并将这些数据写入3个数据统计txt文件中。
#include “pch.h”
#include<stdio.h>
#include<math.h>
#include
int main()
{
FILE fp;
fp = fopen(“down.rgb”, “rb”);
const int width = 256;
const int height = 256;
unsigned char cache[widthheight * 3];
fread(cache, sizeof(unsigned char), widthheight * 3, fp);
unsigned char red[widthheight] = { 0 };
unsigned char green[widthheight] = { 0 };
unsigned char blue[widthheight] = { 0 };
(这一部分是定义数组)
int ri = 0;
int gi = 0;
int bi = 0;
for (int i = 0; i < 3 * width*height; i++)
{
if (i % 3 == 0)
{
blue[bi] = cache[i];
bi++;
}
else if (i % 3 == 1)
{
green[gi] = cache[i];
gi++;
}
else if (i % 3 == 2)
{
red[ri] = cache[i];
ri++;
}
}
(这一部分是计算rgb的数目)
int rtime[256] = { 0 };
int gtime[256] = { 0 };
int btime[256] = { 0 };
for (int i = 0; i < 256; i++)
{
for (int j = 0; j < width*height; j++)
{
if (i == red[j])
rtime[i]++;
if (i == green[j])
gtime[i]++;
if (i == blue[j])
btime[i]++;
}
}
double rfreq[256] = { 0 };
double gfreq[256] = { 0 };
double bfreq[256] = { 0 };
for (int i = 0; i < 256; i++)
{
rfreq[i] = double(rtime[i]) / (width*height);
gfreq[i] = double(gtime[i]) / (width*height);
bfreq[i] = double(btime[i]) / (width*height);
}
(这一部分计算频率)
FILE *R;
FILE *G;
FILE *B;
R = fopen("red.txt", "w");
G = fopen("green.txt", "w");
B = fopen("blue.txt", "w");
fprintf(R, "symbol\tfreq\n");
for (int i = 0; i < 256; i++)
{
fprintf(R, "%d\t%f\n", i, rfreq[i]);
}
fprintf(G, "symbol\tfreq\n");
for (int i = 0; i < 256; i++)
{
fprintf(G, "%d\t%f\n", i, gfreq[i]);
}
fprintf(B, "symbol\tfreq\n");
for (int i = 0; i < 256; i++)
{
fprintf(B, "%d\t%f\n", i, bfreq[i]);
}
(这一部分写入数据)
double hr = 0;
double hg = 0;
double hb = 0;
for (int i = 0; i < 256; i++)
{
if (rfreq[i] != 0)
hr = hr - rfreq[i] * log(rfreq[i]) / log(2);
if (gfreq[i] != 0)
hg = hg - gfreq[i] * log(gfreq[i]) / log(2);
if (bfreq[i] != 0)
hb = hb - bfreq[i] * log(bfreq[i]) / log(2);
}
printf("H(红基色)=%f\n", hr);
printf("H(绿基色)=%f\n", hg);
printf("H(蓝基色)=%f\n", hb);
fclose(fp);
fclose(R);
fclose(G);
fclose(B);
return 0;
}
(这一部分计算熵)
结果如下:
标签:24bitRGB,为例,int,double,height,++,256,width 来源: https://blog.csdn.net/zhaly615/article/details/104884354
本站声明: 1. iCode9 技术分享网(下文简称本站)提供的所有内容,仅供技术学习、探讨和分享; 2. 关于本站的所有留言、评论、转载及引用,纯属内容发起人的个人观点,与本站观点和立场无关; 3. 关于本站的所有言论和文字,纯属内容发起人的个人观点,与本站观点和立场无关; 4. 本站文章均是网友提供,不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属;如您发现该文章侵犯了您的权益,可联系我们第一时间进行删除; 5. 本站为非盈利性的个人网站,所有内容不会用来进行牟利,也不会利用任何形式的广告来间接获益,纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。