在JAVA中,将图像的RGB值转换为二进制的最快方法是什么？

Question

我试图将图像上任何给定像素的RGB值变为12位二进制,每个通道由4位表示.例如,如果一个像素的红色通道位于"255"或二进制"11111111".我希望将其转换为"1111".

我有一个代码工作,但它很慢,我想知道是否有更好的方法来做到这一点.

这就是我所做的.

1. 我使用方法getRGB(x,y)来获取一个像素的RGB值.例如-4278864
2. 我把它转换成3个独立的通道,例如r 190 g 181 b 176
3. 我将它除以16得到4位表示的整数等价
4. 我将数字转换为二进制数.
5. 如果生成的二进制小于4位,则在二进制文件前面添加0.
6. 连接到12位二进制以表示12位颜色输出.例如101110111011

这是我的代码:

import java.awt.Component;
import java.awt.image.BufferedImage;
import java.io.File;
import java.io.IOException;
import javax.imageio.ImageIO;

public class LoadImageApp extends Component {
    private static int[] rgbArray;
    private static double bitPerColor = 4.0;

    public static void main(String[] args) {

       BufferedImage img = null;
       String fileName = "Image.jpg";

       try {
           //Read in new image file
           img = ImageIO.read(new File("src/"+fileName));
       } 
       catch (IOException e){
       }

       if (img == null) {
             System.out.println("No image loaded");
        } 
       else {

                //Get RGB Value
                int val = img.getRGB(500, 500);
                System.out.println("rgbValue from getRGB is: " + val);


                //Convert to three separate channels
                int a = (0xff000000 & val) >>> 24;
                int r = (0x00ff0000 & val) >> 16;
                int g = (0x0000ff00 & val) >> 8;
                int b = (0x000000ff & val);

                System.out.println("rgbValue in RGB is: ");
                System.out.println("a " + a + " r " + r + " g " + g + " b " + b);

                double power = Math.pow(2.0, bitPerColor);
                //Convert each channel to binary
                String r4bit = Integer.toBinaryString((int)(r/(power)));
                String g4bit = Integer.toBinaryString((int)(g/(power)));
                String b4bit = Integer.toBinaryString((int)(b/(power)));



                //Convert concatonate 0's in front to get desired bit count
                int rDifference = (int)bitPerColor - r4bit.length();
                int gDifference = (int)bitPerColor - g4bit.length();
                int bDifference = (int)bitPerColor - b4bit.length();


                for (int i = rDifference; i > 0; i--){
                    r4bit="0"+r4bit;}

                for (int i = gDifference; i > 0; i--){
                    g4bit = "0"+g4bit;}

                for (int i = bDifference; i > 0; i--){
                    b4bit = "0"+b4bit;}

                //Concatonate three channel together to form one binary
                String rgbValue = r4bit + g4bit + b4bit;

                System.out.println("rgbValue in binary is: " + rgbValue);


           }     
       }
}

这一切都可以正常工作.然而,只是读取一个像素,它只是真的,非常难看,而且很慢,只需2-3秒.我希望一次使用代码来读取图像的一部分,但我可以用AGES对它进行成像.

所以任何帮助都将非常感激.

提前致谢.

Answer 1

您的原始代码(十六进制)是0x00rrggbb.你想将其转换为0x00000rgb.这样做:

int rgb24 = ....;
int rgb12 = (rgb24 & 0x00f00000) >> 12 + 
            (rgb24 & 0x0000f000) >> 8  + 
            (rgb24 & 0x000000f0) >> 4;

如果你想"舍入"到最近的颜色而不是截断你可以这样做:

int r = (rgb24 & 0x00ff0000);
int g = (rgb24 & 0x0000ff00);
int b = (rgb24 & 0x000000ff);
r += (r >= 0x00f00000) ? 0x00080000 : 0;
g += (g >= 0x0000f000) ? 0x00000800 : 0;
b += (b >= 0x000000f0) ? 0x00000008 : 0;
int rgb12 = (r & 0x00f00000) >> 12 + (g & 0x0000f000) >> 8 + (b & 0x000000f0) >> 4;

只有当高阶4位不是1111时才会向上舍入(当你冒险溢出时).