java 如何水平翻转图像
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How to flip an image horizontally
提问by Judy Tran
HiI was wondering how to flip and image horizontally, for a practce task I was given a code that reads an image, inverting it to an image indicating it's brightness from 0-5, I had to flip an image.
嗨，我想知道如何水平翻转和图像，对于练习任务，我得到了一个读取图像的代码，将其反转为指示其亮度从 0 到 5 的图像，我不得不翻转图像。
This is my code of my reading an image and drawing it
这是我阅读图像并绘制它的代码
public int[][] readImage(String url) throws IOException
{
// fetch the image
BufferedImage img = ImageIO.read(new URL(url));
// create the array to match the dimensions of the image
int width = img.getWidth();
int height = img.getHeight();
int[][] imageArray = new int[width][height];
// convert the pixels of the image into brightness values
for (int x = 0; x < width; x++)
{
for (int y = 0; y < height; y++)
{
// get the pixel at (x,y)
int rgb = img.getRGB(x,y);
Color c = new Color(rgb);
int red = c.getRed();
int green = c.getGreen();
int blue = c.getBlue();
// convert to greyscale
float[] hsb = Color.RGBtoHSB(red, green, blue, null);
int brightness = (int)Math.round(hsb[2] * (PIXEL_CHARS.length - 1));
imageArray[x][y] = brightness;
}
}
return imageArray;
}
public void draw() throws IOException
{
int[][] array = readImage("http://sfpl.org/images/graphics/chicklets/google-small.png");
for(int i=0; i<array.length; i++)
{
for(int pic=0; pic<array[i].length; pic++)
{
if(array[pic][i] == 0)
{
System.out.print("X");
}
else if(array[pic][i] == 1)
{
System.out.print("8");
}
else if(array[pic][i] == 2)
{
System.out.print("0");
}
else if(array[pic][i] == 3)
{
System.out.print(":");
}
else if(array[pic][i] == 4)
{
System.out.print(".");
}
else if (array[pic][i] == 5)
{
System.out.print(" ");
}
else
{
System.out.print("error");
break;
}
}
System.out.println();
}
}
and this is the code I tried to create to horizontally flip it,
这是我尝试创建的用于水平翻转它的代码，
void mirrorUpDown()
{
int[][] array = readImage("http://sfpl.org/images/graphics/chicklets/google-small.png");
int i = 0;
for (int x = 0; x < array.length; x++)
{
for (int y = 0; y < array[i].length; y++)
{{
int temp = array[x][y];
array[x][y]= array[-x][y];
array[array[i].length-x][y]=temp;
}
}
}
}
I get an error
我收到一个错误
unreported exception java.io.IException;
must be caught or declared to be thrown
回答by eProw
I'd actually do it by this way...
我真的会这样做...
BufferedImage flip(BufferedImage sprite){
BufferedImage img = new BufferedImage(sprite.getWidth(),sprite.getHeight(),BufferedImage.TYPE_INT_ARGB);
for(int xx = sprite.getWidth()-1;xx>0;xx--){
for(int yy = 0;yy < sprite.getHeight();yy++){
img.setRGB(sprite.getWidth()-xx, yy, sprite.getRGB(xx, yy));
}
}
return img;
}
Just a loop whose x starts at the end of the first image and places its rgba value on the flipped position of the second image. Clean, easy code :)
只是一个循环，其 x 从第一张图像的末尾开始，并将其 rgba 值放在第二张图像的翻转位置上。干净，简单的代码:)
回答by Mukul Goel
The function mirrorUpDown() , add a throws IOException there.
函数 mirrorUpDown() ，在那里添加一个 throws IOException 。
Also the function from which you are calling these methods, does that handle exception, does that code enclosed in a try catch block or the function is also set to throw IOException (one of either should be there)
此外，您调用这些方法的函数，是否处理异常，是否包含在 try catch 块中的代码，或者该函数是否也设置为抛出 IOException （其中之一应该在那里）
回答by Alix Martin
How is your image supposed to know it should get it's data from imageArray ?
您的图像应该如何知道它应该从 imageArray 获取数据？
instead, you should access the raster of your image and modify the data in it.
相反，您应该访问图像的光栅并修改其中的数据。
void flip(BufferedImage image) {
WritableRaster raster = image.getRaster();
int h = raster.getHeight();
int w = raster.getWidth();
int x0 = raster.getMinX();
int y0 = raster.getMinY();
for (int x = x0; x < x0 + w; x++){
for (int y = y0; y < y0 + h / 2; y++){
int[] pix1 = new int[3];
pix1 = raster.getPixel(x, y, pix1);
int[] pix2 = new int[3];
pix2 = raster.getPixel(x, y0 + h - 1 - (y - y0), pix2);
raster.setPixel(x, y, pix2);
raster.setPixel(x, y0 + h - 1 - (y - y0), pix1);
}
}
return;
}
回答by Richard Kenneth Niescior
Sorry about posting this here over a year later but it should aid someone at a stage
很抱歉一年多后在这里发布此信息，但它应该会在某个阶段帮助某人
try{
java.awt.image.BufferedImage bi = javax.imageio.ImageIO.read(getClass().getResource("Your image bro.jpg")) ;
int[] h = bi.getRGB(0, 0, bi.getWidth(), bi.getHeight(), null, 0, bi.getWidth());
int [] h1 = new int[h.length];
System.out.println(""+h.length);
for(int j = 0;500>j;j++){
for(int i = 500;i>0;i--){
h1[j*500+(500-i)] = h[(j*500)+(i-1)];
}
}
bi.setRGB(0, 0, bi.getWidth(), bi.getHeight(), h1, 0, bi.getWidth());
}
catch(Exception e){e.printStackTrace();}
Lets break the code down
让我们分解代码
java.awt.image.BufferedImage bi =javax.imageio.ImageIO.read(getClass().getResource("Your image bro.jpg"));
Tries to read the image and stores the read image into the BufferedImage variable bi
尝试读取图像并将读取的图像存储到 BufferedImage 变量 bi
int[] h = bi.getRGB(0, 0, bi.getWidth(), bi.getHeight(), null, 0, bi.getWidth());
int [] h1 = new int[h.length];
instantiate two arrays, h is the original RGB Array and h1 will be the horizontally flipped RGB array.
实例化两个数组，h 是原始 RGB 数组，h1 将是水平翻转的 RGB 数组。
for(int j = 0;500>j;j++){
for(int i = 500;i>0;i--){
h1[j*500+(500-i)] = h[(j*500)+(i-1)];
}
}
Lets look at something in particular more closely
让我们更仔细地看一些特别的东西
h1[j*500+(500-i)] = h[(j*500)+(i-1)];
Images are scanned from position 0;0 to x.length;y.length but it is scanned in a coninual array. Thus we use a psuedo-array to manipulate the flipping of the image. j*500 references the Y values and (500-i) references the x values.
图像从位置 0;0 到 x.length;y.length 扫描，但它以连续阵列扫描。因此，我们使用伪数组来操纵图像的翻转。j*500 引用 Y 值，(500-i) 引用 x 值。
bi.setRGB(0, 0, bi.getWidth(), bi.getHeight(), h1, 0, bi.getWidth());
Finally, the image gets stored back into the BufferedImage variable.
最后，图像被存储回 BufferedImage 变量。
Note that the 500 constant is referencing your x resolution of the image. For example, 1920 x 1080 sized image uses a max value of 1920. The logic is yours to decide.
请注意，500 常量是指图像的 x 分辨率。例如，1920 x 1080 大小的图像使用最大值 1920。逻辑由您决定。
