我要从两个txt文件创建一个bmp图像。第一个是mxn数组:
* * * * * * * * *
m n
c11 c21 .. cm1
...
c1n c2n .. cmn
* * * * * * * *
* * * * * * * *
6 5
.7 .7 .6 1.0 1.2 .1
.9 .3 .7 1.1 .7 .2
1 1.1 1.2 1.3 1.7 .6
.5 .6 .5 .4 .9 .1101
2 .1 .1 .1 2.1 1.1
* * * * * * * *
第二个txt文件是一个色标,就像这样
* * * * * * * * *
min1 max1 r1 g1 b1
min2 max2 r2 g2 b2
...
minx maxx rx gx bx
* * * * * * * *
* * * * * * * *
0 .5 255 128 64
.5 .75 128 255 32
.75 1.25 64 64 225
01.50 5 128 128 0
* * * * * * * *
所以我要从这两个文件中读取。我试图使用StringTokenizer类从第一个txt文件创建一个数组,但我完全迷失了。从这两个文件我要创建一个bmp图像。 有人能以某种方式帮助我吗?
答案 0 :(得分:4)
如果颜色范围是连续的(您的示例缺少1.25-1.5)并且保证覆盖矩阵文件中的所有可能值,我将首先构建一个TreeMap<Double, java.awt.Color>
,使用颜色的最大值文件作为地图密钥。然后,您可以使用TreeMap#ceilingEntry(K)
方法获取任何矩阵值的颜色。例如。如果使用您的测试数据正确填充,ceilingEntry(0.2).getValue()将返回Color(255,128,64)。
您可以更轻松地直接使用java.awt.BufferedImage
进行绘制,然后使用javax.imageio.ImageIO
将缓冲的图像写为BMP文件,而不是将矩阵文件读入数组。
答案 1 :(得分:1)
叹息......当我写下程序时,@ jarnbjo解释了同样的想法。但是这里有一些代码:
import java.awt.Graphics;
import java.awt.Image;
import java.awt.image.BufferedImage;
import java.util.NavigableMap;
import java.util.TreeMap;
import javax.swing.JFrame;
import javax.swing.JPanel;
public class ImageParser {
public static void main(String[] args) {
String dataContent =
"6 5\n" +
".7 .7 .6 1.0 1.2 .1\n" +
".9 .3 .7 1.1 .7 .2\n" +
"1 1.1 1.2 1.3 1.7 .6\n" +
".5 .6 .5 .4 .9 .1101\n" +
"2 .1 .1 .1 2.1 1.1";
String colorContent =
"0 .5 255 128 64\n" +
".5 .75 128 255 32\n" +
".75 1.25 64 64 225\n" +
"01.50 5 128 128 0";
int width = 0;
int height = 0;
BufferedImage image = null;
NavigableMap<Double, Integer> colorMap = new TreeMap<Double, Integer>();
for (String colorLine : colorContent.split( "\n" )) {
String[] colorValues = colorLine.split( " " );
colorMap.put( Double.parseDouble( colorValues[1] ),
Integer.parseInt( colorValues[2] ) << 16 |
Integer.parseInt( colorValues[3] ) << 8 |
Integer.parseInt( colorValues[4] ) );
}
boolean headerParsed = false;
int y = 0;
for( String dataLine : dataContent.split( "\n" ) ) {
String[] dataArray = dataLine.split( " " );
if( !headerParsed ) {
width = Integer.parseInt( dataArray[ 0 ] );
height = Integer.parseInt( dataArray[ 1 ] );
image = new BufferedImage( width, height, BufferedImage.TYPE_INT_RGB );
headerParsed = true;
}
else {
int x = 0;
for( String data : dataArray ) {
Integer rgbValue = colorMap.higherEntry( Double.parseDouble( data ) ).getValue();
image.setRGB( x, y, rgbValue );
x++;
}
y++;
}
}
JFrame frame = new JFrame();
frame.getContentPane().add( new Viewer( image, width, height, 20 ) );
frame.pack();
frame.setVisible( true );
}
static class Viewer extends JPanel {
Image m_image;
int m_width;
int m_height;
int m_zoom;
public Viewer( Image image, int width, int height, int zoom ) {
m_image = image;
m_width = width;
m_height = height;
m_zoom = zoom;
}
@Override
public void paint(Graphics g) {
g.drawImage( m_image, 0, 0, m_width * m_zoom, m_height * m_zoom, this );
}
};
}
答案 2 :(得分:0)
在同一时间(我需要更多的时间大声笑)我只写了代码,以保存到数组中的第一个矩阵文件:(这里是代码,如果你想看看它的工作原理。但实际上我'在你的身上。我会让你知道我会达到什么。谢谢你们......
import java.io.*;<br>
import java.util.StringTokenizer;
public class TokenizerUser4 {
public static double[][] matrix;
public static void main(String[] args) throws IOException {
FileReader reader = new FileReader
("D:\\sonenos\\java\\FlussiIO\\new\\matrix.txt");
BufferedReader br = new BufferedReader(reader);
String line;
int rowIndex = 0;
int counter = 0;
int[] dim = new int[3];
while ((line = br.readLine()) != null){
counter++;
if (counter == 1) {
StringTokenizer dimensioni = new StringTokenizer(line);
//int[] dim = new int[3];
int i = 0;
while(dimensioni.hasMoreTokens()){
dim[i] = Integer.parseInt(dimensioni.nextToken());
//System.out.println(dim[i] + " i=" + i);
i++;
}
matrix = new double[dim[0]][dim[1]];
}
if (counter != 1){
StringTokenizer theLine = new StringTokenizer(line);
int colIndex = 0;
while (theLine.hasMoreTokens()){
String st = theLine.nextToken();
matrix[rowIndex][colIndex] = Double.parseDouble(st);
colIndex = colIndex + 1;
}
rowIndex = rowIndex + 1;
}
}
for (int x = 0; x<dim[0];x++){
for (int y = 0; y<dim[1]; y++){
System.out.print(matrix[x][y] + " ");
}
System.out.println("\n");
}
br.close();
}
}