Creating random color in Java?

Question

I want to draw random colored dots on a JPanel in a Java application. Is there any way to create random colors?

Answer 1

Use a random library:

import java.util.Random; 

Then create a random generator:

 Random rand = new Random(); 

When the colors are divided into red green and blue, you can create a new random color by creating random primary colors:

 // Java 'Color' class takes 3 floats, from 0 to 1. float r = rand.nextFloat(); float g = rand.nextFloat(); float b = rand.nextFloat(); 

Then, to finally create the color, pass the primary colors to the constructor:

 Color randomColor = new Color(r, g, b); 

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You can also create various random effects using this method, for example, create random colors with a lot of attention to certain colors ... skip less green and blue to create a more pink random color.

 // Will produce a random colour with more red in it (usually "pink-ish") float r = rand.nextFloat(); float g = rand.nextFloat() / 2f; float b = rand.nextFloat() / 2f; 

Or to ensure that only "light" colors are generated, you can create colors that are always> 0.5 of each color element:

 // Will produce only bright / light colours: float r = rand.nextFloat() / 2f + 0.5; float g = rand.nextFloat() / 2f + 0.5; float b = rand.nextFloat() / 2f + 0.5; 

There are various other color functions that you can use with the Color class, for example, to make the color brighter:

 randomColor.brighter(); 

An overview of the Color class can be found here: http://download.oracle.com/javase/6/docs/api/java/awt/Color.html

Answer 2

Single line for random RGB values:

 new Color((int)(Math.random() * 0x1000000)) 

Answer 3

If you like pleasant, pastel colors, it is best to use the HLS system.

 final float hue = random.nextFloat(); // Saturation between 0.1 and 0.3 final float saturation = (random.nextInt(2000) + 1000) / 10000f; final float luminance = 0.9f; final Color color = Color.getHSBColor(hue, saturation, luminance); 

Answer 4

Copy this for vibrant pastel rainbow colors.

 int R = (int)(Math.random()*256); int G = (int)(Math.random()*256); int B= (int)(Math.random()*256); Color color = new Color(R, G, B); //random color, but can be bright or dull //to get rainbow, pastel colors Random random = new Random(); final float hue = random.nextFloat(); final float saturation = 0.9f;//1.0 for brilliant, 0.0 for dull final float luminance = 1.0f; //1.0 for brighter, 0.0 for black color = Color.getHSBColor(hue, saturation, luminance); 

Answer 5

If you don't want this to look terrible, I would suggest defining a list of colors in an array, and then use a random number generator to select it.

If you want a truly random color, you can simply generate 3 random numbers from 0 to 255, and then use the Color (int, int, int) constructor to create a new Color instance.

 Random randomGenerator = new Random(); int red = randomGenerator.nextInt(256); int green = randomGenerator.nextInt(256); int blue = randomGenerator.nextInt(256); Color randomColour = new Color(red,green,blue); 

Answer 6

I know this is a little late for this answer, but I have not seen anyone else do it.

As Greg said, you want to use the Random class

 Random rand = new Random(); 

but the difference I'm going to say is simple:

 Color color = new Color(rand.nextInt(0xFFFFFF)); 

And it's that simple! no need to create many different floats.

Answer 7

 import android.graphics.Color; import java.util.Random; public class ColorDiagram { // Member variables (properties about the object) public String[] mColors = { "#39add1", // light blue "#3079ab", // dark blue "#c25975", // mauve "#e15258", // red "#f9845b", // orange "#838cc7", // lavender "#7d669e", // purple "#53bbb4", // aqua "#51b46d", // green "#e0ab18", // mustard "#637a91", // dark gray "#f092b0", // pink "#b7c0c7" // light gray }; // Method (abilities: things the object can do) public int getColor() { String color = ""; // Randomly select a fact Random randomGenerator = new Random(); // Construct a new Random number generator int randomNumber = randomGenerator.nextInt(mColors.length); color = mColors[randomNumber]; int colorAsInt = Color.parseColor(color); return colorAsInt; } } 

Answer 8

You can create a color instance with three floats (r, g, b), each between 0.0 and 1.0: http://download.oracle.com/javase/6/docs/api/java/awt/Color.html#Color ( float,% 20float,% 20float ).

Using the Java Random class, you can easily create a new random color as such:

 Random r = new Random(); Color randomColor = new Color(r.nextFloat(), r.nextFloat(), r.nextFloat()); 

I can not guarantee that everyone will be beautiful, but they will be random =)

Answer 9

Sure. Just generate color using random RGB values. How:

 public Color randomColor() { Random random=new Random(); // Probably really put this somewhere where it gets executed only once int red=random.nextInt(256); int green=random.nextInt(256); int blue=random.nextInt(256); return new Color(red, green, blue); } 

You might want to change the generation of random numbers if you don't like the colors he encounters. I suppose they will be quite dark.

Answer 10

You seem to need light, random colors. Not sure what you mean with light. But if you need random “rainbow colors,” try this

 Random r = new Random(); Color c = Color.getHSBColor(r.nextFloat(),//random hue, color 1.0,//full saturation, 1.0 for 'colorful' colors, 0.0 for grey 1.0 //1.0 for bright, 0.0 for black ); 

Find the HSB color model for more information.

Answer 11

 package com.adil.util; /** * The Class RandomColor. * * @author Adil OUIDAD * @URL : http://kizana.fr */ public class RandomColor { /** * Gets the random color. * * @return the random color */ public static String getRandomColor() { String[] letters = {"0","1","2","3","4","5","6","7","8","9","A","B","C","D","E","F"}; String color = "#"; for (int i = 0; i < 6; i++ ) { color += letters[(int) Math.round(Math.random() * 15)]; } return color; } } 

Answer 12

Here is a way to get a random color:

 private static Random sRandom; public static synchronized int randomColor() { if (sRandom == null) { sRandom = new Random(); } return 0xff000000 + 256 * 256 * sRandom.nextInt(256) + 256 * sRandom.nextInt(256) + sRandom.nextInt(256); } 

Benefits:

• Get an integer view that can be used with java.awt.Color or android.graphics.Color
• Keep a static link to Random .

Answer 13

It would be helpful.

 Random randomGenerator = new Random(); int red = randomGenerator.nextInt(255); int green = randomGenerator.nextInt(255); int blue = randomGenerator.nextInt(255); Color randomColour = new Color(red,green,blue);