How to generate random color names in C#

问题:

I need to generate random color names e.g. "Red", "White" etc. How can I do it? I am able to generate random color like this:

Random randonGen = new Random();
Color randomColor = Color.FromArgb(randonGen.Next(255), randonGen.Next(255), 
randonGen.Next(255));

but I need the names and not all colors generated like this have a known name.

Thanks

回答1:

Use Enum.GetValue to retrieve the values of the KnownColor enumeration and get a random value:

Random randomGen = new Random();
KnownColor[] names = (KnownColor[]) Enum.GetValues(typeof(KnownColor));
KnownColor randomColorName = names[randomGen.Next(names.Length)];
Color randomColor = Color.FromKnownColor(randomColorName);

回答2:

Take a random value and get from KnownColor enum.

May be by this way:

System.Array colorsArray = Enum.GetValues(typeof(KnownColor));
KnownColor[] allColors = new KnownColor[colorsArray.Length];

Array.Copy(colorsArray, allColors, colorsArray.Length);
// get a randon position from the allColors and print its name.

回答3:

Ignore the fact that you're after colors - you really just want a list of possible values, and then take a random value from that list.

The only tricky bit then is working out which set of colors you're after. As Pih mentioned, there's KnownColor - or you could find out all the public static properties of type Color within the Color structure, and get their names. It depends on what you're trying to do.

Note that randomness itself can be a little bit awkward - if you're selecting multiple random colors, you probably want to use a single instance of Random`. Unfortunately it's not thread-safe, which makes things potentially even more complicated. See my article on randomness for more information.

回答4:

Sounds like you just need a random color from the KnownColor enumeration.

回答5:

Copied code from Retrieve a list of colors in C#

CODE:

private List<string> GetColors()
{
    //create a generic list of strings
    List<string> colors = new List<string>();
    //get the color names from the Known color enum
    string[] colorNames = Enum.GetNames(typeof(KnownColor));
    //iterate thru each string in the colorNames array
    foreach (string colorName in colorNames)
    {
        //cast the colorName into a KnownColor
        KnownColor knownColor = (KnownColor)Enum.Parse(typeof(KnownColor), colorName);
        //check if the knownColor variable is a System color
        if (knownColor > KnownColor.Transparent)
        {
            //add it to our list
            colors.Add(colorName);
        }
    }
    //return the color list
    return colors;
}

回答6:

Put the colors into an array and then choose a random index:

class RandomColorSelector
{
    static readonly Color[] Colors = 
        typeof(Color).GetProperties(BindingFlags.Public | BindingFlags.Static)
       .Select(propInfo => propInfo.GetValue(null, null))
       .Cast<Color>()
       .ToArray();

    static readonly string[] ColorNames =  
        typeof(Color).GetProperties(BindingFlags.Public | BindingFlags.Static)
        .Select(propInfo => propInfo.Name)
        .ToArray();

    private Random rand = new Random();

    static void Main(string[] args)
    {
        var colorSelector = new RandomColorSelector();
        var color = colorSelector.GetRandomColor();

        // in case you are only after the *name*
        var colorName = colorSelector.GetRandomColorName();            
    }

    public Color GetRandomColor()
    {
        return Colors[rand.Next(0, Colors.Length)];
    }

    public string GetRandomColorName()
    {
        return ColorNames[rand.Next(0, Colors.Length)];
    }
}

Note that the sample above simply looks up all static properties of the Color type. You might want to improve this by checking that the actual return type of the property is a Color.

回答7:

Or you could try out this: For .NET 4.5

public Windows.UI.Color GetRandomColor()
{
            Random randonGen = new Random();
            Windows.UI.Color randomColor = 
                Windows.UI.Color.FromArgb(
                (byte)randonGen.Next(255), 
                (byte)randonGen.Next(255),
                (byte)randonGen.Next(255), 
                (byte)randonGen.Next(255));
            return randomColor;
}

回答8:

I would build a lookup table. Especially since some colors are up to personal interpretation.

Go through each color value in the Color struct ( http://msdn.microsoft.com/en-us/library/system.drawing.color.aspx ) and map it to the RGB values. Then to convert back, lookup the RGB value to see if it has a named color.

回答9:

There is no way to Randomize an Enumeration, as you want to do, the most suitable solution would pass by setting a List with all the values of the colors, then obtain an integer randomizing it and use it as the index of the list.

回答10:

private string getRandColor()
        {
            Random rnd = new Random();
            string hexOutput = String.Format("{0:X}", rnd.Next(0, 0xFFFFFF));
            while (hexOutput.Length < 6)
                hexOutput = "0" + hexOutput;
            return "#" + hexOutput;
        }

回答11:

Here, I'm generating colors based on profile completed.

public string generateColour(Decimal percentProfileComplete )
{
    if(percent < 50)
    { 
        return "#" + (0xff0000 | Convert.ToInt32(Convert.ToDouble(percentProfileComplete ) * 5.1) * 256).ToString("X6");
    }
    return "#" + (0xff00 | (255 - Convert.ToInt32((Convert.ToDouble(percentProfileComplete ) - 50) * 5.1)) * 256 * 256).ToString("X6");
}

回答12:

I would have commented on Pih's answer; however, not enough karma. Anywho, I tried implementing this and ran into the issue of the same color being generated from multiple calls as the code was called repeatedly in quick succession (i.e. the randomGen was the same and since it is based on the clock = same results ensued).

Try this instead:

public class cExample
{
    ...
    Random randomGen = new Random();
    KnownColor[] names = (KnownColor[]) Enum.GetValues(typeof(KnownColor));
    ...
    private Color get_random_color()
    {
         KnownColor randomColorName = names[randomGen.Next(names.Length)];
         return Color.FromKnownColor(randomColorName);
    }
    ...
}

回答13:

To clear up the syntax errors in the original question, it's

Color.FromRgb, and

(Byte)random2.Next(255)

converts the integer to a byte value needed by Color:

    Random random2 = new Random();
    public int nn = 0x128;
    public int ff = 0x512;
    Color randomColor = Color.FromRgb((Byte)random2.Next(nn, ff), (Byte)random2.Next(nn, ff), (Byte)random2.Next(nn, ff));

回答14:

generate a random number and cast it to KnownColor Type

((KnownColor)Random.Next());