如何生成这个特定的数字序列?

发布于 2024-09-24 22:16:32 字数 590 浏览 5 评论 0原文

显然.net 中没有可用的预定义列表。

我想使用一些标准颜色,例如红、绿、蓝、黄……即由 00 和 FF 分量组成的典型颜色,后面是带有附加 7F 分量的颜色,……

有吗一种检索这些“标准”颜色的方法,还是我必须自己编写 IEnumerable

编辑:这是 RGB 值的可能输出。

请注意集合的顺序很重要,因为 00/FF 枚举必须在添加 80 之前完成,并且 00/80/FF 枚举必须在添加 40/B0 之前完成,所以在。 集合中的顺序并不重要(即 00 FF 00 可能出现在 00 00 FF 之前)。

00 00 00 // start out with 00 and FF components
00 00 FF
00 FF 00
FF 00 00
FF FF 00
FF 00 FF
00 FF FF
FF FF FF
00 00 80 // ok, now add 80
00 80 00
...
80 80 80
FF 00 80
FF 80 00
FF 80 80
80 FF 00
80 FF 80
FF FF 80
...
// now add 40 and B0

Apparently there is no predefined list available in .net.

I'd like to use a number of standard colors, e.g. something like red, green, blue, yellow, ... i.e. the typical colors consisting of 00 and FF components, followed by those with additional 7F components, ...

Is there a way to retrieve these "standard" colors or do I have to write the IEnumerable<Color> myself?

Edit: This is a possible output for the RGB values.

Please note that order of sets matters in that the 00/FF enumeration has to be complete before 80 is added, and the 00/80/FF enumeration has to be complete before adding 40/B0, and so on. The order within a set does not matter (i.e. 00 FF 00 may come before 00 00 FF).

00 00 00 // start out with 00 and FF components
00 00 FF
00 FF 00
FF 00 00
FF FF 00
FF 00 FF
00 FF FF
FF FF FF
00 00 80 // ok, now add 80
00 80 00
...
80 80 80
FF 00 80
FF 80 00
FF 80 80
80 FF 00
80 FF 80
FF FF 80
...
// now add 40 and B0

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も让我眼熟你 2024-10-01 22:16:32

颜色类上有许多预定义的 ARGB 颜色,颜色 结构。这些包含诸如 YellowWhiteGreen 等...

如果您想要用户定义的系统颜色,您可以使用 SystemColors 类,其中包含 < code>ActiveBorder、WindowText 等...


更新:

框架中没有任何内容可以按 ARGB 值对颜色进行排序,因为它没有多大意义

您可以使用Linq 按组件对颜色列表进行排序(OrderBy 扩展方法)。

The Colors class has a number of pre-defined ARGB colors on it, as does the Color struct. These hold things like Yellow, White, Green etc...

If you want the user defined system colors, you can use the SystemColors class which has things like ActiveBorder, WindowText etc...


Update:

There is nothing in the framework that sorts colors by their ARGB values as such, since it doesn't really make much sense

You can use Linq to sort a list of colors by their components (OrderBy extension method).

抱着落日 2024-10-01 22:16:32

也许, SystemColors 类适合您。

Probably, the SystemColors class is the right thing for you.

花心好男孩 2024-10-01 22:16:32

AFAIK“标准”颜色不等于基于您描述的模式的 RGB 值颜色。例如,紫色具有 #B803FF RGB 值(其中类似的“紫红色”具有 #FF00FF )。

AFAIK "Standard" colors are not equal to colors with RGB value based on patter that you described. For example purple color has #B803FF RGB value ( where similar 'fuchsia' has #FF00FF ).

嘦怹 2024-10-01 22:16:32

这是生成序列的相当快的方法:

public static IEnumerable<Color> StandardColors ()
{
    int r = 0;
    int g = 0;
    int b = 0;
    int inc = 0x100;

    yield return Color.FromArgb (0, 0, 0);

    while (true) {
        if (((r | g | b) & inc) != 0) {
            int outR = r == 0 ? 0 : r - 1;
            int outG = g == 0 ? 0 : g - 1;
            int outB = b == 0 ? 0 : b - 1;
            yield return Color.FromArgb (outR, outG, outB);
        }

        r += inc;
        if (r > 256) {
            r = 0;
            g += inc;

            if (g > 256) {
                g = 0;
                b += inc;

                if (b > 256) {
                    b = 0;
                    inc >>= 1;

                    if (inc <= 1) {
                        break;
                    }
                }
            }
        }
    }
}

这当然可以改进。例如,应避免使用单独的 outR/G/B 变量,并且应通过 2*inc 从奇数(基于 inc)值开始递增,以避免必须测试该值是否为之前已经生成了。

使用此测试

static void Main (string[] args)
{
    var colors = StandardColorEnumerator.StandardColors ().Take (15)
        .Concat (StandardColorEnumerator.StandardColors ().Skip (1000).Take (10));
    foreach (var color in colors) {
        Console.WriteLine (color.B + "\t" + color.G + "\t" + color.R);
    }

    Console.ReadKey (true);
}

会生成以下输出:

0       0       0
0       0       255
0       255     0
0       255     255
255     0       0
255     0       255
255     255     0
255     255     255
0       0       127
0       127     0
0       127     127
0       127     255
0       255     127
127     0       0
127     0       127

15      47      191
15      47      207
15      47      223
15      47      239
15      47      255
15      63      0
15      63      15
15      63      31
15      63      47
15      63      63

This is a reasonably fast way to generate the sequence:

public static IEnumerable<Color> StandardColors ()
{
    int r = 0;
    int g = 0;
    int b = 0;
    int inc = 0x100;

    yield return Color.FromArgb (0, 0, 0);

    while (true) {
        if (((r | g | b) & inc) != 0) {
            int outR = r == 0 ? 0 : r - 1;
            int outG = g == 0 ? 0 : g - 1;
            int outB = b == 0 ? 0 : b - 1;
            yield return Color.FromArgb (outR, outG, outB);
        }

        r += inc;
        if (r > 256) {
            r = 0;
            g += inc;

            if (g > 256) {
                g = 0;
                b += inc;

                if (b > 256) {
                    b = 0;
                    inc >>= 1;

                    if (inc <= 1) {
                        break;
                    }
                }
            }
        }
    }
}

This can certainly be improved. For instance, having a separate outR/G/B variable should be avoided, and incrementing should be via 2*inc starting from an odd (based on inc) value to avoid having to test if the value was already generated earlier.

Using this test

static void Main (string[] args)
{
    var colors = StandardColorEnumerator.StandardColors ().Take (15)
        .Concat (StandardColorEnumerator.StandardColors ().Skip (1000).Take (10));
    foreach (var color in colors) {
        Console.WriteLine (color.B + "\t" + color.G + "\t" + color.R);
    }

    Console.ReadKey (true);
}

the following output is generated:

0       0       0
0       0       255
0       255     0
0       255     255
255     0       0
255     0       255
255     255     0
255     255     255
0       0       127
0       127     0
0       127     127
0       127     255
0       255     127
127     0       0
127     0       127

15      47      191
15      47      207
15      47      223
15      47      239
15      47      255
15      63      0
15      63      15
15      63      31
15      63      47
15      63      63
~没有更多了~
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