[英]Generate Color Gradient in C#
我的問題類似於這里的問題 ,除了我正在使用C#。
我有兩種顏色,我有一個預定義的步驟。 如何檢索兩者之間漸變的Color
列表?
這是我嘗試過的方法,但沒有用:
int argbMax = Color.Chocolate.ToArgb();
int argbMin = Color.Blue.ToArgb();
var colorList = new List<Color>();
for(int i=0; i<size; i++)
{
var colorAverage= argbMin + (int)((argbMax - argbMin) *i/size);
colorList.Add(Color.FromArgb(colorAverage));
}
如果你嘗試上面的代碼,你會發現argb
的逐漸增加並不對應於顏色的視覺逐漸增加。
有什么想法嗎?
您必須提取R,G,B分量並分別對它們中的每一個執行相同的線性插值,然后重新組合。
int rMax = Color.Chocolate.R;
int rMin = Color.Blue.R;
// ... and for B, G
var colorList = new List<Color>();
for(int i=0; i<size; i++)
{
var rAverage = rMin + (int)((rMax - rMin) * i / size);
var gAverage = gMin + (int)((gMax - gMin) * i / size);
var bAverage = bMin + (int)((bMax - bMin) * i / size);
colorList.Add(Color.FromArgb(rAverage, gAverage, bAverage));
}
奧利弗的答案非常接近......但在我的情況下,我的一些步進數字必須是負數。 當將步進器值轉換為Color
結構時,我的值從負值變為更高的值,例如-1變為類似於254的值。我單獨設置我的步長值以解決此問題。
public static IEnumerable<Color> GetGradients(Color start, Color end, int steps)
{
int stepA = ((end.A - start.A) / (steps - 1));
int stepR = ((end.R - start.R) / (steps - 1));
int stepG = ((end.G - start.G) / (steps - 1));
int stepB = ((end.B - start.B) / (steps - 1));
for (int i = 0; i < steps; i++)
{
yield return Color.FromArgb(start.A + (stepA * i),
start.R + (stepR * i),
start.G + (stepG * i),
start.B + (stepB * i));
}
}
也許這個功能可以幫助:
public IEnumerable<Color> GetGradients(Color start, Color end, int steps)
{
Color stepper = Color.FromArgb((byte)((end.A - start.A) / (steps - 1)),
(byte)((end.R - start.R) / (steps - 1)),
(byte)((end.G - start.G) / (steps - 1)),
(byte)((end.B - start.B) / (steps - 1)));
for (int i = 0; i < steps; i++)
{
yield return Color.FromArgb(start.A + (stepper.A * i),
start.R + (stepper.R * i),
start.G + (stepper.G * i),
start.B + (stepper.B * i));
}
}
public static List<Color> GetGradientColors(Color start, Color end, int steps)
{
return GetGradientColors(start, end, steps, 0, steps - 1);
}
public static List<Color> GetGradientColors(Color start, Color end, int steps, int firstStep, int lastStep)
{
var colorList = new List<Color>();
if (steps <= 0 || firstStep < 0 || lastStep > steps - 1)
return colorList;
double aStep = (end.A - start.A) / steps;
double rStep = (end.R - start.R) / steps;
double gStep = (end.G - start.G) / steps;
double bStep = (end.B - start.B) / steps;
for (int i = firstStep; i < lastStep; i++)
{
var a = start.A + (int)(aStep * i);
var r = start.R + (int)(rStep * i);
var g = start.G + (int)(gStep * i);
var b = start.B + (int)(bStep * i);
colorList.Add(Color.FromArgb(a, r, g, b));
}
return colorList;
}
使用double而不是int:
double stepA = ((end.A - start.A) / (double)(steps - 1));
double stepR = ((end.R - start.R) / (double)(steps - 1));
double stepG = ((end.G - start.G) / (double)(steps - 1));
double stepB = ((end.B - start.B) / (double)(steps - 1));
和:
yield return Color.FromArgb((int)start.A + (int)(stepA * step),
(int)start.R + (int)(stepR * step),
(int)start.G + (int)(stepG * step),
(int)start.B + (int)(stepB * step));
將這個答案與使用浮點步驟的其他幾個答案的想法結合起來,這里是一個用浮點步進的完整方法片段。 (通過整數步進,我得到了從藍色到紅色的16色漸變中的不對稱漸變色。)
此版本的重要區別:您在返回的漸變序列中傳遞所需的顏色總數,而不是在方法實現中要采取的步驟數。
public static IEnumerable<Color> GetColorGradient(Color from, Color to, int totalNumberOfColors)
{
if (totalNumberOfColors < 2)
{
throw new ArgumentException("Gradient cannot have less than two colors.", nameof(totalNumberOfColors));
}
double diffA = to.A - from.A;
double diffR = to.R - from.R;
double diffG = to.G - from.G;
double diffB = to.B - from.B;
var steps = totalNumberOfColors - 1;
var stepA = diffA / steps;
var stepR = diffR / steps;
var stepG = diffG / steps;
var stepB = diffB / steps;
yield return from;
for (var i = 1; i < steps; ++i)
{
yield return Color.FromArgb(
c(from.A, stepA),
c(from.R, stepR),
c(from.G, stepG),
c(from.B, stepB));
int c(int fromC, double stepC)
{
return (int)Math.Round(fromC + stepC * i);
}
}
yield return to;
}
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