iOS在两种颜色之间的点处查找颜色

Question

我有一个问题:我需要能够采用两种颜色并从中制作出"虚拟渐变".然后,我需要能够在此行的任何位置找到颜色.我目前的做法是:

if (fahrenheit < kBottomThreshold)
{
    return [UIColor colorWithRed:kBottomR/255.0f green:kBottomG/255.0f blue:kBottomB/255.0f alpha:1];
}
if (fahrenheit > kTopThreshold)
{
    return [UIColor colorWithRed:kTopR/255.0f green:kTopG/255.0f blue:kTopB/255.0f alpha:1];
}

double rDiff = kTopR - kBottomR;
double gDiff = kTopG - kBottomG;
double bDiff = kTopB - kBottomB;

double tempDiff = kTopThreshold - kBottomThreshold;

double rValue;
double gValue;
double bValue;

rValue = kBottomR + ((rDiff/tempDiff) * fahrenheit);
gValue = kBottomG + ((gDiff/tempDiff) * fahrenheit);
bValue = kBottomB + ((bDiff/tempDiff) * fahrenheit);

return [UIColor colorWithRed:rValue/255.0f green:gValue/255.0f blue:bValue/255.0f alpha:1];

变量:

• fahrenheit 是一个传递给我的函数的变量,它是我要为其找到颜色的虚拟行上的数字.
• kTopR,, kTopB和kTopG是渐变一端的RGB值.同样为他们的kBottom同行.
• kBottomThreshold并且kTopThreshold是我渐变的终点.

这是我的问题: 当fahrenheit超过渐变的任何一端时,渐变似乎"跳"到不同的值.

我已经包括示例项目,主持我的S3服务器上,在这里.

你真的需要下载项目并在模拟器/设备上试一试,看看我的意思(除非你疯狂聪明,只能通过查看代码来判断)

Answer 1

Swift - 3.0 && 4.0

extension UIColor {
    func toColor(_ color: UIColor, percentage: CGFloat) -> UIColor {
        let percentage = max(min(percentage, 100), 0) / 100
        switch percentage {
        case 0: return self
        case 1: return color
        default:
            var (r1, g1, b1, a1): (CGFloat, CGFloat, CGFloat, CGFloat) = (0, 0, 0, 0)
            var (r2, g2, b2, a2): (CGFloat, CGFloat, CGFloat, CGFloat) = (0, 0, 0, 0)
            guard self.getRed(&r1, green: &g1, blue: &b1, alpha: &a1) else { return self }
            guard color.getRed(&r2, green: &g2, blue: &b2, alpha: &a2) else { return self }

            return UIColor(red: CGFloat(r1 + (r2 - r1) * percentage),
                           green: CGFloat(g1 + (g2 - g1) * percentage),
                           blue: CGFloat(b1 + (b2 - b1) * percentage),
                           alpha: CGFloat(a1 + (a2 - a1) * percentage))
        }
    }
}

用法:-

let colorRed = UIColor.red
let colorBlue = UIColor.blue

let colorOutput = colorRed.toColor(colorBlue, percentage: 50)

结果

Answer 2

感谢@ramchandra-n，我实现了扩展以按百分比从颜色数组中获取中间颜色

extension Array where Element: UIColor {
    func intermediate(percentage: CGFloat) -> UIColor {
        let percentage = Swift.max(Swift.min(percentage, 100), 0) / 100
        switch percentage {
        case 0: return first ?? .clear
        case 1: return last ?? .clear
        default:
            let approxIndex = percentage / (1 / CGFloat(count - 1))
            let firstIndex = Int(approxIndex.rounded(.down))
            let secondIndex = Int(approxIndex.rounded(.up))
            let fallbackIndex = Int(approxIndex.rounded())
            
            let firstColor = self[firstIndex]
            let secondColor = self[secondIndex]
            let fallbackColor = self[fallbackIndex]
            
            var (r1, g1, b1, a1): (CGFloat, CGFloat, CGFloat, CGFloat) = (0, 0, 0, 0)
            var (r2, g2, b2, a2): (CGFloat, CGFloat, CGFloat, CGFloat) = (0, 0, 0, 0)
            guard firstColor.getRed(&r1, green: &g1, blue: &b1, alpha: &a1) else { return fallbackColor }
            guard secondColor.getRed(&r2, green: &g2, blue: &b2, alpha: &a2) else { return fallbackColor }
            
            let intermediatePercentage = approxIndex - CGFloat(firstIndex)
            return UIColor(
                red: CGFloat(r1 + (r2 - r1) * intermediatePercentage),
                green: CGFloat(g1 + (g2 - g1) * intermediatePercentage),
                blue: CGFloat(b1 + (b2 - b1) * intermediatePercentage),
                alpha: CGFloat(a1 + (a2 - a1) * intermediatePercentage)
            )
        }
    }
}

您可以使用它来获得两种或多种颜色之间的中间颜色：

let color = [.green, .yellow, .red].intermediate(percentage: 70)

Answer 3

问题是,你不减去kBottomThreshold从farenheit.

但是,让我们简化.

首先,我们要将输入温度映射到t[0 ... 1]范围内的参数.然后,我们要映射t到的范围内的输出[ kBottomR... kTopR],并在范围内的输出[ kBottomG... kTopG],并在范围内的输出[ kBottomB... kTopB].

UIColor *colorForDegreesFahrenheit(double fahrenheit) {
    double t = (fahrenheit - kBottomThreshold) / (kTopThreshold - kBottomThreshold);

    // Clamp t to the range [0 ... 1].
    t = MAX(0.0, MIN(t, 1.0));

    double r = kBottomR + t * (kTopR - kBottomR);
    double g = kBottomG + t * (kTopG - kBottomG);
    double b = kBottomB + t * (kTopB - kBottomB);

    return [UIColor colorWithRed:r/255 green:g/255 blue:b/255 alpha:1];
}

Answer 4

如果您的渐变比 2 色渐变更复杂，您可以考虑将 CGGradientRef 绘制到临时 CGImageRef 中，并直接从图像缓冲区读取 RGBA 值。

这是我必须用 5 个渐变点和颜色做的事情：

    CGFloat tmpImagewidth = 1000.0f; // Make this bigger or smaller if you need more or less resolution (number of different colors).
    CGColorSpaceRef colorSpace = CGColorSpaceCreateDeviceRGB();

    // create a gradient
    CGFloat locations[] = { 0.0,
        0.35,
        0.55,
        0.8,
        1.0 };
    NSArray *colors = @[(__bridge id) [UIColor redColor].CGColor,
                        (__bridge id) [UIColor greenColor].CGColor,
                        (__bridge id) [UIColor blueColor].CGColor,
                        (__bridge id) [UIColor yellowColor].CGColor,
                        (__bridge id) [UIColor redColor].CGColor,
                        ];
    CGGradientRef gradient = CGGradientCreateWithColors(colorSpace, (__bridge CFArrayRef) colors, locations);
    CGPoint startPoint = CGPointMake(0, 0);
    CGPoint endPoint = CGPointMake(tmpImagewidth, 0);

    // create a bitmap context to draw the gradient to, 1 pixel high.
    CGContextRef context = CGBitmapContextCreate(NULL, tmpImagewidth, 1, 8, 0, colorSpace, kCGImageAlphaPremultipliedLast);

    // draw the gradient into it
    CGContextAddRect(context, CGRectMake(0, 0, tmpImagewidth, 1));
    CGContextClip(context);
    CGContextDrawLinearGradient(context, gradient, startPoint, endPoint, 0);

    // Get our RGB bytes into a buffer with a couple of intermediate steps...
    //      CGImageRef -> CFDataRef -> byte array
    CGImageRef cgImage = CGBitmapContextCreateImage(context);
    CGDataProviderRef provider = CGImageGetDataProvider(cgImage);
    CFDataRef pixelData = CGDataProviderCopyData(provider);

    // cleanup:
    CGGradientRelease(gradient);
    CGColorSpaceRelease(colorSpace);
    CGImageRelease(cgImage);
    CGContextRelease(context);

    const UInt8* data = CFDataGetBytePtr(pixelData);

    // we got all the data we need.
    // bytes in the data buffer are a succession of R G B A bytes

    // For instance, the color of the point 27% in our gradient is:
    CGFloat x = tmpImagewidth * .27;
    int pixelIndex = (int)x * 4; // 4 bytes per color
    UIColor *color = [UIColor colorWithRed:data[pixelIndex + 0]/255.0f
                                     green:data[pixelIndex + 1]/255.0f
                                      blue:data[pixelIndex + 2]/255.0f
                                     alpha:data[pixelIndex + 3]/255.0f];

    // done fetching color data, finally release the buffer
    CGDataProviderRelease(provider);

我并不是说这比上面答案中的“数学方式”更好，当然，生成临时图像会产生内存和 CPU 税。然而，这样做的优点是，无论您需要多少个梯度停止点，代码复杂性都保持不变......