是否可以以编程方式反转图像的颜色？

Question

我想在 iOS 中拍照并反转颜色。

Answer 1

为了扩展 quixoto 的回答，并且因为我有来自我自己的项目的相关源代码，如果您需要使用 CPU 上的像素操作，那么我添加了说明的以下内容应该执行技巧：

@implementation UIImage (NegativeImage)

- (UIImage *)negativeImage
{
    // get width and height as integers, since we'll be using them as
    // array subscripts, etc, and this'll save a whole lot of casting
    CGSize size = self.size;
    int width = size.width;
    int height = size.height;

    // Create a suitable RGB+alpha bitmap context in BGRA colour space
    CGColorSpaceRef colourSpace = CGColorSpaceCreateDeviceRGB();
    unsigned char *memoryPool = (unsigned char *)calloc(width*height*4, 1);
    CGContextRef context = CGBitmapContextCreate(memoryPool, width, height, 8, width * 4, colourSpace, kCGBitmapByteOrder32Big | kCGImageAlphaPremultipliedLast);
    CGColorSpaceRelease(colourSpace);

    // draw the current image to the newly created context
    CGContextDrawImage(context, CGRectMake(0, 0, width, height), [self CGImage]);

    // run through every pixel, a scan line at a time...
    for(int y = 0; y < height; y++)
    {
        // get a pointer to the start of this scan line
        unsigned char *linePointer = &memoryPool[y * width * 4];

        // step through the pixels one by one...
        for(int x = 0; x < width; x++)
        {
            // get RGB values. We're dealing with premultiplied alpha
            // here, so we need to divide by the alpha channel (if it
            // isn't zero, of course) to get uninflected RGB. We
            // multiply by 255 to keep precision while still using
            // integers
            int r, g, b; 
            if(linePointer[3])
            {
                r = linePointer[0] * 255 / linePointer[3];
                g = linePointer[1] * 255 / linePointer[3];
                b = linePointer[2] * 255 / linePointer[3];
            }
            else
                r = g = b = 0;

            // perform the colour inversion
            r = 255 - r;
            g = 255 - g;
            b = 255 - b;

            // multiply by alpha again, divide by 255 to undo the
            // scaling before, store the new values and advance
            // the pointer we're reading pixel data from
            linePointer[0] = r * linePointer[3] / 255;
            linePointer[1] = g * linePointer[3] / 255;
            linePointer[2] = b * linePointer[3] / 255;
            linePointer += 4;
        }
    }

    // get a CG image from the context, wrap that into a
    // UIImage
    CGImageRef cgImage = CGBitmapContextCreateImage(context);
    UIImage *returnImage = [UIImage imageWithCGImage:cgImage];

    // clean up
    CGImageRelease(cgImage);
    CGContextRelease(context);
    free(memoryPool);

    // and return
    return returnImage;
}

@end

这样就为 UIImage 添加了一个类别方法：

1. 创建一个清晰的 CoreGraphics 位图上下文，它可以访问它的内存
2. 向其绘制 UIImage
3. 遍历每个像素，从预乘的 Alpha 转换为未变形的 RGB，分别反转每个通道，再次乘以 Alpha 并存储回来
4. 从上下文中获取图像并将其包装到 UIImage 中
5. 自行清理，并返回 UIImage

Answer 2

使用 CoreImage：

#import <CoreImage/CoreImage.h>

@implementation UIImage (ColorInverse)

+ (UIImage *)inverseColor:(UIImage *)image {
    CIImage *coreImage = [CIImage imageWithCGImage:image.CGImage];
    CIFilter *filter = [CIFilter filterWithName:@"CIColorInvert"];
    [filter setValue:coreImage forKey:kCIInputImageKey];
    CIImage *result = [filter valueForKey:kCIOutputImageKey];
    return [UIImage imageWithCIImage:result];
}

@end

Answer 3

当然，有可能——一种方法是使用“差异”混合模式 (kCGBlendModeDifference)。请参阅this question（以及其他）了解设置图像处理的代码大纲。使用您的图像作为底部（基础）图像，然后在其上绘制纯白色位图。

您还可以手动执行逐像素操作，方法是获取CGImageRef 并将其绘制到位图上下文中，然后在位图上下文中循环遍历像素。

Answer 4

Swift 3 更新： （来自@BadPirate 答案）

extension UIImage {
func inverseImage(cgResult: Bool) -> UIImage? {
    let coreImage = UIKit.CIImage(image: self)
    guard let filter = CIFilter(name: "CIColorInvert") else { return nil }
    filter.setValue(coreImage, forKey: kCIInputImageKey)
    guard let result = filter.value(forKey: kCIOutputImageKey) as? UIKit.CIImage else { return nil }
    if cgResult { // I've found that UIImage's that are based on CIImages don't work with a lot of calls properly
        return UIImage(cgImage: CIContext(options: nil).createCGImage(result, from: result.extent)!)
    }
    return UIImage(ciImage: result)
  }
}

Answer 5

Tommy 的答案就是答案，但我想指出，对于更大的图像来说，这可能是一项非常紧张且耗时的任务。有两个框架可以帮助您处理图像：

1. CoreImage
2. 加速器
   
   值得一提的是来自 Brad Larson 的令人惊叹的 GPUImage 框架，GPUImage 使例程在 OpenGL 2.0 环境下使用自定义片段着色器在 GPU 上运行，具有显着的速度提升。使用 CoreImge，如果有可用的负过滤器，您可以选择 CPU 或 GPU，使用 Accelerator，所有例程都在 CPU 上运行，但使用矢量数学图像处理。

Answer 6

为此创建了一个快速扩展。此外，由于基于 CIImage 的 UIImage 发生故障（大多数库假设设置了 CGImage），我添加了一个选项来返回基于修改后的 CIImage 的 UIImage：

extension UIImage {
    func inverseImage(cgResult: Bool) -> UIImage? {
        let coreImage = UIKit.CIImage(image: self)
        guard let filter = CIFilter(name: "CIColorInvert") else { return nil }
        filter.setValue(coreImage, forKey: kCIInputImageKey)
        guard let result = filter.valueForKey(kCIOutputImageKey) as? UIKit.CIImage else { return nil }
        if cgResult { // I've found that UIImage's that are based on CIImages don't work with a lot of calls properly
            return UIImage(CGImage: CIContext(options: nil).createCGImage(result, fromRect: result.extent))
        }
        return UIImage(CIImage: result)
    }
}

Answer 7

更新到@MLBDG 答案的Swift 5 版本

extension UIImage {
    func inverseImage(cgResult: Bool) -> UIImage? {
        let coreImage = self.ciImage
        guard let filter = CIFilter(name: "CIColorInvert") else { return nil }
        filter.setValue(coreImage, forKey: kCIInputImageKey)
        guard let result = filter.value(forKey: kCIOutputImageKey) as? UIKit.CIImage else { return nil }
        if cgResult { // I've found that UIImage's that are based on CIImages don't work with a lot of calls properly
            return UIImage(cgImage: CIContext(options: nil).createCGImage(result, from: result.extent)!)
        }
        return UIImage(ciImage: result)
    }
}