使用加速器规范化视差图像

Question

我有一个视差图像，我正在使用下面的示例代码对其进行规范化，但速度非常慢。我需要使用一些加速器，比如定制的CIFilter或任何其他技术，但我不知道怎么做？我目前正在用CIContext（）运行代码，它正在CPU上运行（不确定）。有没有办法在GPU上运行它并在没有定制CIfilter的情况下加速？以下是当前代码：

extension CVPixelBuffer {

  func normalize() {

    let width = CVPixelBufferGetWidth(self)
    let height = CVPixelBufferGetHeight(self)

    CVPixelBufferLockBaseAddress(self, CVPixelBufferLockFlags(rawValue: 0))

    let baseAddr = CVPixelBufferGetBaseAddress(self)!
    let floatBuffer = unsafeBitCast(CVPixelBufferGetBaseAddress(self), to: UnsafeMutablePointer<Float>.self)

    var minPixel: Float = 1.0
    var maxPixel: Float = 0.0

    for y in 0 ..< height {
      for x in 0 ..< width {
        let pixel = floatBuffer[y * width + x]
        minPixel = min(pixel, minPixel)
        maxPixel = max(pixel, maxPixel)
      }
    }


    let range = maxPixel - minPixel

    for y in 0 ..< height {
      for x in 0 ..< width {
        let pixel = floatBuffer[y * width + x]
        floatBuffer[y * width + x] = (pixel - minPixel) / range
      }
    }

    CVPixelBufferUnlockBaseAddress(self, CVPixelBufferLockFlags(rawValue: 0))
  }
}

Answer 1

我使用加速框架vDSP向量函数来标准化视差。在gitHub中查看一个经过修改的PhotoBrowse，以获得一个可用的演示。

下面是两个函数中的相关代码

extension CVPixelBuffer {
    func vectorNormalize( targetVector: UnsafeMutableBufferPointer<Float>) -> [Float] {
        // range = max - min
        // normalized to 0..1 is (pixel - minPixel) / range

        // see Documentation "Using vDSP for Vector-based Arithmetic" in vDSP under system "Accelerate" documentation

        // see also the Accelerate documentation section 'Vector extrema calculation'
        // Maximium static func maximum<U>(U) -> Float
        //      Returns the maximum element of a single-precision vector.

        //static func minimum<U>(U) -> Float
        //      Returns the minimum element of a single-precision vector.


        let maxValue = vDSP.maximum(targetVector)
        let minValue = vDSP.minimum(targetVector)

        let range = maxValue - minValue
        let negMinValue = -minValue

        let subtractVector = vDSP.add(negMinValue, targetVector)
            // adding negative value is subtracting
        let result = vDSP.divide(subtractVector, range)

        return result
    }

    func setUpNormalize() -> CVPixelBuffer {
        // grayscale buffer float32 ie Float
        // return normalized CVPixelBuffer

        CVPixelBufferLockBaseAddress(self,
                                     CVPixelBufferLockFlags(rawValue: 0))
        let width = CVPixelBufferGetWidthOfPlane(self, 0)
        let height = CVPixelBufferGetHeightOfPlane(self, 0)
        let count = width * height

        let bufferBaseAddress = CVPixelBufferGetBaseAddressOfPlane(self, 0)
            // UnsafeMutableRawPointer

        let pixelBufferBase  = unsafeBitCast(bufferBaseAddress, to: UnsafeMutablePointer<Float>.self)

        let depthCopy  =   UnsafeMutablePointer<Float>.allocate(capacity: count)
        depthCopy.initialize(from: pixelBufferBase, count: count)
        let depthCopyBuffer = UnsafeMutableBufferPointer<Float>(start: depthCopy, count: count)

        let normalizedDisparity = vectorNormalize(targetVector: depthCopyBuffer)

        pixelBufferBase.initialize(from: normalizedDisparity, count: count)
            // copy back the normalized map into the CVPixelBuffer

        depthCopy.deallocate()
//        depthCopyBuffer.deallocate()

        CVPixelBufferUnlockBaseAddress(self, CVPixelBufferLockFlags(rawValue: 0))

        return self

    }

}

Answer 2

对于您的用例，vImage可能是最好的选择。参见本答案中的选项3。

在核心图像中也有这样做的方法。我会想象使用CIareaMinMax过滤器来获得极值，然后使用一些巧妙的混合来归一化。如果你愿意，我可以详细说明。

Answer 3

像素值为Float值，因此也可以使用vDSP。

vDSP_minv和vDSP_maxv计算极值，并且：

floatBuffer[y * width + x] = (pixel - minPixel) / range

可以替换为vDSP_vasm（需要乘以范围的倒数）。

查看执行此计算的vDSP_normalize可能也很有用：

            m = sum(A[n], 0 <= n < N) / N;
            d = sqrt(sum(A[n]**2, 0 <= n < N) / N - m**2);

            if (C)
            {
                // Normalize.
                for (n = 0; n < N; ++n)
                    C[n] = (A[n] - m) / d;
            }