9

所以我试图尽可能快地模糊图像（即时感觉），因为当我按下模糊按钮时需要更新活动。

我遇到的问题是，我找不到足够快的模糊效果...... 注意：模糊，最好是高斯模糊，根本不需要是最好的质量......

我尝试了以下内容，但需要几秒钟，无论如何，这段代码是否可以在牺牲质量的情况下更快地运行？或者还有其他选择吗？我会研究 GPU 的东西，但这种模糊实际上只是与 UI 相关的一种效果，只有当我按下打开一个大小为小盒子的透明活动时才会发生......

有任何想法吗？

static Bitmap fastblur(Bitmap sentBitmap, int radius, int fromX, int fromY,
    int width, int height) {

// Stack Blur v1.0 from
// http://www.quasimondo.com/StackBlurForCanvas/StackBlurDemo.html
//
// Java Author: Mario Klingemann <mario at quasimondo.com>
// http://incubator.quasimondo.com
// created Feburary 29, 2004
// Android port : Yahel Bouaziz <yahel at kayenko.com>
// http://www.kayenko.com
// ported april 5th, 2012

// This is a compromise between Gaussian Blur and Box blur
// It creates much better looking blurs than Box Blur, but is
// 7x faster than my Gaussian Blur implementation.
//
// I called it Stack Blur because this describes best how this
// filter works internally: it creates a kind of moving stack
// of colors whilst scanning through the image. Thereby it
// just has to add one new block of color to the right side
// of the stack and remove the leftmost color. The remaining
// colors on the topmost layer of the stack are either added on
// or reduced by one, depending on if they are on the right or
// on the left side of the stack.
//
// If you are using this algorithm in your code please add
// the following line:
//
// Stack Blur Algorithm by Mario Klingemann <mario@quasimondo.com>

Bitmap bitmap = sentBitmap.copy(sentBitmap.getConfig(), true);

if (radius < 1) {
    return (null);
}

int w = width;
int h = height;

int[] pix = new int[w * h];

bitmap.getPixels(pix, 0, w, fromX, fromY, w, h);

int wm = w - 1;
int hm = h - 1;
int wh = w * h;
int div = radius + radius + 1;

int r[] = new int[wh];
int g[] = new int[wh];
int b[] = new int[wh];
int rsum, gsum, bsum, x, y, i, p, yp, yi, yw;
int vmin[] = new int[Math.max(w, h)];

int divsum = (div + 1) >> 1;
divsum *= divsum;
int dv[] = new int[256 * divsum];
for (i = 0; i < 256 * divsum; i++) {
    dv[i] = (i / divsum);
}

yw = yi = 0;

int[][] stack = new int[div][3];
int stackpointer;
int stackstart;
int[] sir;
int rbs;
int r1 = radius + 1;
int routsum, goutsum, boutsum;
int rinsum, ginsum, binsum;

int originRadius = radius;
for (y = 0; y < h; y++) {
    rinsum = ginsum = binsum = routsum = goutsum = boutsum = rsum = gsum = bsum = 0;
    for (i = -radius; i <= radius; i++) {
        p = pix[yi + Math.min(wm, Math.max(i, 0))];
        sir = stack[i + radius];
        sir[0] = (p & 0xff0000) >> 16;
        sir[1] = (p & 0x00ff00) >> 8;
        sir[2] = (p & 0x0000ff);
        rbs = r1 - Math.abs(i);
        rsum += sir[0] * rbs;
        gsum += sir[1] * rbs;
        bsum += sir[2] * rbs;
        if (i > 0) {
            rinsum += sir[0];
            ginsum += sir[1];
            binsum += sir[2];
        } else {
            routsum += sir[0];
            goutsum += sir[1];
            boutsum += sir[2];
        }
    }
    stackpointer = radius;

    for (x = 0; x < w; x++) {

        r[yi] = dv[rsum];
        g[yi] = dv[gsum];
        b[yi] = dv[bsum];

        rsum -= routsum;
        gsum -= goutsum;
        bsum -= boutsum;

        stackstart = stackpointer - radius + div;
        sir = stack[stackstart % div];

        routsum -= sir[0];
        goutsum -= sir[1];
        boutsum -= sir[2];

        if (y == 0) {
            vmin[x] = Math.min(x + radius + 1, wm);
        }
        p = pix[yw + vmin[x]];

        sir[0] = (p & 0xff0000) >> 16;
        sir[1] = (p & 0x00ff00) >> 8;
        sir[2] = (p & 0x0000ff);

        rinsum += sir[0];
        ginsum += sir[1];
        binsum += sir[2];

        rsum += rinsum;
        gsum += ginsum;
        bsum += binsum;

        stackpointer = (stackpointer + 1) % div;
        sir = stack[(stackpointer) % div];

        routsum += sir[0];
        goutsum += sir[1];
        boutsum += sir[2];

        rinsum -= sir[0];
        ginsum -= sir[1];
        binsum -= sir[2];

        yi++;
    }
    yw += w;
}

radius = originRadius;

for (x = 0; x < w; x++) {
    rinsum = ginsum = binsum = routsum = goutsum = boutsum = rsum = gsum = bsum = 0;
    yp = -radius * w;
    for (i = -radius; i <= radius; i++) {
        yi = Math.max(0, yp) + x;

        sir = stack[i + radius];

        sir[0] = r[yi];
        sir[1] = g[yi];
        sir[2] = b[yi];

        rbs = r1 - Math.abs(i);

        rsum += r[yi] * rbs;
        gsum += g[yi] * rbs;
        bsum += b[yi] * rbs;

        if (i > 0) {
            rinsum += sir[0];
            ginsum += sir[1];
            binsum += sir[2];
        } else {
            routsum += sir[0];
            goutsum += sir[1];
            boutsum += sir[2];
        }

        if (i < hm) {
            yp += w;
        }
    }
    yi = x;
    stackpointer = radius;
    for (y = 0; y < h; y++) {
        pix[yi] = 0xff000000 | (dv[rsum] << 16) | (dv[gsum] << 8)
                | dv[bsum];

        rsum -= routsum;
        gsum -= goutsum;
        bsum -= boutsum;

        stackstart = stackpointer - radius + div;
        sir = stack[stackstart % div];

        routsum -= sir[0];
        goutsum -= sir[1];
        boutsum -= sir[2];

        if (x == 0) {
            vmin[y] = Math.min(y + r1, hm) * w;
        }
        p = x + vmin[y];

        sir[0] = r[p];
        sir[1] = g[p];
        sir[2] = b[p];

        rinsum += sir[0];
        ginsum += sir[1];
        binsum += sir[2];

        rsum += rinsum;
        gsum += ginsum;
        bsum += binsum;

        stackpointer = (stackpointer + 1) % div;
        sir = stack[stackpointer];

        routsum += sir[0];
        goutsum += sir[1];
        boutsum += sir[2];

        rinsum -= sir[0];
        ginsum -= sir[1];
        binsum -= sir[2];

        yi += w;
    }
}

bitmap.setPixels(pix, 0, w, fromX, fromY, w, h);

return (bitmap);

}

4

2 回答 2

9

尝试将图像缩小 2、4、8... 倍，然后再次放大。那很快。否则在渲染脚本中实现它。

如果你想要的不仅仅是缩放，你可以在 renderscript 中查看这个代码片段。它与另一个答案中给出的模糊相同。相同的算法可以在 Java 中实现，并且是对另一个答案的优化。这段代码模糊了一行。要模糊位图，您应该对所有行调用它，然后对所有列调用它（您需要重新实现它来处理列）。要快速模糊，只需执行一次。如果你想要一个更好看的模糊做几次。我通常只做两次。

做一行的原因是我试图并行化算法，这给出了一些改进并且在renderscript中非常简单。我为所有行并行调用了以下代码，然后对所有列都调用了相同的代码。

int W = 8;
uchar4 *in;
uchar4 *out;    

int N;
float invN;    

uint32_t nx;
uint32_t ny;    

void init_calc() {
  N = 2*W+1;
  invN = 1.0f/N;    

  nx = rsAllocationGetDimX(rsGetAllocation(in));
  ny = rsAllocationGetDimY(rsGetAllocation(in));
}    

void root(const ushort *v_in) {
  float4 sum = 0;    

  uchar4 *head = in + *v_in * nx;
  uchar4 *tail = head;
  uchar4 *p = out + *v_in * nx;    

  uchar4 *hpw = head + W;
  uchar4 *hpn = head + N;
  uchar4 *hpx = head + nx;
  uchar4 *hpxmw = head + nx - W - 1;    

  while (head < hpw) {
      sum += rsUnpackColor8888(*head++);
  }    

  while (head < hpn) {
      sum += rsUnpackColor8888(*head++);
      *p++ = rsPackColorTo8888(sum*invN);
  }    

  while (head < hpx) {
    sum += rsUnpackColor8888(*head++);
    sum -= rsUnpackColor8888(*tail++);
    *p++ = rsPackColorTo8888(sum*invN);
  }    

  while (tail < hpxmw) {
      sum -= rsUnpackColor8888(*tail++);
      *p++ = rsPackColorTo8888(sum*invN);
  }
}

这是垂直模糊：

int W = 8;
uchar4 *in;
uchar4 *out;    

int N;
float invN;    

uint32_t nx;
uint32_t ny;    

void init_calc() {
  N = 2*W+1;
  invN = 1.0f/N;    

  nx = rsAllocationGetDimX(rsGetAllocation(in));
  ny = rsAllocationGetDimY(rsGetAllocation(in));
}    

void root(const ushort *v_in) {
  float4 sum = 0;    

  uchar4 *head = in + *v_in;
  uchar4 *tail = head;
  uchar4 *hpw = head + nx*W;
  uchar4 *hpn = head + nx*N;
  uchar4 *hpy = head + nx*ny;
  uchar4 *hpymw = head + nx*(ny-W-1);    

  uchar4 *p = out + *v_in;    

  while (head < hpw) {
      sum += rsUnpackColor8888(*head);
      head += nx;
  }    

  while (head < hpn) {
      sum += rsUnpackColor8888(*head);
      *p = rsPackColorTo8888(sum*invN);
      head += nx;
      p += nx;
  }    

  while (head < hpy) {
      sum += rsUnpackColor8888(*head);
      sum -= rsUnpackColor8888(*tail);
      *p = rsPackColorTo8888(sum*invN);
      head += nx;
      tail += nx;
      p += nx;
  }    

  while (tail < hpymw) {
      sum -= rsUnpackColor8888(*tail);
      *p = rsPackColorTo8888(sum*invN);
      tail += nx;
      p += nx;
  }
}

这是调用 rs 代码的 Java 代码：

private RenderScript mRS;
private ScriptC_horzblur mHorizontalScript;
private ScriptC_vertblur mVerticalScript;
private ScriptC_blur mBlurScript;

private Allocation alloc1;
private Allocation alloc2;

private void hblur(int radius, Allocation index, Allocation in, Allocation out) {
    mHorizontalScript.set_W(radius);
    mHorizontalScript.bind_in(in);
    mHorizontalScript.bind_out(out);
    mHorizontalScript.invoke_init_calc();
    mHorizontalScript.forEach_root(index);
}

private void vblur(int radius, Allocation index, Allocation in, Allocation out) {
    mHorizontalScript.set_W(radius);
    mVerticalScript.bind_in(in);
    mVerticalScript.bind_out(out);
    mVerticalScript.invoke_init_calc();
    mVerticalScript.forEach_root(index);
}

Bitmap blur(Bitmap org, int radius) {
    Bitmap out = Bitmap.createBitmap(org.getWidth(), org.getHeight(), org.getConfig());

    blur(org, out, radius);

    return out;
}

private Allocation createIndex(int size) {
    Element element = Element.U16(mRS);
    Allocation allocation = Allocation.createSized(mRS, element, size);
    short[] rows = new short[size];
    for (int i = 0; i < rows.length; i++) rows[i] = (short)i;
    allocation.copyFrom(rows);

    return allocation;
}

private void blur(Bitmap src, Bitmap dst, int r) {
    Allocation alloc1 = Allocation.createFromBitmap(mRS, src);
    Allocation alloc2 = Allocation.createTyped(mRS, alloc1.getType());

    Allocation hIndexAllocation = createIndex(alloc1.getType().getY());
    Allocation vIndexAllocation = createIndex(alloc1.getType().getX());

    // Iteration 1
    hblur(r, hIndexAllocation, alloc1, alloc2);
    vblur(r, vIndexAllocation, alloc2, alloc1);
    // Iteration 2
    hblur(r, hIndexAllocation, alloc1, alloc2);
    vblur(r, vIndexAllocation, alloc2, alloc1);
    // Add more iterations if you like or simply make a loop
    alloc1.copyTo(dst);
}
于 2012-11-17T23:03:01.543 回答
7

高斯模糊的精确度很高。只需迭代地平均像素，就可以完成更快的近似。大量模糊图像仍然很昂贵，但是您可以在每次迭代之间重新绘制，以至少提供即时反馈和图像模糊的漂亮动画。

static void blurfast(Bitmap bmp, int radius) {
  int w = bmp.getWidth();
  int h = bmp.getHeight();
  int[] pix = new int[w * h];
  bmp.getPixels(pix, 0, w, 0, 0, w, h);

  for(int r = radius; r >= 1; r /= 2) {
    for(int i = r; i < h - r; i++) {
      for(int j = r; j < w - r; j++) {
        int tl = pix[(i - r) * w + j - r];
        int tr = pix[(i - r) * w + j + r];
        int tc = pix[(i - r) * w + j];
        int bl = pix[(i + r) * w + j - r];
        int br = pix[(i + r) * w + j + r];
        int bc = pix[(i + r) * w + j];
        int cl = pix[i * w + j - r];
        int cr = pix[i * w + j + r];

        pix[(i * w) + j] = 0xFF000000 |
            (((tl & 0xFF) + (tr & 0xFF) + (tc & 0xFF) + (bl & 0xFF) + (br & 0xFF) + (bc & 0xFF) + (cl & 0xFF) + (cr & 0xFF)) >> 3) & 0xFF |
            (((tl & 0xFF00) + (tr & 0xFF00) + (tc & 0xFF00) + (bl & 0xFF00) + (br & 0xFF00) + (bc & 0xFF00) + (cl & 0xFF00) + (cr & 0xFF00)) >> 3) & 0xFF00 |
            (((tl & 0xFF0000) + (tr & 0xFF0000) + (tc & 0xFF0000) + (bl & 0xFF0000) + (br & 0xFF0000) + (bc & 0xFF0000) + (cl & 0xFF0000) + (cr & 0xFF0000)) >> 3) & 0xFF0000;
      }
    }
  }
  bmp.setPixels(pix, 0, w, 0, 0, w, h);
}
于 2012-11-18T01:50:49.487 回答