Android，如何立即模糊位图？

Question

我想尽快模糊一张图片（实时感受），因为在按下“模糊”按钮后需要更新活动。 问题是，我找不到足够快速的模糊... 注意：最好使用高斯模糊，但质量不需要很好...

我尝试了以下方法，但需要几秒钟时间，有没有办法在牺牲质量的情况下使代码运行更快？还有其他替代方案吗？ 我会研究GPU相关的东西，但这个模糊效果只与UI相关，只会在打开一个小框大小的透明活动时发生...

有什么想法吗？

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 <[e-mail]>

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);

}

- rennoDeniro

---

这对我很有效。请参考：https://futurestud.io/tutorials/how-to-blur-images-efficiently-with-androids-renderscript - AndroidDev

Answer 1

尝试将图像缩小2、4、8倍，然后再放大。这很快。否则在renderscript中实现它。

如果你想要更多的功能，你可以查看renderscript中的代码片段。它执行与另一个答案中所给的模糊处理相同类型的操作。该算法也可以在Java中实现，并且是对其他答案的优化。此代码可以将一行进行模糊处理。要对位图进行模糊处理，您应该为所有行调用此代码，然后为所有列进行相同的操作（您需要重新实现它以处理列）。如果您想要快速模糊，请只执行一次。如果您想要更好的模糊效果，请多次执行。我通常只执行两次。

之所以只处理一行，是因为我试图并行化算法，在renderscript中这非常简单并且有些改进。我为所有行并行调用了下面的代码，然后为所有列执行相同的操作。

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);
  }
}

这里是用于垂直模糊的代码：

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;
  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;
  }
}

以下是调用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);
}

- Tobias Ritzau

---

我不太明白如何实现ScriptC_......一部分，我真的不知道它是如何工作的？而且周围的搜索也没有给出太多帮助 :s - rennoDeniro

这些类是自动生成的。将*.rs文件与您的Java文件放在一起。查看此链接以获取更多信息：http://developer.android.com/guide/topics/graphics/renderscript/graphics.html您使用Eclipse还是其他IDE？ - Tobias Ritzau

但是在 ".forEach" 上，我得到了类型不匹配的 u16 :S？有什么想法我做错了什么吗？ - rennoDeniro

我认为你可以把它作为对mBlurScript的参考。 - StackOverflowed

尝试让这个样例工作。除了每个脚本上的forEach_root（index）调用之外，一切看起来都没问题。编译器抱怨找不到那些方法...有什么想法？ - Eagle11

Answer 2

高斯模糊精确计算成本较高。通过迭代平均像素可以更快地进行近似处理。虽然大量模糊图像仍旧需要耗费时间，但你可以在每次迭代之间重新绘制图像以获得即时反馈和漂亮的模糊动画效果。

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);
}

- gordy

---

这段代码多次重新计算相同的总和。如果你想要这种模糊效果，请参考我的答案。 - Tobias Ritzau

请确保在指定半径时使用2的幂次方 - 如8、16或32。 - gordy

我有一个问题，使用32可以得到更好的结果，但是当我获取图像时，模糊似乎只应用于图像的一半:s，如何让它应用于整个图像？ - rennoDeniro

@rennoDeniro，之前有一个bug导致整个图像没有被处理。我刚刚编辑了答案并修复了这个bug，请告诉我模糊是否足够快和模糊。 - gordy

抱歉，这个模糊效果不够高斯化，需要一些时间 :( - rennoDeniro