所以我试图尽可能快地模糊图像（即时感觉）， 因为我按下模糊按钮需要更新活动。
我遇到的问题是，我找不到一个足够快的模糊... 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，这些代码可以在牺牲质量方面更快地运行吗？还是有其他选择吗？ 我会研究GPU的东西，但这种模糊实际上只是与UI相关的效果，只有当我按下打开一个大小为一个小盒子的透明活动时才会发生......
任何想法？
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);
}
答案 0 :(得分：9)
尝试缩小图像2,4,8，...次，然后再将其放大。那很快。否则在renderscript中实现它。
如果您想要比缩放更多，可以在renderscript中查看此代码段。它具有与另一个答案中给出的相同的模糊效果。可以用Java实现相同的算法，并且是另一个答案的优化。这段代码模糊了一行。要模糊位图，您应该为所有行调用此方法，然后对所有列调用相同的方法（您需要重新实现它以处理列）。为了快速模糊，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。
做一行的原因是我尝试并行化算法，这提供了一些改进，并且在renderscript中非常简单。我为所有行并行调用了下面的代码，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);
}
答案 1 :(得分：7)
高斯模糊确实很难做到。通过迭代平均像素可以实现更快的近似。模糊图像仍然很昂贵，但你可以在每次迭代之间重绘，至少给出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);
}