最近,我尝试blur
图像,半径是可变的。我试图实现它myseft,但它看起来很慢。在这个网站上,我得到了一个名为stack blur
的快速模糊方法:
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);
}
实际上,我希望这种模糊方法可以变化模糊(例如:当图像中心的距离减小时,半径也会减小)。但是,我不擅长移位,我修改这些代码太难了! 所以,我想问一种方法来实现像径向模糊这样的代码!我想知道是否有任何模糊库包含径向模糊,灰度,....在Android!如果有人能帮助我,真的很感激!谢谢你!
答案 0 :(得分:4)
由于其对速度的优化,该算法不适合适应变化的半径。如果你采取不同的方法,你仍然可以使用它:
原则是您创建多个临时地图,每个临时地图具有增加的(均匀)模糊半径,然后根据该点的半径应该将它们中的两个混合在一起。假设您准备3个临时地图,一个半径为4,一个为8,另一个为16.现在您希望一个像素的模糊半径为12。你做的是将地图2和3混合约50%。您使用的临时地图越多,质量越好,但3(加上原始的不模糊地图)通常就足够了。
我在我的复合模糊画布中使用这种技术:http://www.quasimondo.com/CompoundBlurForCanvas/CompoundBlurDemo.html - 这可以让你做倾斜移位效果或晕影。
如果您正在寻找看起来更像是相机缩放效果的径向模糊,则必须使用不同的方法。在这种情况下,您首先将位图从笛卡尔映射转换为极坐标映射,然后进行水平模糊,最后将地图转换回笛卡尔空间 - 我的网站上有一个Flash演示和一些源代码,也有更详细的解释:http://www.quasimondo.com/archives/000697.php
答案 1 :(得分:0)
如何使用滑块实现复合模糊:
/// HTML ///
//使用:CompoundBlur.js //
<DIV CLASS="specimenWindow">
<DIV CLASS="specimenImage">
<IMG ID="specImg" WIDTH="150" HEIGHT="150" />
<CANVAS ID="specCanvas" WIDTH="150" HEIGHT="150"></CANVAS>
</DIV>
</DIV>
<SCRIPT TYPE="text/javascript" SRC="js/CompoundBlur.js"></SCRIPT>
/// CSS ///
.specimenWindow {
position: absolute;
width: 150px;
height: 150px;
left: 37px;
top: 96px;
}
.specimenImage {
height: 150px;
width: 150px;
}
#specCanvas {
position: absolute;
top: 27px;
left: 2px;
}
/// JS ///
var specImg = document.getElementById("specImg");
_slider.ontouchend = function() {
compoundBlur();
}
function compoundBlur() {
var lensRead = _slider.getPosition(0, 80);
var rData = getRadialGradientMap( 150, 150, 75, 75, 25, 60 );
compoundBlurImage( "specImg", "specCanvas", rData, lensRead, 1.5, 2, true );
}
所以,只需调整数值即可获得更流畅的模糊效果,当我的滑块完成触摸时,我会让它变得模糊,这在移动设备上更有效,而滑块更改时会不断模糊,否则iPad会获得温暖并开始迅速耗尽电池电量
所以希望能帮助任何人尝试实现快速复合模糊。