如何在iOS上实现框或高斯模糊

Question

我希望能够拍摄一张图像并相对快速地模糊(比如在0.1秒内).图像大小几乎不会大于256 x 256像素.

我是否必须通过每个像素循环并将它们与邻居平均或者是否有更高级别的方法可以做到这一点？

PS:我知道多个盒子模糊可以接近高斯模糊.

Answer 1

对于iOS3.2 +应用程序,我发现了一种非常快速的糟糕方式

  UIView *myView = [self view];
  CALayer *layer = [myView layer];
  [layer setRasterizationScale:0.25];
  [layer setShouldRasterize:YES];

这会将视图光栅化为4x4像素块,然后使用双线性滤波将其缩放...它非常快,如果您只是想在模态视图下模糊背景视图,则看起来很好.

要撤消它,只需将光栅化比例设置回1.0或关闭光栅化.

Answer 2

从i-create-i-created-blurred-text-in-a-iphone-view:

看看Apple的GLImageProcessing iPhone示例.除其他外,它做了一些模糊.

相关代码包括:

static void blur(V2fT2f *quad, float t) // t = 1
{
    GLint tex;
    V2fT2f tmpquad[4];
    float offw = t / Input.wide;
    float offh = t / Input.high;
    int i;

    glGetIntegerv(GL_TEXTURE_BINDING_2D, &tex);

    // Three pass small blur, using rotated pattern to sample 17 texels:
    //
    // .\/.. 
    // ./\\/ 
    // \/X/\   rotated samples filter across texel corners
    // /\\/. 
    // ../\. 

    // Pass one: center nearest sample
    glVertexPointer  (2, GL_FLOAT, sizeof(V2fT2f), &quad[0].x);
    glTexCoordPointer(2, GL_FLOAT, sizeof(V2fT2f), &quad[0].s);
    glTexEnvi(GL_TEXTURE_ENV, GL_TEXTURE_ENV_MODE, GL_MODULATE);
    glColor4f(1.0/5, 1.0/5, 1.0/5, 1.0);
    validateTexEnv();
    glDrawArrays(GL_TRIANGLE_STRIP, 0, 4);

    // Pass two: accumulate two rotated linear samples
    glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_MIN_FILTER, GL_LINEAR);
    glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_MAG_FILTER, GL_LINEAR);
    glEnable(GL_BLEND);
    glBlendFunc(GL_SRC_ALPHA, GL_ONE);
    for (i = 0; i < 4; i++)
    {
        tmpquad[i].x = quad[i].s + 1.5 * offw;
        tmpquad[i].y = quad[i].t + 0.5 * offh;
        tmpquad[i].s = quad[i].s - 1.5 * offw;
        tmpquad[i].t = quad[i].t - 0.5 * offh;
    }
    glTexCoordPointer(2, GL_FLOAT, sizeof(V2fT2f), &tmpquad[0].x);
    glTexEnvi(GL_TEXTURE_ENV, GL_TEXTURE_ENV_MODE, GL_REPLACE);
    glActiveTexture(GL_TEXTURE1);
    glEnable(GL_TEXTURE_2D);
    glClientActiveTexture(GL_TEXTURE1);
    glTexCoordPointer(2, GL_FLOAT, sizeof(V2fT2f), &tmpquad[0].s);
    glEnableClientState(GL_TEXTURE_COORD_ARRAY);
    glBindTexture(GL_TEXTURE_2D, tex);
    glTexEnvi(GL_TEXTURE_ENV, GL_TEXTURE_ENV_MODE, GL_COMBINE);
    glTexEnvi(GL_TEXTURE_ENV, GL_COMBINE_RGB,      GL_INTERPOLATE);
    glTexEnvi(GL_TEXTURE_ENV, GL_SRC0_RGB,         GL_TEXTURE);
    glTexEnvi(GL_TEXTURE_ENV, GL_SRC1_RGB,         GL_PREVIOUS);
    glTexEnvi(GL_TEXTURE_ENV, GL_SRC2_RGB,         GL_PRIMARY_COLOR);
    glTexEnvi(GL_TEXTURE_ENV, GL_OPERAND2_RGB,     GL_SRC_COLOR);
    glTexEnvi(GL_TEXTURE_ENV, GL_COMBINE_ALPHA,    GL_REPLACE);
    glTexEnvi(GL_TEXTURE_ENV, GL_SRC0_ALPHA,       GL_PRIMARY_COLOR);

    glColor4f(0.5, 0.5, 0.5, 2.0/5);
    validateTexEnv();
    glDrawArrays(GL_TRIANGLE_STRIP, 0, 4);

    // Pass three: accumulate two rotated linear samples
    for (i = 0; i < 4; i++)
    {
        tmpquad[i].x = quad[i].s - 0.5 * offw;
        tmpquad[i].y = quad[i].t + 1.5 * offh;
        tmpquad[i].s = quad[i].s + 0.5 * offw;
        tmpquad[i].t = quad[i].t - 1.5 * offh;
    }
    glDrawArrays(GL_TRIANGLE_STRIP, 0, 4);

    // Restore state
    glDisableClientState(GL_TEXTURE_COORD_ARRAY);
    glClientActiveTexture(GL_TEXTURE0);
    glBindTexture(GL_TEXTURE_2D, Half.texID);
    glDisable(GL_TEXTURE_2D);
    glTexEnvi(GL_TEXTURE_ENV, GL_OPERAND2_RGB,     GL_SRC_ALPHA);
    glActiveTexture(GL_TEXTURE0);
    glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_MIN_FILTER, GL_NEAREST);
    glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_MAG_FILTER, GL_NEAREST);
    glDisable(GL_BLEND);
}

Answer 3

您可能想看看 Mario Klingemann 的StakBlur算法。它不完全是高斯分布，但非常接近。