双线性插值 [英] Bilinear interpolation

问题描述

我得到了这个用于通过双线性插值来缩放图像的代码。我知道这有效但我无法弄清楚如果近似像素值是边缘的话（通过边缘我的意思是它在最后一行或最后一列）输入图像中的像素然后我可以我坐标（x + 1，y + 1）的像素，这应该导致数组索引超出范围错误，但没有出现这样的错误为什么？
代码是：

I got this code for scaling image by bilinear interpolation.I know this works but i can't figure out one thing what if the approximated pixel value is an edge(by edge i mean it is in the last row or last column) pixel in the input image then i can i gt a pixel of coordinate (x+1,y+1) ,this should lead to a array index out of range error but no such error occurs why? The code is:

public int[] resizeBilinearGray(int[] pixels, int w, int h, int w2, int h2) {
    int[] temp = new int[w2*h2] ;
    int A, B, C, D, x, y, index, gray ;
    float x_ratio = ((float)(w-1))/w2 ;
    float y_ratio = ((float)(h-1))/h2 ;
    float x_diff, y_diff, ya, yb ;
    int offset = 0 ;
    for (int i=0;i<h2;i++) {
        for (int j=0;j<w2;j++) {
            x = (int)(x_ratio * j) ;
            y = (int)(y_ratio * i) ;
            x_diff = (x_ratio * j) - x ;
            y_diff = (y_ratio * i) - y ;
            index = y*w+x ;

            // range is 0 to 255 thus bitwise AND with 0xff
            A = pixels[index] & 0xff ;
            B = pixels[index+1] & 0xff ;
            C = pixels[index+w] & 0xff ;
            D = pixels[index+w+1] & 0xff ;

            // Y = A(1-w)(1-h) + B(w)(1-h) + C(h)(1-w) + Dwh
            gray = (int)(
                    A*(1-x_diff)*(1-y_diff) +  B*(x_diff)*(1-y_diff) +
                    C*(y_diff)*(1-x_diff)   +  D*(x_diff*y_diff)
                    ) ;

            temp[offset++] = gray ;                                   
        }
    }
    return temp ;
}

推荐答案

原因是 x_ratio 和 y_ratio 计算错误。

考虑最后一行的最后一个像素：

Consider the last pixel of the last row:

i=h2, j=w2 

然后：

x = x_ratio * j = (w-1)/w2 * (w2-1) = (w-1) * (w2-1)/w2 <= w-1
y = y_ratio * i = (h-1)/h2 * (h2-1) = (h-1) * (h2-1)/h2 <= h-1

index = y*w+x <= (h-1)*w + (w-1) < w*h

所以索引总是小于像素的大小数组。

so the index is always less than the size of the pixels array.

但是请注意，这是一个非常脏的黑客，这将导致不准确的结果，特别是对于小图像。

Note however, that this is a very dirty hack, which will result in inaccurate results, especially for small images.

您应该按如下方式计算宽度/高度比率：

You should calculate width/height ratio as follows:

float x_ratio = ((float)w)/w2;
float y_ratio = ((float)h)/h2;

并创建一个将坐标转换为数组索引的函数 - 让它命名为 coord2index 。此函数考虑了超出范围的坐标并实现了所谓的边界选项，模拟图像边界外有像素。

and create a function which converts coordinates to array index - let's name it coord2index. This function takes out-of range coordinates into account and implements a so-called boundary option, which simulates that there are pixels outside the image boundary.

边界的常见选项是：

• 对称 - 图像边界外的像素通过镜像反射边界处的图像来计算。在这种情况下，这可能是最好的可能性。

• symmetric - pixels outside the image bounds are computed by mirror-reflecting the image at the border. This is probably the best possibility in this case.

复制 - 假设图像边界外的像素等于边界处的最近像素。这是最简单的方法。

replicate - pixels outside the image bounds are assumed to be equal to the nearest pixel at the border. This is the simplest way.

循环 - 图像几乎在所有方向上定期重复。用于一些先进的图像处理算法;不适合图像大小调整。

circular - the image is virtually repeated periodically in all directions. Used for some advanced image processing algorithms; not good for image resizing.

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