数字图像处理双线性插值：1.5倍缩放时1.5列的物理矩阵表示

Great question — let's break this down step by step, since bilinear interpolation's handling of subpixel positions is often where folks get tripped up.

1. What "1.5 Column" Means in the Original Physical Matrix #

The original image’s matrix is a discrete grid of pixels, with columns indexed by integers (0, 1, 2, ..., W-1 for an image of width W). Each integer index corresponds to the center of that pixel’s physical area.

A "1.5 column" position is a subpixel location: it sits exactly halfway between the centers of the 1st column (index 1) and 2nd column (index 2) pixels in the original matrix. It’s not a discrete pixel itself, but a continuous point between two existing pixels.

2. How Bilinear Interpolation Uses This Position #

Bilinear interpolation calculates the value at this subpixel point by linearly combining the values of the surrounding pixels. Let’s assume the row position is an integer y for simplicity:

• First, identify the nearest integer columns around 1.5: floor(1.5) = 1 and ceil(1.5) = 2.
• Since 1.5 is exactly halfway between 1 and 2, each column gets an equal weight of 0.5.
• The interpolated value at (y, 1.5) is:

  interpolated_value = 0.5 * I[y][1] + 0.5 * I[y][2]
  

If the row position was also a subpixel (e.g., y=2.3), you’d first interpolate along the columns for rows 2 and 3, then interpolate those two results along the row direction using weights matching the fractional part (0.7 for row 2, 0.3 for row 3).

3. Quick Edge Case Note #

If your subpixel position was closer to an edge (e.g., 0.2 columns), you’d still use the nearest valid columns (0 and 1 here) and adjust weights accordingly (0.8 for column 0, 0.2 for column 1). Most implementations handle boundary positions by clamping to edge pixels or mirroring, but that’s a separate detail.

内容的提问来源于stack exchange，提问作者Satyaki Chatterjee