线性插值基本原理推导

news/2024/4/24 0:57:39/文章来源:https://blog.csdn.net/m0_46926492/article/details/129259857

线性插值基本原理

1. 为什么使用线性插值？
2. 单线性插值
- 2.1 单线性插值推导
3.多线性插值
- 3.1 多线性插值推导注意事项
- 3.2 多线性插值推导
- 3.3 插入坐标越远权重越大

1. 为什么使用线性插值？

在深度学习对图片进行上采样和下采样的时候会应用到线性插值
- 对图片上采样，原始图片(33)范围红色框中的值，会得到(44)框中红色框的值
- 假设目标图片红框坐标为(i,j),那么在原始图片位置 $(i∗34,j∗34)(i*\frac{3}{4},j*\frac{3}{4})$ ,
- 已知 $i = 2, j = 3$ ,所以在原始图片位置 $(1.5, 0.75)$
  - 不是整数，在找原始图片位置时，会自动取整
  - 即，需要使用线性插值，来降低误差

2. 单线性插值

在这里插入图片描述

2.1 单线性插值推导

如图所示，在 $p_0和p_1$ 中间插入 $p$ ，求p点位置
根据斜率公式
$y−y0x−x0=y1−y0x1−x0y=y0+(y1−y0)(x−x0)x1−x0y=(x1−x0)y0+(y1−y0)(x−x0)x1−x0y=x1y0−x0y0+xy1−x0y1−xy0+x0y0x1−x0y=(x1−x)y0+(x−x0)y1x1−x0y=(x1−x)x1−x0y0+(x−x0)x1−x0y1\begin{aligned} \frac{y-y_{0}}{x-x_{0}}&=\frac{y_{1}-y_{0}}{x_{1}-x_{0}}\\ y&=y_{0}+\frac{(y_{1}-y_{0})(x-x_{0})}{x_{1}-x_{0}} \\&y=\frac{(x_{1}-x_{0})y_{0}+(y_{1}-y_{0})(x-x_{0})}{x_{1}-x_{0}} \\&y=\frac{x_{1}y_{0}-x_{0}y_{0}+xy_{1}-x_{0}y_{1}-xy_{0}+x_{0}y_{0}}{x_{1}-x_{0}} \\&y=\frac{(x_{1}-x)y_{0}+(x-x_{0})y_{1}}{x_{1}-x_{0}} \\&y=\frac{(x_{1}-x)}{x_{1}-x_{0}}y_{0}+\frac{(x-x_{0})}{x_{1}-x_{0}}y_{1} \end{aligned}$
假设y对应的值，就是图像中的像素值p，得到
$p=(x1−x)x1−x0p0+(x−x0)x1−x0p1\begin{aligned} p=\frac{(x_{1}-x)}{x_{1}-x_{0}}p_{0}+\frac{(x-x_{0})}{x_{1}-x_{0}}p_{1} \end{aligned}$

3.多线性插值

3.1 多线性插值推导注意事项

双线性插值是三维的，需要从二维计算出某一个坐标轴的差值才可以
- 如果按照单线性插值，直接按照三位图，那么求 $p_{1}$
  - 显然不可能会出现 $y_{1}-y_{1}$ 这个操作
    $y−y1x−x2=y1−y1x1−x2\begin{aligned} \frac{y-y_{1}}{x-x_{2}}&=\frac{y_{1}-y_{1}}{x_{1}-x_{2}} \end{aligned}$

3.2 多线性插值推导

根据单线性插值公式，得出公式
- 我们计算像素值，直接将 $y$ 变成 $p$ 即可
  $y=(x1−x)x1−x0y0+(x−x0)x1−x0y1\begin{aligned} y=\frac{(x_{1}-x)}{x_{1}-x_{0}}y_{0}+\frac{(x-x_{0})}{x_{1}-x_{0}}y_{1} \end{aligned}$

在这里插入图片描述

由单线性插值,对x轴得到【自己转换维度】
$p1=(x2−x)x2−x1p11+(x−x1)x2−x1p21p2=(x2−x)x2−x1p12+(x−x1)x2−x1p22\begin{aligned} p_{1}=\frac{(x_{2}-x)}{x_{2}-x_{1}}p_{11}+\frac{(x-x_{1})}{x_{2}-x_{1}}p_{21}\\ p_{2}=\frac{(x_{2}-x)}{x_{2}-x_{1}}p_{12}+\frac{(x-x_{1})}{x_{2}-x_{1}}p_{22} \end{aligned}$
由单线性插值,对y轴得到
$p=(y2−y)y2−y1p1+(y−y1)y2−y1p2\begin{aligned} p=\frac{(y_{2}-y)}{y_{2}-y_{1}}p_{1}+\frac{(y-y_{1})}{y_{2}-y_{1}}p_{2} \end{aligned}$
将 $p_{1},p_{2}$ 代入化简 $p$
$p=(y2−y)y2−y1((x2−x)x2−x1p11+(x−x1)x2−x1p21)+(y−y1)y2−y1((x2−x)x2−x1p12+(x−x1)x2−x1p22)=(y2−y)(x2−x)(y2−y1)(x2−x1)p11+(y2−y)(x−x1)(y2−y1)(x2−x1)p21+(y−y1)(x2−x)(y2−y1)(x2−x1)p12+(y−y1)(x−x1)(y2−y1)(x2−x1)p22\begin{aligned} p&=\frac{(y_{2}-y)}{y_{2}-y_{1}}(\frac{(x_{2}-x)}{x_{2}-x_{1}}p_{11}+\frac{(x-x_{1})}{x_{2}-x_{1}}p_{21})+\frac{(y-y_{1})}{y_{2}-y_{1}}(\frac{(x_{2}-x)}{x_{2}-x_{1}}p_{12}+\frac{(x-x_{1})}{x_{2}-x_{1}}p_{22}) \\&=\frac{(y_{2}-y)(x_{2}-x)}{(y_{2}-y_{1})(x_{2}-x_{1})}p_{11}+\frac{(y_{2}-y)(x-x_{1})}{(y_{2}-y_{1})(x_{2}-x_{1})}p_{21}+\frac{(y-y_{1})(x_{2}-x)}{(y_{2}-y_{1})(x_{2}-x_{1})}p_{12}+\frac{(y-y_{1})(x-x_{1})}{(y_{2}-y_{1})(x_{2}-x_{1})}p_{22} \end{aligned}$
在图像中，相邻像素点坐标差值为1，即
$x2=x1+1y2=y1+1\begin{aligned} x_{2}&=x_{1}+1\\ y_{2}&=y_{1}+1 \end{aligned}$
将 $x_2,y_2$ 关于 $x_1,y_1$ 的表达式代入p
$p=(y2−y)(x2−x)(y2−y1)(x2−x1)p11+(y2−y)(x−x1)(y2−y1)(x2−x1)p21+(y−y1)(x2−x)(y2−y1)(x2−x1)p12+(y−y1)(x−x1)(y2−y1)(x2−x1)p22=(y2−y)(x2−x)p11+(y2−y)(x−x1)p21+(y−y1)(x2−x)p12+(y−y1)(x−x1)p22\begin{aligned} p&=\frac{(y_{2}-y)(x_{2}-x)}{(y_{2}-y_{1})(x_{2}-x_{1})}p_{11}+\frac{(y_{2}-y)(x-x_{1})}{(y_{2}-y_{1})(x_{2}-x_{1})}p_{21}+\frac{(y-y_{1})(x_{2}-x)}{(y_{2}-y_{1})(x_{2}-x_{1})}p_{12}+\frac{(y-y_{1})(x-x_{1})}{(y_{2}-y_{1})(x_{2}-x_{1})}p_{22} \\&=(y_{2}-y)(x_{2}-x)p_{11}+(y_{2}-y)(x-x_{1})p_{21}+(y-y_{1})(x_{2}-x)p_{12}+(y-y_{1})(x-x_{1})p_{22} \end{aligned}$

3.3 插入坐标越远权重越大

由于p是由 $p_{1}$ 和 $p_{2}$ 得到的
$P1≈(y2−y)(x2−x)p11+(y2−y)(x−x1)p21P2≈(y−y1)(x2−x)p12+(y−y1)(x−x1)p22\begin{aligned} P_{1}&\approx(y_{2}-y)(x_{2}-x)p_{11}+(y_{2}-y)(x-x_{1})p_{21}\\ P_{2}&\approx(y-y_{1})(x_{2}-x)p_{12}+(y-y_{1})(x-x_{1})p_{22} \end{aligned}$
- 根据 $p_{1}$ ,都有系数 $y_{2}-y)$ ,如果 $p_{1}$ 向的值比较大，那么 $x_{2}-x)$ 的值就要大，所以当插入的点离哪一个坐标远，那个坐标的权重越大【插入值的权重越大】