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推导
三角形ABC仿射成为三角形DEF的变换矩阵M
猜测矩阵M=
[
[a1,b1,c1],
[a2,b2,c2]
]
仿射变换的数学联系
对于A点和D点
AX*a1+AY*b1+c1=DX
AX*a2+AY*b2+c1=DY
对于B点和E点
BX*a1+BY*b1+c1=EX
BX*a2+BY*b2+c2=EY
对于C点和F点
CX*a1+CY*b1+c1=FX
CX*a2+CY*b2+c2=FY
求解
对以上数学联系式均取第一条
AX*a1+AY*b1+c1=DX
BX*a1+BY*b1+c1=EX
CX*a1+CY*b1+c1=FX
可以看出这是一个三元一次方程组，我们可以借助scipy库linalg模块的solve函数进行求解，它的用法大家可以参考一下其他资料，这里不多赘述
对数学联系式均取第二条
AX*a2+AY*b2+c1=DY
BX*a2+BY*b2+c2=EY
CX*a2+CY*b2+c2=FY
也是一个三元一次方程组，同样可以借助scipy库linalg模块的solve函数进行求解
验证
根据以上各个点的坐标，分别用opencv的函数，和我们猜想的计算过程，观测对比两个结果是否一致
import numpy as np import cv2 as cv import math AX = 4 AY = 2 BX = 9 BY = 2 CX
= 4 CY = 7 DX = 13 DY = 8 EX = 13 EY = 6 FX = 11 FY = 8 ps1 = np.float32([ [AX,
AY], [BX, BY], [CX, CY] ]) ps2 = np.float32([ [DX, DY], [EX, EY], [FX, FY] ]) M
= cv.getAffineTransform(ps1, ps2) print(f"opencv_M=\n{M}") print() A =
np.array([ [AX, AY, 1], [BX, BY, 1], [CX, CY, 1] ]) b1 = np.array([ [DX], [EX],
[FX] ]) b2 = np.array([ [DY], [EY], [FY] ]) from scipy import linalg r1 =
linalg.solve(A, b1) r2 = linalg.solve(A, b2) r = np.array([ r1.T[0], r2.T[0] ])
print(f"self_M=\n{r}")
运行结果
可以看出，结果是一致的，函数的底层实现原理就是如此
