如果你想要一个通用的解决方案,你可以使用Sympy http://www.sympy.org,它允许您使用符号表达。在下面的代码中表达式K.T.M = m
重新表述为标准线性方程HH.xx = mm
, where xx
是提取未知数的向量T
:
from IPython.display import display
import sympy as sy
sy.init_printing() # LaTeX like pretty printing for IPython
# declaring symbolic variables:
x, y, X, Y, Z, fx, fy, cx, cy = sy.symbols("x y X Y Z f_x f_y c_x c_y", real=True)
x00, x01, x02, x10, x11 = sy.symbols("x00, x01, x02, x10, x11", real=True)
x12, x20, x21, x22 = sy.symbols("x12, x20, x21, x22", real=True)
Tx, Ty, Tz = sy.symbols(" T_x T_y T_z", real=True)
# Building matrices and vectors:
M = sy.Matrix([X, Y, Z, 1])
m = sy.Matrix([x, y, 1])
K = sy.Matrix([[fx, 0, cx, 0],
[0, fy, cy, 0],
[0, 0, 0, 1]])
T = sy.Matrix([[x00, x01, x02, Tx],
[x10, x11, x12, Ty],
[x20, x21, x22, Tz],
[0, 0, 0, 1]])
print("KTM = K.T.M = ")
KTM = sy.simplify(K*T*M)
display(KTM)
print("Vector of Unkowns xx.T = ")
xx = sy.Matrix(list(T.atoms(sy.Symbol)))
display(xx.T)
print("For equation HH.xx = mm, HH = ")
HH = KTM[:2, :].jacobian(xx) # calculate the derivative for each unknown
display(HH)
正如 @Sven-Marnach 已经指出的那样,没有足够的方程来实现唯一的解决方案。由于向量的最后一行KTM
and of m
为 1,则十二个变量只有两个方程。
如果您有多个像素要评估,即多对(m, M)
, 您可以使用Numpy 最小二乘求解器 https://docs.scipy.org/doc/numpy/reference/generated/numpy.linalg.lstsq.html找到解决方案。