如何使用python执行坐标仿射变换?第2部分 [英] how to perform coordinates affine transformation using python? part 2
问题描述
我有与此处所述相同的问题: 如何使用python执行坐标仿射变换?
我试图使用所描述的方法，但由于某些原因，我会收到错误消息. 我对代码所做的更改是为了替换主系统点和辅助系统点.我通过使用不同的origo创建了辅助坐标点.在我正在研究的实际情况下，测量坐标时会出现一些错误.
primary_system1 = (40.0, 1160.0, 0.0) primary_system2 = (40.0, 40.0, 0.0) primary_system3 = (260.0, 40.0, 0.0) primary_system4 = (260.0, 1160.0, 0.0) secondary_system1 = (610.0, 560.0, 0.0) secondary_system2 = (610.0,-560.0, 0.0) secondary_system3 = (390.0, -560.0, 0.0) secondary_system4 = (390.0, 560.0, 0.0)
执行以下命令时出现错误.
*Traceback (most recent call last):
File "affine_try.py", line 57, in <module>
secondary_system3, secondary_system4 )
File "affine_try.py", line 22, in solve_affine
A2 = y * x.I
File "/usr/lib/python2.7/dist-packages/numpy/matrixlib/defmatrix.py", line 850, in getI
return asmatrix(func(self))
File "/usr/lib/python2.7/dist-packages/numpy/linalg/linalg.py", line 445, in inv
return wrap(solve(a, identity(a.shape[0], dtype=a.dtype)))
File "/usr/lib/python2.7/dist-packages/numpy/linalg/linalg.py", line 328, in solve
raise LinAlgError, 'Singular matrix'
numpy.linalg.linalg.LinAlgError: Singular matrix*
可能是什么问题?
问题是您的矩阵是奇异的，这意味着它是不可逆的.由于您正尝试采取相反的方法，因此这是一个问题.您链接到的线程是解决问题的基本方法，但实际上并不是最佳解决方案.您实际要做的不仅仅是解决矩阵的求逆问题，而是解决最小二乘最小化问题，以便为可能有噪声的数据找到最佳仿射变换矩阵.这是您的操作方式:
import numpy as np
primary = np.array([[40., 1160., 0.],
[40., 40., 0.],
[260., 40., 0.],
[260., 1160., 0.]])
secondary = np.array([[610., 560., 0.],
[610., -560., 0.],
[390., -560., 0.],
[390., 560., 0.]])
# Pad the data with ones, so that our transformation can do translations too
n = primary.shape[0]
pad = lambda x: np.hstack([x, np.ones((x.shape[0], 1))])
unpad = lambda x: x[:,:-1]
X = pad(primary)
Y = pad(secondary)
# Solve the least squares problem X * A = Y
# to find our transformation matrix A
A, res, rank, s = np.linalg.lstsq(X, Y)
transform = lambda x: unpad(np.dot(pad(x), A))
print "Target:"
print secondary
print "Result:"
print transform(primary)
print "Max error:", np.abs(secondary - transform(primary)).max()
您的原始矩阵奇异的原因是您的第三个坐标始终为零，因此无法判断该坐标上的变换应为零(零乘以零，因此任何值都可以工作).
打印A的值将告诉您最小二乘已找到的转换:
A[np.abs(A) < 1e-10] = 0 # set really small values to zero
print A
结果
[[ -1. 0. 0. 0.]
[ 0. 1. 0. 0.]
[ 0. 0. 0. 0.]
[ 650. -600. 0. 1.]]
等效于x2 = -x1 + 650, y2 = y1 - 600, z2 = 0，其中x1, y1, z1是原始系统中的坐标，而x2, y2, z2是新系统中的坐标.如您所见，由于系统实际上是二维的，因此最小二乘将与第三维相关的所有项设置为零.
I have same problem as described here: how to perform coordinates affine transformation using python?
I was trying to use method described but some reason I will get error messages. Changes I made to code was to replace primary system and secondary system points. I created secondary coordinate points by using different origo. In real case for which I am studying this topic will have some errors when measuring the coordinates.
primary_system1 = (40.0, 1160.0, 0.0) primary_system2 = (40.0, 40.0, 0.0) primary_system3 = (260.0, 40.0, 0.0) primary_system4 = (260.0, 1160.0, 0.0) secondary_system1 = (610.0, 560.0, 0.0) secondary_system2 = (610.0,-560.0, 0.0) secondary_system3 = (390.0, -560.0, 0.0) secondary_system4 = (390.0, 560.0, 0.0)
Error I get from when executing is following.
*Traceback (most recent call last):
File "affine_try.py", line 57, in <module>
secondary_system3, secondary_system4 )
File "affine_try.py", line 22, in solve_affine
A2 = y * x.I
File "/usr/lib/python2.7/dist-packages/numpy/matrixlib/defmatrix.py", line 850, in getI
return asmatrix(func(self))
File "/usr/lib/python2.7/dist-packages/numpy/linalg/linalg.py", line 445, in inv
return wrap(solve(a, identity(a.shape[0], dtype=a.dtype)))
File "/usr/lib/python2.7/dist-packages/numpy/linalg/linalg.py", line 328, in solve
raise LinAlgError, 'Singular matrix'
numpy.linalg.linalg.LinAlgError: Singular matrix*
What might be the problem ?
The problem is that your matrix is singular, meaning it's not invertible. Since you're trying to take the inverse of it, that's a problem. The thread that you linked to is a basic solution to your problem, but it's not really the best solution. Rather than just inverting the matrix, what you actually want to do is solve a least-squares minimization problem to find the optimal affine transform matrix for your possibly noisy data. Here's how you would do that:
import numpy as np
primary = np.array([[40., 1160., 0.],
[40., 40., 0.],
[260., 40., 0.],
[260., 1160., 0.]])
secondary = np.array([[610., 560., 0.],
[610., -560., 0.],
[390., -560., 0.],
[390., 560., 0.]])
# Pad the data with ones, so that our transformation can do translations too
n = primary.shape[0]
pad = lambda x: np.hstack([x, np.ones((x.shape[0], 1))])
unpad = lambda x: x[:,:-1]
X = pad(primary)
Y = pad(secondary)
# Solve the least squares problem X * A = Y
# to find our transformation matrix A
A, res, rank, s = np.linalg.lstsq(X, Y)
transform = lambda x: unpad(np.dot(pad(x), A))
print "Target:"
print secondary
print "Result:"
print transform(primary)
print "Max error:", np.abs(secondary - transform(primary)).max()
The reason that your original matrix was singular is that your third coordinate is always zero, so there's no way to tell what the transform on that coordinate should be (zero times anything gives zero, so any value would work).
Printing the value of A tells you the transformation that least-squares has found:
A[np.abs(A) < 1e-10] = 0 # set really small values to zero
print A
results in
[[ -1. 0. 0. 0.]
[ 0. 1. 0. 0.]
[ 0. 0. 0. 0.]
[ 650. -600. 0. 1.]]
which is equivalent to x2 = -x1 + 650, y2 = y1 - 600, z2 = 0 where x1, y1, z1 are the coordinates in your original system and x2, y2, z2 are the coordinates in your new system. As you can see, least-squares just set all the terms related to the third dimension to zero, since your system is really two-dimensional.
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