考虑以下两组点。我想找到最佳的 2D 平移和旋转，以对齐数据集蓝色和数据集橙色之间的最大数量的点，如果到另一个数据集中最近邻居的距离小于阈值，则认为该点对齐。

我知道这与“迭代最近点”算法有关，但在这种情况下，情况有点困难，因为并非一个数据集中的所有点都在另一个数据集中，而且有些点可能会变成“误报”(噪音)。

有没有一种有效的方法来做到这一点？

1 Câu trả lời

我遇到了同样的问题并在 comaring the CCD stars observation figures 中找到了解决方案，基本思想是找到两组点的三角形的最佳匹配。

Sau đó tôi sử dụng astroalign bưu kiện计算变换矩阵，并对齐所有点。感谢上帝，它工作得很好。

import itertools
nhập numpy dưới dạng np
nhập matplotlib.pyplot dưới dạng plt
import astroalign as aa

def getTriangles(set_X, X_combs):
    """
    Inefficient way of obtaining the lengths of each triangle's side.
    Normalized so that the minimum length is 1.
    """
    triang = []
    for p0, p1, p2 in X_combs:
        d1 = np.sqrt((set_X[p0][0] - set_X[p1][0]) ** 2 +
                     (set_X[p0][1] - set_X[p1][1]) ** 2)
        d2 = np.sqrt((set_X[p0][0] - set_X[p2][0]) ** 2 +
                     (set_X[p0][1] - set_X[p2][1]) ** 2)
        d3 = np.sqrt((set_X[p1][0] - set_X[p2][0]) ** 2 +
                     (set_X[p1][1] - set_X[p2][1]) ** 2)
        d_min = min(d1, d2, d3)
        d_unsort = [d1 / d_min, d2 / d_min, d3 / d_min]
        triang.append(sorted(d_unsort))
    return triang

def sumTriangles(ref_triang, in_triang):
    """
    For each normalized triangle in ref, compare with each normalized triangle
    in B. find the differences between their sides, sum their absolute values,
    and select the two triangles with the smallest sum of absolute differences.
    """
    tr_sum, tr_idx = [], []
    for i, ref_tr in enumerate(ref_triang):
        for j, in_tr in enumerate(in_triang):
            # Absolute value of lengths differences.
            tr_diff = abs(np.array(ref_tr) - np.array(in_tr))
            # Sum the differences
            tr_sum.append(sum(tr_diff))
            tr_idx.append([i, j])

    # Index of the triangles in ref and in with the smallest sum of absolute
    # length differences.
    tr_idx_min = tr_idx[tr_sum.index(min(tr_sum))]
    ref_idx, in_idx = tr_idx_min[0], tr_idx_min[1]
    print("Smallest difference: {}".format(min(tr_sum)))
    return ref_idx, in_idx


set_ref = np.array([[2511.268821,44.864124],
 [2374.085032,201.922566], 
 [1619.282942,216.089335], 
 [1655.866502,221.127787], 
 [ 804.171659,2133.549517], ])

set_in = np.array([[1992.438563,63.727282], 
 [2285.793346,255.402548], 
 [1568.915358, 279.144544], 
 [1509.720134, 289.434629], 
 [1914.255205, 349.477788], 
 [2370.786382, 496.026836], 
 [ 482.702882, 508.685952], 
 [2089.691026, 523.18825 ], 
 [ 216.827439, 561.807396], 
 [ 614.874621, 2007.304727], 
 [1286.639124, 2155.264827], 
 [ 729.566116, 2190.982364]])

# All possible triangles.
ref_combs = list(itertools.combinations(range(len(set_ref)), 3))
in_combs = list(itertools.combinations(range(len(set_in)), 3))

# Obtain normalized triangles.
ref_triang, in_triang = getTriangles(set_ref, ref_combs), getTriangles(set_in, in_combs)

# Index of the ref and in triangles with the smallest difference.
ref_idx, in_idx = sumTriangles(ref_triang, in_triang)

# Indexes of points in ref and in of the best match triangles.
ref_idx_pts, in_idx_pts = ref_combs[ref_idx], in_combs[in_idx]
print ('triangle ref %s matches triangle in %s' % (ref_idx_pts, in_idx_pts))

print ("ref:", [set_ref[_] for _ in ref_idx_pts])
print ("input:", [set_in[_] for _ in in_idx_pts])

ref_pts = np.array([set_ref[_] for _ in ref_idx_pts])
in_pts = np.array([set_in[_] for _ in in_idx_pts])

transf, (in_list,ref_list) = aa.find_transform(in_pts, ref_pts)
transf_in = transf(set_in)
print(f'transformation matrix: {transf}')

plt.scatter(set_ref[:,0],set_ref[:,1], s=100,marker='.', c='r',label='Reference')
plt.scatter(set_in[:,0],set_in[:,1], s=100,marker='.', c='b',label='Input')
plt.scatter(transf_in[:,0],transf_in[:,1], s=100,marker='+', c='b',label='Input Aligned')
plt.plot(ref_pts[:,0],ref_pts[:,1], c='r')
plt.plot(in_pts[:,0],in_pts[:,1], c='b')
plt.legend()
plt.tight_layout()
plt.savefig( 'align_coordinates.png', format = 'png')
plt.hiển thị()

关于python - 当这些点集包含噪声时如何对齐两组点(平移+旋转)？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/62238853/