Python 基于最近距离的坐标点过滤

我有一个numpy数组，它包含像这样的坐标点

[[ 581  925]
 [ 582  926]
 [ 582  931]
 [ 582  939]
 [ 584  933]
 [ 584  937]
 [ 585  943]
 [ 586  944]
 [ 589  944]]

正如您所看到的，有些点的x坐标相同，但y坐标不同从第一个坐标开始，计算到下一个最近的直接坐标的距离。
例如，可以找到从

[581 925]

到下一个最近坐标的距离。候选坐标为

[582 926]、[582 931]、[582 939]

，因为它们是最接近

[581 925]

的直接坐标

很明显，在这种情况下，

[582 926]

是距离

[581 925]

最近的坐标，我只希望该坐标存在，并删除其他两个候选坐标。所以结果数组应该是

[[ 581  925]
 [ 582  926]
      .
      .
      .
 [ 589  944]]

现在，应该从

[582 926]

开始执行相同的操作，以此类推，直到结束

具有未过滤坐标的轮廓：

带过滤坐标的等高线：

什么是最具python风格（最好是numpy）的方法，时间复杂度最低，因为它是最受关注的问题？

---

注意：线条细化不重要，只需要删除不必要的点/坐标。

我这样做：

要使该方法起作用，首先必须根据相等的x轴值将阵列分成sup组。请参阅详细信息。我将在下面添加代码。数组按相对于x轴的升序排序很重要。如果不是，您可以在数组上应用

np.lextsort

。请参阅了解如何正确应用lexsort。非常感谢为这些帖子提供了精彩的答案

代码：

# Initial array of coordinates

a = np.array([[ 581  925]
     [ 582  926]
     [ 582  931]
     [ 582  939]
     [ 584  933]
     [ 584  937]
     [ 585  943]
     [ 586  944]
     [ 589  944]])

# Following line splits the array into subgroups on the basis of equal x-axis elements
a = np.split(a, np.unique(a[:, 0], return_index=True)[1][1:], axis=0)

# Array after splitting
# [array([[581, 925]]), 
#  array([[582, 926], [582, 931], [582, 939]]), 
#  array([[584, 933], [584, 937]]), 
#  array([[585, 943]]), 
#  array([[586, 944]]), 
#  array([[589, 944]])]

i = 0

# filteredList will initially contain the first element of the array's first sub group
filteredList = np.reshape(np.asarray(a[0][0]), (-1, 2)) # filteredList = [[581 925]]

while not i == len(a) - 1:
    if len(a[i + 1]) > 1:
        # Following line calculates the euclidean distance between current point and the points in the next group
        min_dist_point_addr = np.argmin(np.linalg.norm(filteredList[i] - a[i + 1], axis=1))

        # Next group is reassigned with the element to whom the distance is the least
        a[i + 1] = a[i + 1][min_dist_point_addr]

    # The element is concatenated to filteredList
    filteredList = np.concatenate((filteredList, np.reshape((a[i+1]), (1, 2))), axis=0)
    i += 1

print filteredList

[[581 925]
 [582 926]
 [584 933]
 [585 943]
 [586 944]
 [589 944]]

输出：

# Initial array of coordinates

a = np.array([[ 581  925]
     [ 582  926]
     [ 582  931]
     [ 582  939]
     [ 584  933]
     [ 584  937]
     [ 585  943]
     [ 586  944]
     [ 589  944]])

# Following line splits the array into subgroups on the basis of equal x-axis elements
a = np.split(a, np.unique(a[:, 0], return_index=True)[1][1:], axis=0)

# Array after splitting
# [array([[581, 925]]), 
#  array([[582, 926], [582, 931], [582, 939]]), 
#  array([[584, 933], [584, 937]]), 
#  array([[585, 943]]), 
#  array([[586, 944]]), 
#  array([[589, 944]])]

i = 0

# filteredList will initially contain the first element of the array's first sub group
filteredList = np.reshape(np.asarray(a[0][0]), (-1, 2)) # filteredList = [[581 925]]

while not i == len(a) - 1:
    if len(a[i + 1]) > 1:
        # Following line calculates the euclidean distance between current point and the points in the next group
        min_dist_point_addr = np.argmin(np.linalg.norm(filteredList[i] - a[i + 1], axis=1))

        # Next group is reassigned with the element to whom the distance is the least
        a[i + 1] = a[i + 1][min_dist_point_addr]

    # The element is concatenated to filteredList
    filteredList = np.concatenate((filteredList, np.reshape((a[i+1]), (1, 2))), axis=0)
    i += 1

print filteredList

[[581 925]
 [582 926]
 [584 933]
 [585 943]
 [586 944]
 [589 944]]

---

你能指定你想用哪个距离度量来确定最近的坐标吗？所以，基本上你想计算最近的坐标之间的距离，最近的是指那些距离较小的坐标？没有先计算距离？@Tristan-Euclidean，我将使用

numpy.linalg.norm（a-b）

来计算距离distance@NAmorim不，我想使用这些点绘制轮廓，并想删除那些会阻止我绘制最佳轮廓的点。我将在问题中添加图像。@NAmorim因为有多个点具有相同的x轴值会创建一个厚而锯齿状的轮廓，我只想保留与上一个点最近的点。很高兴看到你终于完成了它@迪瓦卡谢谢！你也帮了很多忙！