根据最近距离过滤坐标点 [英] Filtering coordinate points on the basis of nearest distance

问题描述

我有一个numpy数组，其中包含类似这样的坐标点

I have a numpy array which contains coordinates points like this

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

如您所见，有些点具有相同的x坐标但具有不同的y坐标. 从第一个坐标开始，计算到下一个最接近的即时坐标的距离.
例如，找到从[581 925]到下一个最近坐标的距离.候选对象为[ 582 926], [ 582 931] & [ 582 939]，因为它们是最接近[581 925]的立即坐标.

As you can see there are points which have same x coordinates but different y coordinates. Starting from the first coordinate, distance to the next closest immediate coordinate is calculated.
Like for example, distance from [581 925] to next nearest coordinate is found out. The candidates are [ 582 926], [ 582 931] & [ 582 939] since these are the immediate coordinates which are the closest to [581 925].

很明显，在这种情况下，[582 926]是最接近[581 925]的坐标，我只希望该坐标存在并且要删除其他2个候选坐标.所以结果数组应该是

As it's obvious in this case that [582 926] is the nearest coordinate to [581 925], I only want that coordinate to exist and the other 2 candidate coordinates to be deleted. So the resultant array should be

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

现在应该从[582 926]开始执行相同的操作，依此类推，直到结束.

Now same operation should be performed starting from [582 926] and so on till the end.

具有未过滤的坐标的轮廓:

Contour with unfiltered coordinates:

具有已过滤坐标的轮廓:

Contour with filtered coordinates:

由于这是最令人担忧的事情，因此最省时的方法是最方便的pythonic(最好是numpy)方法?

注意:细化线不是问题，只是要消除不必要的点/坐标.

NOTE: Line Thinning is not of concern, only concern is of removing the unnecessary points/coordinates.

推荐答案

我设法做到了:

要使用该方法，首先必须在相等的x轴值的基础上将数组分为多个sup组.请参考这篇文章以了解如何正确应用lexsort.非常感谢 @Divakar 为这些帖子提供了出色的答案.

For the method to work, first the array has to be split into sup groups on the basis of equal x-axis values. Please refer this post for detailed information. I'll be adding the code below though. It's important that the array is sorted in ascending order with respect to x-axis. If it's not, you can do so by applying np.lextsort on the array. Refer this post to understand how to apply lexsort correctly. A huge thanks to @Divakar for providing awesome answers to these posts.

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

输出:

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

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