从距离到给定点x，y的点排序在我的情况下（x = 0，y = o）

Question

我想排序（最短到最长）数组“ a”（如下所示）到距原点或点（在我的情况下为0,0）的距离，并将其存储到类似的数组类型“ b”或替换数组“ a”

以下给定的点是3d numpy数组

[[[  510.    11.]]

 [[  651.   276.]]

 [[  269.    70.]]

 [[  920.    26.]]

 [[  513.    21.]]

 [[ 1197.   620.]]

 [[  407.   268.]]

 [[  452.    35.]]

 [[  435.     3.]]

 [[  520.    20.]]

 [[ 1151.   499.]]

 [[  104.    26.]]

 [[  754.    28.]]

 [[  263.   111.]]

 [[  731.    12.]]

 [[  972.   200.]]

 [[ 1186.   614.]]

 [[  437.     2.]]

 [[ 1096.    68.]]

 [[  997.   201.]]

 [[ 1087.   200.]]

 [[  913.   201.]]

 [[ 1156.   510.]]

 [[  994.   230.]]

 [[  793.    29.]]

 [[  514.    19.]]]

我找不到有关此类3d np阵列排序的任何有用信息

ps：这些点“ a”是从Goodfeaturestotrack，OPEN CV和python 3.6获得的

以及如何将数组清除为Null类型？

 #this is  clustering algorithm    
    for index in range(len(a): #a is the above matrix 3d np array
#find distance was already defined and is euclidean distance formula
            if findDistance(a[index][0], a[index][1], a[index + 1][0], a[index + 1][1]) < 3: #calculation euclidean distance between ai and ai+1
                c.append(index)
            if findDistance(a[index][0], a[index][1], a[index + 1][0], a[index + 1][1]) > 3: #calculation euclidean distance between ai and ai+1
                if len(c) > 10:
                    cp = np.insert(cp, c, 0)

                    c = [] # should clear c **is this correct ??**

Answer 1

我想方便地使用这种方法来计算几种数组格式的距离……它不是单线的，但它可以工作。 通过搜索“ einsum”和numpy作为关键字，可以在堆栈的其他位置找到有关其实现的详细信息。 导入numpy作为np所需，这只是def并且您需要2个数组

import numpy as np

def e_dist(a, b, metric='euclidean'):

"""Distance calculation for 1D, 2D and 3D points using einsum
: a, b   - list, tuple, array in 1,2 or 3D form
: metric - euclidean ('e','eu'...), sqeuclidean ('s','sq'...),
:-----------------------------------------------------------------------
"""
a = np.asarray(a)
b = np.atleast_2d(b)
a_dim = a.ndim
b_dim = b.ndim
if a_dim == 1:
    a = a.reshape(1, 1, a.shape[0])
if a_dim >= 2:
    a = a.reshape(np.prod(a.shape[:-1]), 1, a.shape[-1])
if b_dim > 2:
    b = b.reshape(np.prod(b.shape[:-1]), b.shape[-1])
diff = a - b
dist_arr = np.einsum('ijk,ijk->ij', diff, diff)
if metric[:1] == 'e':
    dist_arr = np.sqrt(dist_arr)
dist_arr = np.squeeze(dist_arr)
return dist_arr

然后，您可以根据需要对结果进行排序，例如

a = np.random.randint(0, 10, size=(10,2))

orig = np.array([0,0])

    e_dist(a, orig)

    array([  4.12,   9.9 ,   7.07,   6.08,   3.16,  10.63,   8.54,   7.28,   7.21,
             6.08])
    np.sort(e_dist(a, orig))

    array([  3.16,   4.12,   6.08,   6.08,   7.07,   7.21,   7.28,   8.54,   9.9 ,
            10.63])

附录

我应该补充说，您可以使用argsort获得排序后的值，如下所示

np.argsort(e_dist(a, orig))
array([4, 0, 3, 9, 2, 8, 7, 6, 1, 5], dtype=int64)

idx = np.argsort(art.e_dist(a, orig))

closest = a[idx]
array([[3, 1],
       [1, 4],
       [1, 6],
       [6, 1],
       [5, 5],
       [4, 6],
       [2, 7],
       [8, 3],
       [7, 7],
       [7, 8]])

Answer 2

def distance_squared(x1,y1,x2,y2):
    return (x1-x2)**2 + (y1-y2)**2
target_point = 0,0
sorted(a,key=lambda point:distance_squared(target_point[0],target_point[1],*point[0]))