在数组中查找不等于给定点的最近点

Question

我有一个点和一组值。我希望在数组中找到最近的点 y，使 x != y。目前我在 Python 中使用标准的 Scipy KDTree 方法，但这并不仅仅考虑唯一点。是否有任何解决方法或是否需要手动实施？谢谢！

Answer 1

二维数组解法：我回收了大部分与以前相同的方法，但这次它应该适用于您的结构

import math

X = [0,0] # Your cooridinate
x1 = X[0]
y1= X[1]

array = [[1,1],[0,1],[1,0],[-2,2]] # Sample array of data
smallestDistance = 9999 # Make it big so it can be replaced immediately, essentially just a placeholder

for point in array:
    x2 = point[0]
    y2 = point[1]
    separation = math.hypot(x2 - x1, y2 - y1) #Distance equation in easy format
    if separation < smallestDistance and separation!=0:  # Could make this <= instead of < if you want to replace any ties for closest point
        smallestDistance = separation
        closestPoint = point
print(f"Nearest cooridinate = {closestPoint}, separation from {X} is: {smallestDistance}")

一维数组的旧解决方案：我不太确定你的数组结构是什么，但据我所知你有一个一维数组和一个给定的值；您希望数组中最接近原始值的唯一值。如果这是正确的，那么快速而肮脏的解决方案可能类似于：

smallestDifference = 99999999 #Make this some large number
y = array[someindex]
for thing in array:
     difference = abs(thing-y)
          if difference<smallestDifference & difference != 0:  #Must be closer than the last recorded closest point, AND non-zero i.e. a unique point
               smallestDifference = difference #New smallest distance between points
               closestPoint = thing #Record the new closest point
print(f"x = {closestPoint}, |y-x| = {smallestDifference}")

如果这是一个 >1d 数组可能会有点棘手并且可能不是最快的解决方案，因为它必须检查每个点..但可能会完成工作:-)

在数组中查找不等于给定点的最近点

Finding nearest point in an array that is not equal to given point

python

kdtree

nearest-neighbor

scipy