这是使用 RANN的一种方式软件包。该方法类似于此帖子中显示的方法，但是适用于单点集（链接的帖子大约是找到集合A中与集合B中每个点最近的点。

Here's one way, using the RANN package. The approach is similar to that shown in this post, but is adapted for a single set of points (the linked post was about finding the nearest point in set A to each point in set B).

xy <- read.table(text='point x         y
1     1601774   14544454
2     1616574   14579422
3     1608698   14572922
4     1602948   14572990
5     1607355   14573871
6     1615336   14578178
7     1603398   14574495
8     1605153   14570727
9     1606758   14573845
10    1606655   14570953', header=TRUE, row.names=1)

library(RANN)
closest <- nn2(data=xy, k=2)[[1]]

以上，我们将您的单点集 xy 提供给 data 参数，并指定我们要 nn2 来找到每个点两个最近的点（因为最近的点是焦点本身）。 nn2 函数返回包含两个元素的列表：每个 k $ c $的索引的向量（在这种情况下为矩阵） c>最近的点（对于每个查询点）；以及距离的向量（矩阵）。我假设我们对距离不感兴趣，所以在上面我们将结果子集到第一个元素。

Above, we supply your single set of points, xy, to the data argument, and specify that we want nn2 to find the two nearest points to each point (because the nearest point is the focal point itself). The nn2 function returns a list with two elements: a vector (matrix, in this case) of indices of each of the k nearest points (for each queried point); and a vector (matrix) of the distances. I'm assuming we're not interested in the distances, so above we subset the result to the first element.

对于我们的问题，结果是一个两列给出第一列中查询点的索引和第二列中最近点的索引的矩阵。

For our problem, the result is a two-column matrix giving the index of the queried point in the first column and the index of the nearest point in the second.

closest

##       [,1] [,2]
##  [1,]    1    8
##  [2,]    2    6
##  [3,]    3    5
##  [4,]    4    7
##  [5,]    5    9
##  [6,]    6    2
##  [7,]    7    4
##  [8,]    8   10
##  [9,]    9    5
## [10,]   10    8

要获得最近点的坐标矩阵，可以使用：

To get a matrix of coordinates of the nearest points, you could use:

xy[closest[, 2], ]

默认情况下 nn2 使用一棵kd树-您可能要尝试使用 treetype ='bd'。

By default nn2 uses a kd tree - you might want to experiment with treetype='bd'.