来自 3d 数组 R 的 n 个第一个邻居的列表

Question

假设我们有一个 3d 数组：

my.array <- array(1:27, dim=c(3,3,3))

我想创建一个包含前 n 个邻居的列表。

例子：让my.array[2,2,2]=14，所以14的第一个邻居是：

list[14] = [1 to 27] - 14

我也想使用 R、C 或 Matlab 对第二、第三、n 个最近的邻居做同样的事情。

谢谢

Answer 1

基于 cmets，我假设您将“第一最近邻居”定义为欧几里德距离为 1 或以下（不包括自身）的所有单元格，将“第二最近邻居”定义为距离为 2 或以下的所有单元格，等等。您的断言在@evan058's answer 的评论中，“对于（1,1,1）第一级邻居是 2,4,5,10,11,13”，我实际上将其解释为包括直接对角线（距离为 1.414）但不是更远的对角线（在您的示例中，14 将是距离为 1.732 的更远的对角线）。

此函数接受预定义数组 (ary) 或生成数组的维度 (dims)。

nearestNeighbors(dims = c(3,3,3), elem = c(1,1,1), dist = 1)
#      dim1 dim2 dim3
# [1,]    2    1    1
# [2,]    1    2    1
# [3,]    1    1    2
nearestNeighbors(dims = c(3,3,3), elem = c(1,1,1), dist = 1,
                 return_indices = FALSE)
# [1]  2  4 10
nearestNeighbors(dims = c(3,3,3), elem = c(1,1,1), dist = 2,
                 return_indices = FALSE)
#  [1]  2  3  4  5  7 10 11 13 14 19

nearestNeighbors(ary = array(27:1, dim = c(3,3,3)), elem = c(1,1,1), dist = 2)
#       dim1 dim2 dim3
#  [1,]    2    1    1
#  [2,]    3    1    1
#  [3,]    1    2    1
#  [4,]    2    2    1
#  [5,]    1    3    1
#  [6,]    1    1    2
#  [7,]    2    1    2
#  [8,]    1    2    2
#  [9,]    2    2    2
# [10,]    1    1    3
nearestNeighbors(ary = array(27:1, dim = c(3,3,3)), elem = c(1,1,1), dist = 2,
                 return_indices = FALSE)
#  [1] 26 25 24 23 21 18 17 15 14  9

功能：

#' Find nearest neighbors.
#'
#' @param ary array
#' @param elem integer vector indicating the indices on array from
#'   which all nearest neighbors will be found; must be the same
#'   length as \code{dims} (or \code{dim(ary)}). Only one of
#'   \code{ary} and \code{dim} needs to be provided.
#' @param dist numeric, the max distance from \code{elem}, not
#'   including the 'self' point.
#' @param dims integer vector indicating the dimensions of the array.
#'   Only one of \code{ary} and \code{dim} needs to be provided.
#' @param return_indices logical, whether to return a matrix of
#'   indices (as many columns as dimensions) or the values from
#'   \code{ary} of the nearest neighbors
#' @return either matrix of indices (one column per dimension) if
#'   \code{return_indices == TRUE}, or the appropriate values in
#'   \code{ary} otherwise.
nearestNeighbors <- function(ary, elem, dist, dims, return_indices = TRUE) {
  if (missing(dims)) dims <- dim(ary)
  tmpary <- array(1:prod(dims), dim = dims)
  if (missing(ary)) ary <- tmpary
  if (length(elem) != length(dims))
      stop("'elem'' needs to have the same dimensions as 'ary'")
  # work on a subset of the whole matrix
  usedims <- mapply(function(el, d) {
    seq(max(1, el - dist), min(d, el + dist))
  }, elem, dims, SIMPLIFY=FALSE)
  df <- as.matrix(do.call('expand.grid', usedims))
  # now, df is only as big as we need to possibly satisfy `dist`
  ndist <- sqrt(apply(df, 1, function(x) sum((x - elem)^2)))
  ret <- df[which(ndist > 0 & ndist <= dist),,drop = FALSE]
  if (return_indices) {
    return(ret)
  } else {
    return(ary[ret])
  }
}

编辑：更改代码以“轻微”提高速度：使用 256x256x256 数组和 2 距离之前在我的机器上花费了大约 90 秒。现在只需不到 1 秒。即使是 5 的距离（相同的阵列）也需要不到一秒的时间。 未完全测试，请验证是否正确。

编辑：移除函数第 50 行多余的 {。

Answer 2

我认为按照这些思路可以解决问题：

nClosest <- function(pts, pt, n)
{
  # Get the target value
  val <- pts[pt[1], pt[2], pt[3]]
  # Turn the matrix into a DF
  ptsDF <- adply(pts, 1:3)
  # Create Dist column for distance to val
  ptsDF$Dist <- abs(ptsDF$V1 - val)
  # Order by the distance to val
  ptsDF <- ptsDF[with(ptsDF, order(Dist)),]
  # Split into groups:
  sp <-  split(ptsDF, ptsDF$Dist)
  # Get max index
  topInd = min(n+1, length(sp))
  # Agg the split dfs into a single df
  rbind.fill(sp[2:topInd])
}

输出：

> nClosest(my.array, c(1,2,2), 3)
  X1 X2 X3 V1 Dist
1  3  1  2 12    1
2  2  2  2 14    1
3  2  1  2 11    2
4  3  2  2 15    2
5  1  1  2 10    3
6  1  3  2 16    3