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。
示例:让我们得到 my.array[2,2,2]=14,因此 14 的第一个邻居是:
list[14] = [1 to 27] - 14
我还想使用 R、C 或 Matlab 对第二个、第三个、n 个最近的邻居做同样的事情。
谢谢
最佳答案
根据评论,我假设您将“第一最近邻居”定义为欧几里得距离为 1 或以下的所有单元格(不包括自身),“第二最近邻居”为欧几里德距离为 2 或以下的单元格,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1.414),但不是进一步的对角线(在您的示例中,14 将是距离为 1.732 的进一步的对角线)。
此函数接受预定义数组 (ary
) 或构成一个数组的维度 (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 数组和距离 2 之前在我的机器上花费了约 90 秒。现在只需不到 1 秒。即使距离为 5(同一阵列)也只需不到一秒。 尚未完全测试,请验证其是否正确。
编辑:删除了函数第 50 行多余的 {。
关于arrays - 3d 数组 R 中的 n 个前邻居列表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37935323/