本文介绍了在给定2D数字“高度图"的情况下, R中的矩阵,如何找到所有局部最大值?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我有一个带有非负数值的R矩阵.该矩阵实际上是2D高度图,我想在此矩阵中找到所有局部最大值.对于平坦"峰,相邻元素彼此相等(并且它们共同是一个局部最大值),只要我在每个平坦"区域内获得至少一个坐标,我就不会在意什么.

I have an R matrix with nonnegative numeric values. The matrix is effectively a 2D height map, and I want to find all the local maxima in this matrix. For "flat" peaks, where neighboring elements are equal to each other (and they are collectively a local maximum), I don't care what happens as long as I get at least one coordinate within each "flat" region.

是否有任何功能可以有效地做到这一点?显然,我可以手动编写循环以逐个进行每个元素的测试,但是在R中这样做会很慢.我需要对大约一百万个矩阵执行此操作,每个矩阵平均需要884个元素.

Are there any functions to do this efficiently? Obviously I could write the loop manually to go through and test every element individually, but doing that in R would be quite slow. I need to do this for around a million matrices, with an average of about 884 elements per matrix.

理想情况下,会有一个函数将矩阵作为输入并返回一个2列矩阵,其中第1列是行坐标,第2列是列坐标,并且矩阵中的每个局部最大值都包含一行.

Ideally there would be a function that takes the matrix as input and returns a 2-column matrix with column 1 being row coordinates, column 2 being the column coordinates, and one row for each local maximum in the matrix.

允许矩阵边缘上的局部最大值.矩阵外部的区域可以视为高度为零.

Local maxima on the edges of the matrix are allowed. Areas outside of the matrix can be treated as having zero height.

要使用的可复制示例矩阵:

Reproducible example matrix to use:

set.seed(5)
msize <- 20 # Change this to whatever you like
x <- matrix(data=abs(rnorm(msize*2)), nrow=msize, ncol=msize)

推荐答案

raster包中的focal()函数旨在用于此类计算.以下代码返回 all 局部最大值的坐标,包括边缘上的坐标和属于高原"部分的坐标.

The focal() function in the raster package is designed for calculations like this. The following code returns coordinates of all local maxima, including those on edges and those that are parts of "plateaus ".

library(raster)

## Construct an example matrix
set.seed(444)
msize <- 10
x <- matrix(sample(seq_len(msize), msize^2, replace=TRUE), ncol=msize)

## Convert it to a raster object
r <- raster(x)
extent(r) <- extent(c(0, msize, 0, msize) + 0.5)

## Find the maximum value within the 9-cell neighborhood of each cell
f <- function(X) max(X, na.rm=TRUE)
ww <- matrix(1, nrow=3, ncol=3) ## Weight matrix for cells in moving window
localmax <- focal(r, fun=f, w=ww, pad=TRUE, padValue=NA)

## Does each cell have the maximum value in its neighborhood?
r2 <- r==localmax

## Get x-y coordinates of those cells that are local maxima
maxXY <- xyFromCell(r2, Which(r2==1, cells=TRUE))
head(maxXY)
#       x  y
# [1,]  8 10
# [2,] 10 10
# [3,]  3  9
# [4,]  4  9
# [5,]  1  8
# [6,]  6  8

# Visually inspect the data and the calculated local maxima
plot(r)   ## Plot of heights
windows() ## Open a second plotting device
plot(r2)  ## Plot showing local maxima

这篇关于在给定2D数字“高度图"的情况下, R中的矩阵,如何找到所有局部最大值?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

08-20 20:31