如何计算由包含x，y的矩阵定义的两点之间的欧几里得距离? [英] How to calculate Euclidian distance between two points defined by matrix containing x, y?

问题描述

欧几里得距离计算让我很失落.我发现函数dist2 {SpatialTools}或rdist {fields}可以做到这一点，但是它们不能按预期工作.

I am very lost in Euclidean distance calculation. I have found functions dist2{SpatialTools} or rdist{fields} to do this, but they doesn´t work as expected.

我想一个点在笛卡尔系统中有两个坐标，所以[x，y]. 要测量2点之间的距离(按行定义)，我需要2点的4个坐标，所以 点A:[x1，y1] B点:[x2，y2]

I suppose that one point has two coordinates in carthesian system, so [x,y]. To measure distance between 2 points (defined by row), I need 4 coordinates for 2 points, so point A: [x1,y1] point B: [x2,y2]

要点协调:

A[0,1]
B[0,0] 
C[1,1]
D[1,1]

我有两个矩阵: x1 (A和C在那，由行定义)和 x2 (包含B和D).写入矩阵:

I have two matrices: x1(A and C are there, defined by rows) and x2 (contain B and D). Written in matrix:

library("SpatialTools")
x1<-matrix(c(0,1,1,1), nrow = 2, ncol=2, byrow=TRUE)
x2<-matrix(c(0,0,1,1), nrow = 2, ncol=2, byrow=TRUE)

所以我得到

> x1
     [,1] [,2]
[1,]    0    1    #(as xy coordinates of A point)
[2,]    1    1    #(same for C point)

> x2
     [,1] [,2]
[1,]    0    0    #(same for B point)
[2,]    1    1    #(same for D point)

计算之间的欧式距离

A <-> B  # same as x1[1,] <-> x2[1,]
C <-> D  # same as x1[2,] <-> x2[2,]

我假设要获取 EuclidDist :

> x1                           x2                         EuclidDist
     [,1] [,2]                      [,1] [,2]
[1,]    0    1    #A         [1,]    0    0    #B             1
[2,]    1    1    #B         [2,]    1    1    #D             0

我只想获取由[x，y]坐标标识的两点之间的距离的矢量 ，但是，使用dist2我可以获得一个矩阵:

I would like just to obtain vector of distances between two points identified by [x,y] coordinates, however, using dist2 I obtain a matrix:

> dist2(x1,x2)
         [,1] [,2]
[1,] 1.000000    1
[2,] 1.414214    0

我的问题是，根据该矩阵，哪些数字描述了A-B和C-D之间的实际欧几里得距离?我误会了吗?非常感谢您的任何建议或任何解释.

My question is, which numbers describe the real Euclidean distance between A-B and C-D from this matrix? Am I misunderstanding something? Thank you very much for every advice or any explanation.

推荐答案

如果您只想使用矢量，类似的方法将对您有用.

If you just want a vector, something like this will work for you.

尝试这样的事情:

euc.dist <- function(x1, x2) sqrt(sum((x1 - x2) ^ 2))

library(foreach)
foreach(i = 1:nrow(x1), .combine = c ) %do% euc.dist(x1[i,],x2[i,])

这将适用于任何尺寸.

如果不想使用foreach，可以使用一个简单的循环:

If you don't want to use foreach, you can use a simple loop:

dist <- NULL
for(i in 1:nrow(x1)) dist[i] <- euc.dist(x1[i,],x2[i,])
dist

尽管如此，我还是建议foreach(因为对于像这样的各种任务来说非常容易).在软件包的文档中详细了解它.

Although, I would recommend foreach (because it's very easy to for various tasks like this). Read more about it in the documentation of the package.

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