Question

我想运行一个给出两点之间距离的函数。我想计算所有点之间的距离，我该怎么做。我知道可以使用if或for来完成，但我使用它们并不是很好。

我的功能是：

Distance<- function(x,y)    {
  round(sqrt(sum((x - y) ^ 2)),3)
}

我在东部和北部或x＆amp;以上功能：

easting=rbind(609027, 609282, 609501,609497,609405,609704,609718,610277,610530,610573,609875,608947,609865,611105,611169,611243,611388,611598,611339,611310,611212,611150,611358,611626,611763,611887,612043,612134,612160,612539,612857,613062,613154,613303)
northing=rbind(1534293,1534470,1534630,1534848,1534027,1535054,1535315,1535583,1535717,1536254,1536351,1536700,1536746,1536762,1537003,1537261,1537489,1537685,1537838,1538103,1538500,1538812,1539217,1539342,1539627,1539842,1540027,1540357,1540628,1540911,1541623,1541896,1542117,1542494)

如果coords<-as.data.frame(easting,northing)是我的数据集，那么我想计算coords[i,]和coords[j,]之间的距离。其中i，j是数据集中的行。

由于

Answer 1

首先，您需要更改有关如何创建data.frame的详细信息。不要使用easting定义向量northing和cbind，而是使用c。然后使用data.frame，而不是as.data.frame。

easting = c(609027, 609282, 609501,609497,609405,609704,609718,610277,610530,610573,609875,608947,609865,611105,611169,611243,611388,611598,611339,611310,611212,611150,611358,611626,611763,611887,612043,612134,612160,612539,612857,613062,613154,613303)
northing = c(1534293,1534470,1534630,1534848,1534027,1535054,1535315,1535583,1535717,1536254,1536351,1536700,1536746,1536762,1537003,1537261,1537489,1537685,1537838,1538103,1538500,1538812,1539217,1539342,1539627,1539842,1540027,1540357,1540628,1540911,1541623,1541896,1542117,1542494)
coords <- data.frame(easting, northing)

现在，为了使用函数apply，您还需要更改函数，让它接受一个向量作为参数。

Distance<- function(x, y)    {
  round(sqrt(sum((x - y) ^ 2)),3)
}

使用嵌套的for循环

d <- numeric(34^2)
k <- 0
for(i in seq_len(nrow(coords)))
    for(j in seq_len(nrow(coords))){
        k <- k + 1
        d[k] <- Distance(coords[i, ], coords[j, ])
    }

Answer 2

您可以使用dist功能：

df  <- data.frame(easting=easting,northing = northing)
dist(df) # or round(dist(df,upper=T,diag=T),3)

前三行的示例：

round(dist(df[1:3,], upper=T,diag=T),3)

        1       2       3
1   0.000 310.409 581.588
2 310.409   0.000 271.221
3 581.588 271.221   0.000

比较：

round(dist(df[1:3,]),3)

        1       2
2 310.409        
3 581.588 271.221

点矩阵之间的距离，简单如果＆amp;对的

2 个答案: