r - 如何在 Delaunay 三角剖分中设置三角形边的最大长度？

Question

如何从 Delaunay 三角测量中删除大于我需要的距离？

示例数据：

x<-rep(1:12, c(2,2,7,9,10,5,4,6,10,10,9,4))
y<-c(1,2,1,2,1:3,5:8,1:9,1:10,2,7:10,8:11,7:12,3:12,3:12,4:12,5,8:10)
x_plus<-seq(0.2:0.8, by=0.1)
x<-x+sample(x_plus, 78, replace=TRUE)
y<-y+sample(x_plus, 78, replace=TRUE)

绘制地图：

plot(x,y)

使用 tri.mesh() 进行Delaunay 三角剖分- 包（tripack）

my.triangles<-tri.mesh(x,y)
plot(my.triangles, do.points=FALSE, lwd=0.2)
points(x,y, col = "blue", pch=20)

如何仅提取较短的距离？我不需要那些大的，你肯定知道我的意思是什么距离。在 tri.mesh() 函数中是否有一些论点可以做到这一点？或者它可以在它之后完成？

距离甚至存储在这个对象中吗？

my.triangles

triangulation nodes with neigbours:
node: (x,y): neighbours
1: (1.4,1.7) [5]: 2 3 4 11 12 
2: (2,3) [6]: 1 4 7 8 9 11 
3: (3,1.8) [4]: 1 4 5 12 
.
.
.
76: (12.4,8.8) [5]: 68 69 70 75 77 
77: (12.9,9.9) [6]: 70 71 72 75 76 78 
78: (13,11) [4]: 72 73 74 77 
number of nodes: 78 
number of arcs: 221 
number of boundary nodes: 10 
boundary nodes:  1 11 12 45 56 66 74 75 77 78 
number of triangles: 144 
number of constraints: 0 

Answer 1

The documentation (?tri) suggests that the segments are of the form r$tlist[k] -- r$tlist[r$tlptr[k]]: you can remove those beyond some threshold.

r <- tri.mesh(x,y)
k <- seq_len( r$tlnew - 1 )
i <- r$tlist[k]          
j <- r$tlist[r$tlptr[k]]
keep <- i > 0
i <- abs( i[ keep ] )
j <- abs( j[ keep ] )
plot( x, y )
segments( r$x[i], r$y[i], r$x[j], r$y[j], col="grey" )
distances <- sqrt( ( r$x[i] - r$x[j] ) ^ 2 + ( r$y[i] - r$y[j] ) ^ 2 )
threshold <- 2.5  # Choose the threshold manually
i <- i[ distances < threshold ]
j <- j[ distances < threshold ]
segments( r$x[i], r$y[i], r$x[j], r$y[j], lwd = 2 )