r - 在R中找到交点

Question

我有 2 个向量：

set.seed(1)
x1 = rnorm(100,0,1)
x2 = rnorm(100,1,1)

我想将它们绘制为线，然后找到线的交点，如果有多个交点，那么我想定位它们中的每一个。

我遇到了一个类似的问题，并尝试使用 来解决这个问题spatstat，但我无法将包含两个向量值的组合数据框转换为psp object.

Answer 1

如果您实际上只有两个随机数字向量，则可以使用一种非常简单的技术来获得两者的交集。只需找到x1高于 的所有点x2，然后在下一个点低于它，反之亦然。这些是交叉点。然后只需使用相应的斜率来找到该段的截距。

set.seed(2)
x1 <- sample(1:10, 100, replace = TRUE)
x2 <- sample(1:10, 100, replace = TRUE)

# Find points where x1 is above x2.
above <- x1 > x2

# Points always intersect when above=TRUE, then FALSE or reverse
intersect.points <- which(diff(above) != 0)

# Find the slopes for each line segment.
x1.slopes <- x1[intersect.points+1] - x1[intersect.points]
x2.slopes <- x2[intersect.points+1] - x2[intersect.points]

# Find the intersection for each segment.
x.points <- intersect.points + ((x2[intersect.points] - x1[intersect.points]) / (x1.slopes-x2.slopes))
y.points <- x1[intersect.points] + (x1.slopes*(x.points-intersect.points))

# Joint points
joint.points <- which(x1 == x2)
x.points <- c(x.points, joint.points)
y.points <- c(y.points, x1[joint.points])

# Plot points
plot(x1,type='l')
lines(x2,type='l',col='red')
points(x.points,y.points,col='blue')

# Segment overlap
start.segment <- joint.points[-1][diff(joint.points) == 1] - 1
for (i in start.segment) lines(x = c(i, i+1), y = x1[c(i, i+1)], col = 'blue')

Answer 2

这是一个替代段-段交叉代码，

# segment-segment intersection code
# http://paulbourke.net/geometry/pointlineplane/
ssi <- function(x1, x2, x3, x4, y1, y2, y3, y4){

  denom <- ((y4 - y3)*(x2 - x1) - (x4 - x3)*(y2 - y1))
  denom[abs(denom) < 1e-10] <- NA # parallel lines

  ua <- ((x4 - x3)*(y1 - y3) - (y4 - y3)*(x1 - x3)) / denom
  ub <- ((x2 - x1)*(y1 - y3) - (y2 - y1)*(x1 - x3)) / denom

  x <- x1 + ua * (x2 - x1)
  y <- y1 + ua * (y2 - y1)
  inside <- (ua >= 0) & (ua <= 1) & (ub >= 0) & (ub <= 1)
  data.frame(x = ifelse(inside, x, NA), 
             y = ifelse(inside, y, NA))

}
# do it with two polylines (xy dataframes)
ssi_polyline <- function(l1, l2){
  n1 <- nrow(l1)
  n2 <- nrow(l2)
  stopifnot(n1==n2)
  x1 <- l1[-n1,1] ; y1 <- l1[-n1,2] 
  x2 <- l1[-1L,1] ; y2 <- l1[-1L,2] 
  x3 <- l2[-n2,1] ; y3 <- l2[-n2,2] 
  x4 <- l2[-1L,1] ; y4 <- l2[-1L,2] 
  ssi(x1, x2, x3, x4, y1, y2, y3, y4)
}
# do it with all columns of a matrix
ssi_matrix <- function(x, m){
  # pairwise combinations
  cn <- combn(ncol(m), 2)
  test_pair <- function(i){
    l1 <- cbind(x, m[,cn[1,i]])
    l2 <- cbind(x, m[,cn[2,i]])
    pts <- ssi_polyline(l1, l2)
    pts[complete.cases(pts),]
  }
  ints <- lapply(seq_len(ncol(cn)), test_pair)
  do.call(rbind, ints)

}
# testing the above
y1 = rnorm(100,0,1)
y2 = rnorm(100,1,1)
m = cbind(y1, y2)
x = 1:100
matplot(x, m, t="l", lty=1)
points(ssi_matrix(x, m))

Answer 3

延迟响应，但这里是使用包 SP 和 RGEOS 的“空间”方法。这要求 x 和 y 都是数字（或可以转换为数字）。投影是任意的，但 epsg:4269 似乎运作良好：

library(sp)
library(rgeos)
# dummy x data
x1 = rnorm(100,0,1)
x2 = rnorm(100,1,1)

#dummy y data 
y1 <- seq(1, 100, 1)
y2 <- seq(1, 100, 1) 

# convert to a sp object (spatial lines)
l1 <- Line(matrix(c(x1, y1), nc = 2, byrow = F))
l2 <- Line(matrix(c(x2, y2), nc = 2, byrow = F))
ll1 <- Lines(list(l1), ID = "1")
ll2 <- Lines(list(l2), ID = "1")
sl1 <- SpatialLines(list(ll1), proj4string = CRS("+init=epsg:4269"))
sl2 <- SpatialLines(list(ll2), proj4string = CRS("+init=epsg:4269"))

# Calculate locations where spatial lines intersect
int.pts <- gIntersection(sl1, sl2, byid = TRUE)
int.coords <- int.pts@coords

# Plot line data and points of intersection
plot(x1, y1, type = "l")
lines(x2, y2, type = "l", col = "red")
points(int.coords[,1], int.coords[,2], pch = 20, col = "blue")

Answer 4

我需要另一个应用程序的交集，发现 nograpes 的答案无法正常工作：

# another example
x=seq(-4,6,length.out=10)
x1=dnorm(x, 0, 1)
x2=dnorm(x,2,2)

# Find points where x1 is above x2.
above <- x1 > x2

# Points always intersect when above=TRUE, then FALSE or reverse
intersect.points <- which(diff(above) != 0)

# Find the slopes for each line segment.
x1.slopes <- x1[intersect.points+1] - x1[intersect.points]
x2.slopes <- x2[intersect.points+1] - x2[intersect.points]

# Find the intersection for each segment.
x.points <- x[intersect.points] + ((x2[intersect.points] - x1[intersect.points]) / (x1.slopes-x2.slopes))
y.points <- x1[intersect.points] + (x1.slopes*(x.points-x[intersect.points]))

# Joint points
joint.points <- which(x1 == x2)
x.points <- c(x.points, joint.points)
y.points <- c(y.points, x1[joint.points])

# Plot points 
# length(x); length(x1)
plot(x, x1,type='l')
lines(x, x2,type='l',col='red')
points(x.points,y.points,col='blue')

对于这些双正态分布，这些点是关闭的，在这种情况下，尤其是右手边的交点。当 x 轴上的值不是连续整数并且因此连续点的差不为 1 时，就会发生这种情况。我替换intersect.points为x[intersect.points]，这还不够。这是一个遗憾，因为与其他方法相比，该方法相对简单。baptiste 提供的方法效果更好：

m = cbind(x1, x2)
matplot(x, m, t="l", lty=1)
points(ssi_matrix(x, m))

遵循相同的想法，允许连续 x 值 != 1 的差异的更一般的实现是：

intersect.2lines <- function (x, y1, y2){
  above = y1 > y2
  intersect.points <- which(diff(above) != 0) 
  
  y1.diff <- y1[intersect.points+1] - y1[intersect.points]
  y2.diff <- y2[intersect.points+1] - y2[intersect.points]
  x.diff <- x[intersect.points+1]-x[intersect.points]
  
  slope1 = y1.diff/x.diff
  slope2 = y2.diff/x.diff
  intercept1 = y1[intersect.points]-slope1*x[intersect.points]
  intercept2 = y2[intersect.points]-slope2*x[intersect.points]
  x.points = ifelse(slope1==slope2, NA, 
                  (intercept2-intercept1)/(slope1-slope2))
  y.points = ifelse(slope1==slope2, NA,
                  slope1*x.points+intercept1)
  # Joint points
  joint.points <- which(y1 == y2)
  x.points <- c(x.points, x[joint.points])
  y.points <- c(y.points, y1[joint.points])
  return(data.frame(x.points=x.points, y.points=y.points))
}

它是维基百科“给定两条线方程”中给出的公式的实现

现在的结果与巴蒂斯特方法产生的结果相同。