在R中找到交点 [英] finding point of intersection in R

问题描述

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，然后找到线的相交点，如果有多个相交点，那么我想找到它们中的每一个.

I want to plot these as lines and then find the intersection points of the lines, also if there are multiple points of intersection then I want to locate each of them.

我遇到了类似的问题，并尝试使用spatstat解决此问题，但是我无法将包含两个矢量值的组合数据帧转换为psp object.

I have come across a similar question,and tried to solve this problem using spatstat, but I was not able to convert my combined data frame containing both vector values to psp object.

推荐答案

如果您实际上只有两个随机的数字向量，则可以使用一种非常简单的技术来获取两者的交集.只需找到x1在x2之上的所有点，然后在下一个点在其之下，反之亦然.这些是交点.然后只需使用相应的坡度查找该路段的截距即可.

If you literally just have two random vectors of numbers, you can use a pretty simple technique to get the intersection of both. Just find all points where x1 is above x2, and then below it on the next point, or vice-versa. These are the intersection points. Then just use the respective slopes to find the intercept for that segment.

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')

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