在R中找到直线和三角形之间的交点

Question

我试图找到路径和三角形的交点。输出可以是具有一组与三角形相交的路径的经度或纬度点的简单矩阵。这些点可以在图中突出显示。谁能帮我解决这个问题？

# path 01 
longitude <- round(c(48.7188021250007,
                 48.7188133749999,
                 48.7188291249998,
                 48.7188336250004), 8);

Latitude <- round(c (2.39514523661229,
                 2.39512477447308, 
                 2.39472235230961,
                 2.39467460730113), 8);

# path 02
longitude_2 <- round(c(48.71881,
                   48.71882,
                   48.71883,
                   48.71884), 8);

Latitude_2 <- round(c (2.39479,
                   2.3947,
                   2.39469396980779,
                   2.3945), 8);

plot(longitude,Latitude,ylim=range(c(Latitude,Latitude_2)),
xlim=range(c(longitude,longitude_2)), type="l",col="blue")
lines(longitude_2,Latitude_2,col="darkgreen")
title(main = "Two different paths");
legend("right", legend=c("path A", "path B"),
   col=c("blue", "darkgreen"), lty=1:1, cex=0.8, 
   box.lty=0)


# Drawing polygon 

lon <- c(48.71882,48.71883, 48.71884);
lat <- c (2.3945, 2.39478,  2.39479);
x <- cbind(lon, lat)
polygon(x, col='lightblue')

Answer 1

我认为您可以在创建 SpatialLines 和 SpatialPolygons 后使用 rgeos 包中的 gIntersection 函数。

require(sp)
require(rgeos)
longitude <- round(c(48.7188021250007,
                     48.7188133749999,
                     48.7188291249998,
                     48.7188336250004), 8);

Latitude <- round(c (2.39514523661229,
                     2.39512477447308, 
                     2.39472235230961,
                     2.39467460730113), 8);

longitude_2 <- round(c(48.71881,
                       48.71882,
                       48.71883,
                       48.71884), 8);

Latitude_2 <- round(c (2.39479,
                       2.3947,
                       2.39469396980779,
                       2.3945), 8);

lon <- c(48.71882,48.71883, 48.71884);
lat <- c (2.3945, 2.39478,  2.39479);


lines <- SpatialLines(list(Lines(list(Line(cbind(longitude, Latitude)),
                                      Line(cbind(longitude_2, Latitude_2))), 1)))

pol_1 <- SpatialPolygons(list(Polygons(list(Polygon(coords = cbind(lon, lat))),1)))

result <- gIntersection(lines, pol_1)

plot(lines, col = "red")
plot(pol_1, col = "blue", add = T)
plot(result, col = "green", add = T)

或者，如果您想要具有正确缩放比例的相同绘图，您可以使用以下方法重新设计它：

plot(x = longitude,
     y = Latitude,
     ylim = range(c(Latitude,Latitude_2)),
     xlim = range(c(longitude,longitude_2)), 
     type="l",
     col="blue")
lines(longitude_2,Latitude_2,
      col = "darkgreen")
polygon(pol_1@polygons[[1]]@Polygons[[1]]@coords, 
        col = 'lightblue')
title(main = "Two different paths");
legend("right", 
       legend=c("path A", "path B", "Intersect"),
       col = c("blue", "darkgreen", "red"), 
       lty = 1:1, 
       cex = 0.8, 
       box.lty = 0)

lines(result@lines[[1]]@Lines[[1]]@coords, col = "red")
lines(result@lines[[1]]@Lines[[2]]@coords, col = "red")
points(result@lines[[1]]@Lines[[1]]@coords, col = "red")
points(result@lines[[1]]@Lines[[2]]@coords, col = "red")
text(result@lines[[1]]@Lines[[1]]@coords, labels = pasteCols(round(t(result@lines[[1]]@Lines[[1]]@coords),4), sep = " "), cex= 0.7)
text(result@lines[[1]]@Lines[[2]]@coords, labels = pasteCols(round(t(result@lines[[1]]@Lines[[1]]@coords),4), sep = " "), cex= 0.7)

在R中找到直线和三角形之间的交点

Finding intersection points between a line and a triangle in R

r

gps

geospatial

intersection