R以固定的方位角找到线和点之间的距离[重复]

Question

这是我的可重现示例

########################################

library(sf)

# matrix of lon lat for the definition of the linestring
m<-rbind(
  c(12.09136, 45.86471),
  c(12.09120, 45.86495),
  c(12.09136, 45.86531),
  c(12.09137, 45.86540),
  c(12.09188, 45.86585),
  c(12.09200, 45.86592),
  c(12.09264, 45.86622),
  c(12.09329, 45.86624),
  c(12.09393, 45.86597),
  c(12.09410, 45.86585),
  c(12.09423, 45.86540),
  c(12.09411, 45.86495),
  c(12.09393, 45.86471),
  c(12.09383, 45.86451),
  c(12.09329, 45.86414),
  c(12.09264, 45.86413),
  c(12.09200, 45.86425),
  c(12.09151, 45.86451),
  c(12.09136, 45.86471)
)

# define a linestring
ls<-st_linestring(m)

# create a simple feature with appropriate crs
ls<-st_sfc(ls, crs=4326)

# and now again going through the very same 
# definition process for a point

# define a point 
pt <- st_point(c(12.09286,45.86557))

# crate simple feature with appropriate crs
pt<-st_sfc(pt, crs = 4326)

plot(ls)
plot(pt, add=TRUE)

# this is computing the minimum distance from the point to the line
st_distance(ls, pt)


###############

鉴于上面提到的玩具数据集，我需要找到一个合适的方法来计算：

1 - 线的每个顶点到给定点的距离：这可能很容易通过勾股定理的简单应用计算每对点（线顶点与点）之间的距离来完成，即使我对此很怀疑，因为正在使用 crs（即 epsg 4326，以度为单位），所以我可能需要首先将整个数据集转换为另一个参考系统（使用公制单位）...

2 - 以固定方位角（10°、20°、30°、....、360° 从北方）的点和线之间的距离：这就是我真正迷路的地方.. ..

请给我一些帮助，以便正确进行计算，可能使用我现在正在尝试熟悉的“sf”标准

谢谢

Answer 1

感谢你为我指明了正确的方向

为了完整起见，我制定了我在此处发布的最终解决方案

# my reproducible example

library(sf)

# matrix of lon lat for the definition of the linestring
m<-rbind(
  c(12.09136, 45.86471),
  c(12.09120, 45.86495),
  c(12.09136, 45.86531),
  c(12.09137, 45.86540),
  c(12.09188, 45.86585),
  c(12.09200, 45.86592),
  c(12.09264, 45.86622),
  c(12.09329, 45.86624),
  c(12.09393, 45.86597),
  c(12.09410, 45.86585),
  c(12.09423, 45.86540),
  c(12.09411, 45.86495),
  c(12.09393, 45.86471),
  c(12.09383, 45.86451),
  c(12.09329, 45.86414),
  c(12.09264, 45.86413),
  c(12.09200, 45.86425),
  c(12.09151, 45.86451),
  c(12.09136, 45.86471)
)

# define the linestring
ls<-st_linestring(m)

# create a simple feature linestring with appropriate crs
ls<-st_sfc(ls, crs=4326)

# and now again going through the very same 
# definition process for a point

# define the origin point 
pt <- st_point(c(12.09286,45.86557))

# create simple feature point with appropriate crs
pt<-st_sfc(pt, crs = 4326)

plot(ls)
plot(pt, add=TRUE)

# get minimum distance from the origin point to the line
dist_min<-st_distance(ls, pt)

# get cordinates of the origin point
pt_orig<-st_coordinates(pt)

# load library for later use of the function destPoint()
library(geosphere)

# create vector of bearing angles of 10 degress amplitude
b_angles<-seq(0, 350, 10) 

# create empty container for final result as data frame
result<-data.frame(bearing=NULL, distance=NULL)

for(i in 1:length(b_angles)){
  
  result[i,"bearing"]<-b_angles[i]
  
  # calculate destination point coordinates with bearing angle i
  # at fixed safe distance (i.e. 100 times the minimum distance)
  # so that to avoid null intersection in next step calculation
  pt_dest<-destPoint(p=pt_orig, b=b_angles[i],d=dist_min*100)
  
  # define linestring from origin to destination
  b_ls<-st_sfc(st_linestring(rbind(pt_orig, pt_dest)), crs=4326)
  
  # get the intersection point between two features
  pt_int<-st_intersection(ls, b_ls)
  
  # get the distance
  d<-st_distance(pt, pt_int)
  
  result[i,"distance"]<-d
}

我尽可能坚持使用“sf”方法，该方法在执行 st_intersection() 时在 for 循环内给出以下警告：“虽然坐标是经度/纬度，但 st_intersection 假定它们是平面的”

但考虑到我正在使用的短距离，在我看来这是一个可以接受的近似值

顺便说一下，据我了解，它在“sf”包中不存在与 geosphere::destPoint 对应的函数

谢谢