Geosphere package distance from point to polygon - r

Question

I'm trying to use the geosphere package in R to get the distance to a polygon from a set of points that lie outside of that polygon.
The polygon is a shapefile of the Antarctic coastline, found here: https://data.bas.ac.uk/items/e6cf8946-e493-4c36-b4f5-58f7a2ee2a74/ and the points are animal tracking data.
I have tried using the syntax specified in the geosphere documentation (https://www.rdocumentation.org/packages/geosphere/versions/1.5-14/topics/dist2Line)
which is as follows:
dist2Line(p, line, distfun=distGeo)
#my attempt so far:
#libraries
library(rgdal)
library(sf)
library(rgeos)
library(tidyverse)
library(geosphere)
#my points
points <-read.csv("Analyses/example_points.csv") #this is the table included below of 4 example locations.
|ID|LON |LAT |
|--|----------|----------|
|a |-2.515478 |-69.53887 |
|b |-2.601405 |-69.79783 |
|c |-0.153548 |-69.45126 |
|d |26.06987 |-69.55020 |
#my line
line <- <- readOGR('Environmental_Data/COAST/add_coastline_high_res_polygon_v7_5.shp/') #this is the shapefile linked above
#convert points to spatial object
coordinates(points) <- ~LON+LAT
distance <- geosphere::dist2Line(p = points, line = line, distfun = distGEO)
However, I get an error: "Error in .spDistPoint2Line(p, line, distfun) :
Points are projected. They should be in degrees (longitude/latitude)".
The package documentation states that p can be:
"longitude/latitude of point(s). Can be a vector of two numbers, a matrix of 2 columns (first one is longitude, second is latitude) or a SpatialPoints object*" - which is what I'm providing it with. I have seen the same issue encountered on a Reddit post (unanswered) but not on here.
My desired output is as below (distances under distance to coast are made up for now!). I have ~3000 locations I need to find the distance to the coastline for.
ID
LON
LAT
Dist_to_coast (km)
a
-2.515478
-69.53887
40
b
-2.601405
-69.79783
24
c
-0.153548
-69.45126
74
d
26.06987
-69.55020
23
Is there an alternative/better means of doing this?
Thank you.

Answer 1

You have loaded sf, any particular reason for not using sf::st_distance() for the task? Would still need to transform though, as there are 4 sample points vs ~140MB shapefile with ~17000 polygons, points were transformed:
library(ggplot2)
library(dplyr)
library(sf)
coastline <- st_read("add_coastline_high_res_polygon_v7_6.shp/")
p <- readr::read_delim(
"ID|LON |LAT
a |-2.515478 |-69.53887
b |-2.601405 |-69.79783
c |-0.153548 |-69.45126
d |26.06987 |-69.55020" , delim = "|", trim_ws = T) %>%
st_as_sf(coords = c("LON", "LAT"), crs = "WGS84") %>%
# transform points to match crs of the shapefile
st_transform(st_crs(coastline))
# number of different surface polygons
table(coastline$surface)
#>
#> ice shelf ice tongue land rumple
#> 325 37 17233 64
# create a single multipolygon, can take a while;
# you may need to filter first to define any surface types you might want to
# include / exclude ("land" also includes islands)
system.time({
ucoastline <- st_union(coastline)
})
#> user system elapsed
#> 103.40 11.72 116.08
p$dist_ucoastline <- st_distance(p,ucoastline)
# or perhaps select land polygon with max area to
# ignore ice and all the islands:
land_max <- coastline %>%
slice_max(st_area(.))
p$land_max <- st_distance(p,land_max)
ggplot() +
geom_sf(data = st_simplify(ucoastline,dTolerance = 1000), fill = "lightblue", color = NA) +
geom_sf(data = st_simplify(land_max,dTolerance = 1000), fill = "gray70") +
geom_sf(data = p, shape =4, color="red", size = 5) +
theme_bw()
Result:
# convert coordinates back to WGS84,
# geometries to coordinate columns
bind_cols(
st_transform(p, crs = "WGS84") %>% st_coordinates(),
st_drop_geometry(p)
)
#> # A tibble: 4 × 5
#> X Y ID dist_ucoastline[,1] land_max[,1]
#> <dbl> <dbl> <chr> [m] [m]
#> 1 -2.52 -69.5 a 40742. 180479.
#> 2 -2.60 -69.8 b 39750. 157043.
#> 3 -0.154 -69.5 c 6629. 186878.
#> 4 26.1 -69.6 d 45683. 121500.
Created on 2022-11-23 with reprex v2.0.2