计算避开陆地的距离栅格

Calculate distance raster avoiding land

我正在寻找一种方法来计算与点的距离栅格,类似于 raster::distance 所做的,但避开了陆地。所以,就像我可以制作一个点的距离栅格(没有任何限制)一样:

library(sp)
library(raster)


# create cooridnates sp points
col_coord <- data.frame(x      = c(-13.8309),
                        y      = c(28.9942))
coordinates(col_coord) <- ~x+y
col_coord@proj4string <- CRS(projections$WGS84)

#create base raster
baseRaster <- raster(xmn = -80, xmx =55, ymn = -70, ymx = 70, crs = CRS("+proj=longlat +ellps=WGS84"), resolution = 1)

#rasterize spatial point
colraster <- rasterize(col_coord, baseRaster)

#calculate distance
coldist <- raster::distance(colraster, doEdge = T)

#plot
plot(coldist) 

并获取这张图片:

我想制作相同的光栅,但 "shortest distance avoiding an area"(在本例中为陆地)。我知道软件包 gdistance,并且我已尝试遵循 post but this is as far as I've got: (Trans is a transition layer, because it takes a long time to calculate is available here 上的建议。

library(gdistance)
# load transition layer (change directory if needed). This is in another CRS so we transform everything
load("data/Transition_layer.RData")

col_coord <- spTransform(col_coord, Trans@crs)
A <- accCost(Trans, col_coord)
plot(A) #this shows high cost in contintents which is good

# create a raster for water
mapWGS <- rgeos::gBuffer(rworldmap::getMap(resolution = "high"), width = 0)
water <- rasterize(mapWGS, baseRaster, mask = F, field = 1)
table(is.na(water[]))

water[water == 1] <- 0
water[is.na(water)] <- 1
water[water == 0] <- NA
plot(water)

water <- projectRaster(water, crs = Trans@crs)

# change resolution of A to match that of water
A <- resample(A, water)
A2 <- mask(A, water, inverse = F)

plot(A2) # this doesn't make that much sense. 

因此,如果您有任何关于如何从中获得与第一个栅格相似但包含避开陆地的距离的想法,我们将不胜感激!

我认为你可以将 raster::gridDistanceomit 参数一起使用

library(raster)
xy <- data.frame(x=-13.8309, y=28.9942)
r <- raster(xmn=-120, xmx=55, ymn=-70, ymx=70, res=.25)

library(maptools)
data(wrld_simpl)

r <- rasterize(wrld_simpl, r, field= -1)
rr <- rasterize(xy, r, update=TRUE)

x <- gridDistance(rr, 1, omit=-1)
plot(x / 1000000)
points(xy)

地图显示它适用于太平洋,也适用于红海(因为 wrld_simpl 没有巴拿马运河和苏伊士运河)。

但是因为gridDistance的路径必须经过栅格像元的中心,所以估计距离比实际距离长一点。