在R中计算两个经纬度之间的距离，但要避开海岸线

Question

在R中计算两个经纬度之间的距离，但要避开海岸线

4

我正在尝试计算海洋位置和陆地点之间最近的距离，但不通过海岸线。最终，我想创建一个到陆地特征的距离地图。

此地图是使用rdist.earth创建的，是一条直线距离。因此，它并不总是正确的，因为它没有考虑海岸线的曲率。

c<-matrix(coast_lonlat[,1], 332, 316, byrow=T)
image(1:316, 1:332, t(c))

min_dist2_feature<-NULL
for(q in 1:nrow(coast_lonlat)){
diff_lonlat <- rdist.earth(matrix(coast_lonlat[q,2:3],1,2),as.matrix(feature[,1:2]), miles = F)
min_dist2_feature<-c(min_dist2_feature, min(diff_lonlat,na.rm=T))
}

distmat <- matrix( min_dist2_feature, 316, 332)
image(1:316, 1:332, distmat)

陆地特征数据是一个包含xy坐标的二列矩阵，例如：

ant_x <- c(85, 90, 95, 100)
ant_y <- c(-68, -68, -68, -68)
feature <- cbind(ant_x, ant_y)

有没有人有什么建议？谢谢。

- Megan

我相信您想要计算简单多边形内的最短路径？您的数据似乎是一堆坐标，其中一些是海岸线，而另一些则不是。您如何定义多边形，并且如何定义您想要距离的一对点？ - J. Win.

@J.Won。我想计算网格上所有点到要素之间的最短距离，这些点不在海岸线上而是岛屿。[什么是到要素的最短距离]。 - Megan

我认为我还没有完全理解这个问题。当你说“不经过海岸线”时，是指路径必须完全在水中（多边形外部）还是完全在陆地上（多边形内部）？此外，这是矢量数据还是网格数据？如果您能发布用于制作该图像的代码，那将非常有帮助。 - J. Win.

@J.Won。是的，完全在水中（多边形外部）。 - Megan

1个回答

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- J. Win. · Accepted Answer

目前还没有完全进行错误检查，但这可能会让你有所启发。与其使用海岸线，我认为你需要从一个栅格开始，将禁止区域设置为NA。

library(raster)
library(gdistance)
library(maptools)
library(rgdal)

# a mockup of the original features dataset (no longer available)
# as I recall it, these were just a two-column matrix of xy coordinates
# along the coast of East Antarctica, in degrees of lat/long
ant_x <- c(85, 90, 95, 100)
ant_y <- c(-68, -68, -68, -68)
feature <- cbind(ant_x, ant_y)

# a projection I found for antarctica
antcrs <- crs("+proj=stere +lat_0=-90 +lat_ts=-71 +datum=WGS84")

# set projection for your features
# function 'project' is from the rgdal package
antfeat <- project(feature, crs(antcrs, asText=TRUE))

# make a raster similar to yours 
# with all land having "NA" value
# use your own shapefile or raster if you have it
# the wrld_simpl data set is from maptools package
data(wrld_simpl)
world <- wrld_simpl
ant <- world[world$LAT < -60, ]
antshp <- spTransform(ant, antcrs)
ras <- raster(nrow=300, ncol=300)
crs(ras) <- crs(antshp)
extent(ras) <- extent(antshp)
# rasterize will set ocean to NA so we just inverse it
# and set water to "1"
# land is equal to zero because it is "NOT" NA
antmask <- rasterize(antshp, ras)
antras <- is.na(antmask)

# originally I sent land to "NA"
# but that seemed to make some of your features not visible
# so at 999 land (ie everything that was zero)
# becomes very expensive to cross but not "impossible" 
antras[antras==0] <- 999
# each cell antras now has value of zero or 999, nothing else

# create a Transition object from the raster
# this calculation took a bit of time
tr <- transition(antras, function(x) 1/mean(x), 8)
tr = geoCorrection(tr, scl=FALSE)

# distance matrix excluding the land    
# just pick a few features to prove it works
sel_feat <- head(antfeat, 3)
A <- accCost(tr, sel_feat)

# now A still shows the expensive travel over land
# so we mask it out for sea travel only
A <- mask(A, antmask, inverse=TRUE)
plot(A)
points(sel_feat)

似乎正常工作，因为左侧海洋的数值高于右侧海洋，同样当你向下进入罗斯海时也是如此。 enter image description here