将经度纬度坐标（WGS）转换为具有等距轴的网格（在给定区域中） [英] Convert longitude latitude coordinates (WGS) to grid with equidistant axes (in given area)

问题描述

我在拖曳数据集中有很多地理坐标，并且想要运行最近邻居搜索。
我遇到了包'RANN'，函数 nn2（x，y）运行速度非常快。

现在存在这样的问题，那当然在伦敦地区，北方的学位比西方的学位要长一些。 b
我现在的想法是将位置坐标转换为某个网格，其中x方向上的一个步骤与y方向上的一个步骤几乎相同。该地区是伦敦（中心-0.1045,51.489）。

$ b

  library（RANN）
 
 xyunf<  - 结构（c（-0.19117，-0.173862，-0.187623，-0.187623，-0.192366，
 -0.176224,51.489096,51.482442,51.50226,51.50226,51.491632，
 51.495429），.Dim = c（6L，2L），.Dimnames = list（c（1，2，3，
4，6，7），c（Longitude纬度）））
 xyosm < - 结构（c（-0.1966434，-0.1097162，-0.2023061，-0.198467，-0.4804301，
 -0.4286548,51.6511198,51.6134576,51.6042042,51.5186019,51.3757395，
 51.3351355），.Dim = c（6L，2L），.Dimnames = list（NULL，c（lon，
lat）））
 
 res< -  nn2（data = xyunf，query = xyosm，k = 1）
 res $ nn.dists 
 res $ nn.idx 
  

解决方案

如果您阅读了R Spatial Task视图，您可以找到关于Spatial对象的所有信息 - 这些是点，网格，线或多边形可以有一个关联的坐标参照系。

一旦你有了这些，你可以使用 spTransform 在坐标系之间转换。因此，要将经纬度为lat / lon的数据帧从lat-long转换为Ordnance Survey Grid Coordinates：

  coordinates（ptsLL）=〜 Longitude + Latitude＃将一个数据帧转换成SpatialPointsDataFrame 
 proj4string（ptsLL）= CRS（+ init = epsg：4326）＃告诉它是lat-long WGS84 
 ptsOS = spTransform（ptsLL，CRS （+ init = epsg：27700））＃转换为英国网格系统
 ptsOS = pts @ coords 
  

现在关于epsg：27700的一点是，它是一个以米为单位的正方形网格，因此适用于此。如果您需要再次转换为经纬度，请再次 spTransform 。

还有其他方形网格坐标系世界其他地区，所以不要把它用于澳大利亚！

I have a lot of geocoordinates in tow data sets and want to run a nearest-neighbor-search. I came across the package 'RANN' and the function nn2(x,y) runs really fast.

Now there is the problem, that of course in the area of London a degree to the north is a quite a longer way then a degree to the west.

My idea now was to convert the location coordinates to some grid where one step in the direction of x is nearly the same as one step in the direction of y. The area is London (Center -0.1045,51.489). How can I perform this transformation?

library(RANN)

xyunf <- structure(c(-0.19117, -0.173862, -0.187623, -0.187623, -0.192366, 
-0.176224, 51.489096, 51.482442, 51.50226, 51.50226, 51.491632, 
51.495429), .Dim = c(6L, 2L), .Dimnames = list(c("1", "2", "3", 
"4", "6", "7"), c("Longitude", "Latitude")))
xyosm <- structure(c(-0.1966434, -0.1097162, -0.2023061, -0.198467, -0.4804301, 
-0.4286548, 51.6511198, 51.6134576, 51.6042042, 51.5186019, 51.3757395, 
51.3351355), .Dim = c(6L, 2L), .Dimnames = list(NULL, c("lon", 
"lat")))

res <- nn2(data=xyunf, query=xyosm, k=1)
res$nn.dists
res$nn.idx

解决方案

If you read the R Spatial Task view you can find out all about Spatial objects - these are points, grids, lines, or polygons that can have a coordinate reference system associated.

Once you've got those, you can use spTransform to convert between coordinate systems. So to convert a dataframe with lat/lon items from lat-long to Ordnance Survey Grid Coordinates:

coordinates(ptsLL) = ~Longitude+Latitude # turns a dataframe into a SpatialPointsDataFrame
proj4string(ptsLL) = CRS("+init=epsg:4326")  # tells it to be lat-long WGS84
ptsOS = spTransform(ptsLL, CRS("+init=epsg:27700")) # converts to UK Grid system
ptsOS = pts@coords

Now the thing about epsg:27700 is that it is a square grid in metres, so work with that. If you need to convert back to lat-long, spTransform again.

There are other square grid coordinate systems for other parts of the world, so don't use this for Australia!

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