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
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!